Re:A problem involving power series
by surrey-fringe on Thursday October 20, 2016 @01:31AM
(#12749922)
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?
[ Parent
| on HN ]
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.
Re:A problem involving power series
by eeyepieinthesky on Thursday October 20, 2016 @02:08AM
(#12750034)
@cperciva do you ever miss math?
When I see puzzles like this, yes.
context?
Re:A problem involving power series
by blacksmythe on Thursday October 20, 2016 @11:30AM
(#12753182)
Re:A problem involving power series
by surrey-fringe on Thursday October 20, 2016 @01:44AM
(#12749963)
I'm dumb
You could do it for n=0.
Re:A problem involving power series
by The_suffocated on Thursday October 20, 2016 @03:03AM
(#12750242)
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.