Is there a closed form for the following summation? ($k,t \in N, x\in \mathbb R^+$)

$$f(x,k,t)=\sum_{i=0}^{\text{min}(k,t)} {{\binom k i} \over {\binom t i}}\cdot x^i$$


1 Answer 1


Let $g(x,k,t) = f(x^{-1},k,t)$. Then: $$g = {k! \over t!}\sum_{i=0}^{\text{min}(k,t)} {(t-i)! \over (k-i)!} x^{-i}= {k! \over t!}\sum_{j=-\text{min}(k,t)}^0 {(t+j)! \over (k+j)!} x^j$$ Let $I$ be the integral operator with respect to $x$. Then $$I^tg={k! \over t!}x^t\sum_{j=-\text{min}(k,t)}^0 {j! \over (k+j)!}x^{j}$$ $$\frac{t!}{k!}{\partial^k \over \partial x^k}(x^{k-t}I^tg) = \sum_{j=-\text{min}(k,t)}^0 x^{j} = \frac{x-x^{-\text{min}(k,t)}}{x-1}$$ Therefore $$g(x,k,t) = {k! \over t!}{\partial^t \over \partial x^t}\left(x^{t-k}I^k\left({(x-x^{-\text{min}(k,t)}) \over (x-1)}\right)\right)$$ This gives an expression for $f(x,k,t) = g(x^{-1},k,t)$: $$f(x,k,t) = -{k! \over t!}x^2{\partial^t \over \partial x^t}\left(x^{k-t}I^k\left({(x^{\text{min}(k,t)+1}-1) \over (x-1)}\right)\right)$$

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    $\begingroup$ I don't think that a expression involving "$\partial$" and "$I$" qualifies as "closed form", but it is better than nothing. $\endgroup$ Sep 12, 2013 at 23:08

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