I need to write a function for $\sin(x)$ using Padé approximation.
I found here a formula, but I don't know how to achive this.
Thanks in advance.
How to Approximate sin(x) Using Padé Approximation
calculusfunctionsMATLABnumerical methodspade approximation
Related Solutions
Yes, there are formulas that give you an error term depending on the regularity of the function. They are particularly useful for diagonal Pade approximates. This survey http://arxiv.org/pdf/math/0609094v1.pdf provides some useful information.
In practice, one can simply expand both functions, say $\log (1+x)$ and $\frac{x(6+x)}{6+4x}$ into the Taylor series in $|x|\le 1$ and after "cancelling" the corresponding equal terms, you will get an error term. Alternatively, you can set up the function $f(x)=\log{(x+1)}- \frac{x(6+x)}{6+4x}$ and study it's behaviour using derivatives.
In my humble opinion, this is not the way of building a Padé approximant but rather to fit a function over a given range by the ratio of two polynomials of given degrees. Remember that, as Taylor series, a Padé approximant is built around a given point.
Let us show it with $e^x$, the $[2,2]$ Padé approximant of which (built around $x=0$) being $$\frac{1+\frac{1}{2}x+\frac{1}{12}x^2}{1-\frac{1}{2}x+\frac{1}{12}x^2}$$ and now use $$y = a_0 + a_1 x + a_2 x^2 - b_1 x y - b_2 x^2 y$$ with $20$ equally spaced data points $x_i$ between $-1$ and $+1$ with $y_i=e^{x_i}$.
The linear regression would give the following results $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a_0 & +1.00003 & 0.00003 & \{+0.99997,+1.00010\} \\ a_1 & +0.50444 & 0.00235 & \{+0.49943,+0.50946\} \\ a_2 & +0.08385 & 0.00114 & \{+0.08142,+0.08628\} \\ b_1 & -0.49522 & 0.00231 & \{-0.50015,-0.49029\} \\ b_2 & +0.07957 & 0.00104 & \{+0.07736,+0.08178\} \\ \end{array}$$
Moreover, if you speak about least square fit, you must continue with nonlinear regression since what is "measured" is $y$ and not any of its possible transform. This linear regression gives you good estimates to start with; so, now, let us fit the true model $$y = \frac{a_0 + a_1 x + a_2 x^2}{1 + b_1 x + b_2 x^2}$$ This would give $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a_0 & +1.00006 & 0.00003 & \{+1.00000,+1.00012\} \\ a_1 & +0.50900 & 0.00234 & \{+0.50401,+0.51399\} \\ a_2 & +0.08609 & 0.00118 & \{+0.08356,+0.08862\} \\ b_1 & -0.49077 & 0.00228 & \{-0.49562,-0.48591\} \\ b_2 & +0.07761 & 0.00099 & \{+0.07550,+0.07971\} \\ \end{array}$$
For sure, since this is a least square fit, over the range it will be better than the strict Padé approximant. Considering the norms $$\Phi_1=\int_{-1}^{+1} \left(e^x-\frac{1+\frac{1}{2}x+\frac{1}{12}x^2}{1-\frac{1}{2}x+\frac{1}{12}x^2} \right)^2\,dx=1.26 \times 10^{-6}$$
$$\Phi_2=\int_{-1}^{+1} \left(e^x-\frac{a_0 + a_1 x + a_2 x^2}{1 + b_1 x + b_2 x^2} \right)^2\,dx=6.81 \times 10^{-9}$$
Best Answer
Since you know Taylor expansions, let us try a simple way.
You want to make $$f(x)=\frac{\sum_{i=0}^n a_ix^i } {1+ \sum_{i=1}^m b_ix^i }$$ where $m,n$ are predefined degrees. So, write $$f(x)\left({1+ \sum_{i=1}^m b_ix^i }\right)=\sum_{i=0}^n a_ix^i $$ Use the Taylor expansion of $f(x)$ and identify coefficients.
Let us try with $f(x)=e^x$, $m=2$, $n=3$. So, we have $$\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\right)\left(1+b_1x+b_2x^2+b_3x^3\right)=a_0+a_1x+a_2x^2$$ Expanding, grouping terms and setting coefficients equal to $0$ will lead to the following equations $$a_0=1$$ $$1-a_1+b_1=0$$ $$-a_2+b_1+b_2+\frac{1}{2}=0$$ $$\frac{b_1}{2}+b_2+b_3+\frac{1}{6}=0$$ $$\frac{b_1}{6}+\frac{b_2}{2}+b_3+\frac{1}{24}=0$$ $$5 b_1+20 b_2+60 b_3+1=0$$ from which $$a_0=1 \qquad a_1=\frac 25\qquad a_2=\frac 1{20}$$ $$b_1=-\frac35 \qquad b_2=\frac 3{20}\qquad b_3=-\frac 1{60}$$ and then $$\frac{1+\frac{2}{5}x+\frac{1}{20}x^2 } {1-\frac{3 }{5}x+\frac{3 }{20}x^2-\frac{1}{60}x^3 }$$ is the Padé approximant.