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I'm trying to understand analitic continuation. I'm trying to use a Taylor Series to extend

$$\hat{f} (z)= \sum_{n=0}^{\infty} z^n$$

I want to actually compute the series (using mathematica), So I summed up to 100. $$f (z)= \sum_{n=0}^{100} z^n$$

I want to use this definition to make a Taylor Series around the point $ \frac{i}{2}$ and I'm using Mathematica to evaluate the series:

$$\hat{F}(z)= \sum_{n=0}^{\infty} \frac{f^{(n)}(\frac {i}{2})}{n!}(z-\frac {i}{2})^n$$

In theory (actually taking an infinite sum) $\hat{F}(z)$'s radius of convergence should be $\sqrt{5} /2$ around $i/2$. However this is not what I'm finding when I run the code on a computer.

Let:$$ F(z)= \sum_{n=0}^{20} \frac{f^{(n)}(\frac {i}{2})}{n!}(z-\frac {i}{2})^n$$

Also, Let:$$ F_2(z)= \sum_{n=0}^{30} \frac{f^{(n)}(\frac {i}{2})}{n!}(z-\frac {i}{2})^n$$

When I run the code for these two functions $F(z)$ does seem to converge where it is supposed to, however $F_2(z)$ has a smaller radius of convergence (see image below)

1 (above) This is a screen shot of my Mathematica code. The three graph show f,F,F2 and everything that is not in the theoretical radius of convergence is "zeroed" out. In the right most graph, you can see a grey annulus; This is where the series should converge, but it is not.

After messing around with the code, I'm pretty sure the heart of the issue is in the computation of $\frac{f^{(n)}(\frac {i}{2})}{n!}$ changing the upper-limits of the summations affects the values for large enough n.

There seems to be relationship between the upper limit of $f$ and the maximum upper-limit of $F$. Is there a specific formula for the relation?

If I were to do another Taylor series with $F$ to extend the function further, it seems I need to keep reducing the upper-limit. Is there a more efficient when to calculate the series? (maybe some type of series acceleration)

Are there any examples online the show an analytic continuation, with actual values that can be put into Mathematica to generate a graph (I can't actually add infinity numbers)? Or something that explains how the upper-limit of the sum affects the convergence?

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    Yes, its radius of convergence should be $\sqrt{5}/2$ around $i/2.$ What are you seeing when you try a point like $x=-1+i/2$?2017-01-25
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    I just added a screen shot to my post.2017-01-25
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    $f(z)=1/(1-z)$ when $z\ne 1. $2017-01-26

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