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I am very new to Pade' approximation concept, so some detailed derivation for the approximate result of the following function would be very helpful.

The function that I wish to approximate in the limit $x,y\ll1$ is :

$$1-\frac{(1-x^2)\sin^2(\theta)\sin^2(\theta-y)}{(1-(1-x^2)^{1/2}\cos(\theta)\cos(\theta-y))^2}$$

I would like to have an approximation which is correct upto $2^{nd}$ order in both $x,y$.

Sayan
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  • If you are very new with Padé approximants, it is surprising that you are asked for a two variable problem (which is quite unusual). – Claude Leibovici Oct 05 '21 at 02:50
  • Have a look at https://math.stackexchange.com/questions/4065623/how-to-derive-the-multivariate-pade-approximation-for-ln-left-1-fracx – Claude Leibovici Oct 05 '21 at 02:55
  • Hi @claude. Thanks for the answer. I found this expression in a paper and it did not tell how the approximation is made. I did some search over the net and found out that pade approximation could be used to approximate such expression. – Sayan Oct 05 '21 at 06:45
  • It is very simple with one variable. It is already very complicated with two ( they are Chisholm approximant s). With more than two, they are Canterbury approximant. I shall write one for a similar expression. – Claude Leibovici Oct 05 '21 at 06:51
  • It would be great. Another thing that I wanted to know about this approximation is there global ordering. Lets say i want to express the approximation upto second order then what degree should I choose in the numerator and denominator (m/n) ? – Sayan Oct 05 '21 at 07:14

1 Answers1

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A single variable problem.

Assuming that you know Taylor series, building the $[m,n]$ Padé approximant of $f(x)$ is quite simple (I shall make it around $x=0$ and there is no problem to generalize).

So, we look for $$f(x)=\frac{\sum_{i= 0}^m a_i\,x^i}{1+\sum_{i= 1}^n b_i \,x^i }$$ and, on the other side, we have the corresponding Taylor expansion $$f(x)=\sum_{i= 0}^p c_i\,x^i \qquad \text{with} \qquad p>m+n$$ So, write $$\Bigg[1+\sum_{i= 1}^n b_i \,x^i\Bigg]\Bigg[\sum_{i= 0}^p c_i\,x^i\Bigg]=\sum_{i= 0}^m a_i\,x^i$$

Now, as usual, you expand the lhs and identify the coefficients.

Example

SImilar to your problem, consider $$f(x)=1-\frac{\alpha \left(1-x^2\right)}{\left(1-\beta \sqrt{1-x^2}\right)^2}$$ The $[2,2]$ Padé approximant is then $$f(x)=\frac{\left(1-\frac{\alpha }{(\beta -1)^2}\right)+\frac{\alpha (3 \beta -4)-3 (\beta -1)^2 \beta }{4 (\beta -1)^3}x^2 } {1+\frac{3 \beta }{4 (\beta -1)} x^2}$$

Now, we shall compare $$\text{Padé - Taylor}=\frac{\alpha (\beta -2) \beta }{16 (\beta -1)^5}x^6+O\left(x^8\right)$$ Do you see the difference ?

Claude Leibovici
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  • Thanks for the answer. I have a small query, you have written that $p>m+n$, does that mean, if I need an approximation upto say second order of the whole expression then (according to m/n) I can use $1,0$ or $0,1$ or $0,0$ ? – Sayan Oct 05 '21 at 08:34
  • Suppose that the Taylor series starts at $x^3$; then make the $[n+3,n]$. This is my recipe. – Claude Leibovici Oct 05 '21 at 08:37
  • $n$ is the order beyond which you want your series to truncate, right ? – Sayan Oct 05 '21 at 08:43