Coefficients of Partial Fraction Expansion

Coefficients of Partial Fraction ExpansionSwapnil Sunil JainJuly 28 2006

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

$F(s)={\displaystyle \frac{P(s)}{{(s-q)}^r(s-p_1)(s-p_2)\mathrm{\dots }(s-p_i)\mathrm{\dots }(s-p_n)}}$(1)

has a partial fraction expansion given by

$F(s)={\displaystyle \frac{a_0}{{(s-q)}^r}}+{\displaystyle \frac{a_1}{{(s-q)}^{r-1}}}+\mathrm{\dots }+{\displaystyle \frac{a_j}{{(s-q)}^{r-j}}}+\mathrm{\dots }+{\displaystyle \frac{a_{r-1}}{(s-q)}}$

$\mathrm{\hspace{1em}\hspace{1em}}+{\displaystyle \frac{k_1}{(s-p_1)}}+{\displaystyle \frac{k_2}{(s-p_2)}}+\mathrm{\dots }+{\displaystyle \frac{k_i}{(s-p_i)}}+\mathrm{\dots }+{\displaystyle \frac{k_n}{(s-p_n)}}$(2)

where $j=0,1,2,\mathrm{\dots },r-1$ and $i=1,2,3,\mathrm{\dots },n$ and $q\ne p_1\ne p_2\ne \mathrm{\dots }\ne p_n$.

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

$(s-p_i)F(s)={\displaystyle \frac{a_0}{{(s-q)}^r}}(s-p_i)+{\displaystyle \frac{a_1}{{(s-q)}^{r-1}}}(s-p_i)+\mathrm{\dots }+{\displaystyle \frac{a_j}{{(s-q)}^{r-j}}}(s-p_i)+\mathrm{\dots }+{\displaystyle \frac{a_{r-1}}{(s-q)}}(s-p_i)$

$\mathrm{\hspace{1em}\hspace{1em}}+{\displaystyle \frac{k_1}{(s-p_1)}}(s-p_i)+{\displaystyle \frac{k_2}{(s-p_2)}}(s-p_i)+\mathrm{\dots }+{\displaystyle \frac{k_i}{(s-p_i)}}(s-p_i)+\mathrm{\dots }+{\displaystyle \frac{k_n}{(s-p_n)}}(s-p_i)$(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

${\left[(s-p_i)F(s)\right]|}_{s=p_i}=k_i$(4)

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

${(s-q)}^{r}F(s)={\displaystyle \frac{a_0}{{(s-q)}^r}}{(s-q)}^r+{\displaystyle \frac{a_1}{{(s-q)}^{r-1}}}{(s-q)}^r+\mathrm{\dots }+{\displaystyle \frac{a_j}{{(s-q)}^{r-j}}}{(s-q)}^r+\mathrm{\dots }+{\displaystyle \frac{a_{r-1}}{(s-q)}}{(s-q)}^r$

$\mathrm{\hspace{1em}\hspace{1em}}+{(s-q)}^r\left[{\displaystyle \frac{k_1}{(s-p_1)}}+{\displaystyle \frac{k_2}{(s-p_2)}}+\mathrm{\dots }+{\displaystyle \frac{k_i}{(s-p_i)}}+\mathrm{\dots }+{\displaystyle \frac{k_n}{(s-p_n)}}\right]$

$\Rightarrow {(s-q)}^{r}F(s)=a_0+a_1{(s-q)}^1+\mathrm{\dots }+a_j{(s-q)}^j+\mathrm{\dots }+a_{r-1}{(s-q)}^{r-1}+{(s-q)}^r{\displaystyle \frac{A(s)}{B(s)}}$(5)

where we have defined

${\displaystyle \frac{A(s)}{B(s)}}\equiv {\displaystyle \frac{k_1}{(s-p_1)}}+{\displaystyle \frac{k_2}{(s-p_2)}}+\mathrm{\dots }+{\displaystyle \frac{k_i}{(s-p_i)}}+\mathrm{\dots }+{\displaystyle \frac{k_n}{(s-p_n)}}$

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

${\displaystyle \frac{d}{ds}}\left[{(s-q)}^{r}F(s)\right]=a_1+a_2(2)(s-q)+\mathrm{\dots }+a_j(j){(s-q)}^{j-1}+\mathrm{\dots }$

$\mathrm{\hspace{1em}\hspace{1em}}+a_{r-1}(r-1){(s-q)}^{r-2}+{\displaystyle \frac{d}{ds}}\left[{(s-q)}^r{\displaystyle \frac{A(s)}{B(s)}}\right]$(6)

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

${\displaystyle \frac{d^2}{d{s}^2}}\left[{(s-q)}^{r}F(s)\right]=2a_2+\mathrm{\dots }+a_j(j)(j-1){(s-q)}^{j-2}+\mathrm{\dots }$

$\mathrm{\hspace{1em}\hspace{1em}}+a_{r-1}(r-1)(r-2){(s-q)}^{r-3}+{\displaystyle \frac{d^2}{d{s}^2}}\left[{(s-q)}^r{\displaystyle \frac{A(s)}{B(s)}}\right]$(7)

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

${\displaystyle \frac{d^j}{d{s}^j}}\left[{(s-q)}^{r}F(s)\right]=a_j(j)(j-1)(j-2)\mathrm{\dots }(2)(1){(s-q)}^{j-j}+\mathrm{\dots }$

$\mathrm{\hspace{1em}\hspace{1em}}+a_{r-1}(r-1)(r-2)\mathrm{\dots }(r-j){(s-q)}^{r-j-1}+{\displaystyle \frac{d^j}{d{s}^j}}\left[{(s-q)}^r{\displaystyle \frac{A(s)}{B(s)}}\right]$(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

${\left({\displaystyle \frac{d^j}{d{s}^j}}\left[(s-q)F(s)\right]\right)|}_{s=q}=a_{j}j!$(9)

or

$a_j={{\displaystyle \frac{1}{j!}}\left({\displaystyle \frac{d^j}{d{s}^j}}\left[(s-q)F(s)\right]\right)|}_{s=q}$(10)