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I'm looking for some known results for sum of this type but I can't find anything. The sum is defined as: $$S(x,a,b,n)=\sum_{k=0}^n \binom{n}{k} (-1)^{k} f((a(n-k)+bk)x)$$ where $f$ is an arbitrary function and $a$ and $b$ are real and positive constants. For example for the function $f(x)=e^{x}$ the sum can be evaluated with Binomial Theorem and we have: $$S(x,a,b,n)=(e^{ax}-e^{bx})^{n}$$ But for arbitrary function $f$? Thanks.

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  • $\begingroup$ I suspect there is no compact representation of the sum without knowledge of $f$. $\endgroup$ – David G. Stork Dec 19 '18 at 16:43
  • $\begingroup$ There probably isn't a single method for evaluating the sum that works for all functions. If you happen to know a little bit about $f$, you can make certain generalizations, though. For example, if $f$ is linear, then $f((a(n-k)+bk)x)=f(anx)-f(akx)+f(bkx)$. $\endgroup$ – R. Burton Dec 19 '18 at 17:35
  • $\begingroup$ Thanks!! Yes, for an arbitrary function isn't improbable to obtain a closed form. I tried to obtain an Integral representation of the sum via Integral Transform but with no great results. $\endgroup$ – Papemax89 Dec 19 '18 at 17:40

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