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July 30

group isomorphism

I'm a bit uncertain about the definition for group isomorphism. The article says "a bijective group homomorphism". I think this nullifies statements (and their proofs) such as "a group homomorphism is an isomorphism iff it is bijective", and therefore should be changed to something that requires the existence of a sibling homomorphism from the codomain group to the domain one. Unfortunately, i've forgotten the details, but at least that's what i remember from my Algebra 101 class of 9 years ago. -- Jokes Free4Me (talk) 12:28, 30 July 2009 (UTC)

You are right; according to the general definition in category theory, a homomorphism h:G→G' is an isomorphism if and only if there is a homomorphism k:G'→G such that hk=1_G' and kh=1_G. But the equivalence is quite immediate and the various definitions live together... --pma (talk) 12:57, 30 July 2009 (UTC)

Thanks. I was actually thinking about a general isomorphism, and in that base being bijective is not enough. The inverse must also be a morphism. -- Jokes Free4Me (talk) 13:40, 30 July 2009 (UTC)

Indeed, bijectivity is a useful criterion for isomorphisms in special categories; in general, morphisms need not even be functions, in which case bijectivity for them is just meaningless. --pma (talk) 13:58, 30 July 2009 (UTC)

I've heard people distinguish between an "isomorphism into" another group and an "isomorphism onto" another group. With the former, the "other group" is not isomorphic to the domain, but of course it has a subgroup that is. Michael Hardy (talk) 16:44, 30 July 2009 (UTC)

Wouldn't an "isomorphism into" a group just be a monomorphism (i.e. an injective homomorphism)? Maelin (Talk | Contribs) 04:16, 2 August 2009 (UTC)

irreducibility

Eisenstein's criterion mentions that "Then f(x) is irreducible over F[x], where F is the field of fractions of D. When f(x) is primitive [...], it is also irreducible over D[x]." I believe Irred(F[x]) implies Irred(D[x]), since if something is reducible over D[x] it must clearly be immediately reducible over F[x], using the same decomposition. -- Jokes Free4Me (talk) 13:40, 30 July 2009 (UTC)

You are wrong. The polynomial 2x is reducible over the integers but irreducible over the rationals. Algebraist 13:49, 30 July 2009 (UTC)

Oh, right. But then, irreducible polynomial is a bit incomplete and misleading: it only defines irreducibility over fields, and then looks at the ring Z. Maybe such an example as 2X could be added to it, after expanding the definition? Thanks. -- Jokes Free4Me (talk) 14:16, 30 July 2009 (UTC)

The more general definition is given further down the page, at irreducible polynomial#Generalization. That is a rather silly order for things to be in. Algebraist 14:19, 30 July 2009 (UTC)

PS Speaking of the incompleteness of that article, am i right in assuming that 6=2*3 is reducible in Z[X]? -- Jokes Free4Me (talk) 14:16, 30 July 2009 (UTC)

Yes. Algebraist 14:19, 30 July 2009 (UTC)

July 31

Bay window measurement

I'm measuring my backyard and I've ran into a problem with a bay window area. There are three segments, each being 3' 5". The middle segment sticks out from the main exterior wall by 2'. What formula could I use to find the total width of the bay, taking angles into consideration, and then find what degree the angles should be at (for the left and right segments)? 3' 8" * 3 obviously wouldn't work. Here's an image that might help: http://img354.imageshack.us/img354/5492/confusion.jpg

I'd go out and measure it myself, but it's too dark, and it's also a torrential rainfall out there right now.

--69.154.119.41 (talk) 00:15, 31 July 2009 (UTC)

Your best bet is to measure it when the weather gets better, since the abstract math may leave out some stuff like thickness of frames. Also, your diagram is confusing--are the outer segments supposed to be angled towards the wall? If you imagine a right triangle whose hypotenuse is the 3'8" piece of glass, and the side adjacent to the angle at the end runs along the wall, and the opposite side of the angle is the 2' distance from the wall to the middle segment, then the ratio "opposite over hypotenuse" (i.e. 2' / 3'8") is the sine of the angle. Since you know the sine, you can find the angle with your calculator's arcsine function (the arcsine "undoes" the sine function, like squaring undoes a square root). If you want know how much wall space the tilted segment covers, the ratio "adjacent over hypotenuse" is the cosine of the angle. There are a bunch of mathematical relationships ("trig identities") between the sine and the cosine, but conceptually it's simplest in this case to just use a calculator, figure out the sine, use the arcsine button to find the angle, then use the cosine button to get the ratio. The length along the wall will be 3'8"*(cos x) where x is the angle. —Preceding unsigned comment added by 67.117.147.249 (talk) 01:19, 31 July 2009 (UTC)

The floor plan can be viewed as a rectangle and two right triangles: the rectangle is 2' by 3'5". Each right triangle has a 3'5" hypotenuse and a 2' leg. The question is how long the other leg is. So use the Pythagorean theorem:

${\text{other leg}}={\sqrt {(41{\text{ inches}})^{2}-(24{\text{ inches}})^{2}}}\cong 33+1/4{\text{ inches}}.\,$

Thus the whole width of the bay is about

$(33+1/4)+41+(33+1/4)=105+1/2{\text{ inches}}.\,$

Michael Hardy (talk) 02:47, 31 July 2009 (UTC)

Wait: Your diagram says 3'8", but your text says 3'5". For 3'8", change the numbers in what I did above. But use the same method. The bottom line is a bigger number. Michael Hardy (talk) 02:50, 31 July 2009 (UTC)

Picking four true properties of population mean - identified three, but stuck on the last.

Statistics, population mean - question that I'm 75% done with, but need one more:

A study reports that college students work, on average, between 4.63 and 12.63 hours a week, with confidence coefficient .95. Which of the following statements are correct?

MARK ALL THAT ARE TRUE. There are four correct answers. (MARK ALL)

• A. The interval was produced by a technique that captures mu 95% of the time.
• B. 95% of all college students work between between 4.63 and 12.63 hours a week.
• C. 95% of all samples will have x-bar between between 4.63 and 12.63.
• D. The probability that mu is between 4.63 and 12.63 is .95.
• E. 95% of samples will produce intervals that contain mu.
• F. The probability that mu is included in a 95% CI is .95.
• G. We are 95% confident that the population mean time that college students work is between 4.63 and 12.63 hours a week.

I've determined that A, F, and G are TRUE. I'm also sure that D is FALSE (each sample will likely have a different mean and end points).

So that means that one of B, C, and E is true, and the remaining two are false. Any insights on which of B, C, and E is the true property? 70.169.186.78 (talk) 03:02, 31 July 2009 (UTC)

F is equivalent to D. You say D is false, therefore F is not true either. (Igny (talk) 03:37, 31 July 2009 (UTC))

I'm not so sure. See -

• A. The interval was produced by a technique that captures mu 95% of the time. TRUE - That is what "95% confidence" means.
• B. 95% of all college students work between between 4.63 and 12.63 hours a week.
• C. 95% of all samples will have x-bar between between 4.63 and 12.63.
• D. The probability that mu is between 4.63 and 12.63 is .95. FALSE - The population mean will be between the end points of the interval for 95% of all samples. But since each sample will have a different mean, the end points of those intervals will also change. This interval either includes m, or it doesn't. Once I take a sample and compute x-bar, there is no more probability involved.
• E. 95% of samples will produce intervals that contain mu.
• F. The probability that mu is included in a 95% CI is .95. TRUE - As long as we talk about the probability of AN interval, not THIS interval.
• G. We are 95% confident that the population mean time that college students work is between 4.63 and 12.63 hours a week. TRUE: We are trying to estimate the population mean.

-- 70.169.186.78 (talk) 03:50, 31 July 2009 (UTC)

You convinced me as long as you are talking about a random confidence interval, and not random (fixed) mu. (Igny (talk) 04:24, 31 July 2009 (UTC))

A study reports that college students work, on average, between 4.63 and 12.63 hours a week, with confidence coefficient .95. Which of the following statements are correct? The number of hours a particular student works on a particular week depends on which student and which week. You can average over students (giving an average value for each week), or over weeks (giving an average value for each student), or over both (giving one average). It is unclear which of those is the case here. Mentioning confidence indicate that it is unknown population averages rather than known sample averages. A sample of size N has a mean value μ and a standard deviation σ. These sample numbers, N, μ and σ, are known to the authors of the study, but not revealed to you. From these numbers the 95% confidence interval limits for the population average, 4.63 hours and 12.63 hours, are computed, and you can try and compute backwards. The population average is estimated to be close to the sample average, which must have been (4.63+12.63)/2 = 8.63 hours. The sample average is supposed to be normally distributed with a standard deviation equal to the population standard deviation divided by the square root of N. (See Normal distribution). So the difference between the sample average and the population average is normally distributed with mean zero and standard deviation σ/√N. The total width of the 95% confidence interval, 12.63−4.63 = 8 hours, is about four times this standard deviation, so we have 4σ/√N = 8 hours. Assuming that no student works less that 0 hours or more than 40 hours a week, the sample standard deviation is less than 40·√(p(1−p)) where p = 0.22 (such that μ = p·40+(1−p)·0 = 8.63), so σ << 16. (Equality if only the extremes 0 hours and 40 hours are observed). So 4·16/√N >> 8 hours. So N << 64. It must have been a very small study. The decimals on 4.63 and 12.63 hours give an unjustified impression of precision; the population average is given with a huge inaccuracy; and the variation of the observations are untold. So B is false. Bo Jacoby (talk) 20:21, 31 July 2009 (UTC).

B is absolutely false. And I was surprised the first time I saw students interpreting it that way. D is false and F is true if you construe the words the way they're usually construed, but there I have some bones to pick. As someone pointed out, the question is whether we're talking about confidence intervals in general or about THIS interval. That's what "Igny" seems to be missing. F is true of confidence intervals in general, but not of any one particular confidence interval, unless we do some funny and somewhat elaborate revisions of our concept of probability. Where it gets involved is that maybe we SHOULD do some strange and elaborate revisions of the concept, but that's more than I could get into here. E is true, and it's just another way of restating the statement about what the 95% probability means. Michael Hardy (talk) 23:56, 31 July 2009 (UTC)

hypothesis tests

Q2:

For a test of H_o: p = 0.5, the Z-test statistic equals -1.52. Find the p-value for H_a: p < 0.5.

• A. 0.0643
• B. 0.0668
• C. 0.9357
• D. 0.9332
• E. 0.1286
• F. 0.1336

I looked it up in the z-table and found a value of 0.0643 (A) for the area under the standard normal curve. But is this the solution, or is there something more? 70.169.186.78 (talk) 20:03, 31 July 2009 (UTC)

Looks OK. But your use of the letter p makes me wonder about whether it involved a population proportion, and whether you should therefore think about using a continuity correction. Depends on context: if it's an exercise from a textbook, then looking at the exact phrasing of the question might quickly settle the matter. Michael Hardy (talk) 00:01, 1 August 2009 (UTC)

Bilateral statistical hypothesis test

Homework question, sorry. I just can't figure something out.

I'm assuming that I need to do a bilateral statistical hypothesis test.

Info: [...]In 1993-1994, a little more than half (52 %) of foreign exchange students were from Asia.[...]

Assuming that a random sample of 1,346 foreign students contains 756 students from Asia, in 1996.

a) With a significance level (not sure of the term in English) of 5 % can we conclude that the proportion of Asian foreign exchange students, in 1996, is 52 % ?

What I did:

1) H0: µ = 0,52

H1: µ = (not equal) 0,52

2) The significance level is 5 %

3) n = 756 > (or equal) 30

4) Since the size of the sample is n = 756 > (or equal) 30, we can use the normal distribution.

5) If Zx < -1.96 or if Zx > 1.96, the average proposed will be discarded and in consequence the alternative hypothesis H1 will be accepted. If -1.96 < (or equal) Zx < (or equal) 1.96, the average proposed by the null hypothesis H0 will not be discarded.

6) Calculations: http://i31.tinypic.com/xljhc9_th.jpg

However, the standard derivation is given NOWHERE. How am I supposed to calculate Zx ? Is there an error in the manual ? Rachmaninov Khan (talk) 20:02, 31 July 2009 (UTC)

You're running a z-test for proportions. The putative standard deviation is computed as ${\sqrt {\frac {p(1-p)}{n}}}$ (cf. variance of a variable under a binomial distribution). You have the corresponding formula for the statistic z in statistical hypothesis testing, one-proportion z-test. Pallida  Mors 05:48, 1 August 2009 (UTC)

I can't say how much I'm thankful. Thanks again !!! Rachmaninov Khan (talk) 15:08, 1 August 2009 (UTC)

Finding sample size

We want to determine the true average number of drinks University of Michigan students have over a weeklong period. Assume the standard deviation is ~6.3. How many students should be sampled to be within 0.5 drink of population mean with 95% probability?

Is it 610? 70.169.186.78 (talk) 04:39, 31 July 2009 (UTC)

Assuming it's a Normal distribution you can work out p(drinks), and therefore p(drinks-average_drinks > 0.5) for 1 sample.

You could use a computer to calculate (integrate) the same probability for 2 samples. eg integrate over all p(x)p(2d-x) to find p(d) (d=drinks)

And repeat that process to get the probability for 3,4,5 etc samples until p(drinks-mean_drinks > 0.5) is less than 2.5%

Finding the mathematical formula is beyond me. Maybe someone else will know.83.100.250.79 (talk) 22:53, 31 July 2009 (UTC)

See also Standard error (statistics) , Normal_distribution#Estimation_of_parameters 83.100.250.79 (talk) 23:03, 31 July 2009 (UTC)

The answer above misses the point completely and is entirely wrong.

The standard deviation of the sampling distribution of the sample average is σ/√n, where n is the sample size and σ is the population standard deviation, in this case 6.3. Since you've asked for 95%, you get 1.96 out of the normal table (that's the 97.5th percentile, so 2.5% is above that, and another 2.5% below −1.96, so that leaves 95% between ±1.96. So you need

$1.96\cdot {\sigma \over {\sqrt {n}}}=1.96\cdot {6.3 \over {\sqrt {n}}}\leq 0.5.$

That tells you that

${\sqrt {n}}\geq {1.96\sigma \over 0.5}.$

and so on....... Michael Hardy (talk) 00:12, 1 August 2009 (UTC)

Actually it's not entirely wrong - it's a computational method, also the links link to exactly the answer you gave :) 83.100.250.79 (talk) 00:24, 1 August 2009 (UTC)

You wrote "p(drinks-averagedrinks > 0.5)". That that's what was being asked about is entirely wrong. Michael Hardy (talk) 01:00, 1 August 2009 (UTC)

Is this better :

p(drinks_{average after n samples}-average_drinks > 0.5) is less than 2.5%

I can't see the confusion, it's what is being asked? (assuming the distribution is symmetrical)83.100.250.79 (talk) 12:57, 1 August 2009 (UTC)

That's it if you use the absolute value of the difference, not just the difference, and if by "average" you mean the (unobservable) population average. And you're using the word "sample" incorrectly. There's only one sample, which includes n observations. Michael Hardy (talk) 16:45, 1 August 2009 (UTC)

ok buddy, I'll bow to your superior knowledge of naming conventions - but do I actually need the absolute value - that's why I used 2.5% (instead of 5%), and commented that the distribution is symmetrical.83.100.250.79 (talk) 17:28, 1 August 2009 (UTC)

OK, I see what you mean: no absolute value needed when you've got "2.5%" there. Michael Hardy (talk) 17:32, 1 August 2009 (UTC)

oh silly me - I've just realised that they may not know the mean.. 83.100.250.79 (talk) 18:59, 1 August 2009 (UTC)

Clarification: The statement we've derived is that

$\Pr \left(-1.96{\frac {6.3}{\sqrt {n}}}<{\overline {X}}-\mu <1.96{\frac {6.3}{\sqrt {n}}}\right)=0.95,$

But now we want to rearrange it like this:

$\Pr \left({\overline {X}}-1.96{\frac {6.3}{\sqrt {n}}}<\mu <{\overline {X}}+1.96{\frac {6.3}{\sqrt {n}}}\right)=0.95,$

with the unobservable population mean μ in the middle, and only observable quantities as the endpoints of the interval. That observable interval is the 95% confidence interval for the unobservable population mean μ. Michael Hardy (talk) 17:38, 1 August 2009 (UTC)

August 1

computing modulo

I want to compute the remainder when $2^{2^{17}}+1$ is divided by 19. The hint in the book that I have says first compute $2^{17}$ mod 18. How can I do that please? I can't apply Fermat's theorem or Euler's theorem directly, yet as these are the only results proved in the book upto this point I am pretty sure that I am expected to use them only. Thanks.--Shahab (talk) 14:45, 1 August 2009 (UTC)

Try computing the smaller powers of 2 mod 18 first. Michael Slone (talk) 16:30, 1 August 2009 (UTC)

....and observe the pattern that they follow. Michael Hardy (talk) 16:46, 1 August 2009 (UTC)

OK. 2^17=14. What should I do now. I want to get a result mod 19 somehow.--Shahab (talk) 18:35, 1 August 2009 (UTC)

Working mod 19, 2^14=2^-4=16^-1, so you could compute the inverse of 16. Or compute the inverse of 2 and square it twice. Or just divide 16384 by 19 and find the remainder. Algebraist 20:21, 1 August 2009 (UTC)

You now have enough information to apply Fermat's Little Theorem.84.202.41.200 (talk) 22:55, 3 August 2009 (UTC)

Working together word problem

The problem reads:

Abbie paints twice as fast as Beth and three times as fast as Cathie. If it takes them 60 min to paint a living room with all three working together, how long would it take Abbie if she works alone?

Here is what I have:
$A=2B,\,\!\,A=3C$
${\frac {A}{2}}+{\frac {A}{3}}+A=60$
${\frac {11A}{6}}=60$
$A={\frac {360}{11}}$
Which I know is incorrect because I know the answer is actually 1 hour 50 minutes. Where have I gone wrong, and how do I do it correctly? TIA, Ζρς ιβ' ^¡hábleme! 17:02, 1 August 2009 (UTC)

In your first equation you treat A, B, and C as the rates at which they work, and in the second as the times. They can't be both: if you make the rate bigger, you make the time smaller.

The rate is inversely proportional to the time. If Abby can do the job in a hours, she can do 1/a of the job in 1 hour. Beth would then take twice as long: 2a hours; and Cathie would take 3a hours. How much they could do in one hour would be 1/a, 1/(2a), and 1/(3a) for Abby, Beth, and Cathie respectively. So you have 1/a + 1/(2a) + 1/(3a) = 1. That gives a = 11/6 hour = 1 hour and 50 minutes. Michael Hardy (talk) 17:46, 1 August 2009 (UTC)

If you take A to be the rate, then the only thing wrong is the RHS of the second equation. The rate of the three of them working together is indeed A + A/2 + A/3, but the total rate given in the problem is 1 living room per 60 min, which is 1/60 instead of 60 (assuming you're working in minutes). It might be helpful to include the units in your equation. Personally I find that helps avoid mistakes although your mileage may vary. Rckrone (talk) 22:28, 1 August 2009 (UTC)

Two cubes that share three corners

Is it possible for two cubes of the same size to share exactly three corners? NeonMerlin 23:59, 1 August 2009 (UTC)

Yes. Let the vertices of a cube be the eight points in {0, 1}³. Consider the three points

(1, 0, 0)

(0, 1, 0)

(0, 0, 1)

Consider the plane in which they lie to be a mirror. The mirror-image of the cube described here is another cube that shares those same three vertices.

(I wrote a longer reply and it was destroyed by an edit conflict when someone put in the August 2nd heading.) Michael Hardy (talk) 00:11, 2 August 2009 (UTC

You know, you can rescue text lost to an edit conflict - your text is in the textarea at the bottom, you can copy and paste it from there to the top box and submit it. Alternatively, you can copy it from the diff, but that requires a little tidying up afterwards (it is my preferred method, though, since you don't have to search for your text). --Tango (talk) 15:38, 2 August 2009 (UTC)

OK, more long-windedly: The eight corners of one cube are these:

(0, 0, 0)

(1, 0, 0)

(0, 1, 0)

(0, 0, 1)

(1, 1, 0)

(1, 0, 1)

(0, 1, 1)

(1, 1, 1)

The eight corners of the other cube are these:

(2/3, 2/3, 2/3)

(1, 0, 0)

(0, 1, 0)

(0, 0, 1)

(2/3, 2/3, −1/3)

(2/3, −1/3, 2/3)

(−1/3, 2/3, 2/3)

(−1/3, −1/3, −1/3)

Michael Hardy (talk) 03:34, 2 August 2009 (UTC) The three corners of the white triangle are the three vertices that the two cubes share. One corner of each cube—the one that is "inside" the white triangle—is within the interior of the other cube. Michael Hardy (talk) 03:42, 2 August 2009 (UTC)

hallo! I took the liberty of kicking up the cube... pma (talk) 19:40, 2 August 2009 (UTC)

August 2

Image of a set of measure 0

Royden has a problem, number 5.18, which says:

Let g be an absolutely continuous monotone function on [0, 1] and E a set of measure zero. Then g[E] has measure zero.

The article Luzin N property says that every absolutely continuous function on [a, b], f, has the Luzin N property [a, b], which means for $N\subset [a,b]$, mN = 0 implies m(f(N)) = 0.

So, it's true for any absolutely continuous function, not just monotone ones? And, is the more general version much harder to prove?

Thanks StatisticsMan (talk) 01:33, 2 August 2009 (UTC)

Unless I'm doing something wrong, I think it follows pretty quickly from absolute continuity alone. To show that f(E) has measure less than ε, choose a δ that satisfies the absolute continuity property for ε. Since E has measure zero, there exists an open cover A of E consisting of disjoint intervals where the sum of the lengths is less than δ. So then f(A) has measure less than ε and contains f(E). Rckrone (talk) 16:27, 2 August 2009 (UTC)

Oh I guess where monotone comes in is that for an interval I = [x,y], m(f(I)) could be larger than |f(y)-f(x)| for f that's not monotone. I don't think it's much of a problem, since you can probably find a way to break up intervals like that so that that doesn't happen. Alternatively, f can be decomposed into the sum of two monotone absolutely continuous functions. Rckrone (talk) 18:02, 2 August 2009 (UTC)

Yes; or also you may rephrase, clearly equivalently , the definition of absolute continuity this way:

• A function f: I → R is absolutely continuous on I if for every positive number $\varepsilon$, there is a positive number $\delta$ such that whenever a (finite or infinite) sequence of pairwise disjoint sub-intervals [x_k, y_k] of I satisfies

$\scriptstyle \sum _{k}\left|y_{k}-x_{k}\right|\leq \delta$

then

$\scriptstyle \sum _{k}\mathrm {osc} (f,[x_{k},y_{k}])\leq \varepsilon$,

(where $\scriptstyle \mathrm {osc} (f,S)$ denotes the oscillation of f on the set S, that is $\scriptstyle \sup _{S}f-\inf _{S}f$).--pma (talk) 19:22, 2 August 2009 (UTC)

I thought about the fact that an absolutely continuous function is the sum of two monotone functions. I did not think about the fact that they would be absolutely continuous. But, here it is in one of my books. So, assume f is not monotone, then f = f_1 - f_2 where f_1 and f_2 are monotone and absolutely continuous. The result on monotone functions gives f_1[E] and f_2[E] measure 0. Then f[E] is certainly a subset of f_1[E] - f_2[E] = {x - y : x \in f_1[E], y \in f_2[E]}. But, is f_1[E] - f_2[E] also of measure 0? If so, that proves it. If not, ??? StatisticsMan (talk) 18:36, 2 August 2009 (UTC)

Consider that for null subsets N and N' of [0,1], N+N' may have positive measure: take N:=all reals that admit a ternary expansion with only 0 and 1 ; N':=all reals that admit a ternary expansion with only 0 and 2. Clearly, each x in [0,1] is expressable as y+y' with y in N y' in N'. Indeed in this moment I do not see a simplification in decomposing a function as f=f₁-f₂.--pma (talk) 18:53, 2 August 2009 (UTC)

(ec) Your proof is correct; you not have to bother about monotonicity; it never enters. Notice also that for continuous function that is actually a characterization of absolute continuity. Indeed, for a continuous BV function g:[a,b]→R the following are equivalent:

• i. g absolutely continuous;
• ii. for any $\scriptstyle N\subset [a,b]$ with measure zero, g(N) has measure zero;
• iii. for any ε>0 there exists δ>0 such that for any (Lebesgue) measurable $\scriptstyle M\subset [a,b]$, if M has measure less than ε, then g(M) has (exterior) measure less than δ;
• iv. for any measurable $\scriptstyle M\subset [a,b]$ g(M) is measurable.

By (iv), the exterior measure in (iii) is actually the measure. --pma (talk) 18:40, 2 August 2009 (UTC)

(ec) @StatisticsMan: For any interval I = [a,b], for all x in I, f(x) - f(a) ≤ f_1(x) - f_1(a) ≤ f_1(b) - f_1(a) and similarly f(a) - f(x) ≤ f_2(x) - f_2(a) ≤ f_2(b) - f_2(a), so m(f(I)) ≤ |f_1(b) - f_1(a)| + |f_2(b) - f_2(a)|. In other words f can't increase in the interval more than f_1 does or decrease more than f_2 increases, which puts bounds on the measure of f(I). Rckrone (talk) 18:48, 2 August 2009 (UTC)

Good; so if g :=f₁+f₂ you have |f(B)|≤|g(B)|, which reduces the proof to the case of a g monotone. Still, I do not see why monotonicity should make any simpler the proof (your first argument is correct and monotonicity does not enter...)--pma (talk) 19:04, 2 August 2009 (UTC)

The reason I brought it up is that the way my book defines absolute continuity is that for any ε there exists δ such that $\sum {|f(b_{k})-f(a_{k})|}<\epsilon$ for any disjoint intervals (a_k, b_k) with $\sum {(b_{k}-a_{k})}<\delta$, which is also how it's defined on absolute continuity. With other definitions it might not be necessary. Rckrone (talk) 19:14, 2 August 2009 (UTC)

Perfect (I put above a variant of the definition, which it is not necessary to bother about monotonicity in the proof (is this sentence grammatically correct?). Note that the dependence of δ on ε in the modified definition is exactly the same as in the standard one). --pma (talk) 19:24, 2 August 2009 (UTC)

Yeah that makes sense. I probably should've mentioned before what definition I was going by. (it would work if you said "with which" instead of "which") Rckrone (talk) 20:06, 2 August 2009 (UTC)

I read this all and will read it again in a bit. Quickly though, I see you say that for a continuous function, absolute continuity is equivalent to the Lusin N property. However, the article on absolute continuity says it is equivalent to the Lusin N property and bounded variation. "f is absolutely continuous if and only if it is continuous, is of bounded variation and has the Luzin N property." Right? Are you saying it would be fine to say f is absolutely continuous if and only if it is continuous and has the Luzin N property? StatisticsMan (talk) 22:33, 2 August 2009 (UTC)

Okay, I think I am understanding this all. So, for the definition Rckrone is using, which is the same one I am using, monotonicity does matter but the solution is to consider g = f_1 + f_2 and note $m(f(B))\leq m(g(B))$ Then, g is monotone and it works for g so it works for f as well. On the other hand, if we just use the alternate definition that you gave pma, the one involving oscillation, then the same proof works and we don't need to use monotonicity. Is that all right? And, this definition is equivalent, you say clearly :) Let's see if it is to me. First, your definition immediately implies "my" definition since my definition gets a sum which is even smaller than the one involving oscillation. I guess I don't see, right away, the other direction. But, it does seem like it would be true.

Also, is exterior measure also called outer measure? I have heard of the latter but not the former. StatisticsMan (talk) 01:29, 3 August 2009 (UTC)

• Yes: "exterior measure" is less used, but the same as "outer measure". However, for a continuous function f:[a,b]→R the LNP implies that f(M) is measurable for all measurable M, so its outer measure is just the measure.
• "BV and LNP implies AC" is false: if f is a characteristic function of a sub-interval of [a,b] it's BV, discontinuous and maps everything into the null set {0,1}. But you are right, I forgot "BV" above (fixed).
• The definition with "osc" follows at once from the standard definition as Rckrone wrote it: in that situation not only

$\scriptstyle \sum {|f(b_{k})-f(a_{k})|}<\epsilon$,

but in fact for all $\scriptstyle a_{k}\leq a'_{k}\leq b'_{k}\leq b_{k}$ also

$\scriptstyle \sum {|f(b'_{k})-f(a'_{k})|}<\epsilon$

so that

$\scriptstyle \sum _{k}{\mathrm {osc} (f,[a_{k},b_{k}])}=\sup {\big \{}\sum _{k}{|f(b'_{k})-f(a'_{k})|}:a'_{k}\;\mathrm {and} \;b'_{k}\;\mathrm {such\;that} \;a_{k}<a'_{k}<b'_{k}<b_{k}{\big \}}\leq \epsilon$

--pma (talk) 05:56, 3 August 2009 (UTC)

Normal distribution and chi-square test project - Please help

Hello people. As part of my QT (Quantitative Techniques) course, I was required to do an practical exercise of my choice. Before doing that I had to get the idea approved by the teacher. Now the teacher has approved my idea but I have no clue how to go about doing the exercise. My exercise involves normal probability distribution and chi-square test, but I hadn't read up on either of them when I submitted the idea. Please read my idea below (that I submitted to the teacher and got approved) and tell me how I can go about doing it. In particular I don't know how to convert the discrete values of lengths of leaves to the continuous distribution of the normal curve.

"The project involves making measurements on the length of the leaves of Ashoka trees planted in JLT. I shall randomly pluck leaves from Ashoka trees and measure the length of the leaves from the base to the tip. I shall make such measurements for about 200 leaves. After collecting this data I shall plot a graph of the distribution of leaves. According to my expectations, the frequency distribution of the lengths of the leaves should be a normal distribution. I shall use chi square test to gauge the deviation of the distribution from a normal distribution and to determine whether the deviation can be attributed to chance or not. In case it is not, I shall try to find a hypothesis for the deviation from a normal distribution."

Many thanks -- ReluctantPhilosopher (talk) 13:20, 2 August 2009 (UTC)

The lengths of leaves aren't discrete values -- they are continuous. Depending on the scale of your ruler and the lengths of the leaves, your measurements of the lengths will be more or less discrete. For example, if the average leaf is 5 inches and your scale is in quarters-of-an-inch, then your measurements will look more discrete than continuous. If your scale is in millimeters, your measurements will look more continuous than discrete. Wikiant (talk) 14:51, 2 August 2009 (UTC)

Er, my question is, how do I calculate the deviation of the observations from the expected normal distribution. Thanks. ReluctantPhilosopher (talk) 15:22, 2 August 2009 (UTC)

Try the Jarque-Bera test. Wikiant (talk) 15:37, 2 August 2009 (UTC)

Aw thank you so much! That's the kind of thing I was looking for. The article says that it can be further used in a chi-square test also so that means I can stay within the ambit of my "proposal". Thanks again ReluctantPhilosopher (talk) 15:46, 2 August 2009 (UTC)

A follow up question - Is it possible to directly apply the chi-square test? ReluctantPhilosopher (talk) 17:30, 2 August 2009 (UTC)

I'm not sure what you mean by "chi-square test." Every statistical test requires (1) that you construct a test statistic, and (2) that you compare the test statistic to a known distribution. In that sense, any test that uses the chi-squared distribution in step (2) is a "chi-square test." Wikiant (talk) 18:36, 2 August 2009 (UTC)

I apologize for my lack of knowledge about the subject - I am yet to study what chi squared test is and I only have a vague idea what it is used for. Thanks for helping --ReluctantPhilosopher (talk) 16:31, 3 August 2009 (UTC)

Note that the domain of a normal distribution is the entire real axis, while the length of a leave is always a positive number. So take a logarithm of the lengths before testing data for normality (as the logarithm maps the positive semiaxis upon the real axis). Note also the difficulty in defining what it means to randomly pluck a leave; (when is a leave too small to be called a leave?) See multiset and cumulant for defining the deviation of the distribution from a normal distribution. Bo Jacoby (talk) 21:14, 2 August 2009 (UTC).

Those are very good insights. I'll keep those in mind, thanks! --ReluctantPhilosopher (talk) 16:31, 3 August 2009 (UTC)

I imagine you are referring to Pearson's chi square goodness-of-fit test. Whenever I hear "chi-square test", this is the first thing that comes to mind anyhow. I can't give any input as to whether or not it is better (or in what circumstances it might be) than the above-mentioned Jarque-Bera test, though it is presumably what you meant (?) when you told your teacher this was your plan. With Pearson's test there is the problem of choosing the number of "bins" when testing for a continuous distribution, and the Wiki page doesn't seem to give any indication of how to do so; you might search elsewhere for guidelines. You might also check out the Kolmogorov-Smirnov test or the Anderson-Darling test. Nm420 (talk) 20:34, 4 August 2009 (UTC)

0 expected value of a bet = impossible to make money (guaranteed)? why?

If bets are independent and have 0 expected return, does that mean it is IMPOSSIBLE using any strategy to be guaranteed to make money through a series of such bets? What is the proof of this impossibility? (Note: I'm thinking of betting strategies like "martingale"). I'm trying to understand why not just martingale, but any betting strategy is doomed mathematically, so that if it works anywhere, then this is only in places not subject to math. —Preceding unsigned comment added by 82.234.207.120 (talk) 14:18, 2 August 2009 (UTC)

The expected value of a sum of random variables is the sum of the expected values of the variables (that's a basic property of expected values), so if each bet has expected value 0 the expected value of lots of such bets will also be 0. The best you can do is guarantee to get exactly 0 (by hedging your bets precisely), if there is any chance of a positive result then there needs to be a chance of a negative result to balance it. (Note, in real life a bet will generally have an expected value less than zero, so you cannot even guarantee to break even.) --Tango (talk) 14:31, 2 August 2009 (UTC)

That's not actually quite true: a sum of randomly many 0-expectation variables can have positive expectation. E.g. if you have infinite funds and unlimited time, and a house is offering fair bets at evens odds, then the strategy "continue betting until you've won a million dollars" has expected value of a million dollars, even though each bet has expected value zero. The real result is the optional stopping theorem, which requires some additional assumptions. Algebraist 17:05, 2 August 2009 (UTC)

How about if I don't want a million dollars under those conditions, but infinite dollars? Would the strategy "starting at 1 and moving on to the next integer each time you net an additional $1, continue betting until you have netted a win for every positive integer" have infinite expected value? 82.234.207.120 (talk) 23:53, 2 August 2009 (UTC)

No. It takes an infinite time to carry out, and your wealth after an inifinite sequence of plays is not well-defined. Algebraist 00:03, 3 August 2009 (UTC)

Surely the expected value is indeterminate. For the strategy "continue betting until you've won $a or lost $b" the expected gain after any number of steps is zero (even taking into account that you stop upon reaching one of the bounds). Your strategy is the case when a = 10⁶ and b → ∞, but every strategy with finite b has an expected gain of zero. There's some kind of ∞/∞ thing going on. -- BenRG (talk) 00:43, 3 August 2009 (UTC)

No, it's perfectly determinate. The probability that I will be a million dollars up at some finite time is one, so the expected gain is a million dollars. It doesn't work if you have a finite amount of money to lose, or a limited amount of time to play, but I said that already. Algebraist 01:08, 3 August 2009 (UTC)

I see what you mean, but my argument seems valid too. Your expected winnings at any given time are zero. I think there are two incompatible definitions of "expected gain". It reminds me of the balls and vase problem for some reason. -- BenRG (talk) 10:57, 3 August 2009 (UTC)

Yes, your expected winnings at any fixed time is zero, but at a random time can be nonzero, as I said to start with. Since typical betting strategies involve playing for a random time, that's the relevant case. Algebraist 12:15, 3 August 2009 (UTC)

Typical betting strategies usually have a fixed amount of money to lose. Taemyr (talk) 12:33, 3 August 2009 (UTC)

The only surefire way of making money by betting is to be better (oops, pardon the pun) at determining the odds than the house. I know of at least one person who made a living this way. --ReluctantPhilosopher (talk) 15:31, 2 August 2009 (UTC)

Or just better than the other people betting. For something like horse racing, where there is no certain way to determine the odds (unlike roulette, for example), the house just sets the odds based on how people are betting (the initial odds will be based on the actual race, I suppose, but they don't last long). If you know better than everyone else you can, in theory, make money. (You have to know quite a lot better in order to compensate for the profit margin the house adds on to the odds.) --Tango (talk) 15:36, 2 August 2009 (UTC)

There are cases where adding up an infinite number of events each with probability 0 can have a non-zero probability. For example, suppose I am choosing a truly random number from 0 to 1. For any specific value x, the chance that I'll choose x is 0, but the chance of me choosing any number between 0 and 1 should be the sum of these individual probabilities, but is equal to 1. If you bet that I would pick the number x, your expected return would be zero, but if you bet on all the numbers from 0 to 1/2 you would have 50/50 odds. See continuous probability distribution. Rckrone (talk) 17:03, 2 August 2009 (UTC)

You are adding up an uncountably infinite number of events, which I'm not sure is even meaningful to say (although I guess you could add them up by enumerating them with the ordinal numbers). A uniform probability distribution between 0 and 1 is actually a measure on the real numbers between 0 and 1. Measures are, by definition, countably additive, which means that a countably infinite number of 0-probability events cannot possibly add up to a positive value. --COVIZAPIBETEFOKY (talk) 20:34, 2 August 2009 (UTC)

The set of outcomes need not be uncountable. Suppose I choose a random rational number between 0 and 1. The result is the same. Rckrone (talk) 21:08, 2 August 2009 (UTC)

There is no uniform probability distribution on the rationals between 0 and 1, unless you take the unusual step of not requiring countable additivity. Algebraist 21:24, 2 August 2009 (UTC)

Ok point taken, but what if I choose a real number between 0 and 1 at random and you bet on a specific Vitali set? What is your probability of winning? Rckrone (talk) 16:23, 4 August 2009 (UTC)

It must be undefined. I imagine any attempt to determine it empirically would just be unstable - you can only do finitely many trials (countably infinitely many in the limit) and we're working with uncountable sets, so that's perfectly plausible. --Tango (talk) 16:41, 4 August 2009 (UTC)

How do you mean it would be unstable if you test it empirically? If you did N trails, would you ever win? I would think the answer is no. Rckrone (talk) 17:37, 4 August 2009 (UTC)

Vitali sets are unmeasurable, as I'm sure you're well aware. If you're talking about what would actually happen in practice, then (a) it's impossible to specify a specific Vitali set and (b) determining whether a randomly chosen number lies in a specific Vitali set requires knowing that number precisely, i.e. knowing infinite amounts of information. Algebraist 17:46, 4 August 2009 (UTC)

I mean that if you did more and more trials the experimental probability wouldn't stabilise. When you calculate a probability empirically you do a load of trials and calculate successes/total trials, you then do more trials and calculate it again and repeat and repeat until the answer stops changing at the level of precision you want. For something without a well defined frequentist probability the answer would never changing at any level of precision. Of course, Algebraist's point makes mine moot - you couldn't even carry out a trial. --Tango (talk) 18:25, 4 August 2009 (UTC)

Doesn't a "martingale" strategy actually work here -I'm thinking of something like tossing a coin, and either losing or wining the amount bet - that has expectation 0 on constant equal bets. But using the simple 'double the bet' method each time allows the player to win.?

Is that what you were thinking of - because in this case it looks like it is possible to win with enough funds. (I may have made a mistake)83.100.250.79 (talk) 22:08, 2 August 2009 (UTC) Thus there is no proof of the impossibility, but there is proof that a doubling bet method can win; if you want to see it.83.100.250.79 (talk) 22:10, 2 August 2009 (UTC)

You are right, but "enough funds" turns out to be infinite funds. With any finite amount you will find that there is enough chance of losing everything to balance the large probability of winning a tiny amount of money so that you end up with a expectation of breaking even, just as if you had everything you have on a single toss of the coin. --Tango (talk) 22:59, 2 August 2009 (UTC)

August 3

how many ways are there of arranging 8 pairs of socks on a clothesline so that no two socks from a pair are next to each other?

The subject says it all. How many ways are there of arranging 8 pairs of socks on a clothesline, so that no two socks from a pair are next to each other? The clothesline is "open" (it's not circular) and there is no distinction between left and right socks. This is not homework. —Preceding unsigned comment added by 82.234.207.120 (talk) 00:02, 3 August 2009 (UTC)

My guess: 1394852659200

16 * 14!? I'm sure it's not right though, but I think it's a good estimate. 65.184.21.210 (talk) 04:31, 3 August 2009 (UTC)

Inclusion-exclusion. Let n pairs of socks be given. Let's count the wrong arrangements of them: for k from 1 to 2n-1 call A_k the configurations in which there is a pair of socks occupying the k-th and the k+1-th place. We need to count the cardinality of any intersection of r of these sets $A_{i}$: for $1\leq i_{1}<i_{2}<..i_{r}\leq 2n-1$ the set $A_{i_{1}}\cap ..\cap A_{i_{r}}$ is either empty (which happens if and only if $i_{h+1}=i_{h}+1$ for some h), or it has cardinality

$|A_{i_{1}}\cap ..\cap A_{i_{r}}|={\frac {n!}{(n-r)!}}{\frac {(2n-2r)!}{2^{n-r}}}$.

Indeed, in the latter case you can put r pairs of socks in ${\frac {n!}{(n-r)!}}$ ways in the positions $i_{1},i_{1}+1$ (first pair); $i_{2},i_{2}+1$ (second pair),..$i_{r},i_{r}+1$ (r-th pair). The remaining 2n-2r positions are occupied freely by the remaining n-r pairs in ${\frac {(2n-2r)!}{2^{n-r}}}$ ways. This is agreable in that the cardinality only depends on r, and all you need to know is how many non-zero terms are there in the r-th sum of the I-E formula. That is, how many r-subsets of {1..2n-1} with no consecutive elements are there? If you think, there are ${2n-r \choose r}$ of them. Now you can put everything together and write a formula for a(n)= the number of ways n pair of socks can be made in a clothline according to your rule. After computing the first 8 or 10 values of a(k), you can put them in the OEIS [1] and you will surely find more info. --pma (talk) 10:34, 3 August 2009 (UTC)

Spoiler alert: it's A114938. -- BenRG (talk) 10:48, 3 August 2009 (UTC)

;) So, from the computation above I got

$a(n):=\sum _{r=0}^{n}(-1)^{r}{n \choose r}{\frac {(2n-r)!}{2^{n-r}}}$;

the OEIS gives the equivalent

$a(n):=\sum _{r=0}^{n}(-1)^{n-r}{n \choose r}{\frac {(n+r)!}{2^{r}}}$.

--pma (talk) 11:34, 3 August 2009 (UTC)

what is a quarternion?

What is a quarternion?? I'm quite familiar with the concept of complex numbers, their notation in Euler's form, their calculations on complex plane, their extended application in solving geometric problems et cetera. Out of curiosity, I googled if there was a bigger set of numbers that surpassed the concept of complex numbers, and it came up with quarternions. I've read and reread wikipedia's article on quarternions, and it describes their basic calculations and applications to 4 dimensional space quite thoroughly But it seems that the author had written the article on the basis that all the readers are at least familiar with the concept of i, j, and k, and I just can't grasp the concept of the "extra" imaginary parts. How can there be another imaginary number(s) besides the common "i" where j^2 and k^2 both equal -1? I thought that polynomials to the nth degree only had n roots, and as a result, shouldn't the polynomial P(x)=x^2 +1 have only two roots i, -i? And granted that these imaginary numbers j, k do exist, why did mathematicians define them so that ijk=-1? And why was it ever invented (or thought of) anyway? Aren't complex numbers enough? I just can't understand it, and the curiosity's killing me! Can anybody explain these concepts to me in layman's terms (so to speak) so that i can at least understand the basic concept of quarternions? Please enlighten me! Thanks----Johnnyboi7 (talk) 11:23, 3 August 2009 (UTC)

In the sense that you can invent any mathematics you so choose, so long as you can provide a consistent framework for them, why NOT invent such a thing as quaternions, octonions etc.? They are interesting fields of study in their own right. If I recall correctly quarternions were "invented" for the purpose of describing physical quantities in 3 dimensions, so were completely nothing to do with roots of polynomials. Just try to separate out those ideas in your head.

In physics and applied mathematics quarternions came to be almost entirely replaced by vectors in R^3. Quarternion multiplication (of the pur imaginary part) follows the same rules as the cross product in vector multiplication. In fact, in the sense that all hypercomplex numbers are isomorphic to n-dimensional vectors with some extra structure on them, one can define any type of number at all, but only a few of them permit a structure that is both consistent and interesting to study. Zunaid 11:43, 3 August 2009 (UTC)

They are quite popular in computer games and graphics because they don't have any funny points in their representations of rotations, they are much more stable than trying to do it with the usual 3d formulae, see Quaternions and spatial rotation Dmcq (talk) 12:04, 3 August 2009 (UTC)

In a commutative field, a quadratic equation can have only two solutions. Some rings are not commutative fields. A simple example is the integers mod 8. In that ring, the quadratic equation x² = 1 has four solutions. And the same thing happens in various rings of matrices.

The book Visualizing Quaternions by Andrew J. Hanson, published in 2005, treats present-day applications to geometry and engineering, including computer graphics. Michael Hardy (talk) 12:06, 3 August 2009 (UTC)

ijk=-1 doesn't have to be taken as a definition, it can be deduced from some basic assumptions. Just assume there is one extra square root of -1 and that the concept of the modulus of a number extends from complex numbers in the obvious way (and has the same basic properties) and everything else just follows. Consider $(i+j)^{2}=i^{2}+ij+ji+j^{2}=ij+ji-2$ (I'm keeping ij and ji separate because I know what is coming next). The modulus of i+j is ${\sqrt {2}}$, so we know the modulus of ij+ji-2 must be 2, the triangle inequality gives us |ij+ji-2|=<|ij+ji|+2, so ij+ji must be zero, ie. ij=-ji. Then consider (ij)² and you'll find you need to introduce a k and that ijk=-1. So, the quarterions really are the obvious way of extending the complex numbers to have extra square roots of -1. --Tango (talk) 12:43, 3 August 2009 (UTC)

You know the way you can interpret rotation around a circle as multiplication by a complex number? Quaternions are the same way except it's rotation (in some arbitrary direction) around a sphere in 3-space. John Baez has an interesting article about quaternions and their generalization to octonions, sedenions, and so forth. WP's article History of quaternions might also be of interest. 67.117.147.249 (talk) 15:21, 3 August 2009 (UTC)

There are ways of seeing the quaternions which are purely mathematical in nature (mathematically, computations in the quaternions are pointless unless the computations are done for a reason - compare to the complex numbers). Three-dimensional Euclidean space has 3 "axis" and thus any point in this space can be uniquely described by three coordinates. Therefore, (1, 2, 3) or (5, 3.424, 0) are points in three-dimensional Euclidean space. These points can be also written as 1i + 2j + 3k and 5i + 3.424j + 0k, respectively. Think of four-dimensional Euclidean space as simply adding another axis, so that (1, 2, 3, 4) can be written as 1 + 2i + 3j + 4k. The natural question to ask now, is how we can "multiply" two points in 4-dimensional Euclidean space. Certainly we may use real number multiplication so that (for example):

$(1+2i+3j+4k)(0+2i+0j+0k)=2i+4i^{2}+6ji+8ki$

Notice how we have used real number multiplication to define "quaternion multiplication". Now it is necessary to compute $i^{2},ji,andki$. It should be possible to describe these products in terms of i, j and k, only so that the final product is also a quaternion. In general, defining a "nice multiplication on a set" (such as {1, -1, i, -i, j, -j, k, -k} as we have done) and extending it to a multiplication of arbitrary "linear combinations" of elements in the set (1 - 2i + 3j - 9k is an example of a linear combination) defines a Group algebra over a field. Group algebras are very important since they occur in character theory.

There is yet a more mathematical way to view the quaternions. Notice how each complex number can be associated to an "inverse"; that is, given a complex number z, we can find a w, such that:

$zw=wz=1+0i$

We think of w as 1/z, intuitively. Notice also that the complex numbers form a two-dimensional vector space over the real numbers (in milder language, this means that any complex number can be described in terms of 1 and i. The complex numbers 2(1) + 3(i), 5(1) + 7(i) and 0(1) + 0(i) serve as examples). These two facts (together with associativity of multiplication on the complex numbers) implies that the complex numbers forms a finite dimensional (two-dimensional, in fact), associative division algebra (the existence of an inverse for every complex number implies "division" is possible) over the real numbers. A theorem of Frobenius asserts that there is only one other non-trivial mathematical object having this property - the quaternions (the real numbers are a trivial case). This characterizes the quaternions completely. Another consequence is that one cannot move to higher dimensional analogs of the complex numbers (although the octonions seem to constitute such an example, for instance, the multiplication is not associative so it fails to be a higher dimensional analog). Hope this helps. --PST 04:52, 4 August 2009 (UTC)

Notice though, that the quarternions are not a group algebra. The real group algebra constructed from the quarternion group G would be an 8-dimensional algebra with $1\in G$ and $-1\in G$ linearly independent. All four dimensional group algebras are commutative since all the groups of order 4 are commutative. Aenar (talk) 12:27, 5 August 2009 (UTC)

Another lovely complex integral

Hi. I've been knocking my head against the following problem: Prove that $\int _{0}^{\infty }{\frac {\ln(x)}{x^{2}-1}}\,dx={\frac {\pi ^{2}}{4}}$.

The idea (as I see it) is to make some complex function, $f(z)={\frac {\operatorname {L} (z)}{z^{2}-1}}$, where L is some appropriate branch of the complex logarithm, one that restricts to the real natural log on the positive real axis. Then we cleverly choose a contour, avoiding singularities of f, apply the residue theorem, rinse for 5 minutes in cold running water, and solve for our integral. I either keep choosing the wrong contour, and/or the wrong branch or Log, and/or I mess up the calculus. I keep getting imaginary numbers, 90° off from the desired value.

The function has a removable singularity at $z=1$, which is taken care of by setting $f(1)={\tfrac {1}{2}}$, and a simple pole at $z=-1$, with residue $\operatorname {Res} (-1,f)=1-{\tfrac {\pi i}{2}}$. I need a contour that includes a chunk of the positive real axis, and then I can take limits. I also think I should stay away from the pole at -1. My latest attempt is as follows:

Attempt using a weird contour

Let R and r be two positive real numbers, larger and smaller than 1, respectively. Define the contour $\gamma =\gamma _{1}+\gamma _{2}+\gamma _{3}+\gamma _{4}+$, where $\gamma _{1}$ is a straight line from ir to R + ir, $\gamma _{3}$ is a straight line from R-ir to -ir, $\gamma _{2}$ and $\gamma _{4}$ are the obvious (positively oriented) arcs centered at 0. This contour avoids the positive real axis, and thus lies in the domain of the function $\operatorname {L} (z)$, a branch of the logarithm that maps onto $\left\{x+iy\,|\,0\leq y<2\pi \right\}$, discontinuous on $\left[0,\infty \right)$. This branch satisfies the identity $\operatorname {L} (t-ir)={\overline {\operatorname {L} (t+ir)}}+2\pi i$, for $t\in \mathbb {R} ,r\in \left[0,\infty \right)$. (We only care about 0 < r < 1.) Our pole lies inside the contour, and the winding number is 1, so we have: $\int _{\gamma }f(z)\,dz=2\pi i\,\operatorname {Res} \left(-1,f\right)=2\pi i\left(1-{\tfrac {\pi i}{2}}\right)=\pi ^{2}+2\pi i$ This leads to: ${\begin{aligned}\pi ^{2}+2\pi i&=\lim _{R\rightarrow \infty \atop r\rightarrow 0}\int _{\gamma }f(z)\,dz\\&=\lim _{R\rightarrow \infty \atop r\rightarrow 0}\left(\int _{\gamma _{1}}f(z)\,dz+\int _{\gamma _{2}}f(z)\,dz+\int _{\gamma _{3}}f(z)\,dz+\int _{\gamma _{4}}f(z)\,dz\right)\\&=\lim _{R\rightarrow \infty \atop r\rightarrow 0}\left(\int _{\gamma _{1}}f(z)\,dz+\int _{\gamma _{3}}f(z)\,dz\right)\end{aligned}}$ That last line works because $\gamma _{2}$ and $\gamma _{4}$ vanish as R and r go to their limits. The trick now is to express the $\gamma _{3}$ integral in terms of the $\gamma _{1}$ integral. Thus, we parameterize them as $\gamma _{1}(t)=t+ir$, $\gamma _{3}(t)=R-ir-t$, letting t run from 0 to R in both cases. Now: ${\begin{aligned}\int _{\gamma _{3}}f(z)\,dz&=\int _{0}^{R}{\frac {\operatorname {L} (\gamma _{3}(t))}{\left[\gamma _{3}(t)\right]^{2}-1}}{\dot {\gamma _{3}}}(t)\,dt\\&=-\int _{0}^{R}{\frac {\operatorname {L} (R-ir-t)}{(R-ir-t)^{2}-1}}\,dt\\&=-\int _{R}^{0}{\frac {\operatorname {L} (t-ir)}{(t-ir)^{2}-1}}\,(-1)dt\\&=-\int _{0}^{R}{\frac {\operatorname {L} (t-ir)}{(t-ir)^{2}-1}}\,dt\\&=-\int _{0}^{R}{\frac {{\overline {\operatorname {L} (t+ir)}}+2\pi i}{\overline {(t+ir)^{2}-1}}}\,dt\\&=-\int _{0}^{R}{\frac {\overline {\operatorname {L} (t+ir)}}{\overline {(t+ir)^{2}-1}}}\,dt\,-\int _{0}^{R}{\frac {2\pi i}{(t-ir)^{2}-1}}\,dt\\&=-{\overline {\int _{\gamma _{1}}f(z)\,dz}}-\int _{0}^{R}{\frac {2\pi i}{(t-ir)^{2}-1}}\,dt\end{aligned}}$ Ok, whew. Putting all of that together, we get: ${\begin{aligned}\pi ^{2}+2\pi i&=\lim _{R\rightarrow \infty \atop r\rightarrow 0}\int _{\gamma }f(z)\,dz\\&=\lim _{R\rightarrow \infty \atop r\rightarrow 0}\left(\int _{\gamma _{1}}f(z)\,dz+\int _{\gamma _{3}}f(z)\,dz\right)\\&=\lim _{R\rightarrow \infty \atop r\rightarrow 0}\left(\int _{\gamma _{1}}f(z)\,dz-{\overline {\int _{\gamma _{1}}f(z)\,dz}}-\int _{0}^{R}{\frac {2\pi i}{(t-ir)^{2}-1}}\,dt\right)\\&=\lim _{R\rightarrow \infty \atop r\rightarrow 0}\left(2i\cdot \operatorname {Im} \left(\int _{\gamma _{1}}f(z)\,dz\right)-2\pi i\int _{0}^{R}{\frac {dt}{(t-ir)^{2}-1}}\right)\\&=2i\cdot \lim _{R\rightarrow \infty \atop r\rightarrow 0}\left(\operatorname {Im} \left(\int _{\gamma _{1}}f(z)\,dz\right)-\pi \int _{0}^{R}{\frac {dt}{(t-ir)^{2}-1}}\right)\end{aligned}}$

That final expression is pure imaginary, and that's why I've got the blues. If anyone has the patience and understanding to help me figure out what I'm doing wrong, I'll be deeply grateful. I can clarify, if needed, why I think each of those equal signs is true. Thanks in advance. -GTBacchus^(talk) 18:51, 3 August 2009 (UTC)

I'm entirely confused by your choice of contour, and I think that is where your problem is. It seems to me that a better choice is a quarter circle with indent at 0, running along the x-axis from e to r and along the y-axis from e to r. Then the integral round the whole contour is zero by the residue theorem. Now you get (part of) your integral on the real axis, both the curved parts will go to zero as e goes to 0 and as r goes to infinity by easy log estimates, and on the imaginary axis you'll have an imaginary contribution you can ignore and a real part that will be an arctan integral and give you the factor of pi you need. 87.194.213.98 (talk) 19:32, 3 August 2009 (UTC)

Actually, I started out using the quarter-circle you describe, but I ran into my professor partway through the calculation and I asked him about it. He suggested that including the imaginary axis as part of the contour wouldn't work, because of the "- 1" in the denominator. Since then, I have been trying to do it using the real axis + big and little arcs. However, doing it that way I had to avoid -1, which makes it awkward. I'll look again at the quarter-circle contour, and see if I can get it to work. Thank you. -GTBacchus^(talk) 20:04, 3 August 2009 (UTC)

I think your professor must have been confused about the function. The -1 is fine, it only causes problems on the negative real axis which we do not see. Trust me, I'm a doctor. 87.194.213.98 (talk) 22:01, 3 August 2009 (UTC)

Hey! here we shouldn't give medical advices --84.220.119.55 (talk) 12:03, 4 August 2009 (UTC)

With another contour: Success!

That worked out just as you said. Thanks for the advice. I'll trust myself more next time. -GTBacchus^(talk) 15:49, 4 August 2009 (UTC)

Could I hijack slightly and ask for an explanation of: "on the imaginary axis you'll have an imaginary contribution you can ignore." (in 87.194's first response)? Why can we ignore it, that is. I agree we get the answer if you do but I'm not sure why you can. 81.157.165.115 (talk) 21:59, 4 August 2009 (UTC)

I do not know if this is what 87 meant, but we can actually ignore (additive) imaginary contributions because we do know that the result is real: after all the integral of a real function is a real number (even if, as empirical rule, when computing it by the residue theorem it tends to be disrespectfully imaginary) --84.220.119.55 (talk) 22:40, 4 August 2009 (UTC)

Well, I agree that it should be real. However, if it should be real and we show that it has a complex part something has gone rather wrong. It doesn't seem sensible to just ignore the term without showing (or at least spotting a reason) why it'll vanish. 81.157.165.115 (talk) 01:54, 5 August 2009 (UTC)

84 is right, it must vanish, otherwise the residue theorem would be incorrect! 81, look at it this way: the residue theorem tells us a certain equation of the form "some number = some integrals" holds. We are free to take real or imaginary parts of this equation, and taking real parts amounts to ignoring terms we know to be pure imaginary. If the residue theorem is true then the imaginary part of the equation balances too, but this is no concern of ours since we're only interested in the real part. 87.194.213.98 (talk) 14:06, 5 August 2009 (UTC)

August 4

Pseudoscalar

Whilst this is supposedly my area of expertise, I feel I need some advice regarding this article. As you can see from the edit history, I removed the Clifford algebra section, not because it is wrong per se, but for the reason that it was inconsistent with the rest of the article. However, despite a Google scholar search, I'm still not too clear on what a pseudoscalar is, or if there really are two distinct definitions.

If we are to take the statement that "The pseudoscalar is the top-grade element of a Clifford algebra" as truth, then it is wrong to state that it need commute with all other elements and change sign under parity inversion. These are both true if the algebra has an odd underlying basis (number of basis vectors), but neither are true if the underlying basis is even.

Unfortunately, some of the same literature that state one definition seem to suggest that the other is equally general. Thus I am unsure how to proceed.

Given that only odd-dimensional space possesses anything like a pseudoscalar as currently defined in the article, I am slightly suggesting scrapping the current content and replacing it with the Clifford algebra definition (the bit I removed, perhaps wrongly). I am also questioning the article's status on the importance scale as "low", given the number of references in physics publications.

This paper seems to introduce the commuting and parity-inversion properties but in passing for a geometry in which this is true. [[2]]

Please may I have some advice. --Leon (talk) 10:21, 4 August 2009 (UTC)

I don't know that much about Clifford algebras, but the definition of a pseudoscalar as a top-grade element makes sense. Top-grade elements form a 1 dimensional space and are invariant under rotations so they "behave like scalars." The Clifford algebra page also corroborates that definition. In even dimensions it's true that the sign of a top-form element doesn't change when you flip everything, but I don't think there is any element that behaves like what you're asking for (besides 0). Can you link a source that makes these claims about the properties of psuedoscalars that isn't specifically talking about 3D? Rckrone (talk) 17:32, 4 August 2009 (UTC)

Sorry, I simply explained badly in the last paragraph. What I mean is that the current definition of a pseudoscalar (element that is scalar-like but flips under parity inversion) only holds for odd-d space, and thus perhaps should be replaced with the top-grade of Clifford algebra definition. The stuff about the pseudoscalar having the sign-change property with odd-d spaces should be mentioned, especially since we're usually considering 3D space.

Please let me know what you think. --Leon (talk) 18:11, 4 August 2009 (UTC)

AHH, regarding your last question: no. But my point is that those properties appear to be given as defining pseudoscalars, in texts that make no mention of Clifford algebras. --Leon (talk) 18:14, 4 August 2009 (UTC)

What I'm asking is can you give an example of somewhere those properties are used as a general definition? For even dimensions, there are no non-trivial elements that satisfy that definition. I think you're right that the general definition should be given as a top-grade element of a Clifford algebra. Although for physics it's not really necessary to get into that stuff. Rckrone (talk) 18:25, 4 August 2009 (UTC)

Showing a function is right continuous

Let $f:[a,b]\to [0,\infty )$ be Lebesgue measurable. Define the function $\omega$ on $[0,\infty )$ by

$\omega (y)=m(\{x:f(x)>y\})$,

where m is Lebesgue measure.

(a) Prove that $\omega$ is right continuous.

I have been working on this for a while, looking at solutions from others even. I used the fact that $\omega$ is right continuous at y if, and only if, for every monotone decreasing sequence $\{y_{i}\}$ which converges to y, we have $\omega (y_{i})\to \omega (y)$. So, we just let $\{y_{i}\}$ be any such sequence and define

$E_{i}=\{x\in [a,b]:f(x)>y_{i}\},\qquad E=\{x\in [a,b]:f(x)>y\}$.

So, what I want to show is $\lim _{i\to \infty }mE_{i}=mE$. The solution I am looking at has

$\lim _{i\to \infty }mE_{i}=m(\lim _{i\to \infty }E_{i})=mE$

and this confuses me. First of all, my textbook never dealt with limits of sets, so although the right equality seems like it should be true, I am not sure. Obviously, the E_i are getting larger and larger and are getting closer and closer to E. But, this is not rigorous. And, I'm not sure why it is okay (if it is) to move the limit inside the measure. Any ideas? Thanks. StatisticsMan (talk) 16:42, 4 August 2009 (UTC)

Measures are indeed continuous from below in the sense required here. The proof is a simple application of countable additivity. Algebraist 16:52, 4 August 2009 (UTC)

Cool thanks. I have seen it with decreasing sequences I don't remember it for increasing. StatisticsMan (talk) 17:15, 4 August 2009 (UTC)

Solving a system of matrix equations

I have a set of equations of the following form: $X_{i}=ABCCB$, where the product on the right hand side is always a finite sequence of the matrices A B and C. I know the values of all matrices X_i, and I want to solve these for A, B and C. I'm trying to find the right brand of mathematics to deal with this situation, but whatever I google for turns up methods for solving systems of equations with the help of matrices, rather than solving matrix equations like these. I have the following questions. If anybody can point me in the right direction, I'd be much obliged:

1. Assuming all matrices are nxn, and the right hand side always contains a fixed number of matrices, how can I tell whether the problem is over or underdetermined?
2. If there are enough equations for the problem to be overdetermined, is it likely that an analytical least squares solution can be found?
3. What kind of numerical method would be appropriate for this kind of problem? I'm using a kind of genetic algorithm at the moment, but I expect there are less general methods.
4. Is there a way to transform a set of equations like this to the form Ax = b or some other general form?

Thank you. risk (talk) 17:52, 4 August 2009 (UTC)

Well, if somebody has a good clue to this problem, I'm curious to learn it too. In the meanwhile, my obvious remark is that if you have some more information (on the data X_i, or on the form of the equations) things may be different and easier . For instance, if the X_i commute, it could be natural to look for A,B,C commuting too, which changes a lot. In general, a possible approach is maybe to study the map taking (A,B,C) into (X₁, X₂,..) from the point of view of differential geometry and nonlinear analysis, and apply some principle for global invertibility or surjectivity of differentiable mappings (like e.g. Hadamard's theorem: "a C¹ map f:E→E with invertible differential satisfying |Df(x)^-1| ≤ a|x|+b, is a global diffeo"). As to the last point, I would say not: I don't see how one could reduce such a nonlinear problem to linearity. --pma (talk) 21:42, 4 August 2009 (UTC) A further remark: the toy version of your problem: "solving U=AB, V=BA for A and B", already shows that there are nontrivial necessary conditions for the solvability. Assuming that U or V is non-singular (hence both), the equations imply that U and V are similar matrices (U=AVA^-1). This can be checked from their spectral theory, and of course it is also a sufficient condition. Also, some material possibly related with your problem comes back googling "(system) of equations in groups", "equations in matrix groups" and similar combinations.--pma (talk) 08:45, 5 August 2009 (UTC)

What angle do you need to get a certain length of an arc?

Hi there, fellas!

I had nothing to do today, so in my boredom I decided to develop a cool little graphical app in Java. The thing is, there's a whole lot of arcs and circles in it, and trigonometry wasn't my star subject in high school, and by at this point I'm going insane with all the sines, coses and arc-tangents. So now I have a small little problem which is probably ridiculously easy for anyone with an actual brain, but I would really appriciate some help with it.

The problem is that I want to draw an arc of a circle where I can decide exactly how long the arc will be. So this is the problem: say you want to draw an arc that's P pixels long, on a circle that has a radius R pixels. What is the angle (in radians or degrees) of that arc?

As I said, it's probably very easy, but my brain is getting very, very tired, and it's a subject I'm not too comfortable with to begin with. 83.250.236.75 (talk) 21:17, 4 August 2009 (UTC)

You know the circumpherence of a circle or that radius...right? You also know that a pie-wedge taken out of that circle has an arc that's the same fraction of the circumpherence as the angle at the center is a fraction of 360 degrees. So for an arc of length 'L' for a circle of radius 'R', you need an angle that is: A = L * 360 / ( 2 * PI * R ) (in degrees of course). In radians, the 2*pi's cancel nicely so you get the much more elegant A = L / R. SteveBaker (talk) 21:56, 4 August 2009 (UTC)

Right, I figured it was something simple like that :) Thanks! 83.250.236.75 (talk) 22:07, 4 August 2009 (UTC)

(ec) Maybe it is safer to redirect this to the science/RD, specifically to Steve Baker. Oh, but he's already here! Anyway, isn't the number of pixels in a segment, or in an arc, depending on its position and not only on its length? Assuming the arc starts from the X-axis and goes counter-clockwise, and that it is within an angle α not larger than π/2 radiants, I imagine that the number of pixels in it is the same as in its vertical projection on the Y-axis. So, if the ratio s:=(no. of pixels in the arc)/(no. of pixels in the horizontal radius) is less than sqrt(2)/2 my guess is α=arcsin(s). Analogously, compare a short arc starting from the Y-axis with its horizontal projection. In general, I guess one reduces to these cases by adding/subtracting arc segments. --84.220.119.55 (talk) 22:19, 4 August 2009 (UTC)

As I said, I'm not terribly good at this, but Steve's solution worked perfectly. The arcs became the right lengths with his simple formula. 83.250.236.75 (talk) 00:59, 5 August 2009 (UTC)

Yeah - it's really dangerous having me on the math desk...I have this annoying tendancy to want to give answers that people can actually understand. Being a computer graphics geek - I understood perfectly that you meant to use the word "pixel" as a unit of measurement (like "inch" or "centimeter") and not in the sense of "a picture element". Indeed if you had wanted to count the number of picture elements along your arc, life would be difficult and painful and we'd CERTAINLY have to send all of the Mathematicians out to buy more pencils then lock the doors after them. It's a tough question to answer because a lot would depend on which of the 30+ known ways of drawing a circle you were actually using. That's because they tend to hit different numbers of pixels. What '84 is trying to tell us (using "math") is that counting the pixels is non-trivial. Consider that a thin straight (diagonal) line drawn from (x1,y1) to (x2,y2) hits the exact same number of pixels as a much shorter horizontal line drawn from (x1,0) to (x2,0) or a vertical line drawn from (0,y1) to (0,y2) - whichever is greatest. Hence, the number of pixels required to draw a circle is something rather complicated. On an infinitely high resolution display, the number of pixels would simply be four time the radius of the circle - but we don't have infinitely high resolution displays - and that makes the answer horribly complex - and (as '84 does not comprehend) critically dependent on which algorithm you use for drawing your circle. Fortunately, I happened to be here asking about 4D cross-products so none of this uglyness proved necessary!  :-P SteveBaker (talk) 13:25, 5 August 2009 (UTC)

4D analog of a 3D cross-product.

I have a function that takes a 2D vector and returns me another 2D vector that's at right angles to it. I have the standard cross-product function that takes two 3D vectors and returns me another 3D vector that's at right angles to both of the other two...what's the equivelant thing in four dimensions? I presume it needs three 4D vectors as input...but I'm having a mental block as to how to mush them together! (Oh - and is there a general name for these functions?) SteveBaker (talk) 21:50, 4 August 2009 (UTC)

The cross product only exists in 3D. The closest thing in higher dimensions is the wedge product. I've never studied it, though, so I'm not sure if it can be used to do what you want. However, you could easily use the dot product to get a system of simultaneous equations to solve. (Just call the desired vector (x,y,z,w) and dot it with each of the vectors you want it to be orthogonal to.) --Tango (talk) 22:20, 4 August 2009 (UTC)

(edit conflict - verbose answer) In 4D (assuming it's just a normal extension of 3D with another right angle 'squeezed in' somewhere unimaginable) then the cross product of 2 vectors is a plane (since there is a whole flat plane that is at right angles to both vectors).

Taking that plane, and then applying the third vector to it gives a single vector (or 2 vectors pointing in opposite directions to be pedantic) - just as when you took a vector in 2d and got another vector at right angles to it.

So yes in 4D 3 (non colinear) vectors define a fourth vector that is at right angles to all three.

To expand - it's easy to count dimensions (aka degrees of freedom) . Curiously it's an extension of this that allows us to show that in 4D one can rotate a vector around a plane 4(dimensions)=2(dimensional contraints given by the plane)+2(degrees of freedom remaining) - thus a vector constrained to rotate about the plane in 4D has 2 degrees of freedom -just the right amount for rotation.

(By the way the maths is dead simple - just linear algebra, and sins/cosines -even easier still if you've got an automatic linear equation solver to do the donkey work)

So to summarise - to derive the formula - first find the equation of the plane P that satisfies the conditions

V1 is perpendicular to the plane

V2 is perpendicular to the plane

Then use your equation for the right angle in 2D and apply it to the plane with the 3rd vector as a constraint - to give the answer (a vector) 83.100.250.79 (talk) 22:24, 4 August 2009 (UTC)

See Levi-Civita symbol.

2D: $\,y_{i}=\epsilon _{ij}a_{j}$

3D: $\,y_{i}=\epsilon _{ijk}a_{j}b_{k}$

4D: $\,y_{i}=\epsilon _{ijkl}a_{j}b_{k}c_{l}$

The 2D formula is shorthand for

$\,y_{1}=a_{2}$

$\,y_{2}=-a_{1}$

The 3D formula is shorthand for

$\,y_{1}=a_{2}b_{3}-a_{3}b_{2}$

$\,y_{2}=a_{3}b_{1}-a_{1}b_{3}$

$\,y_{3}=a_{1}b_{2}-a_{2}b_{1}$

Bo Jacoby (talk) 22:26, 4 August 2009 (UTC).

In my undergraduate multivariate calculus course (which was like Calc 4), our book had a few exercises that gave an analog for any finite dimension, with a subtitle "Cross products in R^n". As you guessed, you need N-1 vectors in N dimensions and you do it exactly like a cross product. That is, you take a matrix and you put e_1, e_2, ..., e_n (instead of i, j, k) in the top row and you put the N-1 vectors in the rest of the rows. Then, you compute the determinant. The exercises are to show it has various properties. Most are obvious from properties of determinants. But, one is that the vector you obtain is orthogonal to all N-1 vectors. My book is Vector Calculus by Susan Jane Colley. My edition is likely not the newest any more but it's a few exercises at the end of Section 1.6 Some n-dimensional Geometry in my edition. StatisticsMan 01:13, 5 August 2009 (UTC)

I think that Tango is correct about the wedge product thing, but incorrect about the number of vectors required (still two). It's important to remember that oriented line segments (vectors) are dual to oriented plane segments (bivectors, which are what you get with a cross product on vectors though you don't realize it!) in R^{3} but not R^{4}. The wedge product is constructed from the Clifford algebra. The lack of duality is manifest in that the wedge of two vectors from R^{4} spans a six dimensional vector space as opposed the three. Might I ask your purpose in knowing about this? I may be able to recommend some resources.--Leon (talk) 13:10, 5 August 2009 (UTC)

OK - the reason is kinda complicated & messy. <sigh> When you multiply a bunch of nice simple rotation matrices together on some real-world device such as a graphics processor, you get roundoff error. That error isn't particularly important in itself because there is typically a degree of interactivity that tends to correct any total error in the angle rotated through. However, after enough times through the multiplication pipeline, the matrices start to become non-orthonormal...which tends to result in all sorts of nasty skews and scalings that are pretty undesirable. This is a common enough problem in computer graphics and we've learned to "repair" 3x3 rotation matrices that seem to be a little the worse for wear by treating the three rows as vectors, normalizing them and then recalculating the third row to make it be orthogonal to the other two. This results in a matrix that may be rotating by slightly incorrect angles - but at least there is no skew or scale going on. You doubtless have "better" ways to do this - but before you leap in with them, I'll point out that I'm doing this a bazillion times a second on graphics hardware that's very efficient at things like normalizing and doing cross products and the like - and sucks badly at doing any of the things you were just about to suggest!

Well, now I need to do the same thing with some 4D rotation matrices...so I need an analogous function that (given three rows of the matrix) allows me to calculate the fourth row to be orthogonal to the other three. SteveBaker (talk) 13:55, 5 August 2009 (UTC)

August 5

A question about geographical coordinates and significant digits

There are a number of different ways in which a geographical position can be identified using coordinates. For example the location of Mount Whitney in California is listed in different ways by different sources.

The Geographic Names Information System (GNIS) uses two formats:

36.5782684, -118.2934244 (i.e. 36.5782684°N, 118.2934244°W) — Decimal Degrees (DD)

363442N, 1181736W (i.e. 36°34′02″N, 118°17′73″W) — Degrees, Minutes, Seconds (DMS)

The U.S. National Geodetic Survey (NGS) uses the following format:

36 34 42.89133(N) 118 17 31.18182(W)

Others use:

N 36° 05.885 W 115° 05.583 — Degrees, Minutes (DM)

For example when converting DMS to DD format how may decimal places are significant:

36°36′36″ then (36*3600 + 36*60 + 36)/3600 = 36.61°

How can I determine how many decimal places are significant when converting (e.g. from NGS data) to DD format. Is there some general rule I can follow. What about the inverse function (i.e. converting DD to DMS). Assume 36° implies >= 35.5° and < 36.5° etc. --droll [chat] 00:18, 5 August 2009 (UTC)

Find the precision of the initial value and convert it into the new units. For example 36°34′02″ implies uncertainty on the order of 1". Converting 1" to degrees gives you .000277°. You don't want the precision of your new value to be any higher than that of the initial value, so round up your uncertainty to the nearest power of 10 (.001° in this case) and get rid of any digits beyond that one. In this case the result would be 36.567°. Rckrone (talk) 02:59, 5 August 2009 (UTC)

Sorry if being a bit off-topic: When comparing geodesic data from two different sources, it is also convenient to verify that coordinates are expressed with the same reference datum. Pallida  Mors 05:38, 5 August 2009 (UTC)

True, I use decimal degrees with 4 decimal places in any maps and give a selection of maps a user could use. They're not too badly correlated now but in the past you could get to completely the wrong building with different maps. By the way I try to indicate the entrance rather than the body of a building. Dmcq (talk) 09:33, 5 August 2009 (UTC)

By the way I'd convert the precision up rather than down if comparing things, then subtract the two and see if the difference is smaller than what you're checking for. Who is going to actually read the numbers? Normally they'd just go to another computer. Dmcq (talk) 09:42, 5 August 2009 (UTC)

Mandelbrot set

If one iteration is done on the whole Mandelbrot set, is the result the whole of the set or just a subset? --Masatran (talk) 09:43, 5 August 2009 (UTC)

An iteration of what? (Isn't that you mean the Julia set of a map, instead?) --pma (talk) 10:14, 5 August 2009 (UTC)

Did you mean to take test the iterated condition only once - eg using

zn+1 = zn2 + c

Only to go as far as z₁

That I think would be a subset, or more specifically a single quadratic equation.83.100.250.79 (talk) 13:10, 5 August 2009 (UTC)

And what do you mean? I can't understand neither of you. Can you formulate your question precisely?--pma (talk) 14:38, 5 August 2009 (UTC)

See Mandelbrot set first paragraph - using the formula - one iteration

Mathematically, the Mandelbrot set can be defined as the set of complex values of c for which the orbit of 0 under iteration of the complex quadratic polynomial z_n+1 = z_n² + c remains bounded.

The first iteration (for a general point [x,y]) yields a quadratic formula for the boundary condition, the second iteration a quartic, and so on...HappyUR (talk) 15:07, 5 August 2009 (UTC)