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I 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.
Earlier quoted context omitted.
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.
I 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.
Why? It's not bounded towards negative infinity.
Earlier 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.
@cperciva do you ever miss math?
Earlier 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.
@cperciva do you ever miss math?
Earlier quoted context omitted.
@cperciva do you ever miss math?
context?