|
The partial partial fractions form for a simple complex roots is: \[ \frac{p(x)}{(x^2 + ax + b)\phi(x)} = \frac{Ax+B}{x^2+ax+b} + \frac{q(x)}{\phi(x)} \] where the quadratic polynomial $x^2 + ax + b$ has complex roots, that is, $a^2 - 4b < 0$, and where $\phi(x)$ has no factor of $x^2 + ax + b$. Upon clearing it of fractions, this reduces to \begin{equation} \label{ident} p(x) = (Ax+B)\phi(x) + (x^2+ax+b)q(x). \end{equation}
We intend to substitute $x=x_1$ in equation \eqref{ident} where $x_1$ is a (complex) root of $x^2+ax+b$, but we don't do it literally. We observe that upon the substitution $x=x_1$, the last term on the right hand side of \eqref{ident} drops out because $x_1^2+ax_1+b=0$. Furthermore, since $x_1^2 = -ax_1 - b$, we may replace the remaining occurrences of $x^2$ with $-ax_1 - b$ to the extent possible. What I mean is that we replace all even powers $x^{2m}$ by $(-ax_1-b)^m$ and all odd powers $x^{2m+1}$ by $x_1(-ax_1-b)^m$. We simplify the result and repeat the substitution process as necessary until only the zeroth and first powers of $x_1$ remain. In this way, equation \eqref{ident} reduces to: \begin{equation} \label{ident reduced} \gamma x_1 + \delta = \alpha(A,B) x_1 + \beta(A,B), \end{equation} where $\alpha(A,B)$, $\beta(A,B)$, $\gamma$, $\delta$ are real quantities and $\alpha(A,B)$ and $\beta(A,B)$ depend linearly on $A$ and $B$. Equating the imaginary parts of the two sides we obtain $\gamma \Im(x_1) = \alpha(A,b) \Im(x_1)$. But the imaginary part $\Im(x_1)$ of $x_1$ is nonzero (since we are dealing with complex roots) therefore $\gamma = \alpha(A,B)$. Then it follows from \eqref{ident reduced} that $\delta = \beta(A,B)$. We solve the linear system \begin{equation} \label{linear system} \alpha(A,B)= \gamma, \qquad \beta(A,B) = \delta \end{equation} to compute $A$ and $B$.
To compute $A$ and $B$ we clear the fractions: \[ 3x^2 + 2x + 1 = (Ax+B)(x-1) + C(x^2+x+1), \] and expand the $(Ax+B)(x-1)$ term: \[ 3x^2 + 2x + 1 = Ax^2 + (B-A)x - B + C(x^2+x+1). \] Then we substitute $x=x_1$ where $x_1^2 + x_1 + 1 = 0$. As pointed out in the preceding note, there is no advantage in writing $x_1$ for the root. We will write a plain $x$ instead, keeping in mind that $x^2+x+1=0$, thus $x^2 = -x - 1$. We get: \[ 3(-x -1) + 2x + 1 = A(-x-1) + (B-A)x - B, \] and simplify it to \[ -x-2 = (B-2A)x - A - B. \] Equating like powers of $x$ we get \[ B-2A = -1, \quad \text{and} \quad -A-B=-2. \] We conclude that $A=B=1$, and \[ \frac{3x^2 + 2x + 1}{(x-1)(x^2+x+1)} = \frac{x+1}{x^2 + x + 1} + \frac{2}{x-1}. \]
Equating like powers of $x$ we see that: \begin{gather*} c_3(a^2-b) - c_2a + c_1 = A(a^2-ac-b+d)+B(c-a), \\ (c_3 ab - c_2 b + c_0) = Ab(a-c) + B(d-b). \end{gather*}
This linear system may be solved for $A$ and $B$. The coefficients $C$ and $D$ are found in the same way.
To compute the coefficients $B$ and $C$, we substitute for $x$ a root of $x^2+1=0$, thus $x^2=-1$. We obtain $ 1 = (Bx+C)(x-1)(-1+9)$ which expands to $1 = 8\big(Bx^2+(C-B)x -C\big)$. Replacing $x^2$ by $-1$ once more, results in $1 = 8\big(-B+(C-B)x -C\big)$, which simplifies to $ (C-B) x - (B+C) = 1/8$. Equating like powers of $x$ yields \[ C-B=0, \qquad B+C=-\frac{1}{8} \] therefore $B=C=-1/16$.
To compute the coefficients $D$ and $E$, we substitute for $x$ a root of $x^2+9=0$, thus $x^2=-9$. We obtain $ 1 = (Dx+E)(x-1)(-9+1)$ which expands to $1 = -8\big(Dx^2+(E-D)x -E\big)$. Replacing $x^2$ by $-9$ once more, results in $1 = -8\big(-9D+(E-D)x -E\big)$, which simplifies to $ (E-D) x - E - 9D = -1/8$. Equating like powers of $x$ yields \[ E-D=0, \qquad E+9D=\frac{1}{8} \] therefore $D=E=1/80$.
We conclude that: \[ \frac{1}{(x-1)(x^2+1)(x^2+9)} = \frac{1}{20(x-1)} - \frac{x+1}{16(x^2+1)} + \frac{x + 1}{80(x^2+9)}. \]
|
Author: Rouben Rostamian