If the Taylor series for $e^x$ is known, the Taylor Series for $e^{x^2}$ can be found by replacing all $x$ with $x^2$.

What about a Taylor Series known for the following $f(x)=\ln\frac{1+x}{1-x}=2\displaystyle\sum\limits_{n=0}^ \infty \frac{x^{2n+1}}{2n+1}$. Can the Taylor Series of $f(x)=\ln\frac{1+x^2}{1-x^2}$ be found by replacing all $x$ with $x^2$ - making it $2\sum^\infty_{n=0}\frac{x^{4n+2}}{2n+1}$?

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    $\begingroup$ No, because you still have an $x$ in the denominator. This would work for $\ln \frac {1+x^2}{1-x^2}$. $\endgroup$
    – lulu
    Jan 12 '20 at 18:35
  • $\begingroup$ @lulu - whoops! that was a typo! Indirectly you answered my question anyway though. $\endgroup$
    – Burt
    Jan 12 '20 at 18:36
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    $\begingroup$ @joriki seems like you would like me to fix it. will do it now:) $\endgroup$
    – Burt
    Jan 13 '20 at 0:57
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    $\begingroup$ @Burt: Well observed :-) Thank you. By the way, if your question was answered (albeit indirectly), you could post the answer and accept it so that the question doesn't clog up the list of unanswered questions. $\endgroup$
    – joriki
    Jan 13 '20 at 1:00

Yes, this method would work. Because $x$ was replaced with $x^2$ in the original function, you just make that change in the series as well.


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