Taylor and Páde expansion

Next: Approximate the wave equation Up: Forward modeling Previous: Forward modeling

Taylor and Páde expansion

The Taylor expansion of a function at is written as

$\displaystyle f(x+h)=f(x)+\frac{\partial f(x)}{\partial x}h+\frac{1}{2!}\frac{\... ...)}{\partial x^2}h^2+\frac{1}{3!}\frac{\partial^3 f(x)}{\partial x^3}h^3+\ldots.$

(10)

A popular example is

$\displaystyle (1+x)^{\alpha}=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\ldots+\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}x^n+\ldots.$

(11)

Here we mainly consider the following expansion formula:

$\displaystyle (1-x)^{\frac{1}{2}}=1-\frac{1}{2}x-\frac{1}{8}x^2-\frac{1}{16}x^3-\frac{5}{128}x^4-\ldots, \vert x\vert<1.$

(12)

The Páde expansion of Eq. (12) follows from expansion in continuous fractions:

$\displaystyle (1-x)^{\frac{1}{2}}=1-\frac{x/2}{1-\frac{x/4}{1-\frac{x/4}{\ldots}}}$

(13)

I provide an informal derivation:

$\begin{displaymath} \begin{split} y&=(1-x)^{\frac{1}{2}}\Rightarrow x=1-y^2=(1-y... ...-\frac{x/2}{1-\frac{x/4}{1-\frac{x/2}{1+y}}}=\ldots \end{split}\end{displaymath}$

The 1st-order Pade expansion is:

$\displaystyle (1-x)^{\frac{1}{2}}=1-\frac{x}{2}$

(14)

The 2nd-order Pade expansion is:

$\displaystyle (1-x)^{\frac{1}{2}}=1-\frac{x/2}{1-\frac{x}{4}}.$

(15)

And the 3rd-order one is:

$\displaystyle (1-x)^{\frac{1}{2}}=1-\frac{x/2}{1-\frac{x/4}{1-\frac{x}{4}}}.$

(16)

Next: Approximate the wave equation Up: Forward modeling Previous: Forward modeling

2021-08-31