Let $f(x)$ be a real-valued twice continuously differentiable function, and considered the below double sum $$F(t,f(x)):=\dfrac{1}{t}\Big(\sum_{k=0}^{\infty}\sum_{m=0}^{\infty}f(x+(k-m)/\sqrt{n})\dfrac{t^{k+m}}{2^{k+m}k!m!}e^{-t}-f(x)\Big).$$ I want to compute the limit of $F(t,f(x))$ when $t\rightarrow 0$, but I don't even know how to start....

My idea was to write this double sum into two parts, the first part may be our desired answer and the second part may be zero when $t\rightarrow 0$, but I don't know what to do to analyze such a complicated sum...

Any idea?

  • $\begingroup$ Is $n$ on your formula a typo? or, otherwise, who''s that? $\endgroup$ – Ilya Bogdanov May 11 at 15:30
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    $\begingroup$ Is $n$ a constant here? If yes, then what is the problem? The term $k = m = 0$ cancels out, the others converge to $0$ if $k+m>1$, and are constants when $k+m=1$. $\endgroup$ – Mateusz Kwaśnicki May 11 at 15:31
  • $\begingroup$ @MateuszKwaśnicki yes, $n$ is constant here. Would you to give a proof? $\endgroup$ – user157884 May 11 at 15:33
  • $\begingroup$ @IlyaBogdanov let us treat $n$ to be a constant. after I compute the limit of $F(t,f(x))$ when $t\rightarrow 0$, I will then take $n$ to $\infty$, so now it is fixed. $\endgroup$ – user157884 May 11 at 15:34
  • $\begingroup$ why would $F$ even be well defined? $\endgroup$ – jcdornano May 11 at 15:40

$$F(t,f(x))=\dfrac{1}{t}\Big(\sum_{k=0}^{\infty}\sum_{m=0}^{\infty}f(x+(k-m)/\sqrt{n})\dfrac{t^{k+m}}{2^{k+m}k!m!}e^{-t}-f(x)\Big)$$ only the terms $k=0,m=0$, $k=0,m=1$, $k=1,m=0$ survive in the limit $t\rightarrow 0$, the other terms vanishing at least linearly in $t$, $$\lim_{t\rightarrow 0}F(t,f(x))=-f(x)+\tfrac{1}{2}f(x+1/\sqrt n)+\tfrac{1}{2}f(x-1/\sqrt n),$$ where I used that $$\lim_{t\rightarrow 0}\frac{e^{-t}-1}{t}=-1.$$

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  • $\begingroup$ could you be more specific about why only these three terms survive? $\endgroup$ – user157884 May 11 at 15:49
  • $\begingroup$ assuming that the sum converges, the terms with $n+m>1$ vanish at least linearly in $t$ $\endgroup$ – Carlo Beenakker May 11 at 15:55
  • $\begingroup$ oh. okay. Thank you! so after this when I take $n\rightarrow\infty$, the limit is $0$? $\endgroup$ – user157884 May 11 at 15:56
  • $\begingroup$ yes it is indeed, it vanishes as $1/n$. $\endgroup$ – Carlo Beenakker May 11 at 16:01
  • $\begingroup$ Okay Thank you! $\endgroup$ – user157884 May 11 at 16:02

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