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Consider the Taylor expansion of $\tanh$ around $0$. The radius of convergence is finite ($\pi/2$).

Define a $\tanh$-like function a function $f:\mathbb R\to\mathbb R$ such that:

  • $f(0) = 0$;
  • $\lim_{x\to+\infty} f(x) = 1$;
  • $\lim_{x\to-\infty} f(x) = -1$;
  • $f$ is non-decreasing.

Clearly $\tanh$ is a $\tanh$-like function. Other common examples are $x\mapsto \frac{2}{\pi}\arctan x$, or $x\mapsto \frac{2e^x}{1+e^x}-1$.

All these examples are analytic but have a finite radius of convergence when expanded around $0$.

Does it exist an analytic $\tanh$-like function that has infinite radius of convergence around $0$?

I could not find any trivial example that satisfy the requirements to be $\tanh$-like and has infinite convergence radius. My guess would be that such a function does not exist...

ECL
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    I believe that an entire function whose real part is bounded is constant. Thus, this is impossible. – Robbie May 10 '22 at 14:01
  • They're not counterexamples. The claim is true (see e.g. https://math.stackexchange.com/questions/229312/an-entire-function-whose-real-part-is-bounded-must-be-constant). However, I'm happy to accept on further reflection the theorem doesn't immediately yield the result! – Robbie May 10 '22 at 16:40
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    I think the error function might work. https://en.wikipedia.org/wiki/Error_function – my bad! – Robbie May 10 '22 at 16:54
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    @ECL $\sin$ and $\cos$ do not have bounded real parts (it would contradict Picard's little theorem among other things). In fact $\sin,\cos:\Bbb{C}\to\Bbb{C}$ are surjective (easy to prove directly). It is their restrictions to $\Bbb{R}$ which have bounded image. This is a slightly different statement, but it makes all the difference in the conclusion. This is also why while Robbie's initial comment doesn't actually answer the question (though of course the subsequent example of the error function does work). – peek-a-boo May 10 '22 at 19:37
  • @peek-a-boo yep you're right, my bad! – ECL May 10 '22 at 19:58

1 Answers1

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The error function, defined by $$ \text{erf}(z) = \frac{2}{\sqrt\pi}\int_0^z e^{-z^2} dz $$ satisfies your requirements. The first four requirements are clear I think; to show this is entire (and therefore has infinite radius of convergence), see e.g. the answer here.

Robbie
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