I am in trouble with the following problem from the Schaum's outline of Complex variables:
"A function $F(z)$ is represented in $|z-1|<2$ by the series $\sum_0^\infty\cfrac{(-1)^n(z-1)^{2n}}{2^{2n+1}}$. Prove that the value of the function at $z=5$ is $\cfrac{1}{16}$"
My try:
I define $F(z)$ as $F(z)=\sum_0^\infty\cfrac{(-1)^n(z-1)^{2n}}{2^{2n+1}}$
Then $F(5)=\lim_{x\to5}F(x)=\lim_{x\to5}\sum_0^\infty\cfrac{(-1)^n(x-1)^{2n}}{2^{2n+1}}$
Since $\cfrac{(-1)^n(x-1)^{2n}}{2^{2n+1}}=\cfrac{[-(x-1)^2]^n}{4^n\cdot2}=\left[\cfrac{-(x-1)^2}{4}\right]^n\cfrac{1}{2}$
Then if we make $u=\cfrac{(x-1)^2}{4}$ then the sum equals:
$F(x)=\sum_0^\infty\cfrac{(-1)^n(z-1)^{2n}}{2^{2n+1}}=\cfrac{1}{2}\sum_0^\infty\left[\cfrac{-(x-1)^2}{4}\right]^n=\cfrac{1}{2}\sum_0^\infty(-u)^n=\cfrac{1}{2}\cdot\cfrac{1}{u+1}$
$F(x)=\cfrac{1}{2}\cdot\cfrac{1}{\cfrac{(x-1)^2}{4}+1}=\cfrac{1}{2}\cdot\cfrac{4}{4+(x-1)^2}=\cfrac{2}{4+(x-1)^2}$
Therefore $\lim_{x\to5}F(x)=\lim_{x\to5}\cfrac{2}{4+(x-1)^2}=\cfrac{1}{10}$
But the book says that $F(5)=\cfrac{1}{16}$. I think that the reason of the difference radicates in that $F(z)$ is defined in $|z-1|<2$.
Any help is appreciated!
Check the problem again if you made an error transcribing it from the source. I checked your solution and from the comments at least Mindlack came to the same conclusion, so I'd say that we all made the same error is not impossible, but maybe no so likely.
You wrote:
Well, that's how many problems in complex analysis start. You are given a power series that converges inside it's radius of convergence, so a-priori the function defined by the power series is only defined inside some circle (and possibly on the perimeter). It's easy to see that the radius of convergence of the given power series is $2$, so defining it inside $|z−1|<2$ is totally natural.
Then the usual task is to find what the function actually is and it turns out in most cases the function can be extended (and still be holomorphic) to a much bigger domain than the circle. Due to the various uniqueness theorems for holomorphic extensions, it then makes sense to ask "What is $F(5)$?"
So your concern that the problem comes from "$F(z)$ is defined in $|z−1|<2$" is unfounded. Note that when you talk about the limit
$$\lim_{x\to5}\sum_0^\infty\cfrac{(-1)^n(x-1)^{2n}}{2^{2n+1}}$$
this is something that does not exist at all, simply because the sum isn't convergent for any $x \ge 2$, so can't be evaluated near $5$. You (intuitively, I guess) made the right choice to find a closed form of $F(z)=\frac2{4+(z-1)^2}$.
Of course this is how such problems usually work out. You find the closed form of the sum, which happens to be defined and holomorphic except for a few points (in this case $1\pm2i$, which again illustrates that the radius of convergence of the initial power series is $2$), so any holorphic extension of $F(z)$ must be $\frac2{4+(z-1)^2}$, and at $z=5$ this evaluates to $\frac1{10}$.