[Math] why are there two different Pade approximation of delay

Question

There are 2 different second order pade approximations of delay given in internet

What is the difference between these two approximation? Which one is the correct Pade approximation

Answer 1

The first one is not a Padé approximant but the ratio of two Taylor series.

Written as $$e^{-x}=\frac{e^{-x/2}}{e^{x/2}}=\frac{1-\frac{x}{2}+\frac{x^2}{8}+O\left(x^3\right)}{1+\frac{x}{2}+\frac{x^2}{8}+O\left(x^3\right) }$$

This does not have anything to do with the second one which is the $[2,2]$ Padé approximant of $e^{-x}$

To show how is better the Padé approximant, let us compute $$\Phi_1=\int_0^1 \left(e^{-x}-\frac{1-\frac{x}{2}+\frac{x^2}{8}}{1+\frac{x}{2}+\frac{x^2}{8}}\right)^2\,dx\approx 5.033\times 10^{-5}$$ which, for sure, is better than $$\Phi_0=\int_0^1 \left(e^{-x}-\left(1-x+\frac{x^2}{2}-\frac{x^3}{6}\right)\right)^2\,dx\approx 1.375\times 10^{-4}$$ but so worse than $$\Phi_2=\int_0^1 \left(e^{-x}-\frac{1-\frac{x}{2}+\frac{x^2}{12}} {1+\frac{x}{2}+\frac{x^2}{12} }\right)^2\,dx\approx 3.127\times 10^{-8}$$

Edit

Developed at a point $x=a$, for a given function $f(x)$, the $[m,n]$ Padé approximant write $$p_{m,n}=\frac{\sum _{i=0}^m a_i (x-a)^i } {1+\sum _{i=1}^n b_i (x-a)^i }$$ Then, let us write in a first step $$\left(1+\sum _{i=1}^n b_i (x-a)^i \right)\,f(x)=\sum _{i=0}^m a_i (x-a)^i $$ and replace $f(x)$ by its corresponding Taylor expansion $$f(x)= \sum_{k=0} ^ {\infty} \frac {f^{(k)}(a)}{n!} (x-a)^{n}=\sum _{k=0}^{\infty} c_k (x-a)^k$$ which will be truncated to "some" order.

Now, expanding the summations and grouping for a given power of $(x-a)^j$, we have a series of linear equations in $a,b,c$ easy to solve for the $a$'s and $b$'s where the $c$'s are known. All of this is quite simple.

For illustration, let us do it for $f(x)=e^{-x}$ using the $[2,2]$ Padé approximant. We then have $$(1+b_1 x+b_2 x^2)\left(1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}\right)=a_0+a_1 x+a_2x^2$$

Expanding and grouping, this will give $$(1-a_0)+ (-a_1+b_1-1)x+ \left(-a_2-b_1+b_2+\frac{1}{2}\right)x^2+\left(\frac{b_1}{2}-b_2-\frac{1}{6}\right) x^3+\frac{1}{24} (-4 b_1+12 b_2+1) x^4=0$$ Cancelling all coefficients, the gives the equations $$1-a_0=0 \qquad -a_1+b_1-1=0 \qquad -a_2-b_1+b_2+\frac{1}{2}=0$$ $$ \frac{b_1}{2}-b_2-\frac{1}{6}=0\qquad -4 b_1+12 b_2+1=0$$ the solutions of which being $$a_0= 1\qquad a_1=-\frac 12 \qquad a_2=\frac 1 {12} \qquad b_1=\frac 12 \qquad b_2= \frac 1 {12}$$ which is your second expression.

Back to the two formulations given in your post, let us make the long divisions to get $$\frac{1-\frac{x}{2}+\frac{x^2}{8}}{1+\frac{x}{2}+\frac{x^2}{8} }=1-x+\frac{x^2}{2}-\frac{x^3}{8}+\frac{x^5}{64}+O\left(x^6\right)=1-x+\frac{x^2}{2}+O\left(x^\color{red}{3}\right)$$ while $$\frac{1-\frac{x}{2}+\frac{x^2}{12}}{1+\frac{x}{2}+\frac{x^2}{12}}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}-\frac{x^5}{144}+O\left(x^6\right)=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}+O\left(x^\color{red}{5}\right)$$ when compared to the Taylor expansion of $e^{-x}$.

I hope and wish that this makes things clearer for you. If not, just post.