There exists a power series $$f(x) = \sum\limits_{n=1}^\infty a_n x^n = a_0 +a_1 x + a_2 x^2 + \cdots$$ Assume that the radius of convergence of this power series is $\infty$, i.e.,

$$\limsup_{n\rightarrow\infty} \big(|a_n|^{\frac{1}{n}}\big) = 0$$ by Cauchy-Hadamard theorem.

My question :

"Is there a sufficient (or necessary and sufficient) condition that $f(x)=O(1)$ as $x \rightarrow \infty$?"

For example,

the Maclaurin series of $\sin x$, $\cos x$, and $e^{-x}$ are well-defined in $\mathbb{R}$, and $O(1)$ as $x \rightarrow \infty$.

  • 2
    $\begingroup$ I don't know the answer for $\mathbb{R}$, but if you ask the same question on $\mathbb{C}$, the necessary and sufficient condition for the boundedness of a power series with infinite convergence radius is that it is constant $\endgroup$ – Max Jan 5 '18 at 19:52

A necessary and sufficient condition comes from Ramanujan's master theorem.

Let $z = x+ iy$. If and only if there is a holomorphic function $\phi(z)$ for $\Re(z) > 0$ and $|\phi(z)| < C e^{\rho|x| + \kappa|y|}$ for $C,\rho \in \mathbb{R}^+$ and $0<\kappa < \pi/2$ then

$$f(x) = \sum_{n=0}^\infty (-1)^n \phi(n+1)\frac{x^n}{n!}$$ is $O(1)$ for $|\arg(x)| < \kappa$ as $x \to \infty$.

It's a little tricky to prove this, it's rather complicated. Don't have time to find a reference at the moment.

  • $\begingroup$ Could you explain how $f(x)$ is $O(1)$ with you assumption? $\endgroup$ – Doyun Nam Jan 6 '18 at 15:38
  • $\begingroup$ It comes from the isomorphism of spaces (holomorphic functions bounded as I wrote, and the space of $O(1)$ functions) induced by the Mellin transform $\Gamma(1-z)\phi(z) = \mathcal{M}f = \int_0^\infty f(x)x^{-z} \, dx \Leftrightarrow f(x) = \sum_{k=0}^\infty (-1)^k \phi(k+1)\frac{x^k}{k!}$. What I wrote is slightly incorrect though (wrote it a bit too fast), this is a sufficient and necessary for $f(x) = O(1/x)$. We can make $f(x)$ grow like any power of $x$ by noticing if $\phi(z)$ holomorphic for $\Re(z) > a$ (bounded as I wrote, $a<1$) then $f(x) = O(1/x^{1-a})$ $\endgroup$ – user335907 Jan 7 '18 at 0:54
  • $\begingroup$ So, for example, if $f$ is $O(1)$ then there exists a function $\phi$ such that $\phi(z)\Gamma(1-z) = \mathcal{M}f$ where $(-1)^n\phi(n+1)/n! = a_n$. (We take $\mathcal{M}$ to be an extension of the Mellin transform, not simply the integral which could diverge). And if we have $\phi$ we can get $f$. $\endgroup$ – user335907 Jan 7 '18 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.