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+---
+title: "Partial fraction decomposition"
+date: 2021-02-22
+categories:
+- Mathematics
+layout: "concept"
+---
+
+**Partial fraction decomposition** or **partial fraction expansion**
+is a method to rewrite quotients of two polynomials $g(x)$ and $h(x)$,
+where the numerator $g(x)$ is of lower order than $h(x)$,
+as sums 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:
+
+$$\begin{aligned}
+    \boxed{
+        f(x) = \frac{c_1}{x - h_1} + \frac{c_2}{x - h_2} + ... + \frac{c_N}{x - h_N}
+    }
+\end{aligned}$$
+
+The constants $c_n$ 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 this 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 $c_{1,1}, ..., c_{1,m}$.
+
+
+
+## References
+1.  O. Bang,
+    *Applied mathematics for physicists: lecture notes*, 2019,
+    unpublished.