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Let's say I have a Maclaurin series for some function $f(x)$, and to find Maclaurin series $f(x^2)$ can I just substitute $x^2$ for each term Maclaurin series for $f(x)$?

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  • $\begingroup$ Yes, normally it is done . $\endgroup$ – georg Sep 23 '16 at 19:33
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If $$f(x)=\sum_{n\ge0}a_nx^n \text{ for }\vert x \vert < R,$$ then $$f(x^k)=\sum_{n\ge0}a_nx^{nk} \text{ for }\vert x^k \vert < R\text{, i.e.} \vert x \vert <R^{1/k}$$

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Yes, because you are evaluating the function at the point $x^2$ instead of just $x$. All you are really doing is chaning the domain of the function, so that if the original radius of convergence was $r$, the new radius of convergence is $\sqrt r$.

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    $\begingroup$ What if $x^2$ is no longer in the radius of convergence? $\endgroup$ – Zestylemonzi Sep 23 '16 at 19:36
  • $\begingroup$ Thanks for that. I was thinking of the Taylor series expansions for $e^x$, $\sin x$, etc. $\endgroup$ – AlgorithmsX Sep 23 '16 at 19:39
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$${{{{{{\mathbf{\text{Yes}.}}}}}}}$$

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    $\begingroup$ The radius of convergence may differ however. $\endgroup$ – gj255 Sep 23 '16 at 19:34
  • $\begingroup$ Just beat me to saying that. What if $x^2$ is no longer in the radius of convergence? $\endgroup$ – Zestylemonzi Sep 23 '16 at 19:35
  • $\begingroup$ @Zestylemonzi What you call $x$ you can rename as $y^2$ or $z^3$. The radius of convergence will adapt. $\endgroup$ – Von Neumann Sep 23 '16 at 19:37
  • $\begingroup$ Doesn't it make a difference that $x^2 \ge 0$? $\endgroup$ – GFauxPas Sep 23 '16 at 19:38
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    $\begingroup$ @Fourier Transform True, but given the level of the question I think it's worth pointing out in your answer. It's something that can easily be overlooked - i.e just sub in $x^2$ with a value of $x$ that worked previously. $\endgroup$ – Zestylemonzi Sep 23 '16 at 19:42

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