From d33ac5f01a6599406d516edfd45b9938795cea6d Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sun, 21 Feb 2021 20:20:46 +0100
Subject: Add "Partial fraction decomposition" and "Hilbert space"

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+% Partial fraction decomposition
+
+
+# Partial fraction decomposition
+
+*Partial fraction decomposition* or *expansion* is a method to rewrite a
+quotient of two polynomials $g(x)$ and $h(x)$, where the numerator
+$g(x)$ is of lower order than $h(x)$, as a sum of fractions with $x$ in
+the denominator:
+
+$$\begin{aligned}
+    f(x) = \frac{g(x)}{h(x)} = \frac{c_1}{x - h_1} + \frac{c_2}{x - h_2} + ...
+\end{aligned}$$
+
+Where $h_n$ etc. are the roots of the denominator $h(x)$. If all $N$ of
+these roots are distinct, then it is sufficient to simply posit:
+
+$$\begin{aligned}
+    \boxed{
+        f(x) = \frac{c_1}{x - h_1} + \frac{c_2}{x - h_2} + ... + \frac{c_N}{x - h_N}
+    }
+\end{aligned}$$
+
+Then the constant coefficients $c_n$ can either be found the hard way,
+by multiplying the denominators around and solving a system of $N$
+equations, or the easy way by using the following trick:
+
+$$\begin{aligned}
+    \boxed{
+        c_n = \lim_{x \to h_n} \big( f(x) (x - h_n) \big)
+    }
+\end{aligned}$$
+
+If $h_1$ is a root with multiplicity $m > 1$, then the sum takes the
+form of:
+
+$$\begin{aligned}
+    \boxed{
+        f(x)
+        = \frac{c_{1,1}}{x - h_1} + \frac{c_{1,2}}{(x - h_1)^2} + ...
+    }
+\end{aligned}$$
+
+Where $c_{1,j}$ are found by putting the terms on a common denominator,
+e.g.:
+
+$$\begin{aligned}
+    \frac{c_{1,1}}{x - h_1} + \frac{c_{1,2}}{(x - h_1)^2}
+    = \frac{c_{1,1} (x - h_1) + c_{1,2}}{(x - h_1)^2}
+\end{aligned}$$
+
+And then, using the linear independence of $x^0, x^1, x^2, ...$, solving
+a system of $m$ equations to find all $c_{1,1}, ..., c_{1,m}$.
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