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March 17

Elliptic and Parabolic equations

I am still confused what exactly are elliptic and parabolic differential equations, and why exactly are they called so. Could someone explain? deeptrivia (talk) 00:32, 17 March 2009 (UTC)

The operator of an elliptic pde, like the laplacian, has Fourier transform x^2 + ... + y^2 like a circle or ellipse. The operator of a parabolic pde, like the heat operator, has Fourier transform x^2 - y, like a parabola. The operator of a hyperbolic pde, like the wave operator, has Fourier transform x^2 - y^2 like a hyperbola. JackSchmidt (talk) 04:20, 17 March 2009 (UTC)

Two point boundary value problems

I want to get as much information as I can about nonlinear two-point third order boundary value problems (e.g., y ' ' ' + a*y' + b*cos(y) = 0, y(0) =y0, y(1) = y1, y'(1) =yp1). Some things I would like to know are: existence of solutions, uniqueness of solutions, analytical solutions (if possible), approximations, numerical solutions, range of the solution space, etc. Could anyone point me to some references on this? Thanks! deeptrivia (talk) 01:22, 17 March 2009 (UTC)

Surface name?

Does the undulating sheet like curve/surface formed by the equations $\sin(x)cos(y)\$ or $\sin(y)cos(x)\$, etc. have a paricular name? --Leif edling (talk) 07:00, 17 March 2009 (UTC)

You may talk about the equations $\scriptstyle z=\sin(x)\cos(y)\$ or $\scriptstyle z=\sin(y)\cos(x)\$. No, I do not know of particular names. Bo Jacoby (talk) 10:40, 17 March 2009 (UTC).

Oops! Well $\scriptstyle z=\sin(x)\cos(y)\$ was what I meant. --Leif edling (talk) 16:58, 17 March 2009 (UTC)

Chess problem

Consider a game of chess in which Black gives White queen odds: that is to say, the game is played exactly like a standard game of chess, except that Black starts without his queen. Given skillful play by both sides, White will win, of course, but for how many moves can Black delay checkmate? —Preceding unsigned comment added by 75.24.76.213 (talk) 08:39, 17 March 2009 (UTC)

I'd be amazed if this is known. In fact, I don't think anyone has even proved that white can force a win in this situation. Algebraist 13:08, 17 March 2009 (UTC)

On this page (warning: old and hard to load) dedicated to "difficult" computer problems, there was a note on the position where Black has only a Queen. While the page was about 12 years old, it pointed out that while computers could find mates in 12, it was suspected that there might be a mate in 11, but they couldn't prove or disprove it. While I suspect that today's computers could answer this question one way or the other, it shows how hard it is to "prove" a win.

I don't doubt that Queen odds is lost with best play, but it will be some time before it's proven, let alone "Mate in ?" determined. (BTW, the main site of that Queen problem has quite a few interesting positions and commentary.) Baccyak4H (Yak!) 15:38, 17 March 2009 (UTC)

So if it's so hard to solve, it would perhaps make an interesting game. I have seen board games with unequal armies, but never one in which the win condition for the weaker side was simply to survive for a given number of moves. —Preceding unsigned comment added by 75.37.236.230 (talk) 20:04, 17 March 2009 (UTC)

White has a winning position... But finding a winning line is almost as hard as solving chess... For the record, after a minute or two my engine can find nothing but a very comfortable position for white (+11.28, depth of 20 half-moves). Pallida  Mors 15:35, 17 March 2009 (UTC)

It depends on how well each side plays. I don't know whether you can define "equal ability". --PST 23:29, 17 March 2009 (UTC)

We aren't talking about 'equal ability', we're talking about perfect play, which is easy to define. Algebraist 23:38, 17 March 2009 (UTC)

Expected value of length of a chord

If I were to randomly construct a chord in the unit circle, what is the probability that the length of the chord is greater than ${\sqrt {3}}$? What is the expected value of the length of the chord? AMorris (talk)●(contribs) 09:23, 17 March 2009 (UTC)

The answer depends on the random method of construction of the chord. You may for example (A) pick a random point inside the unit circle as the center of the chord, or you may (B) pick a random point on the circumference of the circle as one end point of the chord and pick a random arc between 0 and 360 degrees for defining the other. In case A the chord is greater than $\scriptstyle {\sqrt {3}}$ when the midpoint is inside a circle of radius 1/2, and the probability is 1/4. In case B the the chord is greater than $\scriptstyle {\sqrt {3}}$ when the arc is between 120 and 240 degrees, and the probability is 1/3. So your question is not well posed and the answer is not well defined. Bo Jacoby (talk) 10:28, 17 March 2009 (UTC).

(ec) It depends what you mean by "randomly" - see Bertrand's_paradox. AndrewWTaylor (talk) 10:30, 17 March 2009 (UTC)

Or in other words, it depends on the probability distribution of the random chord. Michael Hardy (talk) 20:22, 18 March 2009 (UTC)

Trapezoid solving an angle

I don't know how to solve for angle MNC. I know that it is an isosceles trapezoid and I know that angle B is 88 degrees. That means that angle A is 88 degrees, and angles C and D are 92 degrees. But I don't know how to solve for angle MNC, could you guys help me out? BTW- M is the midpoint of AD, and N is the midpoint of BC. I just drew the horizontal line in there. Thanks for the help! --71.98.25.121 (talk) 23:31, 17 March 2009 (UTC)

By what you are saying MNCD is also an isosceles trapezoid and the MNC=ABC=88 degrees. Dauto (talk) 00:28, 18 March 2009 (UTC)

Please remember that we do not do homework here, although it is customary to provide guidance or a hint if the OP has shown an effort to solve the problem themselves (as this poster has done). A few helpful comments or directed questions would have been better. ("Give a man to fish...") -- Tcncv (talk) 01:09, 18 March 2009 (UTC)

Yes, thank you for your post. I am aware that you do not do homework here. My assignment included over 40 problems; I only asked about one part of a problem I had already tried but didn't understand. Thank you for your explanation Dauto. --71.98.25.121 (talk) 01:37, 18 March 2009 (UTC)

My apologies, I did not mean to be condescending, and I think you misunderstand me – my note was not directed at your question, but at the way it was answered. Questions such as yours are welcome, especially since you had shown that you had already made a good faith effort at working it yourself (which I acknowledged above). My point was that a helpful hint that allows you to find the solution is preferable to giving the answer outright. At least that is my understanding of our policy here. And yes, Dauto is a regular contributor of much useful information here, such as the detailed analysis in the Probability in Rummikub topic above. -- Tcncv (talk) 02:39, 18 March 2009 (UTC)

Since AM and BN have equal lengths and MAB and NBA have equal angles, the distances of the two points M and N below the line AB must be equal; hence the line MN is parallel to AB. Two lines parallel to each other must both meet a line crossing both of them at the same angle. That should answer your question. Michael Hardy (talk) 20:20, 18 March 2009 (UTC)

Thank you for your responses. I'm guessing Michael Hardy's response is the "correct" one. And I'm sorry, Tcncv, for being so rude. I was having a bad day. Not much of an excuse, but I shouldn't have taken it out on you. --71.117.36.70 (talk) 21:02, 18 March 2009 (UTC)

No Problem. I suspect I am not alone in that I enjoy seeing questions such as yours come along. They add variety and give us amateurs a chance to exercise our minds while making ourselves useful. Thank you. -- Tcncv (talk) 00:23, 19 March 2009 (UTC)

March 18

F-statistic application

In comparing of two models(regressions), using F-statistic, why they should be nested in each other? —Preceding unsigned comment added by 217.219.166.149 (talk) 10:42, 18 March 2009 (UTC)

If the models aren't nested, then there is no way to build a hypothesis test to carry out with F-values or generic ANOVAs, as far as I know. Hence, your F-statistics serve merely as fitness measures, but nothing more than that (you may dischard one model, but not compare one with another if both seem to adjust well).
On the other hand, if models are nested, possibilities of variance analysis for direct comparisons are straightforward, as with the case of significance of additional variables, etc.
There are some ways of comparing non-nested models: for instance, analysis of information measures, such as AIC, etc. I'd like to mention the article Model selection, as it can be of your interest. Pallida  Mors 16:33, 18 March 2009 (UTC)

AIME

Does anybody know where to find solutions for the 2009 AIME? Lucas Brown 42 (talk) 18:25, 18 March 2009 (UTC)

http://www.unl.edu/amc/e-exams/e7-aime/archiveaime.shtml should have them at some point. JackSchmidt (talk) 02:59, 19 March 2009 (UTC)

Prime number problem

You don't need to give me the whole answer, but I need to know how to get started. Find all the prime numbers p with the property that 8p^4-123 is also a prime. Thanks! 76.248.244.232 (talk) 22:43, 18 March 2009 (UTC)

THIS IS PART OF THE IUPUI MATHEMATICS CONTEST AND THE ORIGINAL POSTER IS TRYING TO CHEAT. 131.215.158.184 (talk) 19:50, 20 March 2009 (UTC) -->

March 19

Functional Convergence

In a recent thread, if I understand correctly, pma says that $\textstyle \sum _{1\leq k\leq n}\log \cos(k^{2}{\sqrt {n^{5}}}s)$ converges pointwise to $\textstyle -s^{2}/10$ as n approaces infinity. How would you prove that? Black Carrot (talk) 07:53, 19 March 2009 (UTC)

Won't the limit depend on what branch of log you choose for negative arguments? Algebraist 10:23, 19 March 2009 (UTC)

My apologies: I made a misprint there (now corrected): the change of variables was $t=n^{-5/2}s$, with a minus in the exponent (this is consistent with the line below, that had it). So the term ${\sqrt {n^{5}}}$ is at the denominator, and the argument of log goes to 1 (actually, in that computation it was always positive). Do you see how to do it now?--pma (talk) 12:40, 19 March 2009 (UTC)

Here it is:

• Write the second order Taylor expansion for $\textstyle \log \cos(x)$ at 0, with remainder in Peano form: so, for all $x\in [0,\pi /2)$

$\log \cos(x)=-{\frac {x^{2}}{2}}+o(x^{2})$, as $x\to 0$.

• For any s we only have to consider the integers $n>4s^{2}/\pi ^{2}$. Replace $x={\frac {k^{2}}{\sqrt {n^{5}}}}s={\frac {s}{\sqrt {n}}}\left({\frac {k}{n}}\right)^{2}$ in the expansion above, getting

$\log \cos \left({\frac {k^{2}}{\sqrt {n^{5}}}}s\right)=-{\frac {s^{2}}{2n}}\left({\frac {k}{n}}\right)^{4}+o({\frac {1}{n}})$, as $n\to \infty$, and uniformly for all $1\leq k\leq n$.

• Summing over all $1\leq k\leq n$

$\textstyle \sum _{1\leq k\leq n}\log \cos({\frac {k^{2}}{\sqrt {n^{5}}}}s)=-{\frac {s^{2}}{2n}}\sum _{1\leq k\leq n}\left({\frac {k}{n}}\right)^{4}+o(1)$, as $n\to \infty$.

• Then you may observe that $\scriptstyle {\frac {1}{n}}\sum _{1\leq k\leq n}\left({\frac {k}{n}}\right)^{4}$ is the Riemann sum for the integral of $x^{4}$ on [0,1] (or use the formula for $\scriptstyle \sum _{1\leq k\leq n}k^{4}$) and conclude that the whole thing is $\textstyle -{\frac {s^{2}}{10}}+o(1)$.

Warning: I have re-edited this answer, to make it more simple and clear (hopefully) --pma (talk) 13:40, 19 March 2009 (UTC)

Differential Equation

How should one go about solving this equation.

$y{\frac {dy}{dx}}=xy{\frac {d^{2}y}{dx^{2}}}+x{\bigg (}{\frac {dy}{dx}}{\bigg )}^{2}$

92.9.236.44 (talk) 20:30, 19 March 2009 (UTC)

The right hand side is $x{\frac {d}{dx}}{\bigg (}y{\frac {dy}{dx}}{\bigg )}$. Does that help? —JAO • T • C 20:48, 19 March 2009 (UTC)

Ah yes. It seems to yeild a solution of the form $y={\sqrt {Ax^{2}+B}}$ does that seem correct? —Preceding unsigned comment added by 92.9.236.44 (talk) 21:01, 19 March 2009 (UTC)

Check for yourself - differentiate that a couple of times, substitute everything in and see if the two sides match. If they do, you've got it right, if they don't, you haven't! --Tango (talk) 23:05, 19 March 2009 (UTC)

March 20

First and Second derivative

$h(x)=x{\sqrt {9-x^{2}}}$: I need help finding the first and second derivatives. Im confused with the product and chain rule. —Preceding unsigned comment added by Dew555 (talk • contribs) 02:42, 20 March 2009 (UTC)

If you show what you've done so far, we can see where you may be going wrong. It's policy not to just give answers. However, you could start with the simpler case of $h(x)=x{\sqrt {x}}$, both by the product rule and, to check, by simplifying first to get the overall power of x. If that's OK, what's the derivative of $h(x)={\sqrt {9-x^{2}}}$, using the chain rule? Then put the parts together. →86.132.238.145 (talk) 10:12, 20 March 2009 (UTC)

(Explicit answer by User:Superwj5 removed --131.114.72.215 (talk) 10:18, 20 March 2009 (UTC))

These people will help you if you show you have made some effort to solve your problem. This way they can understand better where is the point where you met difficulties. So please try answering the following:

1. Are you able to apply the product (Leibnitz) formula for the first and second derivative of a function $f(x)=xu(x)$ in terms of$u'(x)$ and $u''(x)$ ? What do you get?

Now put $u(x):={\sqrt {9-x^{2}}}$.

2. Are you able to write $u'(x)$ and $u''(x)$ for this function? What do you get?

3. Are you able to substitute the result of point 2 into the formula of point 1? What do you get?

--131.114.72.215 (talk) 10:14, 20 March 2009 (UTC)

(edit conlict *3) If you are having difficulty with this problem, then try starting with some easier problems. For example, try differentiating the following functions. If you get stuck, let us know where you get stuck:

1. ${\sqrt {x}}$ (Requires power rule.)
2. $(x-1)(2x-3)$ (Requires either product rule or power rule.)
3. $\cos(x)+x^{3}$ (Requires power rule and derivatives of sums.)
4. $\sin(x)(\cos(x)+x^{3})$ (Requires product rule.)
5. $({\sqrt {x}})(x^{3}+x^{2})(\cos(x))$ (Requires product rule twice.)
6. $\sin(x^{4}+2)$ (Requires chain rule.)
7. ${\sqrt {\sin(x^{4}+2)}}$ (Requires using the chain rule twice.)
8. $\cos {\sqrt {\sin(x^{4}+2)}}$ (...and three times.)

After you understand how to apply the product rule and chain rule separately, then try your original problem.

Maybe some other ref-desk-ers can offer help explaining...? Eric. 131.215.158.184 (talk) 10:17, 20 March 2009 (UTC)

Polynomial Roots

Is there a formula or method to find the roots of any polynomial? And if yes, what?The Successor of Physics 09:50, 20 March 2009 (UTC)

there are formulas that give you the roots of quadratics, cubics and quartics, and I've no doubt we have articles on them. For the quintic and higher it's a theorem (see Galois theory) that there's no formula that only involves ordinary arithmetic operations and nth roots. But you can solve the quintic with the ultraradical and there are probably higher degree analogues. Of course if an approximation is good enough, you can solve any polynomial with numerical methods...129.67.186.200 (talk) 10:15, 20 March 2009 (UTC)

The articles in question are Quadratic equation#Quadratic formula, Cubic function#General formula of roots.5B8.5D and Quartic function#The general case, along Ferrari's lines. Note that the last two of these are extremely rarely used in practice, because the roots are almost always simpler when written "the roots of polynomial so-and-so" than when written as whatever comes out of these formulas. And if you need the roots for an application, you'll need to compute an approximation sooner or later anyway, and I wouldn't be surprised if it would often take more computer power to calculate all these radicals to the needed precision than to just go for the roots of the polynomial directly with something like Newton's method. The search for these formulas make an interesting piece of mathematics history though. —JAO • T • C 11:31, 20 March 2009 (UTC)

Indeed. Approximating a radical is no easier than approximating a root of a general polynomial, so if you are going to calculate the root then the radical expressions are useless, it is faster and more precise to approximate the root directly. — Emil J. 13:25, 20 March 2009 (UTC)

To third this opinion: yes, it takes more computation to evaluate an expression in radicals (correctly) numerically than it does to find the root. One major problem is many of these expressions suffer from catastrophic cancellation, so that to get k-bits of accuracy in the final answer, you may need nk-bits of accuracy approximating the radical, where I think it can be very difficult to bound n above. In other words, even with an oracle that spit out the expression in radicals for free, it might still take much much longer than directly applying a root finding technique, especially if the roots are real and not clustered. This is similar to matrix inversion. Even if someone offers you the inverse matrix for free, multiplication by it (correctly) might be more expensive than using a standard linear solver. JackSchmidt (talk) 19:00, 20 March 2009 (UTC)

See solvable group and Abel's impossibility theorem. --PST 02:19, 21 March 2009 (UTC)

Hey, let me tell you something: I HATE APPROXIMATIONS!!!!!!!!!!!!!!!!!!!!The Successor of Physics 17:17, 21 March 2009 (UTC)

Unless the polynomial can be solved in radicals (ie. the polynomial group is solvable), approximations are pretty much all you can have (either that, or really weird notation that doesn't help you much). Even when there is a solution in radicals, it may be extremely complicated, so doesn't help much anyway. In many situations it's best to just not solve the polynomial at all and just work with it as it is, when you do need to solve it approximations are often best. --Tango (talk) 17:44, 21 March 2009 (UTC)

And the radicals don't help you get any more than approximations if you want the answers to be expressed numerically. So you're no worse off for not being able to use radicals. Michael Hardy (talk) 19:26, 21 March 2009 (UTC)

Well, yes, "numerical" essentially implies "approximate" (unless it happens to be a rational number with a small denominator). --Tango (talk) 23:54, 21 March 2009 (UTC)

Good question! While there is not a "formula" in the usual sense, one can nevertheless operate on algebraic roots in a fairly arbitrary manner. For instance one can form polynomials in any finite number of roots of polynomials, by a result due to Leopold Kronecker. There are more modern algorithms, although I do not remember the source. Write to me if you are really interested, and I will try to dig them out. Bounding the roots is more straightforward. To the best of my knowledge, an initial bisection followed by Newton's method generally suffices to isolate the roots. 71.182.216.55 (talk) 02:51, 22 March 2009 (UTC)

Question of Cubes

I am trying to prove that 1³ ± 2 ³ ± ... ± n³=0 for all n of the form 4k-1 or 4k.

To do this, I need what I call start points for the proof. These are n = 11, 12, 15, 16, 19, 20, 23, 24. Once I have these numbers I can easily show that any 16 consecutive cubes can be made equal 0 by a particular arrangement of signs. However, proving the start points cases is something that eludes me. Any suggestions as to the choice of sign in each case for the following expressions:

1³ ± 2³ ± ... ± 11³

1³ ±2 ³ ± ... ± 12³

1³ ± 2³ ± ... ± 15³

1³ ± 2³ ± ... ± 16³

1³ ± 2³ ± ... ± 19³

1³ ± 2³ ± ... ± 20³

1³ ± 2³ ± ... ± 23³

1³ ± 2³ ± ... ± 24³,

in order to make each of them 0? —Preceding unsigned comment added by 143.117.143.13 (talk) 12:34, 20 March 2009 (UTC)

This looks like a perfect task for a computer: finite, numerical, and boring. Algebraist 13:27, 20 March 2009 (UTC)

Yes, just looking at the last case, there are 23 signs which can either + or -. That's 2²³ possibilities, or some 8.4 million. That's well beyond what a person can do, but easy for a computer to handle. StuRat (talk) 14:56, 20 March 2009 (UTC)

Come on, people. How did mathematicians ever get anything done before there were computers? I found one solution for n=12 and think there are no solutions for n=11.

$1^{3}+2^{3}-3^{3}+4^{3}-5^{3}-6^{3}-7^{3}+8^{3}+9^{3}-10^{3}-11^{3}+12^{3}=0$.

I promiss to find another one if the original poster explains how his general solution works. Dauto (talk) 18:24, 20 March 2009 (UTC)

Before computers mathematicians had to do a lot more tedious work. I for one am not masochistic enough to go back to those days without good reason. Algebraist 18:33, 20 March 2009 (UTC)

And those cases for n=11 and n=12 are far simpler. The n=11 case only has 1024 possibilities to try, which, while tedious, is at least possible to do by hand. StuRat (talk) 18:40, 20 March 2009 (UTC)

As for the general solution, it works similarly to the case 1² ± 2² ± … ± n² = 0 which was answered on this desk recently (but I didn't succeed in locating it in the archives): for every n, the sequence of cubes n³, …, (n + 15)³ with signs +−−+−++−−++−+−−+ sums to zero. — Emil J. 18:37, 20 March 2009 (UTC)

For those of you saying it's impossible to do it by hand, I got a solution for n=24, and I used only a hand held calculator

$1^{3}+2^{3}+3^{3}+4^{3}-5^{3}-6^{3}+7^{3}+8^{3}-9^{3}+10^{3}+11^{3}+12^{3}-13^{3}+14^{3}+15^{3}+16^{3}+17^{3}+18^{3}+19^{3}-20^{3}-21^{3}-22^{3}+23^{3}-24^{3}=0$

Dauto (talk) 20:31, 20 March 2009 (UTC)

So how did you do it ? Trying every combo obviously isn't the answer. StuRat (talk) 21:48, 20 March 2009 (UTC)

No, trying every combination would take too long. First, the number of solutions probabily grows quite fast as well and all we need is one solution for each value of n. Second (and more important) there are some heuristics that can speed things up considerably. During the seach, anytime that a partial sum (or subtraction) exceeds the sum of the remaining terms (the ones that were not yet included in the partial sum) we simply skip that branch of the search since no combination of signs of the remaining terms will be able to cancel the partial sum. That saves an incredible amount of time since the terms distribution is quite skewed, as long as we start the search by fixing the signs of the larger terms first and work our way down to the smaller terms. That alone allows the solution for n=12 to be found in just a few minutes. Dauto (talk) 04:49, 21 March 2009 (UTC)

Yes how did you do? Well, some nights ago I was asleep and made some of these computation on my computer, just out of curiosity... so: no solutions for n<12, and only two for n=12: one is Dauto's, the other is it with inverted signs. For the cases n=15 and n=16, we have solutions for free with the 16 signs written by EmilJ, because we can start the sequence either with 0 or 1. To make it short, it came out that there are solutions for any $\scriptstyle n=4k-1,\,n=4k,\,12\leq n\leq 27$. I did not made the program write them though, just the number of solutions. The day after I remembered of the Sloane's OEIS, and made a search with the sequences of the number of solutions relative to sums of squares and cubes: there was neither, so I sent them (after all, it contains such sequences like a(n)=100n+1, so it may well have these ones too). However, the number of solution to $\scriptstyle \pm 1\pm 2..\pm n=0$ was there: [1], with some interesting links. One is to S.R.Finch page, where he finds in a quick and nice way the asymptotics for the number of solutions, by means of the central limit theorem; it seems that the computation extends to the case of sums of r-powers: luckily, giving the same result one gets by the expansion of the integrals.

By the way, note that the 16 signs in the sequence above are those of the expansion of the polynomial $\scriptstyle P(x):=(1-x)(1-x^{2})(1-x^{4})(1-x^{8})=\sum _{j=0}^{15}\varepsilon _{j}x^{j}$. Since $(1-x)^{4}$ divides $\scriptstyle P(x)$, if S is the shift operator, acting on sequences by $\scriptstyle (Sx)_{k}=x_{k+1}$, then the linear operator $\scriptstyle P(S)=\sum _{j=0}^{15}\varepsilon _{j}S^{j}$ contains the fourth discrete difference operator $\scriptstyle (S-I)^{4}$ as a factor: and this is the reason why it killes the sequence of cubes, and in fact, all polynomial sequences of degree <4. Of course, this argument generalizes for any degree r, so we have solutions to $\scriptstyle \pm 1^{r}\pm 2^{r}\pm ..\pm n^{r}$ for all $n=2^{r+1}k$ and $n=2^{r+1}k-1$. It seems difficult to say what's the least n=n(r) for which there is a solution; it could be much less than the bound $n(r)<2^{r+1}$.... --pma (talk) 21:54, 20 March 2009 (UTC)

Here's some Python code to solve this using dynamic programming:

numzerosumcubes = {0:1}
for i in range(1,25):
    replacement = {}
    for sum,frequency in numzerosumcubes.iteritems():
        replacement[sum - i**3] = replacement.get(sum - i**3,0) + frequency
        replacement[sum + i**3] = replacement.get(sum + i**3,0) + frequency
    numzerosumcubes = replacement
    print "Number of solutions for n =",i,": ", numzerosumcubes.get(0,0)

It finds solutions for all n congruent to 0 or -1 mod 4, starting at 15, and outputs the number of solutions for each n. —David Eppstein (talk) 22:11, 20 March 2009 (UTC)

Intrinsic Equations

Given the intrinsic equation of a curve $s=f(\psi )$, how do you find the Cartesian form of the equation? 92.3.124.220 (talk) 22:22, 20 March 2009 (UTC)

Here $\psi$ is the angle between the tangent to the (planar) curve and the x-axis, I suppose. Write it as $\psi =g(s)$. Then you shold find a cartesian parametrization by arc-length ( x(s),y(s) ), where x(s) and y(s) come from the two differential equations $x'(s)=\cos(g(s))$ and $y'(s)=\sin(g(s))$. Check here also. --84.221.81.199 (talk) 09:58, 21 March 2009 (UTC)

March 21

Looking for a reference

A document I'm reading has the following passage (italics are mine):

"...laboratory controls must include the establishment of scientifically sound and appropriate specifications, standards, sampling plans and test procedures to assure that components and products conform to appropriate standards. One example of a scientifically sound statistical sampling and analytic plan is based on a binomial approach (see Table 1: Product Performance Qualification Criteria for the Platelet Component Collection Process). The sampling sizes described in Table 1 will confirm with 95% confidence a < 5% non-conformance rate for pH and residual WBC count, and < 25% non-conformance rate for actual platelet yield.

However, other statistical plans may also be appropriate, such as the use of scan statistics."

Do we have an article on scan statistics that is under another name? If not, does anyone know of a concise introduction to the principles? I see books on Amazon, but I'm looking for a not-too technical overview (i.e. less than ten pages in length). SDY (talk) 19:54, 21 March 2009 (UTC)

There is a short (and IMHO not very good) introduction to scan statistics in the "Guide to the preparation, use and quality assurance of blood components" published by the Council of Europe. Basically, it is binomial statistics with a small twist: instead of considering your samples independently, you look at a "moving window". Say your sample size is n. When you do a new quality control, you look at the set of the n-1 previous quality controls plus your new one. When setting up a scan statistics-based QC program, you need to know (1) the baseline error rate that is considered acceptable, and (2) the error rate at which you want the test to indicate a quality failure. Next, you need to find a combination of a moving window size, and a maximum acceptable number of failed tests within such a window, that result in a low false-positive rate and a high probability of detecting a quality failure. The chapter presents a table with some combinations of window sizes and maximum allowed failures per window. Unfortunately, the accepted error rates in that table are too high for the table to be very useful. In addition to determining a window size and a maximum acceptable number of failures per window, the table requires that you specify a third number, the "universe", corresponding to the number of samples analyzed per year. The false-positive rates in the table are calculated with respect to the "universe", whereas the power of the test to detect quality failure is calculated on a sample-by-sample basis.

Since two consecutive samples are not independent (they share n-1 observations), the maths for calculating false positive-rates becomes quite tricky. The chapter refers to this book for the calculations. However, I've heard from a trustworthy source that the false positive-rates in the table were actually calculated by Monte Carlo simulations, and not by the formulae in the book. The probability of detecting a quality failure (power) as presented in the table, was calculated using the cumulative binomial distribution, and is thus easily checked. If you do so, you will see that there is an error in the bottom row of the table. I've done simulations myself, which have shown that the false-positive rates in the table appear to be correct. --NorwegianBlue ^talk 13:24, 22 March 2009 (UTC)

That helps, thanks. SDY (talk) 15:35, 22 March 2009 (UTC)

Central Limit Theorem & the Gamma Distribution

Hi there refdesk - I'm trying to show that the limit of the gamma distribution under the following integral:

${\frac {\lambda ^{n}}{(n-1)!}}\int _{0}^{\lambda n}x^{n-1}e^{-\lambda x}dx$ tends to 0 as n tends to infinity for any positive real lambda, by using the central limit theorem. By taking the limit as n tends to infinity, we should have a normal distribution as follows:

${\frac {G-{\frac {n}{\lambda }}}{\frac {\sqrt {n}}{\lambda }}}\to N(0,1)$ where G is our gamma distribution, so then $G\to N({\frac {n}{\lambda }},{\frac {n}{\lambda ^{2}}})$, right? Then how do I show that the integral of the normal PDF over $[0,\lambda n]$ tends to 0 as n tends to infinity?

Mathmos6 (talk) 22:48, 21 March 2009 (UTC)Mathmos6

Take $\lambda =1$. Then the integral of the normal pdf tends to 1 as $n\to \infty$. So there is an error somewhere. 71.182.216.55 (talk) 02:31, 22 March 2009 (UTC)

I figured there must be but I can't spot where - i assume it must be my very first bit with G, but I'm not sure where I've gone wrong... Mathmos6 (talk) 02:48, 22 March 2009 (UTC)Mathmos6

I believe that the initial assertion is wrong. I can do it with n! but not (n-1)!. Check your earlier calculations. 71.182.216.55 (talk) 03:08, 22 March 2009 (UTC)

With an n! it's obvious: the quantity given is a probability, so at most 1, so dividing by n makes it tend to 0. Algebraist 03:13, 22 March 2009 (UTC)

I didn't mean to suggest that my modification of the result was deep. But the result as stated is wrong. Would you care to confirm? 71.182.216.55 (talk) 03:23, 22 March 2009 (UTC)

The limit appears to be 1/2 (for λ=1). I can't see anything wrong with the CLT argument, and the integral of the normal tends to 1/2 (not 1 as stated above). Algebraist 03:27, 22 March 2009 (UTC)

Yes, 1/2 is right (I took the normal distribution over (-n,n) rather than (0,n)). Anyway, we are substantively in agreement that the result is definitely not zero. 71.182.216.55 (talk) 03:38, 22 March 2009 (UTC)

OK, we're looking at this integral:

${\frac {\lambda ^{n}}{(n-1)!}}\int _{0}^{\lambda n}x^{n-1}e^{-\lambda x}\,dx$

Now let

${\begin{aligned}u&=\lambda x\\du&=\lambda \,dx\end{aligned}}$

Then as x goes from 0 to λn then u goes from 0 to λ²n, and already I'm wondering if you didn't intend n/λ instead of λn. If what you wrote is what you intended, then the integral becomes

${\frac {1}{(n-1)!}}\int _{0}^{\lambda ^{2}n}u^{n-1}e^{-u}\,du.$

But if you intended n/λ, then the integral becomes

${\frac {1}{(n-1)!}}\int _{0}^{n}u^{n-1}e^{-u}\,du.$

For a Gamma distribution with expected value n and variance n, this integral is the probability that a random variable with that distribution is between the mean and √n standard deviations below the mean. That doesn't go to 0, but maybe it looks more promising than the other thing. Michael Hardy (talk) 03:43, 22 March 2009 (UTC)

I had considered myself whether n lambda was the correct upper limit for the integral but checked and rechecked and it certainly is - perhaps the question was simply written down wrong? Mathmos6 (talk) 04:55, 22 March 2009 (UTC)Mathmso6

Yes, something is wrong, as they say. But what exactly did you check? Anyway, you may clarify this thing a little if you consider what the central limit theorem actually tells you about a sequence of iid random variables $\xi ,\xi _{1},\xi _{2},\xi _{3},...$ with exponential distribution law, which was the topic of your problem, as we may reasonably presume (and I assume that it was an exercise, whose text has been corrupted at some point). If $\xi$ has pdf $\scriptstyle \lambda \exp(-\lambda x)$ supported in $\scriptstyle \mathbb {R} _{+}$, you should find, by the central limit theorem,

${\frac {\lambda ^{n}}{(n-1)!}}\int _{0}^{{\frac {n}{\lambda }}+{\frac {\sqrt {n}}{\lambda }}c}x^{n-1}e^{-\lambda x}dx$ tends to ${\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{c}e^{-{\frac {x^{2}}{2}}}dx$,

as n tends to infinity, for any positive real lambda and all $\scriptstyle c\in \mathbb {R}$. Notice that for $\lambda =1$ and $c=0$ we have again the upper limit n in the integral, and we find again the correct limit 1/2 for the special case considered above --pma (talk) 15:27, 22 March 2009 (UTC)

RMK. It is possible that the original integral had an upper bound $\scriptstyle \lambda _{n}$, defined somewhere; then the definition has been lost, and successively the unintelligible $\scriptstyle \lambda _{n}$ has been wrongly corrected into $\scriptstyle \lambda n$; or that $\scriptstyle \lambda _{n}$ became $\scriptstyle \lambda n$ after a typo, and consequently the definition of $\scriptstyle \lambda _{n}$ was expunged as useless. Still, I can't see where the statement that the limit be 0 comes from. Ignorabimus...  :-(

Oh, but most likely the integral is exactly the one you wrote, only the correct statement is that the limit is 0 for all positive $\lambda$ strictly less than 1 (and it is 1/2 for $\lambda =1$, and it is 1 for all $\lambda$ strictly larger than 1). If so, you just need to use the central limit theorem as written above to conclude. The point is that, no matter what $\scriptstyle c\in \mathbb {R}$ is, we have: $\scriptstyle \lambda n<{\frac {n}{\lambda }}+{\frac {\sqrt {n}}{\lambda }}c$ for all large n, respectively, $\scriptstyle \lambda n>{\frac {n}{\lambda }}+{\frac {\sqrt {n}}{\lambda }}c$ for all large n, according whether we are in the case $0<\lambda <1$ or in the case $1<\lambda$. This allows to make a comparison of integrals, finding limit (superior) =0 in the former case, respectively, limit (inferior) = 1 in the latter. Does this make sense to you? --pma (talk) 16:19, 22 March 2009 (UTC)

March 22

Value of an L-function

I'm interested in the s-derivative at s=0 of the analytic continuation of

$\mu (s)=\sum _{z\in \mathbb {Z} [i]}{\frac {}{}}\!\!^{\prime }{\frac {1}{|z|^{2s}}}.$

Up to an overall gamma function and a constant, by taking the proper Mellin transform this is equal to the integral

$\mu (s)=\int _{0}^{\infty }\theta (x)^{2}x^{s-1}\,dx,$

where θ is a Jacobian theta function. Does this function have a name? (The Hurwitz zeta function initially seems promising, but doesn't quite do the trick.) I'd like to compute $\mu '(0)$, provided it can be done in elementary transcendental functions. 71.182.216.55 (talk) 01:29, 22 March 2009 (UTC)

What is $\theta$ here? Algebraist 01:32, 22 March 2009 (UTC)

Sorry, I made some edits in the mean time. 71.182.216.55 (talk) 01:35, 22 March 2009 (UTC)

Anyway, up to constants, $\theta (x)=\sum _{n}e^{-n^{2}x}$. 71.182.216.55 (talk) 01:40, 22 March 2009 (UTC)

Yeah, so the question is still live, despite recent activity on earlier threads ;-) Is anyone here knowledgable about arithmetic? 71.182.216.55 (talk) 04:08, 22 March 2009 (UTC)

The problem is interesting because it is the zeta function regularization of the determinant of the Laplacian on the flat (square) torus. Despite the fact that the should be the easiest test case of the determinant, it actually appears to me to be non-trivial. (It turns out that disks in symmetric spaces are easier, provided one believes in radial functions.) 71.182.216.55 (talk) 04:43, 22 March 2009 (UTC)

Birthdays

I am wondering: (A) if there are any actual studies or "real" data to answer my question or, if not, (B) if anyone has any relevant ideas or theories to suggest. My question is this. Statistically speaking, is any one day of the year equally likely to be someone's birthday as any other day of the year? In other words ... does a birthday of January 1 come up with equal probability as a birthday of January 2, January 3, January 4, ... and so on ... until December 31? Perhaps stated another way ... does each day of the year actually have a probability of 1/365? (I assume that, at least theoretically, they do ... right?) Are there any data or studies about this? If not, can anyone think of any ideas / reasons / theories as to why one particular birthday might show up with greater (or lesser) frequency than another? For sake of simplicity and convenience in this question, let's ignore the birthday of February 29. Thanks. (Joseph A. Spadaro (talk) 20:15, 22 March 2009 (UTC))

Google gave me this which, concludes, among other things, that conception is more likely to occur in winter and less in summer. That's for people alive in a specific period in a single country, so it may not reflect general trends accurately. The sample also contains people born in many different years, which smooths out effects like the fact (I don't know if this has a significant effect or not) that doctors are less likely to carry out caesareans and such at weekends. Algebraist 20:32, 22 March 2009 (UTC)

Thanks. Yes, I considered both of those issues. First, "winter" in one half of the world is "summer" in the other half of the world ... so the summer/winter distinction should not affect birthday statistics. Also, for example, August 17 might fall on a weekend in one year, but on a weekday in a different year ... so the weekend/weekday distinction also should not affect statistics. I believe? Thanks. (Joseph A. Spadaro (talk) 20:58, 22 March 2009 (UTC))

The world population is heavily hemispherically-skewed, so the seasonal effects will not obviously balance out. Different cultures and climates might have different seasonal effects, though. Algebraist 21:01, 22 March 2009 (UTC)

And the population of the southern hemisphere is heavily skewed towards the equator. There are very few (I think, almost no) people living more than 45 degrees south, compared to a very large number living more than 45 degrees north. --Tango (talk) 21:20, 22 March 2009 (UTC)

In some cultures you might get people aiming to give birth on certain days, and avoid others. They could therefore take various actions to move the birth date of their child, or perhaps lie about it if they failed to move it. StuRat (talk) 03:18, 23 March 2009 (UTC)

March 23

Birthdays Part 2

This is (sort of) a follow-up to my question above. Let us assume that each day of the year is equally likely as any other to be a randomly selected person's birthday. Thus, all of the days of the year have an equal probability of 1/365. (Again, let's "pretend" that February 29 does not exist as a birthday.) If I randomly selected 365 people and, say, placed them in a room ... shouldn't everyone in that room have a different (unique) birthday? In other words, should all 365 days of the year be represented ... or no? Something tells me "no" ... but, why not? Also, if the answer is indeed "no" ... how many people would it, in fact, take to fill a room such that we would insure all 365 birthdays are represented? Thanks. (Joseph A. Spadaro (talk) 04:32, 23 March 2009 (UTC))

The problem of how many people are needed before probably two have the same birthday is the Birthday problem. The problem of how many people are needed before all 365 days appear is the Coupon collector's problem. McKay (talk) 04:38, 23 March 2009 (UTC)

And just to have an idea, it's quite unlikely to get all 365 days among the 365 people: around 1 over 10¹⁵⁵ --84.220.118.44 (talk) 09:01, 23 March 2009 (UTC)

What do mathematicians think about legal laws?

I have studied legal philosophy, but am an amateur math hobbyist. One of my favorite books, talking about elastic collisions (physics) says that nature does the calculation in a split second and adjust each balls momentum to ensure that total kinetic energy of the system is conserved. Its a fun book that toys with the imagination through thought experiments (like the example).

Here goes my question: the laws that govern our society were written by lawyers, not mathematicians. Do mathematicians "see" the ugliness and discontinuity of laws and does it bother them? Petit theft and grand theft differ only by a penny. If someone is adjudged incompetent they avoid criminal so-n-so, if they are not judged incompetent they are 100% open to charges or whatever (these examples are just generalizations, they probably aren't accurate if a lawyer reads the math desk).

I never got an education in math, nor became a mathematician, but I have the heart and brain of a mathematician (i.e. love to see how numbers are involved in every single facet in life!)

It takes extremely in-depth scrutiny and analysis to show why Newton's laws break down at some ultra-deep level, however a small subset of all the laws are surprisingly poorly thought (afterall, legal "loopholes" are simply finding a flaw or else a unimagined implication of the law's use, etc...) Is there a relevant wikipedia page on any of this? Is there an author who wrote a book on the topic? Does he or she or his/her book have a wiki page?

I sincerely value the excellent answers I get from the ref desk here, and it helps keep my hobby going when I finish a book and don't know what to start on next.

Thank you,

PS, I tried to keep this completely not about the examples I used; if I could describe whatever I'm looking for, I could google it, if it exists. After about 30 minutes of looking on wikipedia, the closest thing I have found is metamathematics but its a mathematical dissection of math, not of legal laws. Maybe the info I'm seeking doesn't exist? 71.54.173.193 (talk) 08:45, 23 March 2009 (UTC)

There are a lot of mathematicians (even more if you just count people with math PhDs). So their viewpoints on politics, philosophy, law, etc. will vary quite a bit; I don't think you will be able to find any one viewpoint that 75% of mathematicians subscribe to.

There is a anecdote that may be interesting to you. Kurt Gödel immigrated to the United States during World War II and took a citizenship test in 1947. When the presiding judge asked Gödel if the U.S. could become a dictatorship like Germany had, Gödel started to explain how his analysis of the constitution found a loophole that would allow it to happen. The judge, who knew to expect that sort of thing, cut him off and moved the hearing along. — Carl (CBM · talk) 12:14, 23 March 2009 (UTC)

That's a good point. Derrick Niederman wrote a good book that let me appreciate the way that mathematicians think outside of their field. He wrote how politicians and the public think that cutting pork-barrel spending is important, when in fact the national debt is $10 trillion and gave an analogy of the two (which I wish I could recall). Certainly mathematicians would try and make laws that pulled from their analytical reasoning skills. Since I presume mathematicians had no input into making our laws, the laws are void of mathematicians' input, and mathematician-authors could surely point out plenty of fine examples. If nothing else, I read a few great pages on philosophy--some good stuff is over in that little niche of wikipedia--not at all a bad way to start a Monday morning. 71.54.173.193 (talk) 13:01, 23 March 2009 (UTC)