Wikipedia:Reference desk/Archives/Mathematics/2014 March 13

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March 13

Nonincreasing stars and bars

hello. I'm familiar with stars and bars (combinatorics). But is there a name for the problem where the numbers have to be nonincreasing? That is, you have fixed 'n' and fixed 'r', and we seek non-negative integer solutions $x_{1}$ to $x_{r}$ satisfying $x_{1}+\ldots +x_{r}=n$ where $x_{1}>=x_{2}>=\ldots >=x_{r}$. Is there a name for this? thanks, 122.60.245.210 (talk) 02:36, 13 March 2014 (UTC)

See Partition (number theory), especially the section on restricted partitions. What you're asking for would be "partitions of n with at most r part". See also (sequence A008284 in the OEIS). --RDBury (talk) 16:30, 13 March 2014 (UTC)

In wikipedia article: Multivariate kernel density estimation, Is this a mistake?

Armorjanvan (talk) 09:00, 13 March 2014 (UTC)

In the article "Multivariate kernel density estimation", under "Definition" subsection, it is mentioned that:

---

${\hat {f}}_{\mathbf {H} }(\mathbf {x} )={\frac {1}{n}}\sum _{i=1}^{n}K_{\mathbf {H} }(\mathbf {x} -\mathbf {x} _{i})$

where

• x = (x₁, x₂, …, x_d)^T, x_i = (x_i1, x_i2, …, x_id)^T, i = 1, 2, …, n are d-vectors;
• H is the bandwidth (or smoothing) d×d matrix which is symmetric and positive definite;
• K is the kernel function which is a symmetric multivariate density;
• K_H(x) = |H|^−1/2 K(H^−1/2x).

---

I couldn't find the |H|^−1/2 in any reference, instead, it is just |H|⁻¹, that is: K_H(x) = |H|⁻¹ K(H⁻¹x).

for example, in:

• Multivariate Density Estimation and Visualization P.16, Eq.15
• Lecture Notes on Nonparametrics P.15, Sec. 2.11
• Multivariate Kernel Density Estimation Eq. 3.54

Is this a mistake in the Wikipedia page?

It's not close to my specialty, and it's very possible that there are different conventions. But a naive dimensional analysis of the formula above suggests that H has the dimensions of (co-)variance, and this is consistent with the dimensions of H that appear in several other places in the article. So, again very naively, it looks to me like our article is at least internally consistent, although it may not agree with conventions that appear elsewhere in the literature. Sławomir Biały (talk) 17:43, 13 March 2014 (UTC)

This convention is used in some places, for instance [1], where $\mathbf {H}$ is called the bandwidth matrix. A good rule of thumb is to set $\mathbf {H} \sim {\hat {\mathbf {\Sigma } }}^{1/2}$, where ${\hat {\mathbf {\Sigma } }}$ is the covariance matrix. This transoforms the data to have a unit covariance matrix, so different dimensions are on equal footing with respect to SSE cost functions, etc. --Mark viking (talk) 21:00, 13 March 2014 (UTC)

Find the function

Find the function where $f'(x)$ is equal to $f(x-1)$ . Georgia guy (talk) 19:30, 13 March 2014 (UTC)

f(x)=k⋅a^x where a⋅log(a)−1=0, so a≈1.763222. k is any constant. Bo Jacoby (talk) 20:33, 13 March 2014 (UTC).

Is this the number whose decimal expansion is Sloane's A030797?? Georgia guy (talk) 20:47, 13 March 2014 (UTC)

Yes. And it is this J expression:

  ^@%^:_]1
1.76322

Bo Jacoby (talk) 21:23, 13 March 2014 (UTC).

f(x)=0 202.177.218.59 (talk) 22:45, 16 March 2014 (UTC)

telescoping series

For the analysis in Talk:Mark_and_recapture#Statistical_treatment I need to evaluate the series

$\sum _{N=n+K-k}^{\infty }{\frac {\binom {N-n}{K-k}}{\binom {N}{K}}}$

I am stuck. Your help is appreciated.

The special case K=k is the German_tank_problem#Summation_formula, convergent for K>1 :

$\sum _{N=n}^{\infty }{\frac {1}{\binom {N}{K}}}={\frac {K}{K-1}}{\frac {1}{\binom {n-1}{K-1}}}$

This series is telescoping because:

${\frac {K}{K-1}}\left({\frac {1}{\binom {n-1}{K-1}}}-{\frac {1}{\binom {n}{K-1}}}\right)={\frac {1}{\binom {n}{K}}}$

Bo Jacoby (talk) 20:16, 13 March 2014 (UTC).

You haven't shown that the generalized series above telescopes. In fact, I see no reason why it should. Are you asking about that?--Jasper Deng (talk) 05:54, 14 March 2014 (UTC)

(EC). You are right. I have no proof that it telescopes and I have no proof that it doesn't. The fact that the special case telescopes leaves some hope that the general case telescopes too. Perhaps I should try the case K=k+1. Are there general methods besides trial an error? Is there A Short Table of Sums like Peirce and Foster, A Short Table of Integrals? Bo Jacoby (talk) 08:27, 14 March 2014 (UTC).

Try this: Write S(n, K) for the sum in the K=k case. On other words

$S(n,K)={\frac {1}{\binom {n}{K}}}+{\frac {1}{\binom {n+1}{K}}}+{\frac {1}{\binom {n+2}{K}}}+\dots$

The "next" case, K=k+1, would be

$T(n,K)={\frac {1}{\binom {n}{K}}}+{\frac {2}{\binom {n+1}{K}}}+{\frac {3}{\binom {n+2}{K}}}+\dots$

Then

$T(n,K)+(n-1)S(n,K)={\frac {n}{\binom {n}{K}}}+{\frac {n+1}{\binom {n+1}{K}}}+{\frac {n+2}{\binom {n+2}{K}}}+\dots$

which is

${\frac {K}{\binom {n-1}{K-1}}}+{\frac {K}{\binom {n}{K-1}}}+{\frac {K}{\binom {n+2}{K-1}}}+\dots =KS(n-1,K-1).$

So T can be written in terms of S. Similarly if U is the sum in the K=k+2 case then U can be written in terms of T. Extend recursively to get values for any K-k. The sums are essentially hypergeometric functions at z=1 and presumably the relation above is a form of one of Gauss' contiguous relations. Perhaps there is a closed form expression but, since the degree of the numerators is K-k and there is no obvious cancellation, I'm guessing that a recursive expression is as good as you can do. --RDBury (talk) 08:00, 14 March 2014 (UTC)

Thanks! That is very helpful. The case K−k=0 was

$\sum _{N=n}^{\infty }{\frac {\binom {N-n}{0}}{\binom {N}{K}}}={\frac {K}{K-1}}{\frac {1}{\binom {n-1}{K-1}}}$

and the case K−k=1 is simply

$\sum _{N=n+1}^{\infty }{\frac {\binom {N-n}{1}}{\binom {N}{K}}}={\frac {K}{K-2}}{\frac {1}{\binom {n-1}{K-2}}}$

Bo Jacoby (talk) 13:29, 14 March 2014 (UTC).

You're welcome. Actually I was wrong about there not being a closed form. According to Abramowitz and Stegun the value of ₂F₁(a, b ; c; 1) is

$F(a,b;c;1)={\frac {\Gamma (a)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}}$

which you should be able to plug in to get an expression for the sum. I didn't see this formula in our article on hypergeometric series; I'll double check and add it if it's not there. --RDBury (talk) 16:22, 14 March 2014 (UTC)

See also this book. Count Iblis (talk) 16:22, 14 March 2014 (UTC)