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Comment by cperciva | Hacker News Reader

You could just take the first sequence and set some element to 0 and you've got a new sequence that satisfies the property

No, because if you have f(x) = e^(-x) + x^n / n!, then e^x f(x) is not bounded as x goes to infinity.

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@cperciva do you ever miss math?

When I see puzzles like this, yes.

context?

https://news.ycombinator.com/item?id=35079

"Did you win the Putnam?"

    Yes, I did.

I'm dumb

You could do it for n=0.

No, you couldn't. Suppose $f(x)$ decays like $O(e^{-x})$ and $a_0$ is the constant term in its power series expansion. Then $g(x)=f(x)-a_0$ decays like $O(e^{-x}$ if and only if $f-g$ decays like $O(e^{-x})$, which in turn is equivalent to the condition that $a_0=0$. In other words, you either have $a_0=0$ in the first place (but then you have a circular argument, not a genuine proof), or $g(x)=f(x)-a$ not decaying like $O(e^{-x})$.

Thanks sorry for wasting your time.

If he helped you understand the problem, no time wasted at all!

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