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The Taylor series for $\sin(x)$, centered at any point, converges for all $x$.

The Taylor series for $e^{x}$ and $\cos(x)$ do as well.

Thus, taking an algebraic function of these (without division) forms another 'universally convergent' series:

$\frac{1}{2}(e^{x}-e^{-x}) = \sinh(x)$ converges for all $x$, for example.

And so does $(x^{2}+1)\sinh(x)$.

Note that 'universally convergent', which is a term I invented for the purpose of this question, does not mean that $\exists$ a finite region of convergence when expanding at any $x$ (analytic). It means that $\exists$ a center $x_0$ about which the series converges $\forall x$.

The question: I have not come across any 'universally convergent' functions besides these. Are there any other examples of closed form functions with a universally convergent Taylor series? I ask because such functions are extremely useful algebraically as they can immediately be substituted for their Taylor series in almost any situation.

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    $\begingroup$ $e^{e^x}$? $\frac{\sin x}x$? $\endgroup$
    – bof
    Commented Apr 26, 2015 at 18:55

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In complex analysis, one learns that if $f(z)$ is analytic at $z_0$ (differentiable in a neighborhood of $z_0$) then $f(z)$ can be represented by a Taylor series that converges to $f(z)$ in a circle with radius $r$. What's really nice is that $r$ is the exact distance between $z_0$ and the nearest singularity.

For example you have probably seen that $$\frac{1}{1+x^2} = \frac{1}{1-(-x^2)}= 1-x^2+x^4-\cdots = \sum (-1)^n x^{2n} $$ converges only for $|x|<1$ which is quite odd because this function is infinitely differentiable ($\forall n \forall x\in \mathbb{R}: \exists f^{(n)}(x)$) , but $f(z)=\frac{1}{1+z^2} $ converges in the circle $|z|<1$ in the complex plane, because the distance between $z_0=0$ and $z=i$, the nearest singularity, is $1$.

What if there isn't any singularity (e.g. $\sin z , \cos z , e^z $)? $r=\infty$ , and the Taylor series converges to $f(z)$ for all $z\in \mathbb{C}$.

So every "entire" function, i.e. a function that is (complex) differentiable at every point in $\mathbb{C}$, has a Taylor series that converges to it (for all $z\in \mathbb{C}$)

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  • $\begingroup$ Excellent answer! Thank you! $\endgroup$ Commented Apr 27, 2015 at 20:34
  • $\begingroup$ My pleasure, feel free to ask more questions! $\endgroup$ Commented Apr 28, 2015 at 21:19

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