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Suppose we have a finite number of terms for Taylor expansion of a conditionally convergent function.

For example, $f=\frac1{1-x}$ with expansion $f=\sum_{n=0}^\infty x^n$. This expansion diverges for $|x|\ge1$. On the other hand, having infinite number of series terms, we could re-expand it at a point farther from the pole and do it repeatedly, thus finding a sequence of analytical continuations to any desired point.

The question is: can we do something similar, but with finite number of Taylor coefficients? I.e. can we approximate $f$ for $x:|x|\ge1$ using finite number of Taylor coefficients of expansion at $x=0$?

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No. Once you restricted yourself to a finite number of terms, you have a polynomial. A polynomial $p(z)=\sum_{n=0}^d c_n z^n$ can be expanded in terms of $(z-a)$ for any point $a$ (just by using the binomial formula), but this does not change the polynomial: you still have the same values $p(z)$ at every $z$, just written differently. So, if the polynomial $p$ did not do a good job of approximating $f$ around $a$, the algebraic manipulations won't change that.

The magic of analytic continuation is possible because reshuffling pieces of an infinite series can change its behavior.

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Yes, it can be done. Suppose we have $N$ terms of Taylor expansion for $f(x)$, so we have $$f(x)=\sum_{n=0}^N a(n)x^n+\mathcal O\left(x^{N+1}\right).$$

We are interested in the behavior of the function outside of disk of convergence of the series which we get when $N\to\infty$.

The series converges to $f$, and the same is true for derivatives of $f$: derivatives of the series all converge to derivatives of $f$. This means that as $N$ gets larger, we can have more and more precise expansion coefficients for expansion in any point in the disk of convergence of the original series.

But the derivatives have not identical precision. The higher the derivative, the worse the precision we get from $N$ terms of the original series, because the smaller number of terms is included in their approximation:

$$f^{(k)}(x)=\sum_{n=k}^N a(n)\frac{n!}{(n-k)!}x^{n-k}+\mathcal O\left(x^{N+1-k}\right).$$

To approximate a function using Taylor formula, we don't need to use infinitely precise Taylor coefficients. So, for the point we choose there are some $M<N$ derivatives with sufficient precision to approximate our function to order $M$ by Taylor polynomial. As $N$ gets larger, $M$ also will get larger. NOTE: we don't just rearrange terms of the polynomial: we drop imprecisely computed coefficients of re-expansion, thus reducing the degree of new polynomial, but gaining insight in behavior of the function outside of disk of convergence of original series.

Now, let's do some numerical experiments. Take $f=\frac1{1-x}$ and $N=1000$. Here's how $\Re [f]$ looks:

enter image description here

Find Taylor polynomial for it, expanding the function at $x=0$. The series converges for $x:|x|<1$, so Taylor polynomial is only useful for this domain. Now we want to extend the function to larger domain. We expand our polynomial around $x=-\frac12$, and for $M=100$ see plots of real part of the function:

enter image description here

Here the red part (patchy because it interleaves with blue part) is the original expansion around $x=0$, the oscillations around it are the first indication that the series diverges for bigger $|x|$. I didn't render it outside of disk of series convergence because it isn't useful and would hide the important details. The blue part is the Taylor partial sum with $M$ terms, for expansion at $x=-\frac12$. Clearly, the approximation gives the expected useful domain: $|x+\frac12|<\frac32$. Of course, one has to do some guesstimates to choose proper $M$. But still, it is possible to get the approximation of the function outside of disk of convergence of the original series, using only finite number of series terms.

To better understand precision of such re-expansion, here's a plot of $|f_M(z)-f(z)|$ and $|f_N(z)-f(z)|$, where $f_N$ is $f$ expanded at $x=0$, and $f_M$ is the re-expanded at $x=-\frac12$ polynomial $f_N$:

enter image description here

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  • $\begingroup$ I don't understand this answer. Even you say that "this approximation gives the expected useful domain: $\lvert x+\frac{1}{2}\rvert < \frac{3}{2}$". Why should it be accurate outside the disk of convergence? $\endgroup$ – Giuseppe Negro May 1 '14 at 11:53
  • $\begingroup$ Because it is partial sum of another series, whose approximate coefficients are obtained from partial sum of original series. $\endgroup$ – Ruslan May 1 '14 at 11:55
  • $\begingroup$ I disagree. As the other answer explains, re-expanding a polynomial around another point is just a symbolic manipulation. The polynomial remains exactly the same. So when you take a truncated McLaurin series for $1/1-x$ and re-expand it, you are just manipulating symbols and ultimately doing nothing. $\endgroup$ – Giuseppe Negro May 1 '14 at 11:58
  • $\begingroup$ The core strategy here is to use only precise enough coefficients of re-expansion and drop others. This will lower the polynomial degree, but avoid using imprecise coefficients which would destroy information we're looking for. If we compare these two polynomials, we'll see they're different. $\endgroup$ – Ruslan May 1 '14 at 12:00
  • $\begingroup$ Ok, I didn't get this at first read, thanks for adding that note. I have also had difficulties in interpreting that graph. The flat zone means that the approximation is very good? $\endgroup$ – Giuseppe Negro May 1 '14 at 15:57
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Another option is to use Padé approximant of the given Taylor polynomial. Padé approximants do work for singularities, and moreover, they can go beyond them. And the key property of Padé approximants is that they agree with the Taylor expansion up to $(m+n)$ order, where $m$ and $n$ are orders of numerator and denominator.

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