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I have the following statement -

If $f$ has a Taylor series expansion about zero with radius $R$ , then $g(x) = \displaystyle f\left(\frac{x-1}{2}\right)$ has a Taylor expansion about $X = 1$ of radius $R$.

I would like to get some hint with explanation. Thanks!

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  • $\begingroup$ In this case the radius of convergence of the Taylor series will be $2R$. $\endgroup$ Commented Jun 18, 2016 at 17:16
  • $\begingroup$ Try an example: $f(x) = 1/(1-x)$ has Taylor series with what radius?, then plug in to get $g(x)$ and see what radius that has. $\endgroup$
    – GEdgar
    Commented Jun 18, 2016 at 17:19

1 Answer 1

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Hint: $f(x)=\sum_{n=0}^{\infty}a_n x^n$ and $g(x)=f(\frac{x-1}{2})=\sum_{n=0}^{\infty}a_n (\frac{x-1}{2})^n$.

$g(x)$ has a taylor series about $x=1$ (see brackets) and the radius of convergence can be found by the ratio test, which is $2R$ (see comment Hans Engler).

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