Wikipedia:Reference desk/Archives/Mathematics/2009 May 26

= May 26 =

Statistical fallacies in White Nationalist propaganda?
What statistical or logical fallacies are present in this article, other than misleading bar graphs (where the baseline isn't at zero) and a possible neglect of differences in ambition between the races? (I'm afraid that if no other fallacies can be found, I'll have trouble believing that reverse discrimination is any less of a reality than the authors claim.) Neon  Merlin  02:32, 26 May 2009 (UTC)
 * Would you be equally eager to find fallacies if the propaganda was heading in another direction? Bo Jacoby (talk) 06:58, 26 May 2009 (UTC).
 * The greater the harm that could be caused by falsely accepting a claim, the more carefully it's worthwhile to check it. Neon  Merlin  18:12, 26 May 2009 (UTC)
 * It is a statistical fallacy to scrutinize disagreeable arguments closer than agreeable ones. (Suppose that the disagreeable position is supported by 10 arguments, and the agreeable position is supported by 10 arguments too. Let extraordinary scrutiny discard 5 of the disagreeable arguments and none of the agreeable ones. Then the result is 10-5 in favor of the agreeable position. This is obviously unfair. Perhaps scrutiny of the agreeable position would have discarded 6 arguments or more). Bo Jacoby (talk) 19:17, 26 May 2009 (UTC).
 * There are questions about sample size. It would be good to know how many people fell in each category. Data for Hispanic Females in top 2%ICQ look suspicious, perhaps only one or two people fell in this range. Looking at the graphs of race arranged by ICQ shows that only very few blacks and Hispanics fall in the upper categories. Taking this into account we see a lot of selective presentation in the data - the first graph you see is the top 2% where black income > white income, but if you were to average blacks and whites ignoring ICQ you would get the reverse. There does seem to be an interesting effect in that the very few blacks who fall in the top 2% do especially well. --Salix (talk): 20:18, 26 May 2009 (UTC)
 * Bo is postulating a false dichotomy, and promotes the fallacy that the number, as opposed to the quality, of arguments is relevant.
 * Well, sorry, my example should have been: "Suppose that a disagreeable position is supported by an argument, and an agreeable position is also supported by an argument. Let extraordinary scrutiny discard the quality of the disagreeable argument, while the agreeable one is not scrutinized. Then the result is 1-0 in favor of the agreeable position. This is obviously unfair. Perhaps scrutiny of the agreeable position would have discarded the quality of that argument too". Bo Jacoby (talk) 04:46, 27 May 2009 (UTC).
 * But back to the original question. I don't have the time or inclination to do a full analysis. But a couple of problems are immediately obvious.
 * They separate "Jews" and "Whites". As far as I know, the U.S. census does not separate these - Jews are (overwhelmingly) white.
 * The sample size is 12000. 10% of the population is black, so thats 1200. Assuming IQ as determined by a test is independent of race, 2% of that is 24 - not really statistically significant. For Jews, the number would be 2.4...
 * The whole presentation is assuming a zero sum game and a competition. It's neither.
 * Looking at, I again cannot find any indication that Jews have been sampled as a distinct group.
 * There is a self-selection effect. People who do not answer are dropped.
 * --Stephan Schulz (talk) 20:52, 26 May 2009 (UTC)
 * The main fallacy is that the 'study' completely neglects taking account of any factors other than IQ and race in determining income. Maybe Blacks or Hispanics work harder when they are given educational opportunities. Maybe the see money as the ultimate reward and so gravitate towards jobs that pay well rather than jobs that give them a balanced lifestyle. Maybe Whites perform less well at IQ tests.
 * But that wasn't the first thing I noticed. The first thing I noticed was that the author of the study mis-spelled "affect" (as "effect") in several places. Obviously something that would have been dealt with if he had been given that college place denied him by affirmative action. :-) DJ Clayworth (talk) 21:04, 26 May 2009 (UTC)
 * You mean if they had been given the "priveleges" of an education? ;-) --Stephan Schulz (talk) 21:56, 26 May 2009 (UTC)

Nicer proof of Stirling's approximation
Hi there guys - I was looking for a nice quick (as short as possible!) proof of Stirling's Approximation and I stumbled upon this post here: - can anyone suggest how to show the remainder term which the OP ignores won't affect the result (i.e. his second question at the end of the proof)? I'm intending to (possibly) use this in an exam if such a question turns up, so ideally the shortest possible proof that the result is valid even with the remainder term considered would be brilliant.

Equally, if anyone has any even -shorter- proofs, they'd be extremely well received! :)

Thanks a lot! Mathmos6 (talk) 04:37, 26 May 2009 (UTC)
 * The proof, in a nutshell, says that $$n! = n^ne^{-n}\int_{-n}^{\infty}\exp\left(n(\log(1+\delta/n)-\delta/n)\right)\,d \delta$$ and that $$\int_{-n}^{\infty}\exp\left(n(\log(1+\delta/n)-\delta/n)\right)\,d \delta \approx \int_{-n}^{\infty}\exp\left(\frac{-\delta^2}{2n}\right)\,d \delta \approx \sqrt{2n\pi}$$. The reason the approximation $$\log(1+\delta/n)-\delta/{n} \approx \frac{-\delta^2}{2n^2}$$ works is that the integrand is non-negligible only when $$\delta$$ is on the order of $$\sqrt{n}$$, in which case the higher-order terms of the expansion are negligible (for large n). The fact that n is large was used both here and in the approximation $$\int_{-n}^{\infty}\exp\left(\frac{-\delta^2}{2n}\right)\,d \delta \approx \sqrt{2n\pi}$$. Proving that all of this is indeed valid might require a bit more work. -- Meni Rosenfeld (talk) 12:18, 26 May 2009 (UTC)

Bijection between R and C
Since R, the set of reals and C, the set of all complex numbers have the same cardinality so there must be a bijection between them. Can someone tell me what is it?--Shahab (talk) 06:53, 26 May 2009 (UTC)


 * The complex numbers set $$\mathbb C$$ can be represented as a Cartesian square or reals $$\mathbb C = \mathbb R \times \mathbb R$$ (with appropriate definitions of operators +, –, etc.) so you just need a $$\mathbb R \to \mathbb R \times \mathbb R$$ bijection. This was discussed many times, most recently on May 7. See the archive:       Reference desk/Archives/Mathematics/2009 May 7 CiaPan (talk) 09:03, 26 May 2009 (UTC)

Novel Problem Related To Research
I have a novel problem, but stating it would require stating original research of sorts. Can I ask such a question at all? I am interested only in using someone else as a reference in adding it to a book I am working on.Julzes (talk) 08:24, 26 May 2009 (UTC)

Am I right to assume it won't be a problem as long as I put it here rather than in an article? That seems to be right.Julzes (talk) 09:47, 26 May 2009 (UTC)


 * Questions about mathematics can be asked here, as long as they do not contravene the Reference Desk guidelines. Questions about using or contributing to Wikipedia should be asked at the Wikipedia Help Desk. Your objective "using someone else as a reference in adding it to a book I am working on" is rather unusual, and you may need to expand on that further if you want to receive any useful responses. Gandalf61 (talk) 09:58, 26 May 2009 (UTC)


 * Okay, thank you Gandalf61. I can proceed.  This is a rather unusual type of question from a mathematical perspective because it is related to base-ten specific results, and is asking for the general case.  The objective in using any possible answer I might get is to strengthen an argument concerning the notability (in general, not just per policy here) of the identity (365+1/4)^4=17797577732+7^2/2^8 and the related (365+1/4)^2=3^7*61+9/16, which just look like meaningless coincidences.  The first of these I stumbled upon while trying to answer a birthday coincidence problem, and then I asked the simple question yielding the second on 03/06/09.  It's a strange story.  Anyway, the problem comes first from observing that the sums of the digits of the numerator and denominator of the fractional part of the first identity both equal 13, and then from a relatively simple establishment of the fact that the numbers congruent to 13 modulo 16 are those that can replace 365 to generate the same fractional part.  I thought it was 32 rather than 16 at first, so I noticed a further coincidence involving the final digits of the whole part.  Well, in the same sequence ({32k+13+1/4}) I noted that there are quickly other cases with this group of coincidences (other than all the sevens), and this provoked an even narrower problem (having necessarily fewer solutions).  I immediately found, that is to say, that (13*96+13+1/4)^4 has 96 as the terminal digits of its whole part.  Later I researched to find 672 has the same property as 96 in this regard, and that there are at least no more solutions up to 25000.  I have little programming experience, but I will be attempting to solve the following formal statement of the problem.  I don't think that there is a nice non-enumerative approach to finding results.  My interest is in finding all reasonably small solutions (it's possible to argue heuristically that there ought to be about the same number of solutions on average for each length in digits, and so infinitely many solutions), over all choices of fraction and exponent, and also over all (reasonably small?) choices of base.  I don't want to set a hypothesis.  I would like results, and whether it does or doesn't confirm what I guess could be useful to me.  Here is the formal statement, then:

For what values of the base B, a rational number x in the unit interval, integral power n greater than 1, and integral modulus M does the expression (k*M+S+x)^n have the sum of the digits of the numerator of its fractional part equal to S, where S is the sum of the digits of the denominator of x raised to the nth power (and, necessarily, of the expression given), for all integral k AND have the expression's integral part's terminating (rightmost) digits--the proper number of digits--form M when k=S? (Enumerate solutions up to some Bmax of the choice of base and constrain solutions so that M is also less than some Mmax--arbitrarily at the researcher's discretion). That's about it. I'll check back here periodically to see if anybody has said something I might take interest in or need to give a response to. Thanks.Julzes (talk) 12:15, 26 May 2009 (UTC)


 * The OP repeats the substance of yesterday's post "New Elementary Suggestive Results Found By Birthday Coincidences Related To History". The poster is clearly seeking a collaborator to further OR and possibly a book to be written about an as yet unknown hypothesis.Cuddlyable3 (talk) 09:38, 27 May 2009 (UTC)


 * Clarification: There will be a book late this year. Whether it is mostly only the problem that makes it (without solution) in addition to other material is at issue.  Whether there is confirmation of hypothesis or something more or less is sort of correct.  Hoping to refer to published work unless I actually can describe the process of my own research in the book.  (Little programming experience is hindrance)Julzes (talk) 10:44, 27 May 2009 (UTC)


 * More detail coming in my talk pageJulzes (talk) 16:49, 28 May 2009 (UTC)


 * I was going to reply but I notice tl;dr redirects to Top Level Domain which is just silly so I haven't any reference. Dmcq (talk) 10:22, 27 May 2009 (UTC)


 * There is nothing in the equalities you wrote. Since your question seems to have to do with birthdays, what about redirecting your query to the Entertainment desk? --139.18.116.55 (talk) 17:32, 28 May 2009 (UTC)


 * Well, the book is planned to be rather broad in scope. The question was MOTIVATED by specifics (having to do with birthdays).  Your objection has been noted.Julzes (talk) 12:44, 29 May 2009 (UTC)


 * In the event this has been held as of interest by someone other than myself, I will report on the current state of things. There will not be a book in 2009, and only if I collaborate with a friend will there be one in 2010.  I'm taking my time with a host of matters that need to be discussed and some that need to be researched further.   I have begun programming work, and I can report on some interesting things, and I will do so shortly.Julzes (talk) 03:31, 1 January 2010 (UTC)


 * In regards to the specific base-ten case with S=13, there are only the two solutions given up to a value of 10^12. There are numerous cases with exponent n=2 in a great number of bases, there are a much smaller number of cases for n=3, and for n=4 the base-10 result is not unique, though it is rare.  For each of n=5 and n=6, exactly one class of solutions exists for a reasonable-sized, and n=7 and n=8 yield nothing.  The general-case search done was for all bases up to the minimum of 4096 and the denominator of the result and all multiples up to 2^20 times the minimum possible modulus for M.  The specifics for n=4, 5 and 6 are interesting, but unless someone actually asks I won't expand on the subject right here.Julzes (talk) 09:11, 1 January 2010 (UTC)

Published puzzle
Trying to solve a newspaper teaser, I've converted the wordy description into the following form, but can't get an answer despite a lot of fiddling with it.

Find w so that w/(k+1) is either 10, 11, 12, 13, 14 or 15 for two different values of k (k>1, k≠2, k≠3), and for each value of k

if 13 then w/(k-1) is either 10, 11, 12, 13, 14 or 15

and the four values from (10,11,12,13,14,15) are all different.

There is no stated requirement for k or w to be integral. Can anyone confirm that there is indeed a solution, and if so point me very strongly in the right direction?81.154.107.95 (talk) 08:25, 26 May 2009 (UTC)


 * Attribution? Published where?  —Preceding unsigned comment added by Julzes (talk • contribs) 12:44, 26 May 2009 (UTC)


 * There is obviously no solution, unless one imposes severe restrictions on k. For example, it is impossible for w(2 − k)/(k + 1) to be 10, 11, 12, 13, 14 or 15 for each value of k such that 1 < k < 2 unless there are only finitely many values of k allowed in that interval. The whole thing sounds very odd, are you sure you got the description right? — Emil J. 13:14, 26 May 2009 (UTC)

It was sent to me verbatim, I don't know the source. The scenario is people of different swimming speeds criss-crossing a river and meeting at various distances from the bank. The problem is to find the width of the river whick I've called w, k being the ratios of the speed of the fastest swimmer to that of the other two involved. I'm confident that my formulation is correct.

It would be clearer if "for each value of k" was replaced by "for both values of k" or "for each of the two values of k".81.154.107.95 (talk) 16:10, 26 May 2009 (UTC)

Solved it - w=32, k1=9/7 (generating 14 and 10), k2=19/17 (generating 15 and 13). In each case 1<k<2.81.154.107.95 (talk) 19:41, 26 May 2009 (UTC)

Diffeology
Diffeology is the branch of mathematics which studies diffeological spaces. I have known about the existence of this field for sometime, and have some relatively basic knowledge of it. However, this knowledge is mainly from the perspective of manifold theory. This is reasonable as much of diffeology is essentially a generalization of concepts in manifold theory, but of course there is a lot more to the subject than this. Despite some searching, I have not found any results on diffeology in the literature, except in regards to its relation with manifolds. In essence, I acknowledge that diffeology is very much related to manifold theory, but I am unsure of the extent to which the two subjects are related. For example, I quote one paper:

"Let (M,F) be a foliated manifold. We study the relationship between the basic cohomology Hb(M,F) of the foliation and the De Rham cohomology H(DF) of the space of leaves M/F as a quotient diffeological space. We prove that for an arbitrary foliation there is a morphism from H(DF) to Hb(M,F). It is an isomorphism when F is a Q-foliation."

Forgive me for my ignorance, but is there any research in diffeology currently, outside the realm of manifold theory (i.e research in diffeology for the sake of research rather than for the purpose of applications in manifold theory)? There is one online book on diffeology which can be found rather easily. However, in the one research paper I found on the subject, this was the only book quoted purely on the subject. This suggests strongly that not much active research exists in the field of diffeology, but anyone who can either contradict this or support this is most welcome. In particular, I feel that many important topics are absent in diffeology, and also, I would appreciate the quotation of any research papers on the subject in reputable journals. Thankyou for future comments. -- PS T  13:35, 26 May 2009 (UTC)

Implicit differentiation problem
I am having trouble understanding where I have gone wrong with a question about implicit differentiation. It states: "A curve is definted implicitly by the equation $y^3=2xy+x^2$. Show that $\frac{\mbox{d}y}{\mbox{d}x}=\frac{2(x+y)}{3y^2-2x}$."

Now, then, my method for differentiating $$2xy$$ is:

$$u=2x$$

$$u'=2$$

$$v=y$$

$$v'=1$$

$$\Rightarrow vu'+uv'=2y+2x$$

$$\therefore 3y^2\frac{\mbox{d}y}{\mbox{d}x}=2y\frac{\mbox{d}y}{\mbox{d}x}+2x+2x$$

$$3y^2\frac{\mbox{d}y}{\mbox{d}x}-2y\frac{\mbox{d}y}{\mbox{d}x}=4x$$

$$\frac{\mbox{d}y}{\mbox{d}x}\left ( 3y^2-2y \right )=4x$$

$$\frac{\mbox{d}y}{\mbox{d}x}=\frac{4x}{3y^2-2y}$$

But the mark scheme says:

$$3y^2\frac{\mbox{d}y}{\mbox{d}x}={\color{Red}{2x}}\frac{\mbox{d}y}{\mbox{d}x}+2y+2x$$

$$\Rightarrow \left ( 3y^2-2x \right )\frac{\mbox{d}y}{\mbox{d}x}=2y+2x$$

$$\Rightarrow \frac{\mbox{d}y}{\mbox{d}x}=\frac{2(x+y)}{3y^2-2x}$$

Now, I don't understand how the $$\frac{\mbox{d}y}{\mbox{d}x}$$ gets applied to the $$2x$$ rather than the $$2y$$; any help would be very much appreciated.  It Is Me Here   t / c 16:51, 26 May 2009 (UTC)
 * You should have said $$v' = \frac{dy}{dx}$$ instead of 1. Then adjust your vu' + uv' accordingly, and you should get it right. Staecker (talk) 17:00, 26 May 2009 (UTC)
 * (ec) To differentiate a product you need to differentiate both parts with respect to x. $$\frac{\mathrm{d}}{\mathrm{d}x}(2x)=2$$, as you said, but $$\frac{\mathrm{d}}{\mathrm{d}x}(y)=\frac{\mathrm{d}y}{\mathrm{d}x}$$, not 1. --Tango (talk) 17:02, 26 May 2009 (UTC)


 * Yay, thanks, both of you!  It Is Me Here   t / c 17:33, 26 May 2009 (UTC)

It's the product rule. Note:
 * $$\frac{dy}{dx}$$

is not the thing that "gets applied". The thing that "gets applied" is
 * $$ \frac{d}{dx}. $$

And what it gets applied to is y, so the other factor, 2x, is still there. In the next term, it gets applied to 2x, and the y is still there. So
 * $$\frac{d}{dx}(2xy) = 2x\frac{d}{dx}y + y\frac{d}{dx}(2x) = 2x\frac{dy}{dx} + 2y.$$

Michael Hardy (talk) 19:34, 26 May 2009 (UTC)