Partial fraction decomposition with arbitrary powers

Question

I understand that this is well- discussed in many places. I have even looked at this paper Efficient Recursive Methods for Partial Fraction Expansion of General Rational Functions.

So probably I may write following as $$\frac{1}{(t+a)^{m} (t +b)^{n}}=\sum_{L=1}^{m}\frac{c_{1L}}{(t+a)^{L}}+\sum_{L=1}^{n}\frac{c_{2L}}{(t+b)^{L}};\,\,a,b>0;\,\,m,n>1$$

Please correct me if this is wrong !

I really did not understand how to write $c_{1L}$ and $c_{2L}$ in mathematical forms.

Can someone help me to write $\frac{1}{(t+a)^{m} (t +b)^{n}}$ in general form with the partial fraction decomposition?

Answer 1

It's convenient to do the substitutions $t = s - a$, $b = r + a$, making this $$ \dfrac{1}{s^m (s+r)^n} = \sum_{L=1}^m \frac{c_{1L}}{s^L} + \sum_{L=1}^n \frac{c_{2L}}{(s+r)^L}$$ The principal part of this at $s=0$ is $\sum_{L=1}^m c_{1L}/s^L$. Now by the binomial series $$(s+r)^{-n} = r^{-n} (1+s/r)^{-n} =\sum_{k=0}^\infty {-n \choose k} s^k r^{-n-k}$$ and thus $c_{1L}$ is the coefficient of $s^{m-L}$ here, i.e. $$ c_{1L} = {-n \choose m-L} r^{-n-m+L} = {-n \choose m-L} (b-a)^{-n-m+L} $$ Interchanging $m$ and $n$, $a$ and $b$, $$ c_{2L} = {-m \choose n-L} (a-b)^{-n-m+L} $$

Answer 2

Summing up the terms on the right hand side we have that

$$\sum_{j=1}^m \frac{c_{1j}}{(t+a)^j} + \sum_{k=1}^n \frac{c_{2k}}{(t+b)^k}=\frac{(t+b)^n\sum_{j=1}^mc_{1j}(t+a)^{m-j}+(t+a)^m\sum_{k=1}^nc_{2k}(t+b)^{n-k}}{(t+a)^m(t+b)^n}=\frac{p(t)}{(t+a)^m(t+b)^n},$$

where $p(t)$ is the $m+n-1$ degree polynomial given by the large expression above. Since we wish to have this final expression equal to $(t+a)^{-m}(t+b)^{-n}$, it follows that $p(t)=1$ and so as a polynomial must have equal coefficients. We can then solve for $c_{1j},c_{2j}$ as the solution to the linear system of equations produced, namely

\begin{align*} 1 = & b^n \sum_{j=1}^mc_{1j}a^{m-j}+ a^m\sum_{k=1}^n c_{2k}b^{n-k}\quad&\text{Coefficient equation for constant terms}\\ 0 = & nb^{n-1}\sum_{j=1}^mc_{1j}a^{m-j}+b^n\sum_{j=1}^{m-1}(m-j)c_{1j}a^{m-j}+ ma^{m-1}\sum_{k=1}^n c_{2k}b^{n-k}+a^m\sum_{k=1}^{n-1}(n-k)c_{2k}a^{n-k}\quad&\text{Coefficient equation for first order terms}\\ \vdots\\ 0 = & c_{1,1}+c_{2,1}\quad&\text{Coefficient equation for $(m+n-1)^\text{th}$ order terms} \end{align*}

There are $m+n$ equations, all of them rather complicated , though. Kramer's Rule could solve the system, though even setting up the system would be difficult to do by hand for large $m+n$.

I don't think that there's a simple general formula for those coefficients.