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March 6

is number theory worth studying if one doesn't plan to make a contribution?

I understand that number theory is important for things like encryption (indispensible for fast international banking), and so on. Likewise computer science is important for advances in every industry and for society in general.

But for computer science, there is a reason for ME to learn it, even if I never plan to contribute anything to field of computer science; namely I can program for my own benefit.

my question is, for ME, is there a similar, analogous reason for me to learn number theory, if I never plan to make a contribution to the science; something I will be able to use. Thank you. 82.113.121.94 (talk) 00:03, 6 March 2010 (UTC)

Is it worth taking piano lessons if you're never going to be a professional performer or composer? 66.127.52.47 (talk) 01:26, 6 March 2010 (UTC)

It's fun. That's the main reason for doing any pure maths! --Tango (talk) 02:31, 6 March 2010 (UTC)

If I were to ask an arbitrary person the reason he/she does what he/she is doing, I think the most probable answer would be "money" (though there are definitely many exceptions). If I were to ask a mathematician the reason he/she is doing mathematics, the answer would be totally different. There are jobs in existence that yield a far higher salary than doing mathematics, and at the same time, require little, if not zero thinking.

How do you expect us to know how number theory will benefit you? When I first started doing mathematics, I did it because I liked it; simply because I found the ideas interesting, and wanted to have fun (on that note, I think the main fun of doing pure mathematics is "getting ideas"; especially the excitement after having thought and thought and eventually "proved your intended result").

Go ahead and learn number theory! See if you like it. If you do not, then you do not have to do it. If you do, then that is great, and you can do it for enjoyment (which is for YOU). The small risk that you will not like it is worth it because if you do like it, I am certain that you will not only LIKE it but ENJOY it. Besides, what do you have to lose? PST 04:59, 6 March 2010 (UTC)

Not many people contribute to number theory, the easiest way for most would be via Paypal :) It all depends on what type of computing you're hoping to do but I'd advise studying what they're teaching if you're at university so you get good grades and a nice job. I think the main help something like number theory would give is a general background in solving digital problems with a mathematical bent. Dmcq (talk) 15:30, 6 March 2010 (UTC)

You know, I studied a fair amount of computer science in college and it's not that useful to me as a programmer. I wish I'd studied more logic and statistics. Logic doesn't get much respect in applied areas yet, but it's becoming quite relevant to programming practice as the more interesting new programming languages are being explicitly designed around the Curry-Howard correspondence. And statistics knowledge is very useful for making sense of messy real-world problems. 66.127.52.47 (talk) 15:55, 6 March 2010 (UTC)

Quotient constructions

Why do quotient groups and quotient categories (both have an algebraic nature) differ from from quotient space (which is not algebraic)? For quotient space, we can use an arbitrary partition, but for groups and categories we need the partitions to be the cosets of a normal subgroup or congruence for the morphism of a category.

I think that for any partition G/R of a group G, if there exists a group structure on G/R such that the canonical projection p of each element to its class is a homomorphism, and that the universal property (for any homomorphism h:G->K there's a unique homomorphism j:G/R->K that makes h,j,p commute) is satisfied, then the partition is the cosets of a normal subgroup. It would be really neat if this is so; that'll answer why we can't use arbitrary partitions for groups and categories and other algebraic structures. Money is tight (talk) 00:42, 6 March 2010 (UTC)

It is fairly obvious that this is the case. You don't even need the universal property. The identity of G/R must be a normal subgroup N of G (since the kernel of a homomorphism is always a normal subgroup). Suppose g in G is in some other part A of the partition. Then for every n in N, p(gn)=p(g)p(n)=p(g), so gn is in A. Conversely, if h is in A, then p(g⁻¹h)=p(g)⁻¹p(h)=id, so g⁻¹h is in N. Thus A=gN as desired. Algebraist 00:55, 6 March 2010 (UTC)

The short and (somewhat) informal answer to your first question is this: An arbitrary subset of a topological space can be equipped with a topology such that the corresponding inclusion map is continuous (i.e. a morphism in the concrete category of topological spaces), whereas an arbitrary subset of a group cannot necessarily be equipped with a group structure such that the corresponding inclusion map is a homomorphism (i.e. a morphism in the concrete category of groups). PST 05:26, 6 March 2010 (UTC)

To algebraist: Wow.. I didn't know it was that simple. Thanks. To PST: I was talking about quotient objects but nevertheless your right sub-objects also exhibit the same phenomena; arbitrary subset may not have structures such that inclusion is homomorphism. It's weird hows there's two types of structures: algebraic (includes partial order/lattice, but I think they're called relation structures), and the special subsets: topological and measure spaces. Money is tight (talk) 10:41, 6 March 2010 (UTC)

Inner Product Spaces

Does ZF prove that if an inner product space is such that all absolutely convergent series converge, then the space must be complete? JumpDiscont (talk) 01:22, 6 March 2010 (UTC)

Yes, even for a normed space E, convergence of all absolutely convergent series (i.e. with Σ_k|x_k| < ∞ ) is easily seen to be equivalent to completeness of E. A useful variation, apparently weaker but still equivalent to completeness, is: "convergence of all series with |x_k|  ≤ 2^-k". The latter property passes immediately to the quotient on any closed subspace V, showing that E/V is complete if E is.--pma 08:47, 6 March 2010 (UTC)

I think he means if the axiom of choice is needed in your proof. Money is tight (talk) 10:56, 6 March 2010 (UTC)

No, AC is not needed. It's a plain consequence of these two facts: (1) if a Cauchy sequence in a metric space has a convergent subsequence, then it is convergent to the same limit. (2) Any Cauchy sequence (x_n) in a metric space has a subsequence such that d(x_nk+1 , x_nk) ≤ 2^-k. In (2) you just define by induction the subsequence of indices (n_k).--pma 18:24, 6 March 2010 (UTC)

120 deg?

I need to construct angles of 120 deg on a circle without using a protractor. Is there a way?--79.68.242.68 (talk) 01:35, 6 March 2010 (UTC)

Yes. Assuming you have a pair of compasses, anyway. Set the pair of compasses to the radius of the circle, mark a point anywhere on the circle, then put the point of the compass on that mark and mark a point one radius away on the circle. Move the point of the compass onto that new mark, and mark another point a radius away. The outer two points will be 120 degrees apart. Did that make any sense? If not, I'll draw a diagram... --Tango (talk) 02:35, 6 March 2010 (UTC)

Yes I tried it and it Works! Not sure why but Thanks!--79.76.146.18 (talk) 21:06, 6 March 2010 (UTC)

What you're doing is basically partially constructing an hexagon, which is composed of six equilateral triangle with angles all equal to 60°. Circéus (talk) 20:32, 7 March 2010 (UTC)

Indeed. The key thing is that you get an equilateral triangle when you have a chord with a length equal to the radius of the circle. --Tango (talk) 23:07, 8 March 2010 (UTC)

:::::Yeah it must be =latl triangle, but why is this chord length equal to the circle radius? Just seen it--79.76.188.14 (talk) 18:08, 11 March 2010 (UTC)

sin-1

On my calculator it says sin^-1 on a key. Is this the same as arcsin (and cos-1 as arccos, etc.)? 68.248.236.144 (talk) 02:47, 6 March 2010 (UTC)

Yes. --Tango (talk) 02:52, 6 March 2010 (UTC)

The Inverse trigonometric functions article (arcsin's redirection target) addresses this notation in its lead, pointing out that it may result in confusion between multiplicative inverse and compositional inverse. On a calculator, sin^-1 always means arcsin. In trigonometric identities, such as sin² x + cos² x = 1, sin² x always means (sin x)². To avoid confusion, most people will not write sin^-1 x for (sin x)^-1, although they will write sinⁿ x for (sin x)ⁿ, even where -1 is a possible value for n. What I wonder is, if the compositional use is understood from the context, is sin² x ever used for sin(sin x)? 124.157.249.168 (talk) 23:33, 7 March 2010 (UTC)

Expected minimum from non-parametrized distribution

What is the best way of calculating the expected minimum value in a sample set of size n drawn from an arbitrary distribution, the functional form of which is unknown? To clarify, given a set of N samples from the distribution, how could you go about calculating the minimal value likely to be observed, as a function of n, when performing future samplings of size n from the distribution?

Further details: I am using a non-deterministic minimization protocol, and am performing multiple (independent) trials with different starting values to better locate the global minimum (otherwise unknowable). I would like to gauge how quickly the algorithm converges with increasing number of repeated trials. Using a representative example, I have performed a large number of trials. Just plotting the minimum obtained so far against output order is sub-optimal, because "lucky guesses" near the start of the run skew things. I've seen equations which give the minimum as a function of sample size, but they assume you know the functional form of the underlying distribution. Is there a "mathematically rigorous" way to compute it without the functional form? Thanks. -- 174.21.226.184 (talk) 03:56, 6 March 2010 (UTC)

This seems like an underspecified problem. Consider an unknown distribution on the two-value sample space {I am alive, I am not alive}. Every day I wake up and observe a sample drawn from this distribution, and it's been "I am alive" every day so far. After some thousands of such observations, all the same, should I expect to alive on any particular day (perhaps in the year 3000) with probability 1? If not, then I need a more precise model of how I'm predicting. The article statistical inference might be of some help. 66.127.52.47 (talk) 05:46, 6 March 2010 (UTC)

Granted, a sample of size N may not be representative of the underlying distribution, but what happens if we assume it is? - That is, if we assume that each of the samples in N has been randomly drawn in an unbiased fashion from the (otherwise unspecified) underlying distribution? (I'll note that the alive/dead distribution is time dependent in a way I'm assuming the sampling not to be: the probability of being alive at day n+1 is much greater if the state was "alive" at day n, versus being "not-alive" at day n. Let me be clear that I'm only considering systems where samples are uncorrelated - that is, where the probability of a particular result in trial n+1 is completely independent of the result of trial n, modulo the fact they are drawn from the same population.) -- 174.21.226.184 (talk) 06:20, 6 March 2010 (UTC)

Do you even know what the sample space is? Just how bad is it if you don't actually find the minimum? The point is that with N samples, you will probably miss any "black swan" events whose probability is less than 1/N, so if you also know nothing about the sample space, you're probably hosed. It would probably help if you can describe your actual application, to get more useful answers. There has to be something you know or can infer about the prior distribution. The article hyperprior might help you see some ways to formalize or quantify this. 66.127.52.47 (talk) 06:50, 6 March 2010 (UTC)

I have absolutely zero a priori knowledge of the sample space - all knowledge of it is solely contained within the N test samples. The underlying optimization problem is probably NP-hard (it's a derivative of the protein folding problem), so I'm unlikely to ever know what the true global minimum is, let alone how my non-deterministic minimization protocol samples the underlying search space. I realize that means that I won't be able to predict rare events, which is okay. I'm simply trying to gauge, to the best of my ability, with the knowledge I have, how effective my minimum-finding protocol is, as a function of number of runs. Yes, I may not find the true minimum with N, or even 1 000 000*N samples, but I simply want to know how much increasing the sample size from, say, n ~ N/5 to n ~ N/4 will improve the estimate of the minimum. -- 174.21.226.184 (talk) 16:23, 6 March 2010 (UTC)

Well why do you think the distribution even has a minimum? Suppose it's just the normal distribution, which means the "minimum" is minus infinity. 66.127.52.47 (talk) 21:59, 6 March 2010 (UTC)

If it makes it any easier, just assume that N *is* the underlying distribution. So my question now becomes: assuming a sample of size n drawn from a (total) population of size N, how do you calculate the expected minimum value for some parameter x as a function of sample size n, with the clarification that multiple members of N may share the same value of x? -- 174.21.226.184 (talk) 16:23, 6 March 2010 (UTC)

I'd do the following:

1. Take a random sample of n points from the N you have.
2. Calculate their minimum.
3. Repeat B times.
4. Calculate the average of all B experiments.

-- Meni Rosenfeld (talk) 16:50, 6 March 2010 (UTC)

Well what do you mean by "expected" in "expected minimum value"? Again this seems to suppose some knowledge of a hyperprior of some sort. Otherwise all you know is your n samples (if drawn without replacement) have n/N chance of containing the global minimum. Are we even presuming that the sample space is a set of real numbers (or maybe integers, since this sounds like a combinatorial problem)? Again I'm thinking of a "black swan" distribution: N=1000000, N-1 elements of the sample space are in the interval [0,10], and the Nth element (the "black swan") is -10¹⁰⁰⁰. Either you happen to see the black swan or else you don't. If you don't see it, there is nothing to suggest that it is there at all. You don't even know the expected value of a single draw, since the black swan completely dominates the sum of the possible outcomes. 66.127.52.47 (talk) 22:24, 6 March 2010 (UTC)

I just came across this:

Bayesian_probability#Personal_probabilities_and_objective_methods_for_constructing_priors

It has some interesting discussion. Added: Sunrise problem and the articles it links to are also interesting. 66.127.52.47 (talk) 23:59, 6 March 2010 (UTC)

By "expected minimum" I meant the likely (real-valued) minimum value which would be observed within a subsample limited to size n. To some extent, I was using the phrase "expected minimum" to refer to the expected value of the sample minimum (for the sample of size n), although at this point I'll take any relatively-easy-to-compute "average"-like metric. Let me clarify: Effectively, I have a black box which spits out numbers at me. For all practical purposes, I know nothing about how this black box decides which number to return, except for the fact that each number is *completely independent* of any number previously produced or any number that will be produced, as the black box completely forgets which number it produces immediately after it returns it and maintains no internal state between sessions. I can run the black box N times, and get a list of representative values. (In the "if it makes things easier" formulation, I can say that the black box randomly samples a member from a known population set of size N.) In a hypothetical future, I will produce n numbers from the same black box, for some (as yet to be determined) value of n. Given this data, I wish to know what I can conclude about the properties of this hypothetical sample of size n, without having to actually pull n numbers from the black box. Specifically, I am interested in the minimum value observed in the sample of size n. (That is, if N=300, n=4 and the values of the hypothetical future sample are {0, 6, -2, 10}, the relevant parameter is '-2', the minimum value within the sample set of size n). As the sampling is random, it is expected that this value should vary based on n (specifically, as the size n of the sample increases, the observed minimum should decrease). To be precise, there really isn't going to be a single minimum value for any given n, but rather a probability distribution m(x,n), giving the probability that the observed minimum value within a sample of size n is equal to x. I was hoping for some way of calculating some sort of parameter on this distribution (like the expected value) which describes how the distribution is "centered" on the number line. In writing my question, I was thinking of things like Student's t-test, where meaningful conclusions can be drawn on populations who are characterized by a sample set, rather than a fully-parametrized distribution. -- 174.21.235.250 (talk) 06:34, 7 March 2010 (UTC)

I think I've figured out a way to calculate what I was looking for. Let:

• p(x) = the probability of drawing x
• P(x) = the cumulative distribution function for p(x) (i.e. the probability that a number drawn is less than or equal to x)
• m(x,n) = the probability of having a sample minimum of x in a randomly draw sample of size n
• M(x,n) = the CDF for m(x,n) (i.e. the probability of having a sample minimum less than or equal to x)

By definition, we know that

• m(x,1) = p(x)
• M(x,1) = P(x)

We now can iteratively calculate m(x,n) by hypothesizing drawing an additional number, and adding it to the n-1 numbers already drawn:

• m(x,n) = (probability the minimum so far is greater than x)*(probability of drawing an x) +

(probability the minimum so far is equal to x)*(probability of drawing an x) +

(probability the minimum so far is equal to x)*(probability of drawing higher than x)

• m(x,n) = (1-M(x,n-1))*p(x) + m(x,n-1)*p(x) + m(x,n-1)*(1-P(x))

I would have preferred a non-iterative calculation (and one where I didn't have to explicitly calculate the CDF), but this works for my use, and hopefully is more robust then the average-of-random-samples method. -- 174.21.235.250 (talk) 04:37, 10 March 2010 (UTC)

That tells you the probability that the minimum is in the sample you drew. It tells you nothing about the minimum if the minimum is outside your sample. 66.127.52.47 (talk) 11:19, 12 March 2010 (UTC)

How to determine the angle a line drawn between two concentric arcs makes with the tangent of the outer arc..

(Apologies if this isn't a clear explanation)

In trying to draw something I cam across the following issue...

A line is drawn between two concentric arcs (centered on O)
Define the starting point on the inner arc as  A
Define the end point on the outer arc as B

What is the angle the line AB makes with the tangent to the curve at A or B.

If it helps the distance AB and the radii of the two arcs are known.

Is there a simple solution to this?

The reason I was wanting to know was so as to set the relevant values when performing a transformation on other values.

Yes, express A and B in polar coordinates. That lets you immediately figure out the tangent angles (i.e. 90 degrees away from A and B respectively) and compare them to the vector AB as expressed in polar coordinates. 66.127.52.47 (talk) 14:53, 6 March 2010 (UTC)

That makes sense, but having drawn out a diagram I am still none the wiser... OK Assume for simplicty that A is at Cartesian(0,-r) on the inner arc.... A has polar form (r, -(pi/2) ) ? B has polar form (R, -(pi/2+delta) ) ? AB as a polar vector is (l, theta) ...

r,R and l are known. delta and theta are not known

ShakespeareFan00 (talk) 19:05, 6 March 2010 (UTC)

I don't really see either how polar coordinates are useful, but you can restate the problem as such: you have a triangle formed by A, B and O, you know the three side lengths r, R and I, and you'd like to find 2 of the angles (minus π/2). The formula you'd want is law of cosines. Rckrone (talk) 05:32, 7 March 2010 (UTC)

$c^{2}=a^{2}+b^{2}-2ab\cos(\gamma )\,$

$l^{2}=r^{2}+R^{2}-2*Rr\cos(\gamma )\,$

Solving for gamma:

$l^{2}-r^{2}-R^{2}=-2Rr(\cos \gamma ),$

$\cos(\gamma )={\frac {l^{2}-r^{2}-R^{2}}{-2Rr}},$ - This gives the angle between the end points..

From there the values needed to solve the angles I actually require are obvious :)

I'll post up a full worked out example in the next few days...

Notes

If $\gamma$ is 0 then AB is quite obviously the difference in radii of the two arcs, which is the minimal length l can be

It should also be possible (using cosine or sine) laws to determine the maximum l for

This looks like it might also be useful in respect of another problem, that of approximating a curve using fixed length

lines ShakespeareFan00 (talk) 11:11, 7 March 2010 (UTC)

Hmm , Just had another look at it.

If you consider the triangle OAB and workout angle A then that's the solution desired. ShakespeareFan00 (talk) 14:54, 7 March 2010 (UTC)

statistics-- median in following case

we know that for getting the median of a grouped data with equal class length we first find cumulative frequency(cf) of the classes. then find n/2, the nearest(and greatest) cf. the class satisfying the condition is called median class. then we put the following formula to get median

    median = l+[(n/2 - cf)/f]*h  

where

l is lower class limit

n is the no of observation

cf is the cumulative frequency of the preceding class

f is the frequency of the median class

h is the height or length of the class

SO THE QUESTION IS

What will be the cf(of preceding class) in the formula if the median class is the first class.

THANX --Myownid420 (talk) 16:38, 6 March 2010 (UTC)

0. -- Meni Rosenfeld (talk) 16:53, 6 March 2010 (UTC)

You should really have enough classes so the very first class isn't the one with the median in. Normally one should have at least ten classes, the data probably has rather a long tail compared to its interquartile range if it still has the median in the first class. Dmcq (talk) 17:03, 6 March 2010 (UTC)

I will point out, just as an aside, that the one fatal mistake in statistics is to focus so much on the math that you lose track of common sense. The median is a simple and easy to understand concept (it's the counting midpoint of a set of ordered data points - figure out what n is, count from the left until your reach n/2, and there you are). That's all your equations does, after having taken a couple of spins through the Twilight Zone. Think first, calculate later. --Ludwigs2 21:29, 6 March 2010 (UTC)

March 7

How to do a Recurrence plot

I would like to do a recurrance plot using some time series data I have. I do not understand the deatail of the article's maths. Would anyone like to try to explain to me, a non-mathematician, how to do this simply? Pseudocode if you like, but not computer languages which I don't understand apart from Basic. Thanks 84.13.166.170 (talk) 01:17, 7 March 2010 (UTC)

No answers yet so perhaps I can ask a question of less scope please. In the example Recurrence plot of the Southern Oscillation index, the X axis is from 1880 to 1980, and the Y axis is also from 1880 to 1980.

What determines if a dot is put on the X,Y coordinate or not? Would it mark all years that had the same amplitude as particular years? Thus for 1880 you simply make a mark on the Y axis for years that had the same amplitude as 1880 (within whatever band of tolerance you set)? And then repeat that procedure for all the years along the X axis. Thanks. 89.242.102.148 (talk) 13:29, 7 March 2010 (UTC)

Yes, there is a dot at (X, Y) if the value of the series at time Y is close to that at time X. -- Meni Rosenfeld (talk) 14:38, 7 March 2010 (UTC)

Look at http://www.recurrence-plot.tk/glance.php?show_intro=1 for simple introduction of recurrence plots (a flash animation!).

Simplifying Trig Identities

Given: (cos²x / sin x) + sin x (1)

And

cot²x + 1= csc²x (2)

Then

(1)= cosx cot x + sin x (3)

How to proceed to simplify (1) to the fullest extent? 192.148.117.79 (talk) 05:34, 7 March 2010 (UTC) —Preceding unsigned comment added by 192.148.117.81 (talk) 05:30, 7 March 2010 (UTC)

You don't need the thing you wrote after "and".

${\frac {\cos ^{2}x}{\sin x}}+\sin x=\cos x\left({\frac {\cos x}{\sin x}}\right)+\sin x=\cos x\cot x+\sin x.$

Michael Hardy (talk) 06:06, 7 March 2010 (UTC)

Yes, that's to help as is the simplification after the "then", my question is where to go after (3)? Alternately, ignoring (2) and (3) how to simplify (1) to the fullest extent? 192.148.117.79 (talk) 06:04, 7 March 2010 (UTC)

OK, you're adding fractions, so use a common denominator, which is sin x:

${\frac {\cos ^{2}x}{\sin x}}+\sin x={\frac {\cos ^{2}x}{\sin x}}+{\frac {\sin ^{2}x}{\sin x}}={\frac {\cos ^{2}x+\sin ^{2}x}{\sin x}}={\frac {1}{\sin x}}.$

Michael Hardy (talk) 06:14, 7 March 2010 (UTC)

Undefinability by any "almost-pure identity" formula (of first-order):

Let the domain of discourse be the class of real numbers.

1. Can one prove that there's no definition (even no implicit definition) formulated by a first-order formula which:

• Contains no free variables, and contains logical symbols only (including identity), except for one function-symbol interpreted (in advance) as the Sine function, and another function symbol to be defined by that formula as the Cosine function...?

2. Can one prove that there's no bijection - from the set of (continuous) Arccosine functions - to the set of (continuous) Arctan functions, so that the bijection is induced by a first-order formula which:

• Contains no free variables, and contains logical symbols only (including identity), except for one function-symbol interpreted (in advance) as an Arccosine function, and another function symbol interpreted (in advance) as an Arctan function?

3. Is there any mathematical realm which deals with similar questions?

4. With regard to questions like those mentioned above, is there a specific term for what I've called: "almost-pure identity" formulas (i.e. formulas fulfilling requirements similar to those mentioned above)?

HOOTmag (talk) 09:08, 7 March 2010 (UTC)

The question of definability over the real numbers is exhaustively studied in model theory. It is very sensitive to the language you work in. For example,

cos(x) = sin(x +π/2)

is a perfectly good definition of cosine. If you want to get rid of the π you can do that by defining π as the smallest positive root of the sin function (which you apparently took as a given). If you only want to allow the sin function and equality, but not any of the arithmetical operations, then I have no idea whether it's definable or not. You'd need to work that out on your own; it's unlikely to be in any model theory text because it's not a very natural question from the modern model-theoretic point of view. This is because there's not much point in saying you're studying "real numbers" if you don't include any of the arithmetical operations on them. You're really just studying an infinite set with an infinite-to-one map. — Carl (CBM · talk) 12:04, 9 March 2010 (UTC)

You've probably meant: (x)(cos(x) = sin(π/2-x)), right?

I couldn't figure out what you mean by:

• "there's not much point in saying you're studying 'real numbers' if you don't include any of the arithmetical operations on them".

Of course I do include the arithmetical operations on the real axis, however I don't let my formula include them. To make it clearer why I do think there is much point in asking my original question (whether there is such a formula), notice the difference between a trial to define the cosine function by the sine function, and a trial to define the "ln" function by the "exp" function: In the latter case, there does exist (what I've called): an "almost-pure-identity" formula, namely: "(x)(ln(exp(x))=x)"; Note that this formula involves logical symbols only (except for the symbols for the definiens and the definiendum). Also, notice the difference between a trial to induce a bijection from the set of continuous Arccosine functions to the set of continuous Arctan functions, and a trial to induce a bijection from the set of continuous Arccosine functions to the set of continuous Arcsine functions: In the latter case, there does exist (what I've called): an "almost-pure-identity" formula, namely: "(∃x)(Arcsin(x)=Arccos(x))"; Note that this formula involves logical symbols only (except for the symbols interpreted in advance as the Arcsine and the Arccosine).

Ok, so I see that model theory can't help me. I'm almost convinced that there is a mathematical realm discussing questions similar to mine, but I haven't found it yet.

HOOTmag (talk) 13:47, 9 March 2010 (UTC)

The mathematical realm you are looking for is model theory, it's just that you are imposing strange restrictions on the problem which go a long way towards ensuring that no one has investigated that particular question before. If you don't let your formula use arithmetical operations, your are looking for a formula in the language of the structure consisting of a set of cardinality of continuum with one unary function, and this structure does not include arithmetical operations. This is the meaning of "include" Carl had in mind. If you do that, there's nothing really ensuring that the structure has anything to do with real numbers, it can be an arbitrary set of the same cardinality. Asking whether a function f on the reals is definable using +, ×, and a function g is an interesting question, and therefore a lot of people studied it. Asking whether a function f on a set of particular cardinality is definable from another function g (only) is a much less interesting question.—Emil J. 15:45, 9 March 2010 (UTC)

Anyway, the permutation of R which exchanges π with 2π and leaves other numbers intact is an automorphism of $\langle \mathbb {R} ,\sin \rangle$ (because $\pi ,2\pi \notin \operatorname {ran} (\sin )$ and sin(π) = 0 = sin(2π)), but it is not an automorphism of $\langle \mathbb {R} ,\cos \rangle$. It follows that cos is not first-order definable in $\langle \mathbb {R} ,\sin \rangle$, explicitly or implicitly (and the same holds for second- or higher-order logic, infinitary logic, and in fact pretty much any other logic one can think of).—Emil J. 16:23, 9 March 2010 (UTC)

How about defining surjections by each other? e.g. to prove the undefinability of the function f (where f(x) = x³) - from the function g (where g(x) = x⁵) - using no symbols other than: f,g and the logical symbols? HOOTmag (talk) 17:25, 9 March 2010 (UTC)

The non-surjectivity of sin happened to be convenient in that example, but it's perfectly possible to construct nontrivial automorphisms of surjective functions as well. In this particular case, let h be the bijection such that h(2⁵ⁿ) = 2⁵ⁿ⁺¹ = g(2⁵ⁿ) for every integer n, and h(x) = x for numbers x not of this form. On the one hand, h is a bijection and h(g(x)) = g(h(x)), thus h is an automorphism of $\langle \mathbb {R} ,g\rangle$. On the other hand, 8 = h(f(2)) ≠ f(h(2)) = 32768, thus h is not an automorphism of $\langle \mathbb {R} ,f\rangle$, and f is not definable from g.—Emil J. 18:44, 9 March 2010 (UTC)

To sum up: If I'd like to prove that a given surjection f (on a given domain) is not definable - from a given surjection g (on that domain) - using no symbols other than: f,g and the logical symbols, then I'll have to find a bijection h (on that domain), such that the composition hg is the same as the composition gh, while the composition hf is not the same as the composition fh.

Is this also the case when f has two variables (f being surjective whenever one of its variables becomes a given constant)? Or should I have then to find a corresponding bijection h for every constant which is substituted for a given variable of the function? i.e. have to prove that for every x there is a bijection h and a number y such that hg = gh but h(f(x,y)) ≠ f(x,h(y)), and also to prove that for every y there is a bijection h and a number x such that hg = gh but h(f(x,y)) ≠ f(h(x),y)?

HOOTmag (talk) 20:21, 9 March 2010 (UTC)

They don't have to be surjections, that's completely irrelevant to the issue. In the general case, h has to be an automorphism of the structure, i.e., a bijection of the domain onto itself such that h and h⁻¹ are both homomorphisms (for structures with only functions and no predicates, the condition on h⁻¹ is redundant). For binary functions this means that h is a bijection and h(g(x,y)) = g(h(x),h(y)) for every x, y, but h(f(x,y)) ≠ f(h(x),h(y)) for some x, y. Also, note that this condition is sufficient to prove non-definability, but it is not necessary. It is possible that there is no such automorphism, but f is still not definable from g (one reason being that invariance under isomorphism is a property of a much more general concepts of definability than just first-order definability, as I already mentioned above). Indeed, with more complicated functions it often happens that the structure has no nontrivial automorphism: that's the case of $\langle \mathbb {R} ,+,\cdot \rangle$ for instance.—Emil J. 11:58, 10 March 2010 (UTC)

1. Thank you for your instructive responses, which helped me a lot.
2. In my last posts I've been referring to surjections only (including bijections), and I still am.
3. As you've probably noticed already, I've always been referring to structures containing no predicates.
4. Note that - when referring to non-definablity - I'm interested in proving that no definition is possible, whether explicit or implicit.
5. To sum up what I've learnt from you: even when the functions f,g aren't unary, I simply have to find a bijection h, such that the composition hg is the same as the composition gh, while the composition hf is not the same as the composition fh. Is that correct?
6. How about structures containing more than one function, say, two functions: g,G (and again: no predicates, while f can't be defined from those functions)? What I'm learning from you, is that I simply have to find a bijection h, such that the composition hg is the same as the composition gh, and also the composition hG is the same as the composition Gh, while the composition hf is not the same as the composition fh. Is that correct?
7. What do you mean by "the condition on h⁻¹ is redundant"? Do you mean that once I find such an appropriate bijective homomorphism h - then I'll automatically find out that also h⁻¹ is factually a homomorphism, or you simply mean that even when h⁻¹ is not a homomorphism, still the very existence of the bijective homomorphism h is sufficient for proving non-definability?
8. Is there any 'necessary and sufficient' condition to prove non-definability (for structures containing no predicates)?
9. I think that the most related issue in Wikipedia is this. However, it doesn't deal with non-definability, although it does deal with general structures, including structures which have no arithmetical operations (and no relations).

HOOTmag (talk) 13:47, 10 March 2010 (UTC)

Ad 5: What you write does not make sense. If D is the domain, you have h: D → D and g: D^k → D for some k ≥ 2. The composition gh (or hg if you write composition in postfix order) is thus undefined. You can only compose functions if the codomain of the inner function is the domain of the outer function.

Ad 6: Yes, apart from the composition problem above.

Ad 7: I mean that if h is a bijective homomorphism then h⁻¹ is automatically also a homomorphism (in structures with only functions).

Ad 8: A function (or predicate, it makes no difference) f is not first-order definable in a structure A if and only if there exists an elementary extension $\langle A',f'\rangle$ of the expanded structure $\langle A,f\rangle$ and an automorphism h of A' which does not preserve f'. I think you may even assume that the elementary extension is an ultrapower, but I'm not entirely sure of that.—Emil J. 15:12, 10 March 2010 (UTC)

Thank you again, Emil.

By the composition gh, I of course meant that for every variable of g one substitutes what h returns for that variable. Yes, I know this doesn't fit the regular definition of a composition, however I afforded to use this abbreviation as I was sure you'd figure out what I meant.

Do you think that your claim in #8 can easily be proved by simply using Godel's completeness theorem? If not, then where can I find some information on that (i.e. a proof and the like)? Wikipedia says nothing...

HOOTmag (talk) 17:04, 10 March 2010 (UTC)

I don't recall seeing it formulated in this exact form, but it can be shown easily using standard model theoretic tools. The right-to-left direction follows immediately from the definitions, using the fact that isomorphisms preserve satisfaction. As for the left-to-right direction: first, it is much less messy to work it out for predicates, so let me assume that P is a predicate (the case of f being a function is equivalent to taking its graph as a predicate) undefinable in A. By compactness (or completeness, if you wish), there exists an elementary extension $\langle A',P'\rangle$ of $\langle A,P\rangle$ and sequences ${\vec {a}},{\vec {b}}$ of elements of A' such that $P'({\vec {a}})$, $\neg P'({\vec {b}})$, and $A'\vDash \phi ({\vec {a}})\leftrightarrow \phi ({\vec {b}})$ for every formula in the language L of A. There exists a strongly ω-homogeneous elementary extension A'' of A', hence there exists an automorphism h of A'' such that h(a_i) = b_i for each i. By Robinson's joint consistency theorem, $\langle A'',{\vec {a}},{\vec {b}},h\rangle$ and $\langle A',{\vec {a}},{\vec {b}},P'\rangle$ have elementary extensions whose reducts to $L_{{\vec {a}},{\vec {b}}}$ are isomorphic. These can be combined to a model $\langle A''',P''',h'\rangle$ such that $\langle A''',P'''\rangle$ is an elementary extension of $\langle A',P'\rangle$, and h' is an automorphism of $A'''$ such that h'(a_i) = b_i. The latter guarantees that h' does not preserve $P'''$. You should be able to find the stuff I used in the argument in any model theory textbook, such as Hodges' "A shorter model theory", or the classic Chang&Keisler's "Model theory".—Emil J. 19:18, 10 March 2010 (UTC)

I believe this result, or one very similar, is called "Svenonius' theorem" in the literature. — Carl (CBM · talk) 03:14, 11 March 2010 (UTC)

Thank you for this additional information. HOOTmag (talk) 10:19, 11 March 2010 (UTC)

So, the left-to-right direction isn't so trival; that's interesting!

Back to one of my original questions: Let's take the right-to-left direction, which is trivial - as you've pointed out. As you (and Carl) had indicated, Model Theory uses such tools for proving the impossibility of defining (in a given structure) some given functions / predicates. Intuitively, do you think you can use similar tools for proving the impossibilty of bijecting / surjecting / injecting / mapping (by a first order formula which can be formulated in every language of any structure of a given form) from a given set to another given set (even when their cardinality permits such a correspondence between them), rather than for proving the impossibility of defining (in a given structure) some given functions / predicates?

For example, Let F be the set of all functions f of the form f(x)=x³ⁿ, and let G be the set of all functions g of the form g(x)=x⁵ⁿ, for a natural n. Intuitively, can you think of any tool (e.g. a tool similar to those tools used in Model Theory), which may enable us to prove the impossibilty of surjecting / injecting from F to G, by a first order formula which can be formulated in every language of any structure of the form $\langle \mathbb {R} ,f,g\rangle$? Note that it is just an example, and I'll welcome any simpler example (provided that the functions f,g are total, and the sets F,G have the same cardinality, or any cardinality which permits the very surjection / injection).

HOOTmag (talk) 10:19, 11 March 2010 (UTC)

Here is a much simpler example:

Let S be the set of all polynomials f,g. Note that each one of the function-symbols f,g is to be interpreted as a polynomial. Intuitively, can you think of any tool (e.g. a tool similar to those tools used in Model Theory), which may enable us to prove the impossibilty of surjecting / injecting from S to itself, by a first order formula which can be formulated in every language of any structure of the form $\langle \mathbb {R} ,f,g\rangle$, when ignoring the trivial identity bijection - which of course can be induced by the formula (x)(f(x)=g(x))?

HOOTmag (talk) 16:44, 11 March 2010 (UTC)

Much less interesting? "Interest" is a relative concept: whatever may seem to be boring in your view - may seem to be interesting in other people's view... :)

Back to my original question:

I can ask whether the function ln on the real numbers is definable using logical symbols only - plus a function-symbol interpreted in advance as the exp function; The answer is..."Yes", the implicit definition being: "(x)(ln(exp(x))=x)"; However, when I ask whether the function Cosine on the real numbers is definable using logical symbols only - plus a function-symbol interpreted in advance as the Sin function (on the real numbers), the answer is..."No".

Similarly, I can ask whether any bijection - from the set of continuous Arcsine functions (on the real numbers) - to the set of continuous Arccosine functions (on the real numbers), can be induced using logical symbols only - plus two function-symbols interpreted in advance as the Arcsine function and the Arccosine function; The answer is "Yes", the formula being: "(∃x)(Arcsin(x)=Arccos(x)); However, when I ask whether any bijection - from the set of continuous Arcsine functions (on the real numbers) - to the set of continuous Arctan functions (on the real numbers), can be induced using logical symbols only - plus two function-symbols interpreted in advance as the Arcsine and the Arctan, the answer is..."probably no". My problem is with the "probably"...

HOOTmag (talk) 16:30, 9 March 2010 (UTC)

When EmilJ and I both describe the problem as "not interesting", we're trying to explain why you will find it hard to look up this sort of thing: because, from the point of view of contemporary model theory, the problem is not very interesting. That doesn't stop you from studying it. But it does mean that you won't find much other work on it, and you would need to be able to make a case for your work if you tried to publish it in a model theory journal. — Carl (CBM · talk) 18:58, 9 March 2010 (UTC)

Yeah, I got it, and I realy have an interesting case for my problem (which may seem to be boring in other people's view). HOOTmag (talk) 20:21, 9 March 2010 (UTC)

How many different types of phase-space plot are there?

I'm familar with simple graphs (or "plot" in American-english?) of time against quantity, as in a time series. But I'm not familiar with graphs in what I think is called the phase space.

How many different types of phase-space plot are there please, and what do you do with the data to make them? The simple things you can do with the data must be quite small, so not many different types of plot.

This http://www.springerlink.com/content/5412256211633127/fulltext.pdf?page=1 says "Recurrences were analysed by first return maps, space time separation plots, return time and recurrence time statistics" which lists four or perhaps five different types of plot. Is there anywhere I can see a simple introduction to these plots and how to construct them? Thanks 89.242.102.148 (talk) 15:08, 7 March 2010 (UTC)

See our articles on phase space, Poincaré map and recurrence plot. Space-time separation time plots are described here. Gandalf61 (talk) 14:43, 10 March 2010 (UTC)

Graph of a smooth function is a smooth embedded surface?

Hi all,

is it necessarily true that if you take a smooth function F of 2 variables, defined on some subset of R^2, then the surface given by Z=F(x,y) is necessarily a smooth embedded surface? I can see intuitively why it should be true, but obviously that's hardly a proof.

My axioms for a smooth embedded surface are:

For each point P there exists a map M(u,v) from a subset of R^2 to some open neighbourhood U of P, U in S, such that:

-M is continuous with continuous inverse

-M is smooth (partial derivatives of all orders)

-At each point Q=M(P), the partial derivatives of M w.r.t. u and v are linearly independent.

(I don't know if these are standard, but they're the ones I'm working with!)

Assuming it is true, I haven't got the first clue about how to show this for some general function F. Could anyone help please?

Thanks a lot, 82.6.96.22 (talk) 16:07, 7 March 2010 (UTC)

Yes, it's a general fact: the graph Γ_f of a C^k map f:M→N is a C^k embedded submanifold of M×N, diffeomorphic to the domain; a diffeo h:M→Γ_f is h(x):=(x,f(x)), whose inverse is the restriction to Γ_f of the first projection of the product, sending (x,f(x)) to x. --pma 16:43, 7 March 2010 (UTC)

Fascinating, thankyou :) 82.6.96.22 (talk) 17:25, 7 March 2010 (UTC)

The probability of a normally distributed random variable being a particular value

For example, what is the probability of a normally distributed random variable with mean 3 and standard deviation 2 being exactly 1? One would think that it would be ${\frac {1}{2}}{\Big [}1+\operatorname {erf} {\Big (}{\frac {1-3}{2{\sqrt {2}}}}{\Big )}{\Big ]}-{\frac {1}{2}}{\Big [}1+\operatorname {erf} {\Big (}{\frac {1-3}{2{\sqrt {2}}}}{\Big )}{\Big ]}=0$, so the proability is 0, however one intuitively knows that there is a slight probability of the value being exactly 1 (although the probability is obviously incredibly small) --220.253.247.165 (talk) 23:18, 7 March 2010 (UTC)

It is 0. If your intuition is telling you otherwise, your intuition is seriously mistaken. Algebraist 23:23, 7 March 2010 (UTC)

Why is it 0? And why is it not 0 for a range of values, if the probability for each individual value within that range is 0? --220.253.247.165 (talk) 23:26, 7 March 2010 (UTC)

It is 0 because the random variable will almost surely not be exactly 1. (There is a technical definition of almost surely which is being used here; see the article.) Similarly, for any given value x, the probability is 0 that the random variable will be exactly x. The reason that the probability is not 0 for a range of values is that a range of values contains uncountably many individual values, and you can't add uncountably many probabilities together. This is all pretty counterintuitive, but it follows from the measure-theoretic foundations of probability. —Bkell (talk) 23:36, 7 March 2010 (UTC)

Here's another way to think about it: Consider a random variable that is uniformly distributed on the interval [0, 1]. What is the probability the random variable is exactly 0.5, say? Suppose this probability is some very small but nonzero value ε. Then the probability must also be ε that the random variable is exactly 0.4, and likewise for every other number in the interval [0, 1]. But there are infinitely many numbers in this interval, so when you add up all of those individual probabilities, no matter how small ε is, you will get the nonsensical result that there is an infinite probability that the random variable takes some value in [0, 1]. This is absurd. So there must be a probability of 0 that the random variable is exactly 0.5. Of course, this doesn't mean it's actually totally impossible for the random variable to take that value; after all, it has to take some value in the interval, but beforehand we would have said that such an occurrence had probability 0. So there is a difference between "probability 0" and "impossible", and likewise a difference between "probability 1" and "absolutely certain"—that difference is what is meant by almost surely. Now, this argument doesn't quite work directly for a normal distribution, because it seems that the probabilities of individual values should "vary", and it's possible to imagine that maybe they vary in such a way that their sum adds up to 1. But this isn't possible because there are uncountably many of them. —Bkell (talk) 23:45, 7 March 2010 (UTC)

These results are actually mathematically equivalent to the observation that the "length" of a single point is obviously 0, but a line segment, which is just a bunch of points strung together, has a nonzero length. If you can understand this geometric idea, which may be intuitively paradoxical at first glance, try to apply it to probability in an analogous way. —Bkell (talk) 23:49, 7 March 2010 (UTC)

Check also Continuous probability distribution. --pma 22:30, 8 March 2010 (UTC)

March 8

Catching cheaters in a math class

Okay, so this isn't a math question exactly, but it is specific to math and people who have taught math. I am teaching Calc 1 right now and I assign homework from the textbook, some odds but mostly even. I just found that 9 of 35 students who handed in their last homework copied out of a solutions manual. There is a student solutions manual available with odds, but I'm talking about even problems. The papers of 2 are extremely obvious, word for word, and the other 7 all have several things, each of which by itself would be suspicious, so together make it pretty obvious. Basically they didn't copy as much detail, leaving out words here and there or whatever but it's still the same stuff in the manual and in the same order and all that.

This question is not about how I should handle the situation, like turn in cheaters or give them a 0 or whatever. That is going to be specific to the rules in my university and all that. But, what sort of things could I do to make sure I can catch such cheaters in the future? I want to be able to assign homework from the book because it is way easier than making it up myself. But, I don't want people getting free points. One thought I have, which may or may not work, is asking the author for an errata to the solutions manual. Then I could assign and grade those problems at least occasionally if not regularly. Then any one who copied from the manual would have the exact same mistakes as the manual and that could be a very easy way to at least clue me in to who is cheating. I may need more evidence than that to accuse them. I have emailed one author so I'll see if such an errata even exists.

Thanks for any suggestions/thoughts. StatisticsMan (talk) 16:36, 8 March 2010 (UTC)

One option is to simply not grade homework, so there's no reward for cheating in this way. You could still give smiley faces or something like that to show you appreciate them doing the homework. In this case, their grades are entirely dependent on quizzes and tests, so that doing their homework and learning the material will be rewarded. You could also have them do homework in class, and even on the board, to discourage cheating. This last option might be appropriate for those who you've caught cheating. StuRat (talk) 17:31, 8 March 2010 (UTC)

I agree. Having homework not count towards the final grade makes a lot of sense. It should still be marked and can even still be compulsory (when I was doing a maths degree, you would face disciplinary action if you didn't get at least a C on too many homeworks), but it shouldn't count towards your grade. Homework is very effective as a learning tool, but it is almost entirely useless as an assessment tool (in addition to the difficulty of dealing with cheaters, it is unfair on people that don't understand things straight away - their final grade should describe their ability as they leave the course, not their ability a week after each lecture). If you want to reduce the unfairness on people that are good at the subject but not at exams, you can set project work, which is far harder to cheat on. --Tango (talk) 18:08, 8 March 2010 (UTC)

I teach computer programming, which is rampant with students copying code from websites. I simply set up the grade system for the class such that cheating on the homework will cause you to fail if you don't learn the material. 10% of the grade is in-class work. It is really hard to download code in class when your name is called and you are asked to work something out. 40% is homework. If you cheat and get 100% on the homework, you only guaranteed a 40% in the class. 30% is exams (3 of them at 10% each). 20% is the final. Each semester, I get both extremes. I get students who do great on the homeworks and terrible on the exams - they fail. I get students who do great on the exams, but skip class and never turn in homework. They fail. I don't see why calculus wouldn't be the same. You can cheat on the homework, which will lead to failure on the exam if you don't know the material. -- kainaw™ 17:36, 8 March 2010 (UTC)

Why should students who do well on tests but poorly on homework fail ? Haven't they demonstrated proficiency ? Or do you suspect them of cheating on the tests ?

There is also a fundamental difference in computer programming: it's just about impossible to write a program of any complexity without bugs, which requires a more iterative testing and correction process than just about any other discipline (a typo in a program usually means it won't run at all, while a typo on an English paper is a minor mark down). Therefore, homework (as in writing code) is actually testing something different, their ability to debug, than a test (at least a test on paper). StuRat (talk) 18:28, 8 March 2010 (UTC)

Another idea is to assign homework from a different textbook (or better yet, several different textbooks). Of course, this is a bit more work for you, because you have to make handouts of the homework problems. —Bkell (talk) 19:50, 8 March 2010 (UTC)

Surely it would help if you set them homework questions that you wrote yourself rather than just a) choosing them out of a book or b) choosing them out of a book that has an Answer Book too? 78.146.208.26 (talk) 21:39, 8 March 2010 (UTC)

I agree with the general tenor of responses that homework in a university math class should not be part of the grade, or at least not a significant fraction of the grade. Part of the reason is that collaboration on homework ought to be encouraged (many students learn better in group efforts, provided they actually make an effort), but the line between collaboration and copying is fuzzy and hard to define, much less enforce. Plus you probably have other work to do than grading calculus assignments (of course it's different if the school is willing to pay for a grader; in that case it's a nice thing to help some undergrad have a little income).

I expect you encounter some resistance from the students to this idea; many of them have learned to expect to be able to buffer possible bad results on tests via sufficient grunt work on assignments, whether they understood what they were doing or not. To that I think you just have to tell them that they're not in high school anymore, and this is the way it's done at university. I used to tell my students that homework does count — on the exams. Meaning that if they do the homework, they have a better shot at doing well on the exams. --Trovatore (talk) 21:50, 8 March 2010 (UTC)

At issue is what level of "university" are we discussing. I mentioned computer programming above, which is a Freshman level course. Freshmen are transitioning from high school to college. Homework is a necessity to help in that transition. If I were to use computer architecture, which is a much more advanced course, as an example, then I wouldn't expect as much homework to be involved. -- kainaw™ 22:04, 8 March 2010 (UTC)

It's not that you don't assign homework. You just don't grade it. The idea is that the student is responsible for learning and understanding the material, and then being able to prove that he/she understands it (this notion of "prove" being a bit problematic, but exams are more reliable than homework assignments in this regard). Then the homeworks are provided as an aid for the student to come to that facility, not as an evaluative technique in itself.

Programming is different, though — I'm specifically talking about math classes here. Programming classes need projects, which indeed should be evaluated. If results for the students were the only concern, and instructor time didn't matter at all, you could also have projects for math classes, I guess. I don't know how you would do a project for a cookbook class like calculus, but the idealistic answer is probably "restructure calculus so it's not cookbook anymore". --Trovatore (talk) 22:35, 8 March 2010 (UTC)

Personally I'd like to see more of them working together on a problem during a tutoring session. After all that's what people mostly have to do in life and it trains in cooperative work. Also I believe it helps if they try to explain problems to each other. Dmcq (talk) 09:52, 9 March 2010 (UTC)

I have mandatory homework assignments and have never had students complain about it. Actually I've had students tell me that they really appreciate having to do them. The way to prevent "cheating" is to legalize it. They are allowed to work together, and I don't care if their paper is identical to their friend's. If they're just copying, they might learn something minimal, and then they'll fail the exams anyway. (The exams are weighed heavily.) Everybody says that the way to learn mathematics is to do mathematics. If you really want them to learn mathematics, then make them do mathematics. It's work for me to grade, but you can't beat it for effective learning. Staecker (talk) 13:13, 9 March 2010 (UTC)

Your job as a professor consists of teaching and judging. Do your students look at you as a teacher or as a judge? A good teacher is inspiring and enthusiastic while a good judge is serious and fair and strict. The good teacher makes the students study with pleasure, while the judge threatens them to study. Pick your choice. Bo Jacoby (talk) 14:12, 9 March 2010 (UTC).

Thanks for all the comments. I think homework is very important for students. If I do not assign it, many college freshmen will not do it and will not learn anything. So, it has to be worth something. In fact, I make it somewhat easy to get the points.

I disagree with the comment that it shouldn't be worth much because it's the first time they're learning something. I am fine if they work with others. I am fine if they ask me questions and I will give them a lot of help. I will work through the entire problem with them and ask questions and make comments to try to make sure they understand what I am doing. I am fine if they go to the help room and ask questions. I tell them this. And, I give them most of the credit just for doing it and then only grade a few problems for 30% of the homework points. Then, even if they mess up quite a bit but are sort of on the right track they will get half of that. So, as long as they put in work, they will get a good score on the homework and learn from doing. And, they will learn from my feed back if they look at it. And, I put solutions to some problems on my web page as well so they can learn from that. If they have no idea what they're doing on the homework with all those options for help, then they're not going to know what they're doing when the quiz or test comes around any way so who cares if they lose points on the homework.

I also disagree with legalizing cheating. The fact is, some of these cheaters are people who are doing very well in my class. They have probably had calculus before and do well on tests so they are not ever punished for the cheating if I do not do it myself. On the other hand, if they do very well on homework and then do only alright on everything else, they will still end up passing the class. I would be dishonest if I allowed people to cheat the entire way through the semester and then gave them a C and said they did satisfactory work. Especially when there are other people who don't cheat on the homework and get low grades and then do alright on the tests and those people end up with a D. So, the cheater would literally win in that situation compared to a noncheater. I am not going to allow that to happen. That would be terrible. And, the cheater who passes will then continue the same thing in Calc 2 and Calc 3. And, I guarantee most professors are not going to catch the cheating like I did because most people do not pay attention to detail as much as I do. So, this may be the only time they ever get caught doing this and I need to stop them in hopes that they actually do the right thing for the rest of their college careers. If they don't in other classes, that's not my concern but in my class it is my concern.

The comment about using problems I make up on my own is fine except the point is I don't want to go through that extra work. I assign around 20 to 30 problems a week. I do not want to make up 20 to 30 problems. That's a large part of the point of textbooks. And, the comment about using another textbook is a suggestion a professor here gave me. It may be a good option. I still might have to type up all the problems or something and it's still a large amount of work. He suggested I could just copy the exercise pages and put them on my web page. I guess I could do that but I don't know if that's legal. StatisticsMan (talk) 16:11, 9 March 2010 (UTC)

Well, making up problems (plus TeXing them) is fast, compared to grading them. Unless you have a grader? (Or, a small class?)

One thing I usually did (though I didn't always tell the students about it, at least not "officially" on the syllabus), was have multiple grading formulas, and give each student the grade from the formula most beneficial for that student. So for example the final exam would always count heavily, but midterms and quizzes might count only for those students they helped. If you go with "homework counts only if it helps, and even then not very much", that would considerably reduce the benefit of cheating, while still giving them a direct incentive to do the homework.

Seriously, though, you might consider if it isn't time for them to learn that sometimes you have to do things you don't get directly evaluated on, because they affect things you do get evaluated on. They're legal adults; might as well be now. --Trovatore (talk) 19:19, 9 March 2010 (UTC)

Well, just to knit-pick :) some of my students are not legal adults. One person told me she was 16 and another 17, both graduating from high school early. But, most would be and it still makes sense that they need to learn that lesson. As far as making up problems, if I have to TeX up 20 or 30, that would take me a while. I am a fast typer but I am somewhat of a perfectionist so I take too long on things like that. Right now, I spend about 3 minutes picking out the problems in the book. TeXing would take 15 minutes just to copy them out of some other book and 1 hour or more if I have to actually make up problems, especially word problems. I don't want to add an hour of work time to every week. Especially if I some day am teaching 3 classes at a time, then it is an even greater time. StatisticsMan (talk) 20:45, 9 March 2010 (UTC)

One more data point: at the UK university I teach at, homework for 1st years is compulsory but is worth nothing. I mark it carefully, we go over it together, but the mark they get in their 1st year is based almost entirely on exams. The advantages are obvious: the students know that cheating or copying is a waste of their time, and also the problem sets are a tool for learning things and not somewhere to try and scrape every last mark even if you don't really understand what's happening. The disadvantages are equally clear: some people are strong mathematicians but are not good at written exams, these people may not get the results they deserve.

If you really don't have the time to write your own problems, perhaps you could just make trivial modifications to existing ones (i.e. change a few constants, or even just variable names). It shouldn't cost you too long to solve the new questions, and you'll catch some people so foolish that they copy solutions to the old versions. Tinfoilcat (talk) 16:41, 9 March 2010 (UTC)

I think I like this idea somewhat. Instead of making it 0 points though, I think I will just lower the percentage of points. Perhaps I would make homework 10% of the grade. This is my first time teaching my own course and I'm still learning, so I made homework 25% of the grade. Now, with this situation, it definitely seems like I should have made it lower. Another thing is I want to be clear ahead of time what the punishment for cheating is so that at least some students will be scared from doing it in the first place. StatisticsMan (talk) 18:10, 9 March 2010 (UTC)

Yeah it would probably be helpful to make an announcement now to the class that you've noticed some people are copying their answers from the book and that's really not what you are looking for, and if you see it anymore there will be consequences. You don't have to name names, but make it clear that it's pretty obvious. You might also try to explain why doing the homework for real is important, for whatever that's worth. Rckrone (talk) 19:16, 10 March 2010 (UTC)

Why not set the occasional homework yourself (or one question on each assignment)? Then you could weight the results of these special homeworks (or questions) so that they count more towards the final grade. You will still have the problem that some students will copy from others, or get someone else to do the work for them, but at least they can't just copy from a solution manual. Dbfirs 08:26, 11 March 2010 (UTC)

Ideally all work done by the student ought to be assessed by an objective and independant third-party. Teacher and student then unite with a common purpose. There ought to be "homework" that is marked but does not contribute to final grade, since people do learn and improve with practice and feedback from marks or teacher comments is an essential part of learning. It would be unfare to allow zero practise in a skill they are trying to learn - you are trying to teach them, not merely assess whatever knowledge they came to the course with. 89.243.212.29 (talk) 15:10, 11 March 2010 (UTC)

Easy free software for drawing (phase space) graphs from statistical time-series data?

I would like to draw some graphs to explore what I think is called the phase-space of time series data. I have searched around the internet but have only been able to find software that will draw from a formula only.

Is there any preferably easy to use freeware that people would recommend please? I would like to be able to make simple transformationms of the data in order to show various aspects or kinds of the phase-space. (I'd be interested to know how many different kinds of phase space graphs there are).

Supplemental question: I have R installed, but I have not used it yet. Where should I start in learning to use its graph facilities? Thanks. 78.146.208.26 (talk) 21:37, 8 March 2010 (UTC)

Calculating eigenvector for beginners

I'm trying to write up a description of calculating eigenvectors for a paper to be published in a medical journal - so it has to be VERY light on mathematics. Does anyone here have links to pages that effectively dummy down the calculations so I can get some ideas of how to phrase this with the least amount of matrix mathematics as humanly possible? I'm on revision 23 and I'm still being told it is complicated. -- kainaw™ 23:05, 8 March 2010 (UTC)

You calculate the eigenvalues by calculating the appropriate determinant and solving the polynomial and then you get a system of simultaneous equations to solve for the eigenvectors. So basically you need a simple way to solve polynomials and then a simple way to solve simultaneous equations. What size are the matrices in question? If they are a fixed size (especially if that size is 2 or 3) then there are simple methods of doing that. There is a limit to how simple it can get, though. Can you not tell them to use a computer? --Tango (talk) 23:17, 8 March 2010 (UTC)

Due to the nature of the research, the matrix is always 4x4. I end up with four equations to solve for. I'm a bit flustered because my last attempt was sort of faked. I used a 4x4 matrix with only non-zero values in the upper left 4 positions. So, it was mostly zeros. The reviewers all said it was more confusing. -- kainaw™ 23:31, 8 March 2010 (UTC)

Tango's method is going to be quite painful for a 4x4 matrix (at least the "finding roots of the characteristic polynomial" bit). In practice you would use something like the QR method or the Lanczos method. Do your medics really need to know *how* to calculate eigenvalues, or is it not rather *what* eigenvalues are?213.160.108.26 (talk) 23:33, 10 March 2010 (UTC)

I think I found a way to make this easier for doctors to understand. -- kainaw™ 01:23, 9 March 2010 (UTC)

Don't know if it's appropriate, but you can tell people to use Mathematica or Maple or similar. (Do medical types really need to know how to do it by hand?) For people who don't use this software, they can do them on Wolfram Alpha like so: http://www.wolframalpha.com/input/?i=eigenvectors+of+{{2,2},{4,3}} Staecker (talk) 13:06, 9 March 2010 (UTC)

I'd use just a 2x2 matrix and calculate the eigenvectors and eigenvalues by hand and show what they mean. Plus show a couple of funny cases which don't have two eigenvectors. Then I'd say the 4x4 ones are too much bother to do by hand but can be solved by various different methods using a computer which will be programmed to deal with any quirks. Dmcq (talk) 12:11, 11 March 2010 (UTC)

March 9

Homework question

For what values of x is the sequence Un = $(x-2)^{n}$ convergent to 0?

The answer is obviously 1 < x < 3 but I have no idea how to prove that. Could someone lead me in the right direction? --124.171.116.21 (talk) 04:38, 9 March 2010 (UTC)

Let $z\in \mathbb {R}$ such that $|z|<1$, and let $z_{n}=z^{n}$ for each $n\in \mathbb {N}$. Prove that $z_{n}\rightarrow 0$. Once this has been established, write $z=x-2$, and note that $|x-2|<1$ if and only if $|z|<1$, and therefore, if $1<x<3$, $|z|<1$, $u_{n}=z^{n}$, and thus $u_{n}\rightarrow 0$.

To complete the above proof, you need to establish that $z_{n}\rightarrow 0$, and that if $|r|\geq 1$, and we let $r_{n}=r^{n}$ for each $n\in \mathbb {N}$, the sequence $\{r_{n}\}_{n\in \mathbb {N} }$ does not converge to zero. I have hidden the proof of the former below; if you feel that no furthur attempts at the former will be at least slightly productive, you may see the proof. However, even if you do see the proof of the former, I have not included a proof of the latter (which essentially follows from the line of thinking required to prove the former). If you feel that you cannot solve the latter, even after seeing the hidden proof of the former below, post here again, and I (or another volunteer), will give you additional hints. Hope this helps (and the proof of the former is hidden below). PST 05:36, 9 March 2010 (UTC)

(Click the "show" button at the right to see a proof relevant to the above post or the "hide" button to hide it.)

Result

Let $z\in \mathbb {R}$ such that $|z|<1$, and let $z_{n}=z^{n}$ for each $n\in \mathbb {N}$. Then $z_{n}\rightarrow 0$.

Solution

We must show that for all $\epsilon >0$, there exists $N\in \mathbb {R}$, such that for all $n>N$ ($n\in \mathbb {N}$), $|z_{n}|<\epsilon$ (we can assume without loss of generality that $z\geq \epsilon$). If $\epsilon >0$, let $N=\log _{|z|}(\epsilon )$. Since the function $f(x)=|z|^{x}$ is decreases for $|z|<1$ and $x\geq 1$, we have:

$n>N$

$n>\log _{|z|}(\epsilon )$

$|z|^{n}<\epsilon$ (we assumed without loss of generality that $z\geq \epsilon$)

$|z^{n}|<\epsilon$

$|z_{n}|<\epsilon$

Since the last statement above is true for all $n>N$ and $\epsilon >0$ was arbitrary, the result follows.

PST 05:36, 9 March 2010 (UTC)

Wiktionary does not recognize the word furthur. Did you mean further? -- Meni Rosenfeld (talk) 09:08, 9 March 2010 (UTC)

Of course that is what he meant. Don't be a jerk. StatisticsMan (talk) 18:12, 9 March 2010 (UTC)

Please assume good faith. I was seriously considering the possibility (however remote) that in fact this is an alternative spelling which has not made it to Wiktionary, and did not want to come out a fool if this turned out to be the case.

I should also add that this is the third time I have noticed PST making this mistake, so this being a simple typographical error was out of the question. Either this was, indeed, a correct spelling, or PST mistakenly thought that it was correct. In the latter case I thought he would appreciate noting this error. -- Meni Rosenfeld (talk) 19:38, 9 March 2010 (UTC)

"Your English inflexion is almost perfect, Oberleutnant PST " ;-) --pma 08:48, 10 March 2010 (UTC)

This is no laughing matter. I find it repugnant that StatisticsMan would insult me like this and not bother to follow up on the discussion. -- Meni Rosenfeld (talk) 20:49, 11 March 2010 (UTC)

My excuses, Meni. I still don't know what "jerk" means nor if it has offensive connotations (I am possibly the only one here who would need explanations about English); in any case I'm quite sure SM didn't meant to insult you. As to my comment, it was not addressed to anybody in particular; the episode just reminded me of some old good Bogart style movie. --pma 21:28, 11 March 2010 (UTC)

That is no problem - I just didn't want the lighthearted reference to overshadow how seriously I take the matter.

See also the third definition of jerk. Presumably, StatisticsMan was under the impression that my post was sarcastic and intended to ridicule PST. Usually the word isn't terrible, but in the current context StatisticsMan's comment was incredibly offensive. -- Meni Rosenfeld (talk) 07:38, 12 March 2010 (UTC)

I do not believe that Meni intended to ridicule my use of the English language with his comment, but this conclusion is certainly influenced by the fact that I have interacted with Meni at the reference desk quite often recently. Had I been reading an identical discussion between two people whom I did not know, I might have seen the comment as an insult. I think that this scenario is relevant to the present context.

StatisticsMan has not commented at the reference desk frequently in the recent past (he only contributed to the reference desk five times last month - [1]); of course, that does not necessarily give any indication as to how often he visits the reference desk. Nonetheless, in StatisticsMan's case, I believe that he does not visit the reference desk very frequently either. Therefore, StatisticsMan's failure to follow up on the discussion may simply be because he has not visited the reference desk since posting his most recent comment, in which case I do not think that StatisticsMan can be considered repugnant in the context subsequent to his most recent response. Furthermore, knowing StatisticsMan, I do not think that he would be under the impression that he is "too good" to follow up on this sort of discussion.

That said, I shall emphasize once more that I do not believe Meni intended to ridicule me with his comment; personally, it seems unlikely (if not impossible) that he would behave in that manner, having had some experience interacting with him. StatisticsMan, on the other hand, may not have significantly interacted with Meni in the past, and as I mentioned earlier, if the name "Meni" had not been attached to Meni's comment, I might have interpreted it as an insult (but as I shall clarify once more, I definitely do not think that Meni is a jerk, in that it is very unlikely that his comment is an intended insult).

With regards to the usage of "jerk", I think that sometimes it is considered appropriate for someone to deliver an insult to a "definite troll" (for instance, if a troll came here and vandalized the reference desk, and demanded a response to his behavior, it would be considered OK (by some) to dismissively insult him (I certainly would not object to this use of an insult if the nature of the vandalism were extreme, but that is a different matter in a different context)). Perhaps StatisticsMan decided that Meni's comment was "in the manner of a jerk" and dismissed it with what he considered appropriate. As I said, whether this was appropriate or not is "relative", but I think that anyone who knows Meni Rosenfeld well would see the comment as an incorrect portrayal of his character.

Regarding my English, sometimes I have a habit of making minor misspellings when I type; had I written that comment by hand, I do not think that I would have made that error. Perhaps the first "u" in "further" influences my brain to believe that the "u" repeats, in much the same way as in other words having a "repeating vowel", and perhaps this influence is more significant while I type. I don't really know. But thanks to Meni for noting that; I might have noticed it in the past when editing my typed text, but I never really knew that I was repeating the mistake (it would be nice if Wikipedia had a "spell-check" like Microsoft Word; perhaps I should paste my text into Microsoft Word and run "spell-check", and then paste the corrected text into Wikipedia ;)). PST 09:41, 12 March 2010 (UTC)

My post was intended to be as diametrically opposite to "insulting" or "jerky" as possible, in the same way that "you are wrong" is inferior to "I think you may be wrong". The former is arrogant and rude and assumes the speaker is infallible while the target is error-prone (unless, of course, the speaker is an authority on the subject).

Prior to StatisticsMan's comment, it never crossed my mind that my post could have been interpreted as an insult. Furthermore, if I had been asked "do you think your post could be interpreted as an insult?" my answer would have been a definite "no". I would not be able to conceive of anything more polite than presenting the evidence I have that a mistake has taken place, offering a correction and acknowledging that I might be wrong myself.

Now that I know it can be interpreted negatively, I have certainly noted to myself how to avoid such problems in the future (primarily "don't be so polite nobody would believe you are that polite and assume you are rude instead"). I will also be happy to hear any additional insight on what I should have said instead in this particular situation.

I believe that StatisticsMan, even without deep familiarity with me, had enough information that with a little WP:AGF, he should not have concluded that I am a jerk.

"[maybe] because he has not visited the reference desk since posting his most recent comment" - this is factually false, as he has posted here since my reply to him. Obviously he has no obligation to visit here at any frequency, but I, for one, when I refer to someone as a jerk (not very often), I sure don't wait >2.5 days to check back if he has something to say for himself. Or, if I anticipate a period of unavailability, I don't enter heated debates to begin with.

What browser do you use? Firefox has a spellchecker. -- Meni Rosenfeld (talk) 11:31, 12 March 2010 (UTC)

Alternatively, if you are OK with 1/n → 0 and with the sandwich theorem, you may prove that if 0 ≤ a < 1 then 0 ≤ aⁿ ≤ C/n with C := a/(1-a). Start from the Bernoulli inequality with x := (1-a)/a .--pma 09:38, 9 March 2010 (UTC)

Divisibility In Digit-by-Digit Inter-Base Conversion

I've been trying to get my head around the meaning of some empirical results that don't seem to have an obvious explanation. Up to some very large number at least, there is no number that generates only numbers not divisible by any of {2, 3, 5} when one takes its representations in bases 2 through 5 and translates these into greater bases up to 6 (Ten numbers output for each input). Can anyone prove the impossibility of finding such a number, explain reasonably precisely why the first such number would be so inordinately large that it may not be computable, or demonstrate an example which does give ten numbers relatively prime to 30?Julzes (talk) 08:40, 9 March 2010 (UTC)

What's wrong with the number 1? Dmcq (talk) 09:45, 9 March 2010 (UTC)

20821 - outputs are 5321791, 285282577, 6348063151, 80542674037, 267781, 1973131, 10134727, 94531, 328147, 58633; outputs modulo 30 are 1, 7, 1, 7, 1, 1, 7, 1, 7, 13. Gandalf61 (talk) 11:51, 9 March 2010 (UTC)

Ok, that means my most recent program was in error. Anyway, can anyone find a number that generates 10 primes or suggest to me a stronger sieve than simply checking numbers congruent to 1 modulo 60? I know why the first such number is likely to be huge. In fact, just checking numbers like mentioned has another program that hopefully is not also defective up to around 190 billion without finding such a number.Julzes (talk) 16:36, 9 March 2010 (UTC)

Well, if anyone comes up with anything profound on this, please let me know.

Off-topic slightly, but here is a brief description of the granddaddy of coincidences that I discovered since the last time I posted anything of the nature here: List the primes that translate as primes twice in going from base 2 to base 10 and also translate twice as primes in going from base 3 to base 10. The 4th of these is the first to also generate a prime once in going from base 4, the 44th is the first to do it twice, both begin with the digits 234 in base 10, and the tenth on the list begins with 365. Furthermore, the 4th on the list is more special in that it translates twice as primes from base 5, does not translate once as a prime again until base 20 (allowing 'digits' greater than 9, and the tautology at base ten isn't counted) at which base it translates 5 times as a prime, and then it also does it 4 times at base 22 and twice at base 25 (not at bases 21, 23, or 24) to boot. Don't ask me how I found it.Julzes (talk) 03:42, 10 March 2010 (UTC)

March 10

How do you prove that $x$ generates the multiplicative group of $GF(p)[x]/f(x)$ if $f(x)$ is a primitive polynomial?

How do you prove that $x$ (and any element in $GF(p)[x]/f(x)$ other than the additive identity) generates the multiplicative group of $GF(p)[x]/f(x)$ if $f(x)$ is a primitive polynomial?

If it is fairly easy to prove, I'd like to give it a try, but could use some hints. On the other hands, if it is difficult, could someone point me to a complete proof? Thanks. --98.114.146.242 (talk) 02:57, 10 March 2010 (UTC)

Well, x generates the field because it is a root of a primitive polynomial... However, for example, if the multiplicative group of the field has size 8 and z is a generator, then z⁴ cannot possibly be a generator, because z⁴ will have order 2. So it will not be true that "any element in $GF(p)[x]/f(x)$ other than the additive identity" will generate the multiplicative group. In general, a cyclic group of size n has only φ(n) generators. — Carl (CBM · talk) 03:04, 10 March 2010 (UTC)

You're right. I confused a special case with the general case. --98.114.146.242 (talk) 03:28, 10 March 2010 (UTC)

Project Euler problem

In this problem, how many simulations of random walks would you expect to have to do in order to get a result accurate to 6 decimal places? (and yes, I am aware this isn't a good way to solve the problem). 149.169.164.152 (talk) 07:11, 10 March 2010 (UTC)

That's easier to estimate once you've done a few simulations. You'll get very rough estimates of the mean $\mu$ and standard deviation $\sigma$. If you then do n simulations and take their average, the result will be distributed normally with expectation $\mu$ and standard error $\sigma /{\sqrt {n}}$. Let $\mu =m\cdot 10^{e}$, $1<m<10$. To be highly confident the result is accurate to 6 s.f., you'll want the SE to be about $10^{e-6}$, so $n=\sigma ^{2}10^{12-2e}\approx {\frac {\sigma ^{2}}{\mu ^{2}}}10^{13}\approx {\frac {{\hat {\sigma }}^{2}}{{\hat {\mu }}^{2}}}10^{13}$. If the distribution of the experiment result is roughly exponential, expect the required number to be in the trillions. -- Meni Rosenfeld (talk) 08:27, 10 March 2010 (UTC)

Oh, decimal places, not s.f.. Then what you really want is an SE of $10^{-7}$, which means $n\approx {\hat {\sigma }}^{2}10^{14}$, which is much higher (order of $10^{20}$, maybe?) -- Meni Rosenfeld (talk) 08:35, 10 March 2010 (UTC)

Is C^∞ dense between the measure preserving transformations?

My problem is:

Suppose we have a diffeomorphism T which preserves the Lebesgue measure and is C³. Is it possible to find a T' which is C³-close to T, preserves the Lebesgue measure and is C^∞? Usually density arguments make use of convolution but I guess this would destroy the property of preserving the measure. How can I do?--Pokipsy76 (talk) 10:39, 10 March 2010 (UTC)

At least in the case of a C³ diffeo T:M→M on a C^∞ compact connected differentiable manifold M with a volume form α, I think this follows quite immediately by a celebrated theorem by Jürgen Moser [2], that states that if α and β are two volume forms with the same total mass on such an M, there exists a C^∞ diffeo f:M→M such that β=f^*(α), and in fact f can be chosen arbitrarily C^∞ close to the identity provided α and β are suitably close. Now, take a C^∞ diffeo S which is C³ close to T. Then S^*(α) is a volume form close to α and has the same total mass. By Moser theorem with β=S^*(α) there exists a C^∞ diffeo f:M→M close to the identity such that S^*(α)=f^*(α), so (Sf ^-1)^*(α)=α, which mans that Sf ^-1 is a C^∞ diffeo C³ close to T which preserves α, as you wanted. I strongly suggest you to read Moser's beautiful proof, that may give you several hints. It has been generalized to various cases (manifold with boundary, non-compact, &c). You should work out the details of what I sketched, but I think it's ok and I hope it's of help. It is possible that there is a more direct proof, but I guess in any case one has to pass close to some of Moser's proof's lemmas. --pma 21:15, 11 March 2010 (UTC)

Thank you very much, I think this article will be useful.--Pokipsy76 (talk) 13:10, 12 March 2010 (UTC)

Another thing. Let F:R×R^m→R^m be a time-dependent field with with vanishing divergence (wrto the x variable), and such that its integral flow

$\scriptstyle \partial _{t}g(t,x)=F(t,g(t,x))$

$\scriptstyle g(0,x)=x$

is defined at time 1 for any initial value x.

Then it's clear that the map at time 1, T(x):=g(1,x) is a diffeomorphism that preserves the Lebesgue measure. Conversely, any measure preserving diffeo is of this form -I don't have a reference here but it is a standard result (not completely elementary; it uses Moser's theorem and a few facts about fibrations); I am pretty sure that you can do it for any T of class C³, like the ones you are treating. Of course, such a map T can be very easily approximated by a C^∞ measure preserving diffeo: just take the integral flow of a regularized field with null divergence (convolution now works well, because it gives a fiels with null divergence). So if you find a ready reference, the approximation should follow very easily. Hope this help. --pma 22:13, 12 March 2010 (UTC)

'Extension' of a linear transformation?

Say I have a linear transformation of R², $J$, and a linear transformation of R⁴, $A=\left({\begin{array}{c|c}J&0\\\hline 0&K\end{array}}\right),$ is it fair to call A an 'extension' of J, since, considering R² a subset of R⁴, J and A bring about the same transformation on this subset? Is there a proper term for this sort of situation? Thanks, Icthyos (talk) 17:03, 10 March 2010 (UTC)

You could say that J the restriction of A to R². This is term has pretty general application, see Domain of a function#Formal definition. I doubt the specific situation you describe has a name.--RDBury (talk) 03:03, 11 March 2010 (UTC)

Thanks, that makes sense. Restriction will do nicely! Icthyos (talk) 11:48, 11 March 2010 (UTC)

Actually A an extension of J (or J a restriction of A, as said above) is fair, exactly with the justification you wrote, and is indeed common language for linear operators also. Note that you can say the same even if there's only one zero-blok in the matrix (which one: NE or SW? ;-) ). If you want to describe the diagonal decomposition, you may also say that J is a factor of A, or that A splits into the direct sum of J and K, and similar expressions. --pma 16:46, 11 March 2010 (UTC)

Well, as it stands, A is an extension of both J and K - if only the NE block is zero, it remains an extension of J, but no longer for K, and if only the SW block is zero, it is still an extension of K, but not of J, correct? I suppose I'll just pick my favourite expression and run with it, thanks! Icthyos (talk) 21:57, 11 March 2010 (UTC)

Complex Analytic Functions

Let's say I have a function f : C --> C, which is infinitely differentiable in a neighbourhood of $z_{0}$. I define a new sequence of polynomial functions ${f_{n}}$ given by

$f_{n}(z):=\sum _{k=0}^{n}{\frac {f^{(k)}(z_{0})}{k!}}(z-z_{0})^{k}.$

I want to consider the new sequence of functions ${g_{n}}$ given by $g_{n}(z):=|f(z)-f_{n}(z)|.$ I want to know how to investigate the limit of the sequence ${g_{n}(z_{0})}$ as n tends to infinity. I think this is called the pointwise convergence. Ultimately I'd like to investigate the uniform convergence. But I want some explicit methods. A nice worked example would be great. Let's say $z_{0}=0$ and $f(z)=e^{\sin(z)}$. I know that $f(z)$ is complex analytic, a.k.a. holomorphic, and so I know that it's equal to it's power series. I just use this as a familiar non-trivial example for the explicit computations. •• Fly by Night (talk) 19:53, 10 March 2010 (UTC)

it's not quite clear if you are talking of an f defined everywhere on C, but only C^∞ in a nbd of 0. If so, it is a rather strange assumption, and you should be aware that there is no relation at all between f and the f_n in the large (i.e. out of the neighborhood Ω you are talking of). Besides, it is not clear if you mean real or complex differentiability. In the first case, you can have a C^∞ function with f(x)≠0 for all x≠0, and all the f_n identically 0. If you mean infinitely differentiable in the complex sense, then complex differentiability (that is f holomorphic) is sufficient (it's equivalent) ad implies that f_n converges to f uniformly on all closed disks centered at z₀ and contained in Ω. You can't have uniform convergence on C even if f is holomorphic on C (entire) unless f itself is a polynomial (in which case the sequence of the f_n stabilizes and the uniform convergence is trivially true). For instance, the function you wrote is bounded and nonconstant, whereas a nonconstant polynomial is not. So the uniform distance is infinite. --pma 16:36, 11 March 2010 (UTC) (PS: above I meant: bounded on R, sorry)

I gave an explicit example: $z_{0}=0$ and $f(z)=e^{\sin(z)}$. I also said that I know that $f(z)$ is complex analytic, a.k.a. holomorphic, and so I know that it's equal to it's power series. I just used this as a familiar non-trivial example for the explicit computations. •• Fly by Night (talk) 21:47, 12 March 2010 (UTC)

Ok, then for this particular entire function the polynomials f_n converge to f uniformly on every disk, but do not converge uniformly on C, for the reason I wrote (it is bounded on R whereas the f_n are not. It was not clear to me what you were asking exactly. Is it all clear now? --pma 22:50, 12 March 2010 (UTC)

March 11

Quick question on integration by parts and non-integer values of the Gamma function

Hi all,

Just briefly:

Could anyone explain to me why $\Gamma ({\frac {-1}{2}})=\int _{0}^{\infty }t^{\frac {-3}{2}}\exp(-x)$ is non-singular? Or, more specifically, when I integrate by parts, I get $-2t^{\frac {-1}{2}}\exp(-x){\Big \vert }_{0}^{\infty }-2\int _{0}^{\infty }t^{\frac {-1}{2}}\exp(-x)=-2t^{\frac {-1}{2}}\exp(-x){\Big \vert }_{0}^{\infty }-2\Gamma ({\frac {1}{2}})$; but the first term, unless I'm being stupid here (most likely) goes to 0 at infinity, but not at 0. However, $\Gamma ({\frac {-1}{2}})=-2\Gamma ({\frac {1}{2}})$ is a well-documented result (indeed on the Wiki page here), which would imply $t^{\frac {-1}{2}}\exp(-x){\Big \vert }_{x=0}=0$, no? Exp(0)=1, so the behaviour is like ${\frac {1}{\sqrt {t}}}$ as $t\to 0$. What have I done wrong?

Thanks very much! Otherlobby17 (talk) 02:21, 11 March 2010 (UTC)

The integral formula only works when the real part of the argument is greater than 0. So if you try to apply it with -1/2 then it's not surprising you get inconsistent results.--RDBury (talk) 02:53, 11 March 2010 (UTC)

Precisely, the integral formula holds for poisitive arguments of the gamma function, that is exponents strictly greater than -1; more generally, it gives Γ(z) for any complex z with Re(z)>0. --pma 16:16, 11 March 2010 (UTC)

Length of a function

Say I have a function f(x). If I know the lower and upper x bounds, is there a way, using calculus, I could find the length? I know how to do it for a linear function (distance formula), but what about a really curvy and irregular function? For example, if I have a function f(x)=3x³-6x²+x, how would I find how long it is between the x coordinates 7 and 15? Thanks 69.210.140.77 (talk) 03:26, 11 March 2010 (UTC)

I think you want arc length. Algebraist 03:30, 11 March 2010 (UTC)

And the relevant formula is $\int _{a}^{b}{\sqrt {1+(f'(x))^{2}}}\,dx.$ -- Meni Rosenfeld (talk) 08:35, 11 March 2010 (UTC)

Note that the linked article is somehow misleading, for it report the (correct) definition as total variation of f:[a,b]→X, ( X,d being a metric space) and then gives a (correct) integral formula for the curve $\scriptstyle x\mapsto (x,f(x))$, where f is a function (thus not the same role as the f of the definition!). That is, it is the "length of the graph curve". The integral formula corresponding to the definition given in the link, for the length (variation) of a C¹ curve f:[a,b]→Rⁿ with respect to any norm ||.||, is:

$l(f,[a,b])=\int _{a}^{b}||f^{'}(t)||dt$,

that also holds for absolutely continuous f (the integral is then in the Lebesgue sense). I will rectify that thing (or somebody else) as soon as I'm more free.--pma 15:55, 11 March 2010 (UTC)

Why doesn't Cantor's diagonal argument apply to rationals?

Resolved

Hi, I know that it is settled that Cantor's diagonal argument does not apply to integers, and that there exists a bijective map between integers and rationals numbers (so there is the "same number" of both). If it takes aleph₀ digits to represent all real in (0,1), and aleph₀ digits to represent all rationals in (0,1) then why doesn't the diagonal argument apply to rationals? 0¹⁸ (talk) 15:57, 11 March 2010 (UTC)

because, of course, not every infinite sequence of digits represents a rational number (try to follow the analogous argument, and it will stop in the middle before reaching any contradiction) --pma 16:00, 11 March 2010 (UTC)

Because the number constructed in the diagonalization argument (as it is usually presented) is guaranteed to be a real number, but not guaranteed to be a rational number. So the number constructed is a real number not in the initial sequence, but may not be a rational number not in the original sequence. The fact that the rationals are countable shows there is no way to patch up the argument to make it output a rational number regardless of what the input sequence is. — Carl (CBM · talk) 16:01, 11 March 2010 (UTC)

Okay, I follow your logic. I appreciate the prompt responses. I guess then what I wonder why this isn't an injection from the integers (indexed by j) into the decimals of base b:

d_i = [j mod_b bⁱ - j mod_b b^{i -1}] / b^i-1

where mod_b is the base b modulo (sorry, I don't recall how to make the minus sign without a math tag). 0¹⁸ (talk) 17:22, 11 March 2010 (UTC)

BTW, I think this is just saying, why can't I write the digits backwards in base b (which I realize I can not but I don't know why). 0¹⁸ (talk) 17:32, 11 March 2010 (UTC)

How can that produce an infinite decimal? —Bkell (talk) 17:58, 11 March 2010 (UTC)

Understanding your comment is what I want to do. How many digits can an integer have? If the answer is finite, I'd think you could tell me the length. I must be missing something. 0¹⁸ (talk) 18:44, 11 March 2010 (UTC)

For any finite length, there is an integer with that many digits. But there is no integer that has infinitely many digits. --Trovatore (talk) 18:55, 11 March 2010 (UTC)

Just to see if I understand: if there were an integer with infinitely many digits, it would be infinite and infinity is not in I. Is that right? 0¹⁸ (talk) 19:15, 11 March 2010 (UTC)

Well, almost. A string of infinitely many digits doesn't really represent "infinity", though. A string of infinitely many digits is pretty much just a string of digits — in the usual context, it doesn't represent anything but itself. (In the context of p-adic numbers, things are a bit different, but they're not relevant here.) --Trovatore (talk) 19:24, 11 March 2010 (UTC)

Your argument does prove that the set of real numbers (more precisely, those between 0 and 1) that have a finite decimal expansion is a countable set, which is true. —Bkell (talk) 19:10, 11 March 2010 (UTC)

Oh, and by the way, if I'm understanding it correctly, your map is an injection, as you claim, which shows that the set of real numbers is at least countably infinite. But it is not a surjection, because there are real numbers that are not the image of any integer under your map (for example, ${\sqrt {2}}$). —Bkell (talk) 19:16, 11 March 2010 (UTC)

Thanks. This is not quite what I was originally asking but another related question I had was how do we know that pi and sqrt(2) have decimal representations (even allowing for infinite digits)? I can see how we can say that they are within delta for any delta larger than zero, but not that we really can write them down that way. 0¹⁸ (talk) 19:41, 11 March 2010 (UTC)

One first has to define "decimal expansion". But basically it's a function that takes n and returns the nth decimal digit. So your question comes down to, how do we know that if we fix n then the nth decimal digit of π exists? The answer is to put a grid on the real line starting at 0 and incrementing by a step size of 10^-n. Then π will be immediately between some pair of adjacent points in the grid, and the one on the left will tell you first nth decimal digits. — Carl (CBM · talk) 19:55, 11 March 2010 (UTC)

Okay, so what is n a member of, and how large does n have to be for the pi to not be arbitrarily close, but actually represented by by the decimal expansion? 0¹⁸ (talk) 20:11, 11 March 2010 (UTC)

n is any positive integer, and the decimal expansion has to match for any positive integer (the part before the decimal point needs to match too). -- Meni Rosenfeld (talk) 20:25, 11 March 2010 (UTC)

As for the second part of your question, there is no positive integer n such that pi is actually equal to the first n digits of the decimal expansion of pi. The decimal expansion of pi goes on forever; if you truncate it at some point, you get an approximation of pi, but not the true value. If you're familiar with infinite series, it's the same idea: Truncating a (convergent) infinite series after finitely many terms gives an approximation to the true value, but (generally) it is not exactly equal to the true value. —Bkell (talk) 20:34, 11 March 2010 (UTC)

(backdent) sorry, my question was how do we know we can represent an irrational real with an infinite string of decimals and not just get arbitrarily close. The answer I was told is that we can generate a string of digits until the cows come home but it will still just come close. I'm still confused. 0¹⁸ (talk) 02:33, 12 March 2010 (UTC)

If two real numbers are arbitrarily close to each other, then they are at distance 0, and so they are equal. You make a decimal expansion one digit at a time; the resulting infinite decimal expansion determines some real number, and that real number is arbitrarily close to the number you started with, so the infinite decimal expansion determines the same number you started with.

The most common mistake is to confuse the finite initial segments of the decimal expansion with the decimal expansion itself. To define the decimal expansion, all you have to do is define each of its initial segments. However, the overall decimal expansion of an irrational number will have an infinite number of digits. Thus each of the finite initial segments of the expansion will be unequal to the irrational number, but the entire infinite expansion gives back exactly the irrational number you started with. — Carl (CBM · talk) 02:42, 12 March 2010 (UTC)

I'd just add here that the reason it "gives that back" is that that's what it's defined to give back. The denotation of a decimal representation is, by definition, the limit of the denotations of the finite truncations.

It's not an arbitrary definition — it's the only reasonable definition that causes the decimal representations to denote real numbers. This is what those who refuse to believe that 0.999... is really exactly 1 are missing: The real numbers are the real, underlying Platonic objects we're trying to get at; the decimal representations are just names for them. If they didn't name real numbers, they wouldn't be doing the job we set them; the limit definition is the only reasonable one under which they do name real numbers. --Trovatore (talk) 02:50, 12 March 2010 (UTC)

Okay, so I asked about an injective map from the integers to the reals and was told it is not surjective because the algorithm could never generate enough digits--there simply are not enough. Then I asked you about generating the digits and you gave me a method that can generate any finite number of digits but clearly can not generate infinitely many digits (the bins are countable after all). So there must be an additional step that I am missing for how you get from an irrational number to an exact decimal representation. 0¹⁸ (talk) 04:35, 12 March 2010 (UTC)

As was mentioned, a decimal representation is a function which takes a positive integer as an argument and returns a digit (I'm focusing on the part after the dot for simplicity). Given any real x, the corresponding function is given by $f(n)=\lfloor 10^{n}x-10\lfloor {10^{n-1}x}\rfloor \rfloor$. So if you take $x=\pi$ and $n=10$ you get $f(10)=5$; for $n=1000000$ you get $f(1000000)=1$. In practice we can only expect to calculate f for finitely many digits "until the cows come home", but in theory the function f is capable of producing a digit for any n, however large. In other words we have provided a "recipe" to obtain any digit of $\pi$ we want. Thus, the decimal expansion given by f describes $\pi$ exactly. -- Meni Rosenfeld (talk) 07:52, 12 March 2010 (UTC)

So when I build a number using the integers, I can say the same thing. For any fixed (finite) n, I can build the number, regardless of the size of n. There must be something that allows the decimal expansion to go past this. 0¹⁸ (talk) 17:15, 12 March 2010 (UTC)

You seem to be confusing two different functions:

• Given a real number, produce a decimal expansion
• Given an integer, use that as a finite decimal expansion of some real number

The first bullet is talking about a function whose domain is the entire set of real numbers, and whose range consists of decimal expansions. The second is talking about a function whose domain is the integers and whose range is included in the set of real numbers. These are not the same function. It is true that every real number has a decimal expansion (which may be infinite), but this is not the same as saying that there is a way to produce all these decimal expansions with a function whose domain is just the integers.— Carl (CBM · talk) 12:59, 12 March 2010 (UTC)

Carl, I agree with everything you say. I would especially emphasize that I am confused. My question is why are these different? See my above response to Meni Rosenfeld. 0¹⁸ (talk) 17:44, 12 March 2010 (UTC)

I don't know if what I'm about to say is going to make any more sense than what has already been said, but I'm going to say it anyway. Let's consider ${\sqrt {2}}$. Do you agree that this number is irrational? If so, then you must agree that it cannot have a finite decimal representation (for if it did we could multiply this finite decimal expansion by some sufficiently large power of 10 to get an integer, and then make a fraction that equals ${\sqrt {2}}$ by writing this integer over the power of 10 that we used). So there is no single integer that can be interpreted as the decimal expansion of ${\sqrt {2}}$, because every integer has only finitely many digits. Agreed? Therefore, ${\sqrt {2}}$ is an example of a real number that is not in the range of the map you defined.

That being said, it is possible to find arbitrarily good approximations of ${\sqrt {2}}$ that are in the range of your map—take the decimal expansion of ${\sqrt {2}}$ and truncate it after some number of digits. So the range of your map is dense in the real numbers, but that doesn't mean it actually includes all the real numbers (it doesn't). The rational numbers are dense in the real numbers (given any real number x and some allowable error ε > 0, no matter how small, we can find a rational number q that approximates x to within ε—though note that a different choice of ε may require a different choice for q), but there are many real numbers that are not rational. It is the same with the range of your map. Okay so far?

What we have said so far is that not all real numbers can be represented with finite decimal expansions (for example, ${\sqrt {2}}$, and in fact even simple fractions like 1/3 if we are using base 10), but every real number can be approximated arbitrarily well by a finite decimal. More precisely, if there is an allowable error ε > 0 specified ahead of time, we can find a finite-decimal approximation q to any real number x so that |x − q| < ε.

So what hope do we have of representing a real number like 1/3 with a decimal expansion? We must resort to an infinite decimal expansion, but we need to define what such a thing means. Let's assume we're working in base 10, for simplicity (this argument easily generalizes to any base), and let's consider just the part of the decimal expansion after the decimal point. Suppose an infinite decimal expansion has digits 0.d₁d₂d₃…. The value of this decimal expansion is defined to be

$\sum _{i=1}^{\infty }{\frac {d_{i}}{10^{i}}},$

which is an infinite series that is in turn defined as the limit of the partial sums:

$\sum _{i=1}^{\infty }{\frac {d_{i}}{10^{i}}}=\lim _{n\to \infty }\sum _{i=1}^{n}{\frac {d_{i}}{10^{i}}}.$

For example, 1/3 is 0.333…. If you replace d_i by 3 in the series above, you get a geometric series, and you can calculate that the sum of the geometric series is indeed 1/3.

Now, how do we know that every infinite decimal expansion results in a real number, and that every real number can be represented with an infinite decimal expansion (i.e., as the sum of an infinite series like these)? The answer to the first question is that the real numbers are complete: every Cauchy sequence (roughly, every sequence of points that are getting closer and closer together) converges to a limit, i.e., a real number. The answer to the second question is essentially that this procedure is how the real numbers are constructed from the rationals (actually, usually the construction involves something a little subtler, like Dedekind cuts; but it is safe to think of the real numbers as precisely the set of "limit points" of Cauchy sequences of rational numbers).

So, finally, here is the distinction between Carl's two bullet points above. The first idea is producing a decimal expansion from a real number. Unless the real number happens to have a finite decimal expansion, this process will produce an infinite sequence of digits, and the decimal expansion of the real number is the entire infinite sequence—not the finite pieces at the beginning, but the entire thing. (If you don't have all of the digits of the decimal expansion of 1/3, you don't have the decimal expansion of 1/3.) The infinite sequence of digits is an exact representation of the real number, but you need all infinity of them for it to be exact. The second idea is using a given integer as the (finite) decimal representation of some real number. This works fine for every real number that can be represented with a finite decimal expansion, but it cannot give you 1/3. It can give you real numbers that are very close to 1/3, but not 1/3 itself.

One last comment: Carl mentioned earlier that "If two real numbers are arbitrarily close to each other, then they are at distance 0, and so they are equal." This is true, but it requires careful interpretation. What it means is this: If two given real numbers x₁ and x₂ are such that for every ε > 0 the inequality |x₁ − x₂| < ε is satisfied, then x₁ = x₂. The important thing to notice is that x₁ and x₂ are fixed at the beginning—we aren't changing them for different values of ε—and they are real numbers, not sequences of real numbers. For example, let's consider the number 1/3 again, and try to find a number q in the image of your map that is "arbitrarily close" to 1/3. Well, q = 0.33333 is pretty close, and it will satisfy |1/3 − 0.33333| < ε for any ε ≥ 0.000003333…; but any value of ε less than this will cause the inequality to be false. So 0.33333 is not "arbitrarily close" to 1/3 in the sense that Carl meant. You can try to get closer by adding some more 3's to the end of q, but no matter how many 3's you add (as long as you add only finitely many), there will be some value of ε that is small enough (yet still positive) to make |1/3 − q| ≥ ε. So there is no such thing as a single number "arbitrarily close" to 1/3 in the image of your map, and therefore 1/3 is not in the image of your map.

Was this helpful? —Bkell (talk) 19:05, 12 March 2010 (UTC)

Bkell, that was very helpful. I think you answered my question but I am not sure. Lets see if you agree with my understanding. Consider these statements:

(1) Using the map I proposed, one can generate a string of digits of any finite length or a string larger than any finite length.

(2) Using the algorithm proposed above for coming up with digits for reals one can produce a string of digits of any finite length or a strong larger than any finite length.

(3) Any method that can produce only finite length strings of digits will never get exactly to a map to the reals.

(4) There is some map (not mentioned above) that gives infinite length strings of digits given a real value in its domain.

(5) statement 4 is false. Instead, reals are the set of all values generated by 1 or 2 plus the limit points of all Cauchy sequences that can be generated from this set.

I am quite sure of (1) and (3). If one of these is wrong, I'd really like to know. I think you are saying 1,2,3, and 5 are true. Is that right? If not, I am not sure about 2,4, and 5 and commenting on them would be very helpful for me. 0¹⁸ (talk) 19:31, 12 March 2010 (UTC)

Here are my comments:

1. It is certainly true that your map can generate a string of digits of any finite length. The second half of your statement depends heavily on interpretation. On the one hand, it is true that if I give you any finite length (say, n), then your map can produce a string of digits that is longer than that (for example, of length n + 1). But it is not true that your map can produce a string S of digits for which it can be said that "S is longer than any finite length." If S is finite, then it has some length n, and then it is not longer than length n + 1. So saying that "S is longer than any finite length" is the same as saying that "S has infinite length." Your map cannot produce strings of infinite length, because no integer has infinitely many digits. Most mathematicians would interpret "longer than any finite length" in the second way, that is, to mean "of infinite length"; under this interpretation, the second half of your statement is false.
2. It is true that the decimal expansion of a real number may be a finite sequence of digits or an infinite sequence of digits (that is to say, a sequence longer than any finite length). (It is often useful to consider the decimal expansion of any real number to be an infinite sequence of digits, by adding infinitely many zeroes at the end of the finite sequences. This way we don't have to consider two different cases. So, when I refer to an infinite decimal expansion below, you should consider this to include finite decimal expansions as a special case.) However, in your statement you need to be careful about what you mean by "algorithm" and "produce" (or "generate"). An algorithm, by definition, must terminate in a finite amount of time, so it is impossible for an algorithm to produce infinitely much output (such as the entire infinite decimal expansion of ${\sqrt {2}}$). Now, there is an algorithm that will produce, in a finite amount of time, any desired digit in the decimal expansion of ${\sqrt {2}}$, and for most practical purposes this is just as good as having the entire decimal expansion (since, in a finite amount of time, we could only use finitely much of the infinite decimal expansion anyway). But, strictly speaking, this algorithm does not actually generate the infinite decimal expansion itself. The idea of an infinite decimal expansion is a purely conceptual thing, just like any infinite set. There is no way to "produce" or "generate" an infinite set in "the real world." But, if you accept the idea of the natural numbers, for example, as an infinite set (an infinite set that exists conceptually in its entirety, rather than a more "real-world," "algorithmic" view of them as an unending list that is "produced" by a "process"), you should also accept the idea of an infinite decimal expansion. We accept that real numbers have infinite decimal expansions on a conceptual level, and there are algorithmic ways to symbolically manipulate these decimal expansions (or at least finite portions of them), but I can't actually present you with an infinite decimal expansion itself. The existence of the infinite decimal expansion of a real number does not depend on the existence of an algorithm to produce this expansion.
3. Yes, it is true that any function (I am using the word "function" instead of your word "method" to avoid the algorithmic connotations) that produces only finite strings of digits cannot produce the decimal expansions of all real numbers, because some (in a certain sense, most) real numbers do not have a finite decimal expansion.
4. Yes, there does exist a map from the set of real numbers to the set of infinite sequences of digits, such that every real number is mapped to its decimal expansion. There are some sticky points here, though. The first annoyance is that real numbers with finite decimal expansions can be represented in two ways (for example, 0.999… is exactly equal to 1.000…), so this map is not unique without some additional restrictions. The second difficulty is that this map may not be computable—there are uncomputable numbers for which no algorithm exists to compute their decimal expansions. (But their decimal expansions still do exist—remember, the existence of the decimal expansion does not require the existence of an algorithm to produce it). The third snag, from an algorithmic point of view, is the question of how to specify an arbitrary real number as "input" to this function without simply specifying its infinite decimal expansion in the first place. So the map you refer to in statement (4) exists, but it exists only in the mathematical sense of existence; it doesn't exist in a tangible or implementable sense.
5. Well, as I understand statement (2), you can say directly that the reals are the set of numbers whose decimal expansions are produced by (2), without having to say anything about Cauchy sequences or limit points at all; but that's a rather circular statement, because I understand (2) to refer directly to the set of decimal expansions of the real numbers. So here's what I think is a better statement: The set of real numbers is the set of all rational numbers that have finite decimal expansions, plus the limit points of Cauchy sequences of such numbers. (Often the reals are thought of as limit points of rationals, but the subset of rationals having finite decimal expansions is also dense in the reals.) This does not imply that statement (4) is false.

In summary, there is a difference between saying "every real number has an infinite decimal expansion" and saying "there is an algorithm to produce the infinite decimal expansion of any real number." Perhaps this is the core of the issue? —Bkell (talk) 23:15, 12 March 2010 (UTC)

Bkell, in the last paragraph you hit the nail on the head. Thank you. 0¹⁸ (talk) 23:48, 12 March 2010 (UTC)

What is the 98th percentile in a Normal distribution with μ=100 and σ=15?

What is the 98th percentile in a Normal distribution with μ=100 and σ=15?
It has been years since I went to school, so I have almost completely forgotten that which I knew about mathematics.

If the above question does not make sense, then please tell me how I should have said it, and then answer. ;-) --89.8.104.253 (talk) 17:22, 11 March 2010 (UTC)

Do you want the value such that 98 percent fall below that or the value such that 98 percent should not be that far from the mean, regardless of if they are too high or too low? 0¹⁸ (talk) 17:24, 11 March 2010 (UTC)

I want the value such that 98 percent fall below that value. --89.8.104.253 (talk) 17:32, 11 March 2010 (UTC)

Okay, then use a normal-table and look for about 0.98 in the table. You will find that it is between two points. This will tell you how many SD from the mean this value is. In this case, it is between 2.05 and 2.06. So lets say it is at about 2.055. Then the mean is 100, and the SD is 15, so this value is at 100 + 15 * 2.055 = 130.825, which is close to the correct answer (within 0.1). If you want the (more) exactly right answer, you have to use some software that has exact tables. 0¹⁸ (talk) 17:37, 11 March 2010 (UTC)

No, your answer was precisely what I wanted! Thank you! :-) --89.8.104.253 (talk) 18:31, 11 March 2010 (UTC)

Software typically doesn't "have" exact tables, it computes answers on the fly. -- Meni Rosenfeld (talk) 21:25, 11 March 2010 (UTC)

There is a formula that gives results to within about machine epsilon for all IEEE doubles. It's half way between a table (huge file) and a calculation (say, something based on Newton's method or numerical quadrature). 0¹⁸ (talk) 02:20, 12 March 2010 (UTC)

"98 percentile" means the value which 98 percent of the observations fall. This is a typical highschool math question. See our article on standard score. You can also use one of calcultors listed in the links. --Kvasir (talk) 17:39, 11 March 2010 (UTC)

"standard score" was helpful. Thank you! :-) --89.8.104.253 (talk) 18:39, 11 March 2010 (UTC)

A follow-up question: I seem to remember that in highschool and college, we used to have a small booklet with all the formulas, distributions and tables, that we would ever need in mathematics, probability and statistics. We were required to bring the booklet at exams (so it was not a cheat sheet). Q: What is this kind of booklet called in english? And is there such a booklet available in one of the Wikimedia Foundation projects? --89.8.104.253 (talk) 18:30, 11 March 2010 (UTC)

"Mathematical handbook" comes to mind, but the items I've seen with that name (such as this) are heavier than you describe. "Formula booklet" should do fine.

Wikibooks has Collection of Mathematical Formulas ("work in progress" is an understatement) and Engineering Tables. I also found this. Don't waste ink on the numerical tables - they do it all with computers now. This Wolfram Alpha query is a fairly impressive example.

Wikipedia also has many lists, such as List of probability distributions and Lists of integrals. -- Meni Rosenfeld (talk) 20:20, 11 March 2010 (UTC)

Thank you! :-) --89.8.104.253 (talk) 20:48, 11 March 2010 (UTC)

March 12

little project

I've been studding acase related to prime numbers and how to generate them and iam very convinced it is not possible to follow acertain pattern to generate infinte set of aprime numbers, so i came up with this postulate.Lets, p0, be aprime number≥7 and c(n) are constants generated by acertain pattern or afunction where; {P0+c(n)}; is aset of prime numbers,Now,the maximum prime number, [p0+c(max)]<[P0×P0].It means that any pattern will fail to exceed[p0×p0].I did lots of experiments about this.I want to know if my project is rite and if it can be proven?Husseinshimaljasimdini (talk) 00:05, 12 March 2010 (UTC)

You're going to have to be more precise about what you mean by "a certain pattern or a function." For example, I could define a function c(n) to be

c(n) = [the (n + 5)th prime number] − 7.

Then, for example, c(0) = 11 − 7 = 4, c(1) = 13 − 7 = 6, and so on. This is a perfectly reasonable definition of a function. If I do this, and let p0 = 7, then of course p0 + c(n) is a prime number for every nonnegative integer n. —Bkell (talk) 00:59, 12 March 2010 (UTC).

MY point here is that every pattern no matter what it was; or how it defined would fail under this condition, —Preceding unsigned comment added by Husseinshimaljasimdini (talk • contribs) 10:19, 12 March 2010 (UTC) As amatter of fact to be more specific ,let the set p0+c(0)=p1,.....p0+c(n)=pn,are positve prime numbers < [P0×P0].and my example is 41+2=43+4=47+6=53+.......this pattern fails until 1601+80=41×41.Husseinshimaljasimdini (talk) 10:09, 12 March 2010 (UTC)

The problem is still that you haven't clarified what you consider to be a "pattern". Your example is apparently Euler's trinomial n² - n + 41 which gives distinct primes for n = 1 to 40: 41, 43, 47, 53, ..., 1601. It's mentioned at Formula for primes#Prime formulas and polynomial functions. Most primes from 41 to 1601 are not included in the sequence so it seems you don't require consecutive primes. A sequence with larger jumps can easily end at a larger prime. 41 + 420n is prime for n = 0 to 5: 41, 461, 881, 1301, 1721, 2141. For an example where the number of primes is as large as the initial prime, 5 + 6n is prime for n = 0 to 4: 5, 11, 17, 23, 29. These were simple linear functions. If we allow arbitrary functions then there is no limit as Bkell showed. PrimeHunter (talk) 13:37, 12 March 2010 (UTC).

O!i see. i like these answers i appreciate you. I have also noticed something i`d like to include here,can we say or state that prime numbers can be devided generally into tow types, the first one is the prime numbers that satisfy the following way;for istance; 43.17=731, now if we reverse this number we will get 137 which is aprime;or 17.2=34,which leades to 43 and so on.The second type is the primes that stay primes when we reverse the digits such as 113,101..and so on.Husseinshimaljasimdini (talk) 14:58, 12 March 2010 (UTC)

That is one way to pick a subset of the primes. See List of prime numbers for many others. Palindromic primes are those which read the same backwards (usually with base-10 implied) such as your 101. Emirps (also called reversible primes) are primes which become another prime when read backwards (again, base-10 usually implied) such as your 113. I (Jens Kruse Andersen) happen to be the discoverer of the largest known emirps [3] and several other computational emirp results.[4][5][6][7][8][9] Primality of a number is not a base-dependent property. Studying special properties of the decimal digits of numbers is usually considered recreational mathematics. PrimeHunter (talk) 16:25, 12 March 2010 (UTC)

How many types of numbers are there?

I've known about complex numbers for a while, but I recently discovered the article on dual numbers. I was wondering, are there any other types of numbers? If so, what are they and how many are there? Furthermore, is it possible to have a "complex dual" number (i.e. a number in the form a + bi + cε?--203.22.23.9 (talk) 02:57, 12 March 2010 (UTC)

Well, the thing is, there's no agreed general definition of what it means to be a "number". So you have things like Cayley numbers that may not really strike you as all that "numerical". That makes this really a language question — how many classes of mathematical objects are called such-and-such numbers?

Ones you might be interested in looking at: hyperreal numbers, surreal numbers, transfinite ordinal numbers and cardinal numbers. --Trovatore (talk) 03:09, 12 March 2010 (UTC)

(edit conflict)The concept of number is a moving target and so it makes no sense to count how many types of number there are. See natural number, positive number, rational number, real number, prime number, square number. Lots of mathematical constructions generalize numbers, and some of them are even called numbers. Elements of groups, rings, and fields generalize numbers but are often not called numbers. Bo Jacoby (talk) 03:20, 12 March 2010 (UTC).

Another thought - Is there a set of numbers which can be used to solve equations such as |x| = -1 , in the same sense that the complex numbers are used to solve equations such as x² = -1?--203.22.23.9 (talk) 03:33, 12 March 2010 (UTC)

I can't make any sense of it, but that doesn't prove you shouldn't think about it. (Actually I did think about it, when I was a kid -- a point closer to a given point than that point is! But absolute values are not really analogous to quadratics -- they don't have algebraic properties that obviously generalize. Again, this is not a proof. Feel free to work on it; let us know if you get anywhere!) --Trovatore (talk) 03:41, 12 March 2010 (UTC)

Try Split-complex number. Staecker (talk) 12:23, 12 March 2010 (UTC)

(First off, no one has mentioned the quaternions yet.) If you're going to allow |x| = −1, then your "absolute value" operation is no longer a norm. Norms have very useful properties—you can think of a norm as a generalized absolute value that can be applied to objects other than standard numbers. But I suppose there could be such a thing as a "signed norm" that is allowed to be negative; there are already concepts like signed measures and complex measures, which extend the idea of a measure (essentially a length, area, or volume) by allowing it to be negative or complex, respectively. —Bkell (talk) 03:55, 12 March 2010 (UTC)

Something similar is considering spaces that have "norms" that aren't positive-definite, like Minkowski space. In that context some vectors will satisfy |x|² = -1. But that also breaks from the standard definition of what a norm is. Rckrone (talk) 06:13, 12 March 2010 (UTC)

If |x|² = -1 then |x| = i (or -i), what value for x leads to that? --203.22.23.9 (talk)

Minkowski space describes events in space and time. We can choose some coordinates and describe a point in the space as w = (t,x,y,z) where t describes the point in time and x, y, z describe the position in space. The Minkowski norm of this point is |w|² = -t² + x² + y² + z² (compare this to the usual Euclidean norm which would be |w|² = t² + x² + y² + z²). So for example the point w = (1,0,0,0) has |w|² = -1. The Minkowski norm is useful in special relativity because the value remains unchanged when we change our reference frame with a Lorentz transformation (similar to how the Euclidean norm doesn't change when we rotate the coordinate system), even though it doesn't meet all the criteria of the definition of a norm since |w|² can be negative. Rckrone (talk) 07:16, 12 March 2010 (UTC)

Absolute value of a quaternion

Resolved

What's the absolute value of a quaternion?
Absolute value of a complex number:
$|a+bi|={\sqrt {a^{2}+b^{2}}}$
Absolute value of a quaternion:
$|a+bi+cj+dk|=??$ --220.253.247.165 (talk) 06:55, 12 March 2010 (UTC)

${\sqrt {a^{2}+b^{2}+c^{2}+d^{2}}}$ In general the abs of a hypercomplex number is the sqrt of the number multiplied by its conjugate. Zunaid 07:16, 12 March 2010 (UTC)

Domain of a function

f(x)=√(x-4)÷x-2 then find the domain of a given function —Preceding unsigned comment added by 121.52.145.148 (talk) 07:26, 12 March 2010 (UTC)

I think it's {do, your, own, homework}. --Trovatore (talk) 08:17, 12 March 2010 (UTC)

Also, please be more clear when defining your function. It is difficult to determine whether the function f is defined by $f(x)={\sqrt {\frac {x-4}{x-2}}}$, $f(x)={\frac {\sqrt {x-4}}{x-2}}$, $f(x)={\sqrt {\frac {x-4}{x}}}-2$, or $f(x)={\frac {\sqrt {x-4}}{x}}-2$. If you use brackets to clarify the order of operation, we may be able to guide you towards the solution. We cannot give you the solution, of course, as this will not help you to solve any other problems of a similar nature. PST 08:36, 12 March 2010 (UTC)

How about assuming there is already a proper number of parens in appropriate places? May be there is no need to complicate things, if they are not obviously simplified too much. --CiaPan (talk) 12:12, 12 March 2010 (UTC)

Statistical significance when discussing a population

I understand the concept of statistical significance when dealing with sampling. E.g., if you draw a sample of 1000 American household's incomes for two years you can get a probability that the population income differs between the two years, and hence determine whether any statistically significant change in household income can be found.

But often social scientists use the concept of statistical significance when discussing a population, not a sample drawn from a population. For example, it can be claimed that there is a significant downward trend in GDP growth, or that there is a significant correlation between health and GDP per capita. But how can you talk of significance when you already know the population? I understand that you can use R2 to determine the fit of a regression, but is statistical significance a meaningful concept in this case? What's the intuition? Jacob Lundberg (talk) 17:51, 12 March 2010 (UTC)

Actually, you don't know the population. For example, I may have data on GDP from the beginning of time to the present, but that is not the whole population because it does not include the future. Wikiant (talk) 17:57, 12 March 2010 (UTC)

Jacob Lundberg, you have to postulate that the population is a draw from an infinite set of unrealized populations. Deming has a paper on this topic from his time at the Census. 0¹⁸ (talk) 18:34, 12 March 2010 (UTC)

Here is a link to the article at jstor [10]. 0¹⁸ (talk) 19:06, 12 March 2010 (UTC)

feedback rating

Feedback ratings tend to show how many buyers were satisfied with their purchase and not only do buyers rely on this statistic but auction houses rely on it too in deciding whether an item was defective that was returned. While 4,999 buyers that were satisfied with a food product versus only one that died will still have a 99.98% (4,999/5,000) rating while the risk of death may be ignored. What type of feedback rating calculation regardless of complexity can overcome this dilemma and give a much lower feedback rating to reflect the significance of a defective product so that potential buyers will be appropriately warned? 71.100.11.118 (talk) 22:59, 12 March 2010 (UTC)

Feedback systems tend to be game-able toys (and it's in the venue's interest for them to be game-able) but the article trust metric and its references might give you some ideas. It's silly in general to try to compress complex multi-dimensional data into a single numerical score, although that doesn't stop many from trying. 66.127.52.47 (talk) 23:54, 12 March 2010 (UTC)

Any rating system is going to have a lot of subjectivity from the raters. But let's ignore some of that; there is still a lot of subjectivity from whoever designed the rating system. From a purely mathematical point of view, you're probably going to want some kind of weighted average, but what weights do you assign? If "satisfied" is given a positive weight of 10, say, what weight should "death" be given? If it's given a weight of, say, −1000, then mathematically you're saying that 100 satisfied customers balance out one dead customer. Is that the correct value to put on death? I don't know; it doesn't seem right. On the other hand, if death is given a weight of −∞, then (under typical rules for arithmetic with ∞) your weighted average will be −∞ even if you have six billion satisfied customers and just one customer who died in a freak accident. That doesn't seem right either. So, if you're going to use a weighted-average approach, you're going to have to place numerical values on the relative worths of satisfaction and death, which is certainly going to involve a subjective judgement. —Bkell (talk) 00:03, 13 March 2010 (UTC)

Feedback ratings aren't usually used to assess safety. There are usually government standards to satisfy for that sort of thing, eg. the kitemark system in the UK. You are right that giving one number like that doesn't account for different levels of satisfaction. Often the survey will let people choose from "Very poor, poor, satisfactory, good, very good" or similar and then the number you see is everyone that said "satisfactory" or better and they don't give a breakdown. This kind of loss of information is inevitable if you want to reduce things to a single number. --Tango (talk) 00:22, 13 March 2010 (UTC)

I've discovered an interesting problem with weighted feedback. Many people will give the maximum rating or the minimum rating, but very few choose any value in between. I suspect that people who would otherwise give a low rating actually give the minimum rating, to maximize their impact on the overall rating, and others do something similar at the high end. To deal with this, I would look at previous ratings by each responder, and "normalize" them. Do they always give a 0 or 100 ? Then maybe those ratings should actually count as a 30 or 70. StuRat (talk) 02:30, 13 March 2010 (UTC)

Actually lots of people give responses in the middle. It is (on a scale of 1-9) numbers around 3 and 7 that tend to be ignored. --Tango (talk) 02:58, 13 March 2010 (UTC)

You might be interested in the idea of 'quality adjusted life year' by NICE which has to deal with the problem of putting a price onto people's lives. If one in 5000 people dies who would otherwise live a healthy life your talking about a cost of more than a million pounds, so you'd only need a gain of a few hundred pounds for everyone else to easily offset it according to that. However in fact a death rate of 1 in 5000 would get people in a tizzy if you announced a new product a free HD television and cable for life and said there was a downside that unfortunately one in 5000 of those given the free HD would die of an electric shock. Dmcq (talk) 08:41, 13 March 2010 (UTC)

March 13

Maclaurin expansion

Resolved

Question 6 on the homework is in two parts

(a) Obtain the first four terms of the Maclarin expansion for ln(1+x)
(b) Use the expansion from part (a) to show that:
$\ln(c+x)=\ln(c)+{x \over c}-{x^{2} \over 2c^{2}}+{x^{3} \over 3c^{3}}-{x^{4} \over 4c^{4}}+...$

Part (a) is a piece of cake: $\ln(1+x)=x-{x^{2} \over 2}+{x^{3} \over 3}-{x^{4} \over 4}+...$
Part (b) would also be a piece of cake if I worked it out the typical way (i.e. finding all the derivatives, powers and factorials etc.) and indeed I have already solved it that way, however the question specifically says Use the expansion from part (a) to solve it. It's possibly a mistake in the question. This is what I have so far:
$\ln(c+x)=\ln(1+(x+c-1))=x+c-1-{(x+c-1)^{2} \over 2}+{(x+c-1)^{3} \over 3}-{(x+c-1)^{4} \over 4}+...$

I could probably expand the brackets to continue but you just know I'm never going to rearrange it to the correct form (after all, where the is that ln(c) going to come from?!)--220.253.247.165 (talk) 00:45, 13 March 2010 (UTC)

$c+x=c(1+x/c)$ —Bkell (talk) 01:12, 13 March 2010 (UTC)

That's only helpful if the OP is allowed to use the rule for logs of products. That's always the problem with this type of question - guessing that you are and aren't allowed to take as given. --Tango (talk) 01:24, 13 March 2010 (UTC)

Thanks Bkell - That doesn't help the slightest bit :P Seriously, how is that useful?--220.253.247.165 (talk) 02:52, 13 March 2010 (UTC)

log(c+x)=log(c(1+x/c))=log(c)+log(1+x/c), you can then use your expression for log(1+x) and change all the x's to x/c's. Job done. --Tango (talk) 02:55, 13 March 2010 (UTC)

THANK YOU!!! Problem solved :)--220.253.247.165 (talk) 04:08, 13 March 2010 (UTC)

Statistics - Conditioning

My stats course recently covered conditioning for finding an expectation, probability, etc., but I really don't understand it. When do we use it and why and how is it useful? Please recommend any good resources. I don't understand my prof or textbook. —Preceding unsigned comment added by 142.58.129.94 (talk) 02:36, 13 March 2010 (UTC)

Have you ever watched a televised poker tournament? They commonly have little cameras mounted around the edge of the table or somewhere so that the television audience can peek at the players' hands. Often there is a chart on the side of the screen listing the players and the percentage probabilities that each one will hold the best hand at the end. For example, the chart might say Jones 42%, Smith 9%, Brown 31%, Fernandez 18%, indicating that Jones has a 42% probability of holding the best hand at the end, etc. These probabilities are calculated based on the knowledge of the cards the players hold and (in a variant like Texas hold-em) the community cards that have been revealed. (The probabilities ignore such aspects of the game as bluffing; a player may hold the best hand but still lose if he decides to fold.)

In the hypothetical situation above with four players, before any cards are dealt or revealed, each player has a 25% probability of holding the best hand at the end. But as more knowledge is obtained about the cards, these probabilities change—if a player is dealt three aces, her probability of holding the best hand at the end will probably rise, whereas a player who is dealt a two, a six, and a nine of different suits will probably see his probability fall.

This is the idea of conditional probability. The probability that a person in a four-player poker game will be dealt the best hand, given no other information, is 25%. But the probability that she will be dealt the best hand, given that she has been dealt three aces, is quite a bit higher than 25%.

Of course, an example like poker is an overly complicated example. It's simpler to think about, say, the roll of two fair six-sided dice. Let's say one of the dice is white and the other is red, so that we can tell them apart. There are 36 possible rolls of the two dice; each of these rolls is equally likely. Let's shake the dice in an opaque plastic cup and then slam the cup upside-down on the table—we've rolled the dice, but we can't see what we rolled yet. What is the probability we rolled at least an 11? Well, there are three possible ways that could happen: white 5, red 6; white 6, red 5; or white 6, red 6. So the probability we rolled at least an 11 is 3/36, or 1/12. Now let's say we reveal only the white die, and we see it's a 5. Given that information, what is the probability that we rolled at least an 11? Well, there are only six possibilities for what we rolled (we know the white die was a 5, but the red die could be anything from 1 to 6). Out of those possibilities, two of them give us at least 11. So, given the fact that the white die was a 5, the probability that we rolled at least 11 is 2/6, or 1/3, which is significantly higher than 1/12. Our partial knowledge about the situation changed the probability—that's conditional probability. Now suppose instead that when we revealed the white die we saw that it was a 3. Then the conditional probability that we rolled at least 11, given the fact that the white die is a 3, is 0; do you see why? —Bkell (talk) 04:16, 13 March 2010 (UTC)

Stats -- algebraic manipulation of V(X) and E(X), etc.

I can do simple things but can't solve them when it's really complicated. Could anyone please recommend any resources that have tips or practice problems for doing manipulations with V(X), E(X), etc.? —Preceding unsigned comment added by 142.58.129.94 (talk) 02:38, 13 March 2010 (UTC)

Symmetric linear solve

Given $AE+EA=B$ with all three matrices symmetric, is there an explicit form for E? (If it and A aren't known to commute, of course.) --Tardis (talk) 03:55, 13 March 2010 (UTC)

You want to invert the linear endomorphism $\scriptstyle T_{A}X:=AX+XA$ defined on the linear space of all symmetric square matrices of order $n$ . One case where $\scriptstyle T_{A}$ is certainly invertible (both from matrices to matrices and from symmetric matrices to symmetric matrices), and you have an inversion formula is when A is a perturbation of the identity. Note that $\scriptstyle T_{A}^{n}X=\sum _{j=0}^{n}{n \choose j}A^{j}XA^{n-j}$ (because $T_{A}$ is the sum of a left multiplication operator and a right multiplication operator, and they commute so you can apply the binomial expansion formula). From $\scriptstyle \|T_{A}^{n}\|$ you can compute the spectral radius of $T_{A},$ as $r(T_{A})=2r(A)$. Therefore if $A=I-H$ and $r(H)<1/2$ you have $T_{A}=I-T_{H}$, with $r(H)<1$ and $(T_{A})^{-1}$ has the geometric series expansion. This gives E as a double series, $\scriptstyle E=\sum {n \choose j}H^{j}XH^{n-j},$ summing over all pairs (j,n) (with 0≤j≤n) --pma 10:21, 13 March 2010 (UTC)

simple question.

Is there away to find out how many prime numbers before agiven one?like there are 4 prime numbers before 11.Husseinshimaljasimdini (talk) 06:45, 13 March 2010 (UTC)

Just counting them is one way. Prime-counting function#Algorithms for evaluating π(x) mentions some more sophisticated methods. Algebraist 06:59, 13 March 2010 (UTC)

is there any proof that current compression couldn't be much better?

Is there anything to show that in the future, people won't have computing power coupled with algorithms that will allow them to collapse something that is a 9 megabyte compressed file today, and actually just have it as a 50 kilobyte compressed file?

I don't mean compression of white noise, of random data - obviously impossible (pigeonhole principle). I mean the kinds of data people actually use and compress.

Thank you. 82.113.121.167 (talk) 11:06, 13 March 2010 (UTC)