## Pade approximants: a case study (ii)

In the previous article we have seen how to calculate Padé approximations on the example of

$f(z) = \frac{\log(1-z)}{z}$

## The study (ii)

We are now going to see how efficient Padé’s ansatz is and compare it to the well known series representation $f(z) = -\sum\limits_{k=0}^\infty \frac{z^k}{k+1}$. We also focus on the “diagonal” Padé’ $P^n_n(z)$, which seems to be one of the two popular choices (the other is $P^n_{n+1}(z)$).

Example 1 (inside of Taylor r.o.c): convergence at $z = \frac{3}{4}$. Here the value is

$f\bigl(\frac{3}{4}\bigr) = -\frac{4\log(4)}{3}\approx -1.84839$

Since three quarters lies inside the radius of convergence for the series, we can directly compare how fast the individual approaches to $f$ converge to the exact value:

The numbers on the horizontal axis (0, … ,10) denote the order of the approximation $n$.  Here and below, the black dots represent the exact value, green squares stand for the diagonal Padé approximation $P^n_n\bigl(\frac{3}{4}\bigr)$, and blue diamonds correspond to the $n$-th Taylor polynomial evaluated at the point of interest. At this resolution, already the third Padé does not differ visually from the exact value, whereas the series still deviates significantly!

Example 2 (at the edge of Taylor r.o.c): consider $z = -1$. The series is rather famous

$f(-1)=-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\dots=-\log 2\approx -0.693147$

and known to be slowly convergent. Have a look how the Padés outperform the Taylors:

Example 3 (Padé converges where Taylor explodes): consider $z = -2$. At this point the Taylor series does not work anymore, however the Padé approximants do converge to the correct value

$f(-2)=\frac{\log 3}{2}\approx -0.549306$

Impressive, isn’t it? We will come back to that property in a later post…
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