Approximaths

Padé approximant of 1/(1-x)

Written by

approximaths

in

In this post we will have a look at the Padé approximants of \frac{1}{1-x}. The Maclaurin series of \frac{1}{1-x} is:

\displaystyle 1 + x + x^2 + x^3 + x^4 + x^5 + \cdots

which converges for x \in ]-1,1[. We can calculate the corresponding P(1,1)

\displaystyle \frac{A_0 + A_1 x}{1 + B_1 x} = 1 + x + x^2 + \cdots
\displaystyle = (1 + B_1 x)(1 + x + x^2 + \cdots)
\displaystyle = 1 + x + x^2 + \cdots + B_1 x + B_1 x^2 + B_1 x^3 + \cdots
\displaystyle = 1 + (1 + B_1) x + (1 + B_1) x^2 + (1 + B_1) x^3 + \cdots

Keeping only degrees up to 2:

\displaystyle = 1 + (1 + B_1) x + (1 + B_1) x^2

This implies:

\displaystyle A_0 = 1
\displaystyle A_1 = 1 + B_1
\displaystyle 1 + B_1 = 0

Therefore, A_0 = 1, A_1 = 0 and B_1 = -1 and the P(1,1) is:

\displaystyle \boxed{P(1,1)(x) = \frac{1}{1-x}}

This is an exceptional result since the Padé approximant P(1,1) is equal to the function it is supposed to approximate. This result is very attractive since it suggests a way to solve very hard problems using series up to some terms. Then using Padé approximation we may hope to recover the exact solution (or at least a sufficient approximation for a specific application).

Let’s imagine that we would like to solve the following differential equation:

\displaystyle y' = y^2 \text{ with initial condition } y(0) = 1

This differential equation can be solved exactly since it is separable. The solution is y(x) = \frac{1}{1-x}. Let’s pretend for a moment that solving this equation is a very difficult problem because we don’t know how to solve separable differential equations.

A possible approach is to consider a solution of the form:

\displaystyle y(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots

Then we have:

\displaystyle y(x)' = a_1 + 2 a_2 x + 3 a_3 x^2 + \cdots
\displaystyle y(x)^2 = (a_0 + a_1 x + a_2 x^2 + \cdots)^2
\displaystyle = a_0^2 + a_0 a_1 x + a_0 a_2 x^2 + \cdots
\displaystyle + a_0 a_1 x + a_1^2 x^2 + \cdots
\displaystyle + a_2 a_0 x^2 + \cdots
\displaystyle = a_0^2 + 2 a_0 a_1 x + (a_1^2 + 2 a_0 a_2) x^2 + \cdots

We can write the differential equation keeping only terms up to two:

\displaystyle y(x)' = y(x)^2
\displaystyle a_1 + 2 a_2 x + 3 a_3 x^2 = a_0^2 + 2 a_0 a_1 x + (a_1^2 + 2 a_0 a_2) x^2

we obtain:

\displaystyle a_1 = a_0^2
\displaystyle 2 a_2 = 2 a_0 a_1
\displaystyle 3 a_3 = a_1^2 + 2 a_0 a_2

Setting a_0 = 1 (since y(0) = 1) implies a_1 = a_2 = a_3 = 1. An approximation of the solution to the differential equation is therefore:

\displaystyle 1 + x + x^2 + x^3

As presented above, the corresponding P(1,1) = \frac{1}{1-x} which is the exact solution to the differential equation. In fact, the diagonal sequence of Padé approximants (P(1,1), P(2,2), … P(n,n)) recovers \frac{1}{1-x}. This is, of course, a special case.

So, for this differential equation, we have the following pattern:

Comments

Leave a comment

More posts