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Cally’s Trials – OST Cracked

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How to show $|f(n)| \leq K e^{ -|n|}$ for some $K$ and $n \in \mathbb{Z}$?

Let $f : \mathbb{Z} \to \mathbb{C}$ be a function such that $f(n+1) = e^n f(n)$ for all $n \in \mathbb{Z}$, and $f(0) = 1$.
Prove that $|f(n)| \leq K e^{ -|n|}$ for some $K$ and $n \in \mathbb{Z}$.
I tried to write $f(n) = f(0) + \int_0^n f'(t) dt$ and then use that $f'(n+1) = e^n f'(n)$ and $f'(0)=1$. However, I can’t get the answer.

A:

I took as my base the notes of a colleague who did not study calculus on this subject. This work for $f$, $f(n):=\sum_{k=-\infty}^\infty a_ke^{ikn}$ with $f(0)=1$ and $a_k:=\dfrac{1}{2\pi}\int_{ -\pi}^\pi f(x)e^{ -ixk}dx$, $k\in\mathbb Z$.
It is $$a_{k+1}=a_k e^k,\;\;\;k\in\mathbb Z$$ which leads to $$a_k=a_0e^{k\log(a_0)}\qquad k\in\mathbb Z$$
If $a_0=0$, then $a_k=0$; otherwise the integral a_k=a_0\sum_{n=0}^\infty\left(\dfrac{a_0
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Cally’s Trials – OST Cracked ((NEW))