How do I use Padé approximation to transform a given function?
Pade approximation is the best approximation in mathematics, compared to Taylors approximation & others. I have learnt this approximation transform d given function to another equivalent function but in rational form.
Pls my challenge is to perform the transformation example $e^x = 1 + x + x^2/2! + ...$ (From Maclaurin expansion) in Padé is $$R(x) = \frac{1 + 2x/3 + x^2/6}{ 1 - x/3}???$$ Please help?
To be short, consider a function $f(x)$ which has a Taylor expansion $F(x)$ and you want to build the Pade approximant as $\frac{P_m(x)}{Q_n(x)}$ where $$P_m(x)=\sum_{i=0}^m a_i x^i$$ $$Q_n(x)=1+\sum_{i=1}^n b_i x^i$$ are polynomials of degrees $m$ and $n$. So, you could write $$F(x)=\frac{P_m(x)}{Q_n(x)}\implies F(x) Q_n(x)=P_m(x)$$ and identify the coefficients.
For your specific case $$F(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}$$ $$P_2(x)=a_0+a_1x+a_2x^2\qquad , \qquad Q_1(x)=1+b_1 x$$ So, grouping terms and stopping as soon as all coefficients appeared $$F(x) Q_n(x)-P_m(x)=(1-a_0)+x (-a_1+b_1+1)+x^2 \left(-a_2+b_1+\frac{1}{2}\right)+\left(\frac{b_1}{2}+\frac{1}{6}\right) x^3$$ Cancelling terms, we then have $$1-a_0=0$$ $$-a_1+b_1+1=0$$ $$-a_2+b_1+\frac{1}{2}=0$$ $$\frac{b_1}{2}+\frac{1}{6}=0$$ Then $a_0=1$, $a_1=\frac 23$, $a_2=\frac 16$, $b_1=-\frac 13$ which is your formula.
Making it more general and using the same procedure, you should find that $$a_0=f(0)\qquad a_1=\frac{3 f'(0) f''(0)-f(0) f'''(0)}{3 f''(0)}\qquad a_2=\frac{3 f''(0)^2-2 f'''(0) f'(0)}{6 f''(0)}$$ $$b_1=\frac{f'''(0)}{3 f''(0)}$$
