Pade approximant of infinite order

Question

The Pade approximant states that you can approximate a function $f(x)$ by a rational function $R(x)$ of a given order. My question is, if the order of $R(x)$ goes to infinity, does $R(x)$ approach $f(x)$ exactly?

Answer 1

If you write the Maclaurin series of the rational function $R(x)$ you'll obtain the terms of the original function up until the specified order, since $R(x)$ is constructed so that $R(0)=f(0), R'(0)=f'(0) \cdots R^{(m+n)}(0)=f^{(m+n)}(0)$. If it is constructed to have the exact same Maclaurin expansion, then obviously it becomes $f(x)$ however this defeat the purpose of the Padé Approximation, which is used to simplify functions in the design of Control Systems for example.

Answer 2

Guess how this wonderful plot is generated? Surely enough, while applying pade approximation of a given system, where delay function is located at the feedback path. As it can be seen, the dominant pole of the system is unmoved because of the approximation, but, higher order pole-zero pair moves towards infinity with increasing order of pade approximation. Sure enough, all the non-dominant poles are coupled with zeros, which signifies, system response is unaltered because of the approximation.

NB: I am talking from control system/transfer point of view. OC may pardon me for that.