# Coefficients of Partial Fraction Expansion

Coefficients of Partial Fraction Expansion Swapnil Sunil Jain July 28 2006

Coefficients of Partial Fraction Expansion Let us start with the assumption (or rather a Lemma) that any rational proper function $F(s)$ of the form

 $\displaystyle F(s)=\frac{P(s)}{(s-q)^{r}(s-p_{1})(s-p_{2})...(s-p_{i})...(s-p_% {n})}$ (1)

has a partial fraction expansion given by

 $\displaystyle F(s)=\frac{a_{0}}{(s-q)^{r}}+\frac{a_{1}}{(s-q)^{r-1}}+...+\frac% {a_{j}}{(s-q)^{r-j}}+...+\frac{a_{r-1}}{(s-q)}$ $\displaystyle\qquad+\frac{k_{1}}{(s-p_{1})}+\frac{k_{2}}{(s-p_{2})}+...+\frac{% k_{i}}{(s-p_{i})}+...+\frac{k_{n}}{(s-p_{n})}$ (2)

where $j=0,1,2,...,r-1$ and $i=1,2,3,...,n$ and $q\neq p_{1}\neq p_{2}\neq...\neq p_{n}$.

First, we determine the coefficient $k_{i}$. In order to do so, we multiply both sides of equation (2) by $(s-p_{i})$ which then gives us

 $\displaystyle(s-p_{i})F(s)=\frac{a_{0}}{(s-q)^{r}}(s-p_{i})+\frac{a_{1}}{(s-q)% ^{r-1}}(s-p_{i})+...+\frac{a_{j}}{(s-q)^{r-j}}(s-p_{i})+...+\frac{a_{r-1}}{(s-% q)}(s-p_{i})$ $\displaystyle\qquad+\frac{k_{1}}{(s-p_{1})}(s-p_{i})+\frac{k_{2}}{(s-p_{2})}(s% -p_{i})+...+\frac{k_{i}}{(s-p_{i})}(s-p_{i})+...+\frac{k_{n}}{(s-p_{n})}(s-p_{% i})$ (3)

If we then let $s=p_{i}$, all the terms on the R.H.S drop out except the one containing the coefficient $k_{i}$ and we get

 $\displaystyle\Big{[}(s-p_{i})F(s)\Big{]}{\Bigg{|}}_{s=p_{i}}=k_{i}$ (4)

Now, in order to determine the coefficient $a_{j}$, we multiply both sides of (2) by $(s-q)^{r}$ which yields

 $\displaystyle(s-q)^{r}F(s)=\frac{a_{0}}{(s-q)^{r}}(s-q)^{r}+\frac{a_{1}}{(s-q)% ^{r-1}}(s-q)^{r}+...+\frac{a_{j}}{(s-q)^{r-j}}(s-q)^{r}+...+\frac{a_{r-1}}{(s-% q)}(s-q)^{r}$ $\displaystyle\qquad+(s-q)^{r}\Bigg{[}\frac{k_{1}}{(s-p_{1})}+\frac{k_{2}}{(s-p% _{2})}+...+\frac{k_{i}}{(s-p_{i})}+...+\frac{k_{n}}{(s-p_{n})}\Bigg{]}$ $\displaystyle\Rightarrow(s-q)^{r}F(s)=a_{0}+a_{1}(s-q)^{1}+...+a_{j}(s-q)^{j}+% ...+a_{r-1}(s-q)^{r-1}+(s-q)^{r}\frac{A(s)}{B(s)}$ (5)

where we have defined

 $\displaystyle\frac{A(s)}{B(s)}\equiv\frac{k_{1}}{(s-p_{1})}+\frac{k_{2}}{(s-p_% {2})}+...+\frac{k_{i}}{(s-p_{i})}+...+\frac{k_{n}}{(s-p_{n})}$

Then if we take the derivative of the above equation with respect to $s$ and we obtain

 $\displaystyle\frac{d}{ds}\Big{[}(s-q)^{r}F(s)\Big{]}=a_{1}+a_{2}(2)(s-q)+...+a% _{j}(j)(s-q)^{j-1}+...$ $\displaystyle\qquad+a_{r-1}(r-1)(s-q)^{r-2}+\frac{d}{ds}\Big{[}(s-q)^{r}\frac{% A(s)}{B(s)}\Big{]}$ (6)

If we again take the derivative of both sides of the above equation with respect to $s$ we get

 $\displaystyle\frac{d^{2}}{ds^{2}}\Big{[}(s-q)^{r}F(s)\Big{]}=2a_{2}+...+a_{j}(% j)(j-1)(s-q)^{j-2}+...$ $\displaystyle\qquad+a_{r-1}(r-1)(r-2)(s-q)^{r-3}+\frac{d^{2}}{ds^{2}}\Big{[}(s% -q)^{r}\frac{A(s)}{B(s)}\Big{]}$ (7)

If we keep taking derivatives this way until we have taken the derivative $j$ times, we arrive at

 $\displaystyle\frac{d^{j}}{ds^{j}}\Big{[}(s-q)^{r}F(s)\Big{]}=a_{j}(j)(j-1)(j-2% )...(2)(1)(s-q)^{j-j}+...$ $\displaystyle\qquad+a_{r-1}(r-1)(r-2)...(r-j)(s-q)^{r-j-1}+\frac{d^{j}}{ds^{j}% }\Big{[}(s-q)^{r}\frac{A(s)}{B(s)}\Big{]}$ (8)

If we then let $s=q$, all the terms on the R.H.S drop out except the one containing the coefficient $a_{j}$ which yields

 $\displaystyle\Bigg{(}\frac{d^{j}}{ds^{j}}\Big{[}(s-q)F(s)\Big{]}\Bigg{)}{\Bigg% {|}}_{s=q}=a_{j}j!$ (9)

or

 $\displaystyle a_{j}=\frac{1}{j!}\Bigg{(}\frac{d^{j}}{ds^{j}}\Big{[}(s-q)F(s)% \Big{]}\Bigg{)}{\Bigg{|}}_{s=q}$ (10)
Title Coefficients of Partial Fraction Expansion CoefficientsOfPartialFractionExpansion1 2013-03-11 19:26:06 2013-03-11 19:26:06 swapnizzle (13346) (0) 1 swapnizzle (0) Definition