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Which is the radius of convergence of Taylor series of $f(x)=\frac{1}{1+x^2}$ around $x_0=0$? I am unable to write down the analitycal expression of all its derivatives. Why is it finite even if $f(x)<\infty$ for all $x\in \mathbb{R}$?

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  • $\begingroup$ Are you familiar with the geometric series formula involving $\frac{1}{1-x}$? $\endgroup$
    – anon
    Jan 20, 2013 at 22:32
  • $\begingroup$ @anon, yes, $\frac{1}{1-x}=\sum_n x^n$ $\endgroup$
    – glassy
    Jan 21, 2013 at 16:26

3 Answers 3

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One possibility (which does not require complex analysis) is to use the geometric series, as suggested by anon. We know that for $|r|<1$ $$1+r+r^2+\ldots=\sum_{k=0}^\infty r^k=\frac{1}{1-r}.$$

If we use this with $r=-x^2$, we have $$\frac{1}{1+x^2}=\frac{1}{1-(-x^2)}=\sum_{k=0}^\infty (-x^2)^k=\sum_{k=0}^\infty (-1)^kx^{2k},$$ which is the Taylor series around $x_0=0$ (also known as the Maclaurin series). From the properties of the geometric series, we know that this series converges iff $|-x^2|<1$. This is equivalent to $|x|<1$, so the radius of convergence is one.

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  • $\begingroup$ @Did: The "only if" part is also true with my wording. The terms of a convergent series always tend to zero, and if the quotient has magnitude $\geq 1$ the terms do not approach zero. $\endgroup$
    – Mårten W
    Jan 21, 2013 at 21:41
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    $\begingroup$ Thank you. Can you explain me why it must be $|x|<1$ even if $|f(x)|$ is finite for all $x$? $\endgroup$
    – glassy
    Jan 22, 2013 at 10:15
  • $\begingroup$ @user59051: $|f(x)|$ is indeed finite for all real valued $x$, but as I pointed out in my other answer there are singularities at $x=i$ and $x=-i$. For a complete understanding of what is going on, one needs to know a bit of complex analysis. $\endgroup$
    – Mårten W
    Jan 22, 2013 at 10:30
  • $\begingroup$ What would happens to the function around $x=\pm1$ if we only use real analysis? I guess nothing's special? $\endgroup$
    – Ooker
    Sep 13, 2017 at 19:02
  • $\begingroup$ @Ooker: The function itself will work just fine for all $x\in\mathbb{R}$, but the Maclauring series will not converge if $|x|\geq 1$. $\endgroup$
    – Mårten W
    Sep 13, 2017 at 21:44
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The radius of convergence is the distance to the nearest singularity of the function.

Your function has singularities at $x=\pm i$, so the Maclaurin expansion (Taylor series at zero) of $f(x)=\frac{1}{1+x^2}$ has radius of convergence $r=1$ (the distance from zero to $\pm i$).

If the Taylor series expansion is about $x_0=1$, then the radius of convergence would be $$\min_{x=\pm i}|1-x|=\sqrt{2}.$$

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  • $\begingroup$ Since my undergraduate days I wish to see a proof of this fact by pure real analysis arguments, that is, one that does not resort to complex analysis. $\endgroup$ Jan 21, 2013 at 4:45
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$f(\sqrt{-1}) = 1/0$

$f(-\sqrt{-1}) = 1/0$

so the radius of convergence is blocked by these explosions.

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