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I appreciate that Pade approximants are often nicer than Taylor series; I know that if you take a Pade approximant of order $M/N$ it corresponds loosely to a Taylor approximation of order $M+N$.

Soft question, and apologies if this is deemed a bad question:

If I take $M+N$ as fixed, how do I make a decision on the ratio between $M$ and $N$? I expect this will depend on the function at hand, or perhaps on the computing system we are designing the approximant for, but in general what are the rules of thumb for this? I have tried to research this but have seen no way to make the decision, yet I feel the decision must be important somehow.

FShrike
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  • I would assume that the $N$ is chosen depending on how many poles the function has. Some sources (https://rdrr.io/rforge/pracma/man/pade.html) also suggest that the "errors are smallest when the degrees of numerator and denominator are the same or when the degree of the numerator is one larger than that of the denominator.". So maybe something like $N \approx \text{number of poles}$, $M \approx N$? – SV-97 Nov 05 '21 at 19:54
  • Could you motivate your assumption on $N$ and the poles? It isn't immediately clear to me @SV-97 – FShrike Nov 05 '21 at 19:55
  • My thinking was this: a rational function has poles at the zeros of it's denominator and a polynomial of degree $n$ has precisely $n$ (complex) roots. So if we want to accurately represent a certain number of poles then $N$ is bounded below by the number of poles. I'm not entirely sure if some other factor should be considered here though. – SV-97 Nov 05 '21 at 20:08
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    I forgot to tell you that you can make the Padé approximant of a long polynomial. – Claude Leibovici Nov 06 '21 at 15:33
  • @ClaudeLeibovici interesting; I assume by that you mean a Padé approximation can make a high degree polynomial (say degree 8) into a rational polynomial of lower degree, say [4,4], with presumably small error – FShrike Nov 06 '21 at 15:37
  • Exactly true. Make a polynomial of your choice and send it to me (for the fun). – Claude Leibovici Nov 06 '21 at 15:42
  • $$1+x+2x^2-3x^3-4x^5+6x^6\approx\frac{55+73x+112x^2-59x^3}{55+18x-16x^2+86x^3}$$(+1) for the kindness sir, @ClaudeLeibovici – FShrike Nov 06 '21 at 16:09
  • Perfect. Enjoy, as much as I do ! Cheers :-) – Claude Leibovici Nov 06 '21 at 16:25
  • Suppose that you want to solve $1+x+2x^2-3x^3-4x^5+6x^6=\pi$ and you know that the solution of interest is close to $-\frac 12$. Build the $[2,2]$ approximant around. $x= -\frac 12$. Solve the quadratic to get $x \sim -0.665065$ while the solution is $x=-0.665057$. – Claude Leibovici Nov 07 '21 at 10:16

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I give you my personal solution.

Suppose that you expand your function $f(x)$ around a given point $x=a$ and that the first term corresponds to a degree $p$. Then I choose the $[p+n,n]$ Padé approximant. This exactly coincides with @SV-97 comment.

If you look at this question of mine, which never recieved any answer, you will see one example and a similar concern.

Claude Leibovici
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  • So you would choose more terms in the numerator than denominator? I feel like I have misunderstood your answer as it doesn’t appear to coincide with Sv-97. “The first term corresponds to a degree $p$”; I’m not sure what you mean: the first non-zero term of the function’s Taylor series? – FShrike Nov 06 '21 at 09:19
  • @FShrike. Suppose that $f(x)=x \text{erf}(x) \sinh (x)$; the first term of Taylor series is $\frac{2 x^3}{\sqrt{\pi }}+O\left(x^4\right)$. So, I should use the $[m+3,m]$ Padé approximant. Is it better explained ? – Claude Leibovici Nov 06 '21 at 09:29
  • Gotcha! Am I right to assume you use the [numerator,denominator] convention? – FShrike Nov 06 '21 at 09:31
  • @FShrike. Yes, that's it. – Claude Leibovici Nov 06 '21 at 09:36
  • I obtained the Pade' approximant of: $$\frac{7395-34840x-113056x^{2}}{4640-23680x-56192x^{2}}$$On your exercise... it fits the curve, but without the help of a calculator would have been nightmarish to find. Am I doing something wrong? The quadratic solution is then $x\approx0.147$ and $x\approx-0.77$, and shifting these by $1/2$ does not give me the right answer – FShrike Nov 07 '21 at 11:19
  • @FShrike. Your Padé approximant seems to be wrong. It should be $$\frac{ \frac{51}{32}-\frac{37227 \left(x+\frac{1}{2}\right)}{17378}+\frac{260761 \left(x+\frac{1}{2}\right)^2}{34756}} {1+\frac{18988 \left(x+\frac{1}{2}\right)}{8689}+\frac{14556 \left(x+\frac{1}{2}\right)^2}{8689} }$$ – Claude Leibovici Nov 08 '21 at 04:45