Commit ece54d73 authored by Erik Strand's avatar Erik Strand

Add pade approximant exposition

parent 939034e2
---
title: Problem Set 10 (Functions)
---
## 1
{:.question}
Find the first five diagonal Padé approximants [1/1], ..., [5/5] to $$e^x$$ around the
origin. Remember that the numerator and denominator can be multiplied by a constant to make the
numbers as convenient as possible. Evaluate the approximations at $$x = 1$$ and compare with the
correct value of $$e = 2.718281828459045$$. How is the error improving with the order? How does that
compare to the polynomial error?
For a fixed function $$f$$ and integers $$N \geq 0$$ and $$M \geq 0$$, the Padé approximant
is the function
$$
[N/M]_ f(x) = \frac{\sum_{n = 0}^N a_n x^n}{1 + \sum_{m = 1}^M b_m x^m}
$$
that matches as many terms as possible of the Taylor series of $$f$$. (We can define the approximant
about any point, but without loss of generality we'll only consider the origin.) There are $$N + M +
1$$ parameters in the formula above ($$a_0$$, $$\ldots$$, $$b_N$$, and $$b_1$$, $$\ldots$$,
$$b_M$$), so in general this means we can match $$f$$ up to order $$N + M + 1$$. (Though by
coincidence, or otherwise, we may end up matching some higher order terms as well.)
We can write the resulting system of equations explicitly by expanding the Taylor series of the
approximant to order $$L = N + M$$ (since this gives us $$N + M + 1$$ terms). To make the equations
simpler to write I will introduce a constant $$b_0 = 1$$.
$$
\frac{\sum_{n = 0}^N a_n x^n}{\sum_{m = 0}^M b_m x^m} = \sum_{l = 0}^L c_l x^l
$$
Then we multiply by the denominator.
$$
\begin{aligned}
\sum_{n = 0}^N a_n x^n &= \left(\sum_{m = 0}^M b_m x^m \right) \left( \sum_{l = 0}^L c_l x^l \right) \\
&= \sum_{m = 0}^M \sum_{l = 0}^L b_m c_l x^{m + l}
\end{aligned}
$$
By setting the different powers of $$x$$ equal, this gives us $$L + 1 = N + M + 1$$ equations.
$$
\begin{cases}
a_n = \sum_{m = 0}^{\min(n, M)} b_m c_{n - m} & \text{for } 0 \leq n \leq N \\
0 = \sum_{m = 0}^{\min(n, M)} b_m c_{n - m} & \text{for } N < n \leq L = N + M
\end{cases}
$$
We know what $$c_0$$, $$\ldots$$, $$c_L$$ are, since they must match the Taylor series of $$f$$. (In
particular, $$c_l = f^{(l)} / l!$$.) So by solving this system of equations we can determine all of
the unknown parameters. In particular, the last $$M$$ equations (i.e. for $$N < n$$) can be solved
to determine $$b_1$$, $$\ldots$$, $$b_M$$. Then the first $$N + 1$$ equations immediately give
values for $$a_0$$, $$\ldots$$, $$a_N$$.
The first step can be written as a system of $$M + 1$$ equations.
$$
\begin{bmatrix}
1 & 0 & 0 & \cdots & 0 & 0 \\
c_{N + 1} & c_{N} & c_{N - 1} & \cdots & c_{N - M + 2} & c_{N - M + 1} \\
c_{N + 2} & c_{N + 1} & c_{N} & \cdots & c_{N - M + 3} & c_{N - M + 2} \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
c_{N + M - 1} & c_{N + M - 2} & c_{N + M - 3} & \cdots & c_{N} & c_{N - 1} \\
c_{N + M} & c_{N + M - 1} & c_{N + M - 2} & \cdots & c_{N + 1} & c_{N} \\
\end{bmatrix}
\begin{bmatrix}
b_0 \\
b_1 \\
b_2 \\
\vdots \\
b_{M - 1} \\
b_M
\end{bmatrix}
=
\begin{bmatrix}
1 \\
0 \\
0 \\
\vdots \\
0 \\
0
\end{bmatrix}
$$
Note that if $$N + 1 < M$$, then some upper diagonal chunk of the matrix will be zero, corresponding
to those entries where the subscript of $$c$$ would be negative.
Finally, there's no guarantee that this matrix will be nonsingular. (For example, consider the case
where $$N = M$$ and all derivatives of $$f$$ up to $$N + M$$ are zero. Then all rows of the matrix
except for the first will be zero.) In such a situation, one can try to throw out degenerate
equations, and pull new ones from higher order terms (i.e. consider $$L > N + M$$). However it's
possible that all additional rows will be degenerate; in this case, the approximant is legitimately
underdetermined. This means it can match the function exactly, and you can reduce $$M$$ until your
system is nonsingular.
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