I am currently starting to read papers about Padé approximation of the matrix exponential $\exp(A)$, namely $\exp(A) \approx P_{n,m}(A)Q^{-1}_{n,m}(A)$

I am now seeking for a good motivation behind Padé approximation.

The problem with doing numerical approximation of $\exp(A)$ by taylor series is - neglecting efficiency - that there the values could be very different in the floating point arithmetic, especially very big and of the same size too so that catastrophic cancellation is possible, making the whole method unreliable. This problem still remains with Padé and there is even an additional problem, namely how well-conditioned the "inverse part" of the rational approximation is.

Can someone give me a good motivation and an intuition why Padé really can do better numerically and why it pays off to even consider inversion?

My ideas so far: $P_{n,m}(A) \approx \exp(\frac{A}{2})$, $Q_{n,m}(A) \approx \exp(-\frac{A}{2})$ and taylor of $\exp(A)$ is good for small $\|A\|$. Is this factor $2$ the only motivation? The scaling and squaring benefit is valid for both Taylor and Padé approximation.

  • 1
    $\begingroup$ This question on MO discusses the scalar case, but not the matrix one. $\endgroup$ – Antonio Vargas Oct 23 '15 at 14:28
  • $\begingroup$ A comment stated what I wanted to know. Other ideas are welcome too. $\endgroup$ – mdot Oct 23 '15 at 14:39
  • $\begingroup$ I don't know much about Pade of matrices, but in general Pade approximants are really useful for approximating a function beyond the radius of convergence of its power series. $\endgroup$ – Ron Gordon Oct 23 '15 at 19:20
  • 1
    $\begingroup$ $f$, $g$ being analytical with sufficiently large domain, $f(A)$ and $g(A)$ have equal values at $A$ if they as scalar functions are equal at the eigenvalues of $A$ and in the case of multiplicities, if the derivatives coincide up to the dimension minus one of the eigenspace. Similar should hold if $g$ is an approximation of $f$. The Padé approximation has usually a larger interval of "good" approximation than the Taylor series of the same error order. $\endgroup$ – LutzL Apr 4 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.