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This should be a very simple exercise requiring some algebraic manipulation, but I can not seem to get it right! Its statement is the following: If $f$ ($\mathbb{C}\to\mathbb{C}$) is a power series centred at the origin $z=0$, then $f$ has a power series expansion around any point of its disc of convergence.

So let $z_0$ be a number in the disc of convergence of $f$. Then $$f(z)=f(z-z_0+z_0)=\sum_{n=0}^\infty a_n(z-z_0+z_0)^n = \sum_{n=0}^\infty a_n \sum_{k=0}^n {n\choose k}(z-z_0)^{n-k}z_0^k.$$ Now I should rearrange this to get $(z-z_0)^n$ and some new coefficients $b_k$ not depending on $z$ and I don't know how to do it!

Any help will be appreciated!

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  • $\begingroup$ Just combine all the coefficients of $(z-z_0)^m$ coming from all $(n,k)$ such that $n-k=m$. But that doesn't solve the problem: the hard part of this problem is proving that the resulting power series actually converges to $f$ in a neighborhood of $z_0$. $\endgroup$ Sep 1, 2017 at 18:42
  • $\begingroup$ Possible duplicate: math.stackexchange.com/questions/2188444/… $\endgroup$
    – Alex R.
    Sep 1, 2017 at 18:48

1 Answer 1

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Unfortunately I don't know how to use LaTeX.

It is worth noting that, by the general binomial expansion, the range of k should be from 0 to infinity.

Instead of having the power of (z-z_0) as (n-k) and the power of (z_0) as k, have the power of (z-z_0) as k and the power of (z_0) as n-k. Then, after some argument about rearranging the summation signs and then letting everything under the k sum, apart from the (z-z_0)^k, being b_k, some complex sequence, we should be very nearly done.

Let the radius of convergence of the power series centred at 0 be R. Explaining that the new radius of convergence, R', is related to R by the formula r'=r-|z_0|, although relatively clear, is still required.

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  • $\begingroup$ Throw \$ around your math, like \$z_0\$ which comes out as $z_0$. $\endgroup$
    – JKreft
    Oct 26, 2018 at 8:39

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