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I'm trying to find some function $g(k)$ such that $$\sum_{k=0}^{\infty} g(k) \frac{(n \lambda)^k}{k!} = 0 $$ The textbook says that there is only one solution, that is $g(k)=0$ for all $k$. But I cannot see why it is so. It is also constrained that $g(k)$ depends on $k$ alone and does not depend on $\lambda$ or $n$. $n$ is a positive integer, $\lambda$ is a positive real. My intuitive feeling is that $g(k)$ can take alternating positive or negative values such that the summation is zero, but I cannot prove how. Any ideas ? Or is $g(k)=0$ the only solution ?

EDIT: The above must hold true for all $\lambda$.

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Isn't there a condition that it should be true for any $\lambda$? Besides, I don't see the point of having $n\lambda$. Just call it $x$. Unless there is something you are not telling us here, but then you should. If it's just for one value of $\lambda$ and $n$, it's easy to construct a counter example. (Think Taylor expansion of $\sin$ or $\cos$.) – Raskolnikov Mar 20 '11 at 20:16
$n$ doesn't seem relevant to the question. I think you are missing a quantifier here: don't you want the identity to hold for all $\lambda$? – Qiaochu Yuan Mar 20 '11 at 20:56
Thanks @Raskolnikov and @Qiaochu Yuan, I've edited the question. – satish Mar 20 '11 at 21:52

a convergent power series of the form $$f(x)=\sum_{k=0}^{\infty}\frac{g(k)}{k!}x^k$$ represents the zero function iff $g(k)=0$ for all $k$ (here $g(k)=f^{(k)}(0)$). this can be zero for given values of $x$ as noted in the comments e.g. $$ \sin(\pi/2)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(\pi/2)^{(2k+1)} $$ corresponding to a not-identically-zero $g(k)$ and $\lambda n=\pi/2$.

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Thanks yoyo, I somehow couldn't relate it power series and Taylor series, but is clear now. And the $sin(\pi/2)$ example was good. I think i got the point. – satish Mar 20 '11 at 21:54

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