4 ms·
This problem is interesting but not quite air-tight. You could just take the first sequence and set some element to 0 and you've got a new sequence that satisf
by surrey-fringe 10y ago
This problem is interesting but not quite air-tight. You could just take the first sequence and set some element to 0 and you've got a new sequence that satisfies the property. Is he actually asking whether there's not another A between 0 and 1 such that the series converges for a_n = C(-A)^n?
- cperciva 10y agoYou 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.
- surrey-fringe 10y agoI'm dumb
- eeyepieinthesky 10y ago@cperciva do you ever miss math?
- cperciva 10y agoWhen I see puzzles like this, yes.
- Chris2048 10y agocontext?
- blacksmythe 10y agohttps://news.ycombinator.com/item?id=35079 https://news.ycombinator.com/item?id=35079 "Did you win the Putnam?" Yes, I did.
- petters 10y agoYou could do it for n=0.
- The_suffocated 10y agoNo, 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 $O(e^{-x})$.