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September 2

find scalar values a, b and c

Given $\mathbf {A} ={\begin{bmatrix}2&-5\\3&1\\\end{bmatrix}}$, find scalar values a, b, c (NOT ALL ZERO) for which $a\mathbf {I} +b\mathbf {A} +c\mathbf {A} ^{2}=\mathbf {O}$
I get these 4 simultaneous equations in 3 unknowns:
$a+2b-11c=0$
$-5b-15c=0$
$3b+9c=0$
$a+b-14c=0$
What do I do now? Wikinv (talk) 01:41, 2 September 2010 (UTC)

Check your initial work. I believe one of the four equations is wrong. Next, take a look at Simultaneous equations for various techniques for solving these. One approach is to rearrange one of the equations to isolate one of the unknowns (i.e., $a=...$) and then substitute the results into the next equation. After the second iteration, you should have a value for one of the unknowns. Repeat the process until you've solved them all. -- Tom N (tcncv) talk/contrib 02:09, 2 September 2010 (UTC)

After attempting to solve it myself, I found that there's a gotcha in those equations. The problem has a solution, but that solution may not be unique. I assume this is homework, so I will not give you too obvious a hint. Write back if you need additional help. -- Tom N (tcncv) talk/contrib 02:33, 2 September 2010 (UTC)

Fixed the equation. I was aware that there are multiple solutions, indeed that is why I don't know how to solve it.--Wikinv (talk) 07:05, 2 September 2010 (UTC)

Although, having fixed the equation, the solution becomes quite easy by inspection, but is there an analytic way of solving it?--Wikinv (talk) 07:09, 2 September 2010 (UTC)

Just realised that the second and third equations are in fact exactly the same. Furthermore, it is evident that there is an infinite number of solutions (that's why it was so easy to solve by inspection!)--Wikinv (talk) 07:14, 2 September 2010 (UTC)

Observe that if (a,b,c) is a nonzero solution to the equation $\scriptstyle a\mathbf {I} +b\mathbf {A} +c\mathbf {A} ^{2}=\mathbf {O}$ where $\scriptstyle \mathbf {A}$ is any square matrix, and k is a nonzero number, then (ka,kb,kc) is a nonzero solution too. Bo Jacoby (talk) 05:15, 2 September 2010 (UTC).

In general, the equation $a\mathbf {I} +b\mathbf {A} +c\mathbf {A} ^{2}=\mathbf {O}$ has a one-dimensional space of solutions (a, b, c) for any 2x2 matrix $\mathbf {A}$. To see this, let

$\mathbf {A} ={\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\\end{bmatrix}}{\text{ and }}\mathbf {A'} ={\begin{bmatrix}a_{22}&-a_{12}\\-a_{21}&a_{11}\\\end{bmatrix}}$

Then

$\mathbf {A'A} =\det(\mathbf {A} )\mathbf {I}$

but

$\mathbf {A'} =\mathrm {tr} (\mathbf {A} )\mathbf {I} -\mathbf {A}$

so

$(\mathrm {tr} (\mathbf {A} )\mathbf {I} -\mathbf {A} )\mathbf {A} =\det(\mathbf {A} )\mathbf {I}$

$\Rightarrow \det(\mathbf {A} )\mathbf {I} -\mathrm {tr} (\mathbf {A} )\mathbf {A} +\mathbf {A} ^{2}=\mathbf {0}$

so

$(a,b,c)=(\det(\mathbf {A} ),-\mathrm {tr} (\mathbf {A} ),1)$

or (as Bo says) any multiple of this. Gandalf61 (talk) 08:54, 2 September 2010 (UTC)

Alternatively, just take the coefficients of the characteristic polynomial. —Preceding unsigned comment added by 203.97.79.114 (talk) 09:11, 2 September 2010 (UTC)

See Gaussian elimination. Properly used, it works also for systems of linear equations that have no solutions, multiple solutions, and\or redundant equations. I don't know if our articles cover the details involved, but if not, any introductory linear algebra book will. -- Meni Rosenfeld (talk) 09:17, 2 September 2010 (UTC)

Numerical analysis notation

I was reviewing my numerical analysis textbook, and stumbled across a notation I didn't remember, concerning propagation of errors. For background, the question is, assuming that $f:\mathbb {R} \rightarrow \mathbb {R}$ is a differentiable function, and that ${\overline {x}}$ is an approximation of $x\in \mathbb {R}$, what is an upper bound on the uncertainty $|f({\overline {x}})-f(x)|$? Now, the mean value theorem directly gives that there is a $\xi$ between $x$ and ${\overline {x}}$ such that $|f({\overline {x}})-f(x)|=|f'(\xi )||{\overline {x}}-x|$. We know that $|{\overline {x}}-x|\leq M$, but we do not know an upper bound on $\textstyle |f'(\xi )|$.

Now, what the book does here is that it replaces the unknown $\textstyle f'(\xi )$ with the known $f'({\overline {x}})$. The idea, presumably, is that since $|{\overline {x}}-\xi |$ is small, then so is $|f'({\overline {x}})-f'(\xi )|$. The book tacitly makes this assumption, mind you, without requiring a single fact about $\textstyle f'$—not even that it be continuous! Of course, it is not then necessarily true that $|f({\overline {x}})-f(x)|\leq |f'({\overline {x}})|M$, and the book does not claim so, but replaces the $\leq$ sign with a $\lesssim$ sign, telling the reader to read it as "less than or approximately equal to"—I would say it seems reasonable to interpret this as "unequality that almost certainly holds unless the function is too irregular". It feels kind of sloppy though. Is it just my book or is this common notation? Is this a case of "since we're dealing with applications, we don't have to care about badly-behaving functions"? 85.226.206.114 (talk) 07:57, 2 September 2010 (UTC)

Trying to find an estimate for $\textstyle f'(\xi )$ gets you back to the original problem but with derivative of the function. So this estimate will be in terms of $\textstyle f''$ and if you try it estimate this it will be in terms of $\textstyle f'''$ etc. But each time you're multiplying the estimate by $|{\overline {x}}-x|$ which is assumed to be small. So I think the interpretation of $\lesssim$ is "Less than equal up to a quantity that is small compared to the other quantities." You can construct examples where the derivatives get very large or infinite so that the error term is still large even with the $|{\overline {x}}-x|$ factors, but these aren't encountered often in practice. The $\lesssim$ notation is a bit vague perhaps but if you want more precise notation try a different numerical analysis book, there is no shortage of them.--RDBury (talk) 15:05, 2 September 2010 (UTC)

All right, thanks! 85.226.205.5 (talk) 08:42, 3 September 2010 (UTC)

Drawing 45-45-90 triangles on spheres

Can you draw a 45-45-90 triangle on a sphere so all three sides have whole-number lengths? 20.137.18.50 (talk) 16:31, 2 September 2010 (UTC)

You cannot draw a 45–45–90 triangle on a sphere at all. The sum of angles of any triangle on a sphere is strictly more than 180°, see spherical geometry.—Emil J. 16:36, 2 September 2010 (UTC)

You can, however, draw an isosceles right triangle on a sphere so that all three sides have whole-number lengths. Will that do? -- ToE^T 16:53, 5 September 2010 (UTC)

Doesn't this require the radius of the sphere to be an integer multiple of 2/π?—Emil J. 12:27, 6 September 2010 (UTC)

Only for 90-90-90 equilateral right spherical triangles. For isosceles right spherical triangles with other leg : hypotenuse rations, other restrictions on the radius will hold. -- 124.157.234.26 (talk) 06:39, 7 September 2010 (UTC)

Also of interest here might be The Right Right Triangle on the Sphere, a 2008 paper from The College Mathematics Journal in which:

The question explored here is whether having a 90°- angle is the most fruitful analogue in spherical geometry to right triangles in Euclidean geometry. A strong case is made for the property of a triangle with one angle equal to the sum of the other two.

I don't have access to this paper beyond its abstract, and am not aware of a related Wikipedia article. -- 124.157.234.26 (talk) 03:53, 7 September 2010 (UTC)

September 3

Standard deviation

Hi all! In physics we're doing a bit of stats and I noticed in the standard deviation formula they divide by N-1 rather than just N. I asked my teacher and he said he didn't get it either, and look it up on Wikipedia or something like that, so here I am. I tried looking at your articles Standard_deviation and Bessel's correction, but that didn't really help because I don't have a university-level stats background :/ Can someone who does explain why you divide by N-1, in simpler terms? I'm OK with (and even expect you to) dumb the concept down a little --cc —Preceding unsigned comment added by 76.229.208.208 (talk) 01:58, 3 September 2010 (UTC)

As I understand it, the N-1 come in because you are trying to estimate the actual standard deviation based on sample data. If you put N in the denominator it turns out that the estimate will, on average, be too low. So a correction factor is built into the formula so that the estimate will average to the actual value if the experiment is repeated many times. When the correction factor is added it works out the same as using N-1 in the denominator instead of N. It has been noted here before though, if your sample is small enough that it actually makes a difference then your sample size is too small.--RDBury (talk) 03:46, 3 September 2010 (UTC)

See the Wikipedia article on unbiased estimator, which has the explanation you're looking for. --173.49.14.153 (talk) 04:20, 3 September 2010 (UTC)

If you knew the population (actual) mean rather than estimating it and used that to get the squared differences then N would be correct. However using the sample (estimated) mean makes the sum of the squared differences slightly smaller. In fact the sum of the squared differences from the population mean is equal to the sum of the squares of the differences from the sample mean plus N times the square of the difference between the population mean and the sample mean. This itself gives you an estimate of the probable difference between the population and sample mean so the workings out in the article is just using this to get an estimate of the sum of squared differences from the population mean. A finickety point is that it is only the expression without the square root that is unbiased, the estimated standard deviation from taking the square root is biased but I would worry even less about that than using N instead of N-1 in the denominator. Dmcq (talk) 07:57, 3 September 2010 (UTC)

Maybe it won't hurt to mention also that unbiasedness may be slightly over-rated, at least by non-statisticians. See my paper on this: "An Illuminating Counterexample", American Mathematical Monthly, Vol. 110, No. 3 (March, 2003), pp. 234–238. Michael Hardy (talk) 18:47, 4 September 2010 (UTC)

Random variables

Hello mathematicians! Can you please help me solve this. It's not homework, it's actually work work. Say $S$ is the amount of money I make per "event" and $E$ is the number of events per year. Let's also say that $S$ has a lognormal distribution and $E$ is a poisson distribution (the parameters for $S$ can be estimated from some data and let's assume that the parameter for $E$ is known).

A) Then the total money I make from these events in one year is $P=SE$. Is there an analytic distribution function for $P$ ?

B) Will the following monte-carlo methods work to determine a distribution for $P$ :

1) sample a random value from $E$, say $e$, then sample $e$ values of $S$ and add them up - repeat this many times; or

2) sample a random value from $E$, say $e$, and sample a random value of $S$, say $s$, and then use $es$ - and repeat this many times.

What is the difference between these two methods? What other possible numerical methods can I use to determine $P$ ? Thanks very much. --Mudupie (talk) 17:32, 3 September 2010 (UTC)

I'll assume that the events don't all make the same amount of money, but rather that each makes an independent contribution drawn from some distribution. Then $P\neq SE$. In fact there isn't even an S, there are iid random variables $S_{1},\ S_{2},\ \cdots ,S_{E}$, and $P=\sum _{i}S_{i}\;\!$. So it's clear that you can't sample the distribution of P with method 2 - you'll get a different distribution which has a much higher variance. You can use method 1, though.

You may know that if X and Y are iid then $\mathbb {V} [X+Y]=\mathbb {V} [X]+\mathbb {V} [Y]=2\mathbb {V} [X]$ while $\mathbb {V} [X+X]=\mathbb {V} [2X]=4\mathbb {V} [X]$. If it seems that E being random makes a difference, think what happens when $\lambda$ is large - then E is roughly constant.

If finding the expectation and variance of the distribution suffices, you have $\mathbb {E} [P]=\mathbb {E} [E]\mathbb {E} [S]$, and if I'm not mistaken $\mathbb {V} [P]=\mathbb {E} [E^{2}]\mathbb {E} [S]+\mathbb {E} [E]\mathbb {V} [S]-\mathbb {E} [E]^{2}\mathbb {E} [S]^{2}$. This holds no matter what are the distributions of E and S, as long as everything is independent. -- Meni Rosenfeld (talk) 18:56, 4 September 2010 (UTC)

Thanks very much Meni! That was very useful information. I have one follow up question for now. I'm trying to understand how to derive the expectation of P. I guess the following equation holds but I don't understand why: $\mathbb {E} [S_{1}+...+S_{E}]=\mathbb {E} [S_{1}+...+S_{\lambda }]$, where λ is just the expectation of E. I "get" that it makes sense but I don't know the actual theoretic reason. Can you please explain? --Mudupie (talk) 23:09, 4 September 2010 (UTC)

$\mathbb {E} [S_{1}+...+S_{\lambda }]$ only makes sense when λ is an integer, so it's not useful to talk about it. What I did is to write $P_{k}=\sum _{i=1}^{k}S_{i}$ and $P=P_{E}=\sum _{k=0}^{\infty }I(E=k)P_{k}$. Then finding $\mathbb {E} [P]$ is just some algebraic manipulations. -- Meni Rosenfeld (talk) 11:20, 5 September 2010 (UTC)

Thanks again mate! I managed to arrive at the expression for E[P] using your approach. I'll try to do the variance one as well and come back here if I get stuck. --Mudupie (talk) 09:41, 6 September 2010 (UTC)

Formula images

In every maths page on wikipedia I notice the formulae are images not text. How do you create these? On Mac? Thanks for any replies.86.147.12.111 (talk) 18:05, 3 September 2010 (UTC)

See Help:Displaying a formula. —Bkell (talk) 18:27, 3 September 2010 (UTC)

Thank you86.147.12.111 (talk) 19:42, 3 September 2010 (UTC)

Also, when you see a page with such formulas, if you click on "edit", you'll see how they are created. Michael Hardy (talk) 18:51, 4 September 2010 (UTC)

Homogeneous polynomials

The symmetric degree 4 homogeneous polynomial in two variables: x⁴ + x³y + x²y² + xy³ + y⁴ can be written (x⁵−y⁵)(x−y)⁻¹ for x≠y. What is the analogous expression for the symmetric degree 4 homogeneous polynomial in 3 variables: x⁴ + x³y + x³z + x²y² + x²yz + x²z² + xy³ + xy²z + xyz² + xz³ + y⁴ + y³z + y²z² + yz³ + z⁴ ? Bo Jacoby (talk) 22:28, 3 September 2010 (UTC).

First, just to be consistent with the terminology, these are called the complete homogeneous symmetric polynomials. The expression you're looking for follows from the properties of Schur polynomials.

$s_{4}(x,y,z)={\frac {1}{\Delta }}\;\det \left[{\begin{matrix}x^{6}&y^{6}&z^{6}\\x&y&z\\1&1&1\end{matrix}}\right]$

which turns out to be the complete symmetric polynomial. Here Δ is the product of the differences (x−y)(x−z)(y−z).--RDBury (talk) 04:33, 4 September 2010 (UTC)

Thank you very much! Bo Jacoby (talk) 06:10, 4 September 2010 (UTC).

No problem but please be civil. —Preceding unsigned comment added by 114.72.252.111 (talk • contribs)

It is plainly obvious from the edit history that User:Bo Jacoby did not make the uncivil comment you are referring to, per [1]. I have removed the IP's offending comment. --Kinu ^t/_c 05:19, 5 September 2010 (UTC)

September 4

Need a better parametric test for a sine-wave distribution.

Lunar cycle theorists postulate cyclic fish activity associated with major and minor lunar periods. This creates a theoretical sine-wave centered around the average catch/activity rate. Usually this is illustrated by sine-wave-like imagery with peaks at the top and the negative wave-half flipped as data for both majors and minor and in-between hours can only be positive integer values.

I have a large data set of fish catches, over 10,000 angling hours, with the majority of catches 0 or 1 per hour but a few hours ranging up to a few catches of 9 or 10 per hour. I believe the data to be parametric. Using one way AOV (Statistix 9) I have a P small enough to declare my overall data collection highly significant by scientific standards. However, breakdowns of the data and subsets by type of water fished and fish size show several dependent variables effect results and produce less than significant data sets using AOV. However, these sub-sets still show peaks near the major and minor hours as predicted by the lunar cycle postulates.

Is/are there statistical method/methods that focus on this tendency to have peaks in the right places rather than just seeking sufficient differences between means and SDs? Please provide enough detail so that I might apply your input.

Also please comment if you feel my use of oneway AOV is inappropriate for this type of data. I'm long out of touch with academic sources of statistical guidance.

19:24, 4 September 2010 (UTC)~ —Preceding unsigned comment added by RManns (talk • contribs)

The observed nonnegative integer number of fish catched within an hour, n, has a poisson distribution with some nonnegative real parameter λ. So λ has a gamma distribution with mean value n+1 and variance n+1. The first transformation of data is thus to add 1 to all the observed number of catches, for you are interested in λ rather than in n. The sine-like wave should be nonnegative too, so you should find A,B,ω,φ to minimize

Σ((n(t)+1)−Ae^{B sin(ωt+φ)})²/(n(t)+1)

where this weighted sum of squares is over all the observations, and t is the time coordinate. Bo Jacoby (talk) 07:12, 5 September 2010 (UTC).

Mr. Jacoby,

thanks for your response, but I'm too long away from school and very stupid when it comes to statistics. I just rely on a Statistix Program to do the math. If I make a modified data set by adding 1 to all, what is the appropriate test called? Also, to do this data conversion is the one added to the zero values or only the positive integers? R Manns RManns (talk) 03:04, 7 September 2010 (UTC)

You are welcome. Yes, the one is also added to the zero values. If you catch no fish within an hour, (n=0), then you cannot conclude that the average number of fish catched per hour, (λ), is zero. You might have been unlucky. But you do get some information regarding λ, that λ≈1±1. This should not change the name of the tests you want to perform. I do not know the Statistix Program. Bo Jacoby (talk) 07:07, 7 September 2010 (UTC).

Unclear as to why you are doing this analysis. Do you want to advise people whether they should fish during a particular phase of the moon? If so, your 'bottom line' could be a prediction of the relative success they should experience if they select a particular phase of the moon. (E.g. 'fish during the full moon and you will catch 50% more'). The p-value is not very strong guidance for this kind of prediction. It's also peculiar that, when you attempt to factor out other variables you discover that your p-value becomes less impressive. If you are successful in removing other factors from the problem, then any relationship to the phase of the moon (if there really is one) ought to show up more clearly in your data. It is possible that your removal of the other factors is hurting your value of N, so the data is less informative. EdJohnston (talk) 00:37, 8 September 2010 (UTC)

When to use Kendall / Spearman correlations instead of Pearson's?

Statistical software package I use offers, in addition to Pearson product-moment correlation coefficient, also Spearman's rank correlation coefficient and Kendall tau rank correlation coefficient. I am trying to find an explanation of why one would want (or when one can...) use S/K instead of P. I found a bunch of descriptions, little different from our wiki pages - they go into math; but I don't care about the theory as much as for application (when to use which).

I am guessing that there are times you want to use P, and times you want to use S/K, and if you use incorrect one you'll get a misleading result (It seems that for P, both variables should be normally distributed (how can I test if this is true?). S/K do not have this assumption (doesn't it make them better by default...?)). How to determine which one do you want to use?

In particular, I am looking at some data that seems not significant under P, but more so under S/K. What does that mean? Is the correlation in the data I am looking at statistically significant or not? --_{Piotr Konieczny aka Prokonsul Piotrus| talk} 21:06, 4 September 2010 (UTC)

Well, in short, Pearson correlation coefficient tells us if there is a linear relationship between two variables and the other two tell us if there is a monotonous relationship between those variables. Thus low Pearson correlation coefficient with high Spearman or Kendal tau correlation coefficient indicate that there might be a monotonous relationship between two variables, but it is not linear. For example, it could be that one variable is proportional to the the cube of the other. You might wish to look at a scatter graph to find out more. --Martynas Patasius (talk) 22:10, 4 September 2010 (UTC)

Just FYI, the word in English is monotonic :-). Of course it could also be monotonous.... Trovatore (talk) 09:02, 8 September 2010 (UTC)

Spearman's rank correlation looks for linear relationships between the ranks. If I recall correctly, Spearman's and Pearson's are equal to each other in cases in which the data consist of distinct ranks, but when you have ranks to work with rather than the raw numbers getting ranked, then there's a simpler formula for computing them. Michael Hardy (talk) 00:02, 5 September 2010 (UTC)

Ditto to above, and also: You certainly want normality if you are using Pearson's correlation coefficient; you can refer to the article normality test for how to test the normality of data. If normality is suspect, go with Kendall's or Spearman's coefficients. Nm420 (talk) 20:17, 6 September 2010 (UTC)

Thanks again 18:54, 7 September 2010 (UTC) —Preceding unsigned comment added by RManns (talk • contribs)

What is the test of statistical significance for nominal variables?

Let's say I want to see if a nominal variable (country, or certain groupings of ones) and ratio one (geographical or population size, for example) are statistically significant. What test should I use? Correlation is out, because it is not useful for categorical variables, right...? --_{Piotr Konieczny aka Prokonsul Piotrus| talk} 21:39, 4 September 2010 (UTC)

You could do a multiple comparisons ANOVA. Michael Hardy (talk) 00:03, 5 September 2010 (UTC)

September 5

Question on polynomials

Hey guys, how are ya all? ANyway, I'm stuck on a question and your help would be appreciated.

The question is this: Let k be a natural number and let r be a real number such that |r| < 1. Prove (by induction on k) that for any polynomial P of degree k there's a polynomial Q of degree k s.t. Q(n+1)r^(n+1) - Q(n)r^n = P(n)r^n.

The hint is this: Consider differences of succesive terms for n^k r^n, and use the inductive hypothesis.

OK. So that's the question and hint. I've been trying hard at this question since I've woken up but to no avail. So I've been trying for 5 hours. I really would appreciate an answer. Like for example I've tried it when P(x) = x^2, r = 1/2 and I've found that Q(x) = -2x^2 - 4x - 6 and it works. The reason I did this is 'cause it helps me to sum the series n^2 2^(-n). But I'm really stumped on this one. Help??? Thanks guys ... I've worked out about 7 pages of rough work so please don't say I'm lazy and I want my homework done for me. This isn't homework just independent study and it's for my own benefit but I'd really like a decent hint or an answer please. —Preceding unsigned comment added by 114.72.228.12 (talk) 05:02, 5 September 2010 (UTC)

Also you don't have to use the hint if you don't want to. Thanks guys ...

I've just run through the argument but only have a mess (though I think it's a correct mess). The point you might be missing (which is basically the hint restated) is that you don't want to evaluate what Q is if P = x^k; it suffices to show you can deal with x^k (in much the same way as the grounding case of the induction) and then the left overs are of less degree so can be absorbed into the rest of P and dealt with by Inductive Hypothesis. Also I don't see that you need |r|<1 (though you're in trouble if r=1). Hope that helps. 95.150.22.63 (talk) 14:17, 5 September 2010 (UTC)

Following from 95.150.22.63's bit about induction, I think it is pretty clear that we can rewrite as rQ(n+1)r^n - Q(n)r^n = x^k r^n. We ignore the r = 0 case, since it's rather meaningless, and so we divide through by r^n. Q is a polynomial of degree k, so let us express it as

$Q(n)=\sum _{i=0}^{k}a_{i}n^{i}$, so our equation is $r\sum _{i=0}^{k}a_{i}(n+1)^{i}-\sum _{i=0}^{k}a_{i}n^{i}=x^{k}$

Then expand the first term by the binomial theorem, and then equating coefficients of n^i for i = 0, ..., k, you get k+1 linear equations to solve for the k+1 coefficients a_0, ..., a_k. To calculate, you can start from a_k and work back, a_k = 1/(r-1) always, for example (a_i can be solved for directly once you know the values of a_{i+1}, ..., a_k). In any case, you have k+1 linear equations for k+1 unknowns, and now you want to be sure that for all |r|<1, these have a unique solution. Invrnc (talk) 14:31, 5 September 2010 (UTC)

Jensen's inequality?

Is this a case of Jensen's inequality, or some other inequality?

$(\mathbb {E} (|a-c|^{2}))^{1/2}\leq (\mathbb {E} (|a-b|^{2}))^{1/2}+(\mathbb {E} (|b-c|^{2}))^{1/2}$

—Preceding unsigned comment added by 130.102.158.15 (talk • contribs) 08:14, 5 September 2010 (UTC)

That's Minkowski's inequality, with $p=2$. —Bkell (talk) 15:43, 5 September 2010 (UTC)

Thanks! —Preceding unsigned comment added by 130.102.158.15 (talk) 22:32, 5 September 2010 (UTC)

Limit Question

I know it's possible to reason out that $\lim _{(x,y)\rightarrow (0,0)}\exp {\tfrac {-1}{x^{2}+y^{2}}}=0$, but is there any way to do it algebraically? Thanks. --Basho: banana tree (talk) 20:42, 5 September 2010 (UTC)

For every ε>0, you can choose an appropriate δ>0 (you will be able to express this value in terms of ε) such that for every point (x,y) within a distance of δ of (0,0), $|\exp {\tfrac {-1}{x^{2}+y^{2}}}|<\epsilon$. This is formally how you demonstrate a limit. Rckrone (talk) 22:34, 5 September 2010 (UTC)

(edit conflict) You are interested in how the function behaves as (x,y) tends towards (0,0). But there are many ways in which a point (x,y) can tend towards (0,0). Let's assume that (x,y) follows a line towards (0,0). In other words x = r⋅cos(θ) and y = r⋅sin(θ), where θ is a fixed parameter and r is the variable along the line. We consider:

$\lim _{(x,y)\to (0,0)}e^{\left({\frac {-1}{x^{2}+y^{2}}}\right)}=\lim _{r\to 0}e^{\left({\frac {-1}{(r\cos \theta )^{2}+(r\sin \theta )^{2}}}\right)}=\lim _{r\to 0}e^{\left({-1/r^{2}}\right)}=0.$

Notice that the penultimate expression is independent of θ and does not depend upon the sign of r since 1/(+r)² = 1/(−r)². — Fly by Night (talk) 23:04, 5 September 2010 (UTC)

Note that this is not a proof. There are functions which converge along every line towards the origin, and yet do not converge at the origin.--203.97.79.114 (talk) 09:53, 6 September 2010 (UTC)

That's the whole point of a limit. The function may not be defined at the origin, but you calculate the limit of the function as you tend towards the origin. In this example the function is indeterminate at the origin; but the limit exists and is well defined. — Fly by Night (talk) 12:59, 6 September 2010 (UTC)

I didn't say such functions were undefined; I said they did not converge. Consider the function ${\tfrac {xy}{x+y^{3}}}$. The limit as $(x,y)\rightarrow (0,0)$ along any line is $0$. The limit as $(x,y)\rightarrow (0,0)$ along the curve $x=y^{2}$ is $1$. The limit in this case does not converge, yet if you only considered approaching along a line, you would think it does. —Preceding unsigned comment added by 203.97.79.114 (talk) 13:18, 6 September 2010 (UTC)

I don't think that example works. Surely if we have $x=y^{2}$ then ${\tfrac {xy}{x+y^{3}}}={\tfrac {y}{1+y}}$ and then $\lim _{y\rightarrow 0}{\tfrac {y}{1+y}}=0$ ? (However, Meni's example below illustrates the point).Gandalf61 (talk) 13:36, 6 September 2010 (UTC)

Woops. ${\tfrac {xy^{2}}{x^{2}+y^{4}}}$ works if you want something elementary. Or Meni's, as you say. --203.97.79.114 (talk) 13:48, 6 September 2010 (UTC)

[ec] 203's point is that only looking at straight lines is insufficient; you need it to work for every curve. The canonical example is

$f(x,y)={\begin{cases}1&x>0,y=x^{2}\\0&{\textrm {otherwise}}\end{cases}}$

This function has no limit at the origin; but along any line, it converges to 0 at the origin.

Of course, in the OP's case, this objection can be considered a nitpick, since you have demonstrated that the function depends only on r, so finding the limit for a single curve suffices. -- Meni Rosenfeld (talk) 13:27, 6 September 2010 (UTC)

Strictly speaking, you cannot prove this statment (or any other statement involving limits) algebraically because it is a statement of analysis, not of algebra. The result depends on the topology that you use - with the usual topology on R² the statement is true, but with the discrete topology (for example) it is false. The choice of topology does not affect the algebraic properties of R² - therefore the result cannot be derived from algebraic properties alone. Gandalf61 (talk) 08:34, 6 September 2010 (UTC)

True, but we often teach an algebraic approach to limits that relies on analysis only for the fact that certain basic functions (addition, division where defined, etc) are continuous. Presumably that's what was being asked for.--203.97.79.114 (talk) 13:29, 6 September 2010 (UTC)

So what do we think the OP is looking for ? What would a proof of the OP's statement look like under this algebraic approach to limits ? Presumably it would not involve any δs, εs, open intervals or neighbourhoods ? Gandalf61 (talk) 13:51, 6 September 2010 (UTC)

I think it would be something like $\lim _{(x,y)\rightarrow (0,0)}\exp {\tfrac {-1}{x^{2}+y^{2}}}=\exp \lim _{(x,y)\rightarrow (0,0)}{\tfrac {-1}{x^{2}+y^{2}}}=\exp(-\infty )=0$. -- Meni Rosenfeld (talk) 14:55, 6 September 2010 (UTC)

dL

In physics you express uncertainty as dL or delta-L. WHat does this have to do with derivatives? 76.229.214.25 (talk)` —Preceding undated comment added 23:00, 5 September 2010 (UTC).

because if y=f(x), then dy=f '(x)dx is an approximate computation of the uncertainty of y based on the uncertainty of x. Bo Jacoby (talk) 08:13, 6 September 2010 (UTC).

September 6

Proving e converges

How would you go about proving that $\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}$ converges? Or, say, that $2<\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}<3$, then that $2.70<\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}<2.72$ and so on, zooming in as far as you like? It Is Me Here ^{t / c} 17:54, 6 September 2010 (UTC)

You can use the binomial theorem, $\left(1+{\frac {1}{n}}\right)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}{\frac {1}{n^{k}}}$. It's easy to show that the kth term is less than $1/k!$ so the expression is less than $\sum _{k=0}^{\infty }{\frac {1}{k!}}$. In fact the kth term converges to $1/k!$, and it should be possible to bound the differences to make the entire expression converges.

There should be a result that states conditions under which $\lim _{n\to \infty }\sum _{k=0}^{\infty }a_{n,k}=\sum _{k=0}^{\infty }\lim _{n\to \infty }a_{n,k}$, but I can't remember or find the details. -- Meni Rosenfeld (talk) 18:17, 6 September 2010 (UTC)

The dominated convergence theorem gives convergence here. There's probably a separate name for dominated convergence applied to discrete measure spaces, but if so I don't know it. Algebraist 18:28, 6 September 2010 (UTC)

Also, you can show that the sequence $(1+1/n)^{n}$ is (strictly) increasing: use the inequality of arithmetic and geometric means with choice of $n+1$ numbers $x_{1}=x_{2}=\dots =x_{n}:=1+1/n$ and $\textstyle x_{n+1}:=1.$ To show that it is bounded, note that the argument above also holds more in general for the sequence $(1+c/n)^{n}$ for any real $c$, showing it is increasing as soon as $n+c>0.$ Since for n>|c|

$\left(1+{\frac {c}{n}}\right)^{n}\left(1-{\frac {c}{n}}\right)^{n}=\left(1-{\frac {c^{2}}{n^{2}}}\right)^{n}\leq 1$

and both sequences on the LHS are positive and increasing, they are bounded. So this actually gives the convergence of $(1+c/n)^{n}$ for all real $c$ together with the bounds: in particular for $c=1$ you should obtain, for all $n$

$\left(1+{\frac {1}{n}}\right)^{n}<e<\left(1+{\frac {1}{n}}\right)^{n+1}.$

Note however that the argument via the dominated convergence theorem for series shown by Algebraist also holds when $c$ is more generally a complex number, or a matrix, or even an element in a Banach algebra. Also, it can be shown that the convergence of the series is much faster. --pma 02:46, 7 September 2010 (UTC)

The Space of Oriented Geodesics

Consider the a 2-sphere is 3-space, centred at the origin. The geodesics are given by the great circles. For each great circle we get a unique plane, namely the plane containing the great circle (and passing through the origin). Given a plane passing through the origin we get a unique line passing through the origin, namely the line perpendicular to the given plane passing through the origin. So each great circle gives a line through the origin. The same procedure works in reverse too: a line through the origin gives a plane through the origin (the plane passing through the origin that is perpendicular to the line). A plane through the origin gives a great circle (the circle generated by intersecting the sphere with the plane). It follows that the space of unoriented geodesics is exactly RP², i.e. the real projective plane. Now consider the space of oriented geodesics of the sphere. This is exactly the sphere itself. Given a point p of the sphere, we get an unoriented geodesic by intersecting the sphere with the plane passing through the origin that is perpendicular to the chord joining the origin to p. Let us denote this intersection by C. Assume that g : S¹ → S² gives a smooth parametrisation of C, with parameter θ. We can give an orientation to C by insisting that

${\text{Volume}}\left\{\left[\,p,\,g(\theta ),\,g'(\theta )\,\right]\right\}>0\ {\text{ for all }}\ \theta .$

Clearly, if we replace p by −p then we must reverse the orientation of C, i.e. replace θ with –θ, in order to maintain the inequality above. This means that there is a one-to-one correspondence between the oriented geodesics of the 2-sphere and the 2-sphere itself. My question is this:

• What other manifolds M have the property that there is a one-to-one correspondence between the oriented geodesics of M and M itself?

— Fly by Night (talk) 18:14, 6 September 2010 (UTC)

If two sets have the same cardinality, then there is a one-to-one correspondence between them. The cardinality of the set of real numbers is c. The cardinality of the set of points in a manifold is also c. The cardinality of the set of oriented geodesics in a manifold is also c. So all manifolds M have the property that there is a one-to-one correspondence between the oriented geodesics of M and M itself. You did not require the correspondance to be continuous. Bo Jacoby (talk) 23:37, 6 September 2010 (UTC).

Could you please show how you prove that the space of oriented geodesics on M has the same cardinality as M? — Fly by Night (talk) 00:57, 7 September 2010 (UTC)

The set of all points in an infinite plane has the same cardinality as the set of all points in a finite line segment, namely, c.[2] What is the one-to-one correspondence between an infinite plane and a finite line segment? — Fly by Night (talk) 00:10, 7 September 2010 (UTC)

Well, finding a specific one is a little complicated. Not too bad, but not something you can draw on a napkin like the one between the rationals and the natural numbers.

The easiest approach to giving an explicit bijection is probably to find injections both ways, and then apply the method used to prove the Schröder–Bernstein theorem. Finding an injection from the line segment to the plane is trivial; you can get the reverse injection by something such as interleaving the decimal representations of the x- and y- coordinates of a point, after projecting the plane into a finite square. --Trovatore (talk) 00:39, 7 September 2010 (UTC)

Thanks Trovatore. Do you think this cardinality argument shows that there is a bijective correspondence between the space of oriented geodesics on M and M itself? I mean, how can Bo say that the space of oriented geodesics on M has the same cardinality as M? Geodesics are differential invariants, whereas cardinality is much weaker. Making small bumps on the sphere may stop is having any closed geodesics at all. — Fly by Night (talk) 00:46, 7 September 2010 (UTC)

N.B. I meant to say that there is a bijective correspondence between the 2-sphere and its space of oriented geodesics. And my question should have been: What other manifolds M have the property that there is a bijective correspondence between the oriented geodesics of M and M itself? — Fly by Night (talk) 00:10, 7 September 2010 (UTC)

"how can Bo say that the space of oriented geodesics on M has the same cardinality as M?" An oriented geodesic is defined by the two points (P, P+dP). The cardinality of pairs of points is c when the cardinality of points is c. Bo Jacoby (talk) 06:25, 7 September 2010 (UTC).

Why not try a stronger condition that just bijection, such as Homeomorphism, or Diffeomorphism both of which are exhibited by the sphere? This leads to a more general question of when we can give the space of all geodesics a manifold structure?--Salix (talk): 09:33, 7 September 2010 (UTC)

It seems like there is a bit of existing work on the space of geodesics [3] has a survey article. There are several results about the topological structure of the space, including conditions for it to be a manifold, but falls short of answering Fly by Night's question.--Salix (talk): 10:07, 7 September 2010 (UTC)

That's really interesting. Thanks Salix. This is more like what I had in mind, i.e. does the space of oriented geodesics have a manifold structure and is it the same as the original manifold. I know I didn't say as much but I think you hit the nail on the head. Thanks again! — Fly by Night (talk) 16:41, 7 September 2010 (UTC)

big-O?

I've got some calc homework and I need to solve a problem using the big-O. Unfortunately I forgot how (it's been a long labor day weekend ;) and your article is kind of written more for the person looking something up that they'd learnt years ago than me, just learning it now. I'm not going to give the actual problem because it would be kind of pointless if I didn't figure out how to do it myself, but can someone show me how in simple terms (using any problem)? Thanks 76.229.163.32 (talk) 19:59, 6 September 2010 (UTC)

If you won't give an explicit problem then you must be seeking a general solution. In that case; you'll find everything you need in the formal definition section. — Fly by Night (talk) 20:43, 6 September 2010 (UTC)

September 7

What does this expression count?

Hi. I'm working on a combinatorial problem, and I've encountered a lot of expressions with form similar to the following: ${\tbinom {9}{4}}-{\tbinom {8}{3}}+{\tbinom {7}{2}}-{\tbinom {6}{1}}+{\tbinom {5}{0}}$. Can anyone think of a situation where this counts something? If I can think about a real-world representation of it, I might be able to count that same quantity with a different technique, and thus establish some kind of identity.

Thanks in advance for any ideas. -GTBacchus^(talk) 02:06, 7 September 2010 (UTC)

You can often simplify that kind of expression mechanically with WZ theory, if that's of any interest. See the A=B book (online) linked in the references to that article. For your concrete example I'd put a few terms into OEIS and see what comes out. 67.122.211.178 (talk) 06:11, 7 September 2010 (UTC)

For example, you have this. -- Meni Rosenfeld (talk) 09:18, 7 September 2010 (UTC)

It looks like some form of the Inclusion-exclusion principle to me. 198.161.238.19 (talk) 18:02, 7 September 2010 (UTC)

What about the number of subsets S of {1,...,9} with card(S)=5 and such that max(S) has the same parity of 9?--pma 20:05, 7 September 2010 (UTC)

Good grief, not another homework question

If f is continuous on [0,2], and f(0) = f(2), prove that there is a real number x ∈ [1,2] such that f(x) = f(x-1).

Intuitively I know this should be true, but that hasn't helped me out much. The only approach I can think of is to show that you can find infinitely many pairs of numbers such that f(a) = f(b), and that a-b will range from 2 to 0, but this hasn't helped. Can anyone help me? 74.15.136.172 (talk) 23:02, 7 September 2010 (UTC)

Consider the function g on [1,2] with g(x)=f(x)-f(x-1). Algebraist 23:10, 7 September 2010 (UTC)

Got it, thanks! 74.15.136.172 (talk) 00:17, 8 September 2010 (UTC)

Dixon's Method of Factorization

Perhaps I'm being slow here, but under 'Method' on Dixon's factorization method, why is it the case that N = gcd(a − b, N) × gcd(a + b, N)? Surely N -divides- gcd(a − b, N) × gcd(a + b, N), since gcd(A,BC) divides gcd(A,B)gcd(A,C), but why must the equality hold here? Is it something to do with the method of computing a and b? Or perhaps I'm missing something?

Cheers, 92.40.243.116 (talk) 23:24, 7 September 2010 (UTC)

September 8

Σ and ʃ

How are summation and integration are related? That is to say, if i can only perform a sum of a function from a to b how can I find the area, and if I can only integrate a function along [a, b] how can I find the sum of it (or it's sequence)? 24.92.78.167 (talk) 02:34, 8 September 2010 (UTC)

Two relations are the definition of the integral via Riemann integral#Riemann sums and the Integral test for convergence of series. -- 58.147.53.113 (talk) 05:21, 8 September 2010 (UTC)

Atmospheric carbon scenario difference projections again

I'm moving this back from the archives in hopes that someone else can give it a shot balancing it against File:Extreme-weather-cost.gif: What is the optimal amount of money to spend on climate change mitigation which will minimize total financial losses?

There is a calc problem that I don't understand at Wikipedia:Reference_desk/Archives/Science/2010 August 27#What is the wind-water-solar climate change mitigation scenario atmospheric carbon projection? [copied below] Why Other (talk) 22:22, 27 August 2010 (UTC)

From what I can tell, since there seems to be a lot of jargon there that would be better understood by a atmospheric scientist than a mathematician, the model being used is that the atmosphere is a giant tank of some fluid where some impurity is being added at a given rate while at the same time it's being removed at a rate proportional to its concentration. The tank is assumed to be well mixed, meaning you don't have to worry about the concentration not being the same in different parts of the tank. This is a fairly standard problem in ODE's and the solution is a straightforward use of separation of variables. I hope that helps with the mathematical aspect of the model at least.--68.40.56.142 (talk) 15:04, 28 August 2010 (UTC)

Thanks! Why Other (talk) 03:24, 30 August 2010 (UTC)

In Jacobson, M.Z. (2009) "Review of solutions to global warming, air pollution, and energy security" Energy and Environmental Science 2:148-73 doi 10.1039/b809990c and Jacobson, M.Z. and Delucchi, M.A. (November 2009) "A Plan to Power 100 Percent of the Planet with Renewables" (originally published as "A Path to Sustainable Energy by 2030") Scientific American 301(5):58-65 what is the projected atmospheric carbon over time for their preferred wind-water-solar program?

What year do they start subtracting carbon and when do they reach 350 ppm?

I have asked also here, but I have been having better luck here at WP:RDS. Why Other (talk) 02:04, 27 August 2010 (UTC)

Per Dr. Jacobson, this is related to Eqn. 3 in http://www.stanford.edu/group/efmh/fossil/ClimRespUpdJGR%201.pdf

"[calculate] the time-dependent change in CO2 mixing ratio from a given anthropogenic emission rate, [and with that] the time-dependent difference in mixing ratio resulting from two different emission levels by subtracting results from the equation solved twice. Note that chi in the equation is the anthropogenic portion of the mixing ratio (this is explained in the text) and units of E need to be converted to mixing ratio. The conversion is given in the paper."

This almost might be ready for the math reference desk. Why Other (talk) 22:18, 27 August 2010 (UTC)

Maybe something like the scenario proposed by James Hansen's Alternative Scenario paper? ~AH1^(TCU) 18:38, 28 August 2010 (UTC)

Interesting, but the assumptions in that paper don't involve adjusting the CO2 emissions rate, which is what I am trying to do.

This is from a response to a pointer to this question I put on the Math Reference Desk:

"... This is a fairly standard problem in ODE's and the solution is a straightforward use of separation of variables....--68.40.56.142 (talk) 15:04, 28 August 2010 (UTC)"

I wish I had more experience with ODEs. Why Other (talk) 03:27, 30 August 2010 (UTC)

The separation of variables article describes it some, and a little more explanation is here. Basically the ODE is something like

${dy \over dx}=ay+b$

Separating variables gives you

${dy \over ay+b}=dx$

Integrating both sides,

${1 \over a}{\log(ay+b)}+C_{1}=x+C_{2}$

Exponentiating both sides, rearranging terms, and renaming some constants,

$y={ke^{ax}-b \over a}$

where $k=e^{a(C_{2}-C_{1})}$ or something like that. It's possible that I messed up that calculation somewhere (it's late and I haven't done this in a while) but it's a basic ODE technique.

That said, that linear model sounds oversimplistic for something as complicated as the earth/atmosphere system. See for example clathrate gun hypothesis. 67.119.3.248 (talk) 08:50, 30 August 2010 (UTC)

Can anyone take it further than that? I know compared to File:Extreme-weather-cost.gif this is "merely" an optimization problem in probability distributions, but I'm sure it has few enough degrees of freedom to have a singular optimum solution.

If you want to consider a less tractable, but still important problem: how much is knowing the answer worth? Why Other (talk) 05:40, 8 September 2010 (UTC)