Wikipedia:Reference desk/Archives/Mathematics/2014 October 26

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October 26

Gambler's delight

Hi, a race involves n horses with respective decimal odds x1, x2, ... xn. If 1/x1 + 1/x2 + ... + 1/xn < 1 then a punter can guarantee to make money by betting k/x1 on the first horse, k/x2 on the second, and so on, for any fixed multiple k. My question is whether 1/x1 + 1/x2 + ... + 1/xn < 1 is the only scenario where a betting strategy exists which is certain to make a profit. I feel that it probably is, but if there's an easy way to prove it, I can't see it. 86.160.82.218 (talk) 03:48, 26 October 2014 (UTC)

Let x₁, ..., x_n be odds for which such a strategy exists. Let k₁/x₁, ..., k_n/x_n be bets that guarantee a profit. Let k = min(k₁, ..., k_n) (your minimum possible winnings). Then k > k₁/x₁ + ... + k_n/x_n ≥ k/x₁ + ... + k/x_n, and dividing by k gives 1/x₁ + ... + 1/x_n < 1. I think that works. -- BenRG (talk) 06:39, 26 October 2014 (UTC)

Neat, thanks! 86.160.82.218 (talk) 14:18, 26 October 2014 (UTC)

Stable distribution question

First, am I correct in asserting that if I have the characteristic function for a probability distribution with $X$ as a random variable of said distribution, and want the characteristic function of the distribution followed by ${\frac {X}{n}}$, I simply take the nth root of the first characteristic function?

Second, if I take a mean average of variables from a stable distribution with $\alpha <1$, am I right in thinking that the scale parameter of the distribution of my sample mean increases as I increase the size of my sample? If so, does this mean that my estimate of the distribution mean---of course, the mean doesn't exist, but my estimator for the mean does---gets more variable as I increase the sample size?--Leon (talk) 11:36, 26 October 2014 (UTC)

Equivalence of conditions for a positive operator on complex Hilbert space

Hi,

On the page for positive operators, multiple definitions of such an operator are given. In particular it states that requiring a bounded linear operator on a complex Hilbert space to satisfy

$\langle Tx,x\rangle$ is (real and) non-negative for all $x$

coincides with the notion of self-adjoint elements having a spectrum in the positive reals. I've seen proofs that the $\langle Tx,x\rangle \geq 0$ condition implies $\sigma (T)\subset [0,\infty )$, but none have been elegant. I also haven't seen a proof of the other direction (but I assume this is true).

Does someone know an elegant proof of this? That is, on a complex Hilbert space, for a bounded linear self-adjoint operator $T$,

$\langle Tx,x\rangle \geq 0\iff \sigma (T)\subset [0,\infty )$?

Cheers,

Neuroxic (talk) 11:37, 26 October 2014 (UTC)

I don't know if there is an elementary proof of this equivalence, but it follows from the spectral theorem (our article seems to be unclear on details, see Rudin's "Functional analysis" for a clearer statement). Sławomir Biały (talk) 14:19, 26 October 2014 (UTC)

Indeed, Rudin does use spectral theory machinery very nicely. Thanks for the reference, even if it may not be elementary. Neuroxic (talk) 07:19, 27 October 2014 (UTC)

a polynomial based question

Let f(x)=x^5+ax^4+bx^3+cx^2+d such that f(1)=1,f(2)=2,f(3)=3,f(4)=4, and f(5)=5, then 'd' is equal to which of the following four numbers - A. 100, B. 0, C. -100, D. -120 — Preceding unsigned comment added by 223.176.59.120 (talk) 17:29, 26 October 2014 (UTC)

Welcome to Wikipedia. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. --Kinu ^t/_c 17:55, 26 October 2014 (UTC)

Did you misstate/misread the problem? I'm pretty sure this has no solution. -- BenRG (talk) 23:55, 26 October 2014 (UTC)

The answer is -120 (though, your polynomial will be of form x^5 + ax^4 + bx^3 + cx^2 + dx + e, and the solution will be for what is here "e"). Here's a hint: if y is a fixed point of a polynomial P, then y is a root of P(x) - x. You are given 5 fixed points, so you know the roots of f(x) - x, from there, it is easy to calculate f(x) - x by multiplying binomials; the rest is straight forward. (Or, for a simpler calculation: observe that adding x does not change the product of the roots of a polynomial, that the constant term is the product of the negative of the roots, and that you know the roots of f - x...)Phoenixia1177 (talk) 05:08, 27 October 2014 (UTC)

The answer is E. All of the above (vacuously). -- Meni Rosenfeld (talk) 07:49, 27 October 2014 (UTC)

It seems highly likely the op mis-typed and forgot the first power term - but even going with what is written, your joke doesn't work well, all of the above would not follow vacuously.Phoenixia1177 (talk) 08:41, 27 October 2014 (UTC)

Why not?

If $f(x)=x^{5}+ax^{4}+bx^{3}+cx^{2}+d,\ f(1)=1,\ f(2)=2,\ f(3)=3,\ f(4)=4$ and $f(5)=5$, then $d=100,\ d=0,\ d=-100$ and $d=-120$.

The consequent here is of course absurd, but the statement is still true because the antecedent is impossible. -- Meni Rosenfeld (talk) 19:13, 27 October 2014 (UTC)

This is one of those problems that come up in competitions, where there is a slow and laborious approach and a quick and elegant approach. In this case the slow and laborious approach is to write down and solve 5 simultaneous equations; the quick and elegant approach as pointed out by Phoenixia1177 is to notice that f(x) - x has five obvious roots, and the constant term in f(x) is the product of these roots give or take a sign. Gandalf61 (talk) 16:27, 27 October 2014 (UTC)