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$$f(x)=\dfrac{1}{1-x}=\sum\limits_{n=0}^{\infty}x^n$$ After a bit of experimenting with geometric series, it seems the radius of convergence is restricted because the function blows up at $x=1$.

If I do the power series about a different point, say $x=100$, the radius of convergence increases to 99. The farther I pick the point, the larger the radius of convergence!

Extrapolating this, can I say the power series of continuous functions like $\sin x, \cos x, e^x$ etc should have infinite radius of convergence?

AgentS
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    @JMoravitz: But if you plug $x=10$ into the series expansion of $\frac{1}{1-x}$ about the point $x=100$, it converges. Did you click on the link in the question? – TonyK Sep 09 '19 at 14:35
  • If $x$ were $10$, then the sum converges if we did the power series about any point greater than $9$, right? – AgentS Sep 09 '19 at 14:35

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No. Consider $f(x)=\dfrac{1}{1+x^2}$. This is a $C^\infty$ function defined everywhere in the real line. It has no singularity but the radius of convergence at $x=0$ is $1$. The explanation lies in the complex plane. The radius of convergence is the distance to the closest singularity.

lhf
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  • cool example! ty:) do you mean when $x=i$ it blows up? – AgentS Sep 09 '19 at 14:36
  • @rsadhvika, exactly. It's the standard example. – lhf Sep 09 '19 at 14:36
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    See also https://math.stackexchange.com/questions/3302167/radius-of-convergence-iff-distance-to-the-nearest-singularity and https://math.stackexchange.com/questions/2362626/who-said-the-shortest-trajectory-between-two-realities-is-complex – lhf Sep 09 '19 at 14:37
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    Just to clarify: $\sin x$, $\cos x$ and $e^x$ do have an infinite radius of convergence. – TonyK Sep 09 '19 at 14:38
  • I don't know complex analysis :( if we consider complex singularities too, can we say the radius of convergence is infinite if the function doesn't blow up ? – AgentS Sep 09 '19 at 14:38
  • Oh I think you've answered my question already - radius of convergence is the distance to the closest singularity – AgentS Sep 09 '19 at 14:39
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    @rsadhvika: If the function is infinitely differentiable in the complex plane, then yes, its radius of convergence is infinite. – TonyK Sep 09 '19 at 14:39
  • @TonyK right! Im just watching this short vid on why $\dfrac{1}{1+x^2}$ has such small radius of convergence. Surprising how complex numbers kick in XD https://www.youtube.com/watch?v=z7-vejtX7vo – AgentS Sep 09 '19 at 14:48
  • @TonyK I know all polynomials are infinitely differentiable even though the tails become $0$. Isn't $\dfrac{1}{1-x}$ also infinitely differentiable ? – AgentS Sep 09 '19 at 14:54
  • @rsadhvika: not at $x=1$. – TonyK Sep 09 '19 at 15:26
  • Ohk saying "infinitely differentiable" means "infinitely differentiable everywhere" gotcha! Playing in desmos for $1/(1+x^2)$, if we imagine $i$ on y axis at $y=1$, then the power series about $a$ will have the radius of convergence $\sqrt{1+a^2}$. This is because the circle cannot go beyond the points $z=\pm i$ in complex plane. Hope I got this right. Thanks again:) – AgentS Sep 09 '19 at 18:13