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If I know that there is a Taylor series for a function

$f\left(x\right)=\sum_{i=0}^{\infty}c_i\left(x-x_0\right)^i$

am I always justified in saying that this can be re-expressed as a series of the form

$f\left(x\right)=\sum_{i=0}^{\infty}b_i x^i$

where $b_i$ may depend on the $c_i$ and $x_0$?

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2 Answers 2

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I believe the answer is no, that conclusion is not always justified. $f(x)=\frac{1}{x}$ has a Taylor series expansion about $x_0 =1$, which can be gotten from the series $\frac{1}{1-x} = \sum_{i=o}^{\infty} x^i$. But $f$ has no expansion about $x_0 = 0$, because it does not exist there.

If $0$ is in the interval of convergence for your original function, I believe the conclusion would be justified.

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Yes.

If we say that $$f_n(x)=\sum_{i=0}^nc_i(x-x_0)^i$$ then we can expand this via the binomial theorem using binomial coefficients. Therefore, every term of $c_i(x-x_0)^i$ can be expressed as $$c_i(x-x_0)^i=c_i\sum_{j=0}^ia_jx^j$$ where $a_j$ represents the binomial coefficients and terms involving multiplication by some powers of $x_0$. Substituting this into our original equation gives $$f_n(x)=\sum_{i=0}^n\left(c_i\sum_{j=0}^ia_jx^j\right)$$ which can then of course be written in the form of some series $$f_n(x)=\sum_{k=0}^nb_kx^k$$ where the $b_k$s take all of the above coefficients into account.

However, as Van Latimer pointed out, this is not the same as expanding $f(x)$ as a MacLaurin series around $x=0$. Plugging $x=0$ into $f_n(x)$ will merely get you the value of $f(x_0)$, in the approximation - not the value of $f(x_0)$, because $f$ may not be defined at $x=0$. Your answer may bear no relation to the actual value of $f(x)$ at $x=0$ - if there is one.

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    $\begingroup$ The upper limit on your first, third and last sums should be $\infty$. More important, you don't take convergence considerations into account. If the third sum is not convergent, then you can't re-arrange the terms as you suggest. $\endgroup$
    – B. Goddard
    Jun 23, 2017 at 17:49

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