1
$\begingroup$

Can we get closed form for $$\sum_{k=0}^m \left(-\frac12\right)^k \binom{2m}{m-k}k^p,\quad p\in\mathbb{N}\,?$$ In Concrete Mathematics Knuth describes Gosper's algorithm and its Zeilberger's extension, but they both fails on this partial hypergeometric sum. $-1/2$ can be replaced to $x$, but it doesn't help (I tried to differentiate it for any advances).

Hypergeometric function is not a closed form, however. Because this sum is a hypergeometric function yet (multiplied by binomial coeff.). It will be very cool to find closed form for $p=0$ or $p=1$. But if I understood correctly Knuth, there is no closed form.

Any ideas?

$\endgroup$
  • $\begingroup$ Running the summation in Mathematica for various $p$, these partial sums can be expressed as $x\binom{2m}{m-1}{_{p+1}F_p}(2,2,\ldots,2,1-m;1,1,\ldots,1,2+m;-x)$.for $p>0$ and $\binom{2m}{m}{_2 F_1}(1,-m;1+m;-x)$ for $p=0$. However, while more compact, this is essentially the same summation as in the problem statement. $\endgroup$ – Semiclassical Jun 8 '15 at 3:23
  • $\begingroup$ @Semiclassical, I used Mathematica and Maple, and you are completely right $\endgroup$ – Michael Galuza Jun 8 '15 at 4:47
  • $\begingroup$ To be honest, I would be shocked if there was a 'nice' expression for any of these generalized hypergeometric series. A more modest goal (and one which is often the case) is to find if there are specific values for which such series can be expressed in closed-form. $\endgroup$ – Semiclassical Jun 9 '15 at 17:06
  • $\begingroup$ @Semiclassical, why do you think so? $\endgroup$ – Michael Galuza Jun 9 '15 at 17:31
  • 1
    $\begingroup$ I think what you might want is at sciencedirect.com/science/article/pii/S0012365X03002061 Lemma 2 and then the Sterling numbers can be translated to hypergeometric terms via sciencedirect.com/science/article/pii/S0377042796001677 I don't know what size "p" 's you are interested in. $\endgroup$ – rrogers Dec 22 '15 at 19:08
4
+100
$\begingroup$

Let us write $$S_{m,p}(x)=\sum_{k=0}^m\binom{2m}{m-k}x^kk^p$$ You may try the generating function (ordinary for $m$, exponential for $p$) $$g(x,y,z)=\sum_{m,p}S_{m,p}(x)y^m\frac{z^p}{p!}.$$

First, for $p=0$ we have $$g(x,y,0)=\sum_m S_{m,0}(x)y^m=\sum_{0\le k\le m} \binom{2m}{m-k}x^ky^m$$ We can use $$\sum_m\binom{2m}{m-k}y^k=\frac1{\sqrt{1-4y}}\left(\frac{1-\sqrt{1-4y}}{1+\sqrt{1-4y}}\right)^k=\frac{1}{\sqrt{1-4y}}\frac{(1-\sqrt{1-4y})^{2k}}{4^ky^k}$$ and we get $$g(x,y,0)=\sum_mS_{m,0}(x)y^m=\frac1{\sqrt{1-4y}}\frac{4y}{4y-x(-1+\sqrt{1-4y})^2}.$$

For arbitrary $p$ $$g(x,y,z)=\sum_{m,p}\sum_{k=0}^mx^kk^p\binom{2m}{m-k}\,y^m\frac{z^p}{p!}=\sum_m\sum_{k=0}^m x^k \mathrm e^{kz}\binom{2m}{m-k}y^m=g(x\mathrm e^{z},y,0).$$

There are no hypergeometric functions. The result for $m$ and $p$ fixed is the coefficient of $y^mz^p$ multiplied by $p!$ in $g(x,y,z)$, it is written $$p!\left[y^mz^p\right]g(x,y,z).$$

Probably not what you hoped, but it is not likely to be improved into a less hypergeometric form.

$\endgroup$
  • $\begingroup$ $g(x, y, 0) = 0$, isn't it? Or you assummed summation for all integer $m$, $p$, and $0^0 = 1$? $\endgroup$ – Michael Galuza Jun 12 '15 at 11:41
  • $\begingroup$ Yes, I use $0^0=1$. $\endgroup$ – Tom-Tom Jun 12 '15 at 11:45
  • $\begingroup$ I'm not sure that it is closed form, but it's definitely new point of view. $\endgroup$ – Michael Galuza Jun 12 '15 at 11:56
2
$\begingroup$

One method that canbe used is the following.

Starting with $\delta = x D$ where $D = \frac{d}{dx}$ then $\delta^{p} x^{k} = k^{p} x^{k}$ for which \begin{align} S_{m} = \sum_{k=0}^{m} \binom{2m}{m-k} \, k^{p} \, x^{k} = \delta^{p} \, \sum_{k=0}^{m} \binom{2m}{m-k} \, x^{k} = \delta^{p} \, \sum_{k=0}^{m} \binom{2m}{k} \, x^{m-k}. \end{align} Since \begin{align} \sum_{k=0}^{m} \binom{2m}{k} \, y^{k} = (1+y)^{2m} - \binom{2m}{m+1} \, y^{m+1} \, {}_{2}F_{1}(1, 1-m; m+2; - y) \end{align} then \begin{align} S_{m} = \delta^{p} \left[ x^{m} \left(1 + \frac{1}{x} \right)^{2m} - \binom{2m}{m+1} \frac{1}{x} \, {}_{2}F{1}\left(1, 1-m; m+2; - \frac{1}{x} \right) \right]. \end{align} Calculating the derivatives may lead to what is desired.

$\endgroup$
  • $\begingroup$ You're right, but it's what I want. Maple or Mathematica simplify this series to hypergeometric function. But it's just a hypergeomtric function, nothing more. By "closed form" I meant expression with binomial coefficients and smth. similar, not a special functions. $\endgroup$ – Michael Galuza Jun 3 '15 at 8:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.