Partial Fractions Done Right

The Case of Repeated Real Roots

The partial partial fractions form for a real root $a$ of multiplicity $r>1$ is: $\frac{p(x)}{(x-a)^r\phi(x)} = \frac{A_1}{x-a} + \cdots + \frac{A_{r-1}}{(x-a)^{r-1}} + \frac{A_r}{(x-a)^r} + \frac{q(x)}{\phi(x)}.$

We clear it of fractions $$\label{ident} p(x) = A_1(x-a)^{r-1} \phi(x) + \cdots + A_{r-1}(x-a) \phi(x) + A_{r}\phi(x) + (x-a)^r q(x),$$ then substitute $x=a$ to get $p(a) = A_r \phi(a),$ from which we compute $A_r$.

To compute $A_{r-1}$, we differentiate \eqref{ident} with respect to $x$: \begin{align*} p'(x) &= A_1 \big[ (r-1)(x-a)^{r-2} \phi(x) + (x-a)^{r-1} \phi'(x) \big] \\ &\mbox{} \phantom{+} \cdots \\ &\mbox{} + A_{r-1} \big[ \phi(x) + (x-a) \phi'(x) \big] \\ &\mbox{} + A_r \phi'(x) \\ &\mbox{} + \big[ r(x-a)^{r-1} q(x) + (x-a)^r q'(x) \big], \end{align*} then substitute $x=a$. All the terms involving $x-a$ go away and we are left with $p'(a) = A_{r-1} \phi(a) + A_r \phi'(a),$ from which we can compute $A_{r-1}$ because $A_r$ is known. The coefficients $A_{r-2},\ldots,A_1$ are computed similarly through repeated differentiations of \eqref{ident}.

The calculation of the unknowns $A_r$ through $A_1$ calls for $r-1$ derivatives of the identity \eqref{ident}. The multiplier $(x-a)^r$ of the undetermined polynomial $q(x)$ guarantees that even after $r-1$ differentiations, all occurrences of $q(x)$ and its derivatives will continue having multipliers of $(x-a)^k$, with $k\ge1$, therefore they will drop out after the $x=a$ substitution. The point is, there is no need to differentiate the remainder term; it does not affect the values of $A_1,\ldots A_r$.
Compute $A_1$, $A_2$ and $A_3$ in the partial partial fractions expansion $\frac{4x^2 + 2x + 1}{(x-1)^3(x^2+x+1)} = \frac{A_1}{x-1} + \frac{A_2}{(x-1)^2} + \frac{A_3}{(x-1)^3} + \frac{q(x)}{x^2+x+1} .$
We clear this of fractions to obtain the polynomial identity \begin{multline} \label{ident-ex1} \tag{*} 4x^2 + 2x + 1 = A_1(x-1)^2(x^2+x+1) + A_2(x-1)(x^2+x+1) \\ \mbox{} + A_3(x^2+x+1) + (x-1)^3 q(x). \end{multline} Substituting $x=1$ we get $A_3 = 7/3$.

Next, we differentiate the identity \eqref{ident-ex1} to get \begin{multline*} 8x + 2 = A_1 \big[ (2(x-1)(x^2+x+1) + (x-1)^2(2x + 1) \big] \\ \mbox{} + A_2 \big[ (x^2+x+1) + (x-1)(2x+1) \big] + A_3(2x+1) + \big( (x-1)^3 q(x) \big)'. \end{multline*} Substituting $x=1$ yields $10 = 3A_2 + 3A_3$ whence $A_2 = 10/3 - A_3 = 10/3 - 7/3 = 1$. Note that we haven't bothered expanding the derivative of the $(x-1)^3 q(x)$ in view of the previous remark.

Differentiating the identity \eqref{ident-ex1} once more, we obtain \begin{multline*} 8 = A_1 \big[ 2(x^2+x+1) + \cdots \big] + A_2 \big[ (2x+1) + (2x+1) + 2(x-1) \big] + 2A_3 + \big( (x-1)^3 q(x) \big)''. \end{multline*} In the multiplier of $A_1$ we have elided the details of those terms that have $x-1$ multipliers because they will vanish once we substitute $x=1$. We see that $8 = 2A_3 + 6A_2 + 6A_1$, whence $A_1 = 4/3 - A_3/3 - A_2 = -4/9$.

In conclusion: $\frac{4x^2 + 2x + 1}{(x-1)^3(x^2+x+1)} = - \frac{4}{9(x-1)} + \frac{1}{(x-1)^2} + \frac{7}{3(x-1)^3} + \frac{q(x)}{x^2+x+1} .$

Test Yourself

1. Show that: $\frac{x^2+x+1}{(x+1)^3} = \frac{1}{x+1} - \frac{1}{(x+1)^2} + \frac{1}{(x+1)^3}.$
2. Show that: $\frac{1}{(x^2-1)^2} = - \frac{1}{4(x-1)} + \frac{1}{4(x-1)^2} + \frac{1}{4(x+1)} + \frac{1}{4(x+1)^2}.$

Author: Rouben Rostamian