A problem involving power series
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A problem involving power series
1–10 of 17 posts
Re: A problem involving power series
#2This 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?
Re: A problem involving power series
#3This 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?
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
#4This 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?
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.
I'm dumb
Re: A problem involving power series
#5This 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?
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.
@cperciva do you ever miss math?
Re: A problem involving power series
#6This 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?
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.
You could do it for n=0.
Re: A problem involving power series
#7Earlier quoted context omitted.
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.
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 $O(e^{-x})$.
Re: A problem involving power series
#8Earlier quoted context omitted.
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 l…
Thanks sorry for wasting your time.
Re: A problem involving power series
#9Earlier quoted context omitted.
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 l…
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Re: A problem involving power series
#10I think the user Av's answer (using Liouville's theorem) is correct, and a very simple argument, though it does use the fairly hefty machinery that goes into "a bounded analytic function is constant".
EDIT: as tome kindly points out, e^x f(x) need not remain bounded as x goes to -infinity, or indeed any other direction of infinity on the complex plane.