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Revision difference : partial fractions of expressions
Version 19 Version 18
Let \,$R(z) = \frac{P(z)}{Q(z)}$\, be a {\em fractional expression}, i.e., a quotient of the polynomials $P(z)$ and $Q(z)$ such that $P(z)$ is not divisible by $Q(z)$. \,Let's restrict to the case that the coefficients are real or complex numbers. Let $R(z) = \frac{P(z)}{Q(z)}$ be a {\em fractional expression}, i.e., a quotient of the polynomials $P(z)$ and $Q(z)$ such that $P(z)$ is not divisible by $Q(z)$. \,Let's restrict to the case that the coefficients are real or complex numbers.
If the distinct complex zeros of the denominator are $b_1, b_2, ..., b_t$ with the multiplicities $\tau_1, \tau_2, ..., \tau_t$ ($t \ge 1$), and the numerator has not common zeros, then $R(z)$ can be decomposed uniquely as the sum If the distinct complex zeros of the denominator are $b_1, b_2, ..., b_t$ with the multiplicities $\nu_1, \nu_2, ..., \nu_t$ ($t \ge 1$), and the numerator has not common zeros, then $R(z)$ can be decomposed uniquely as the sum
$$R(z) = H(z)+ $$R(z) = H(z)+
\sum_{j=1}^t\left(\frac{A_{j1}}{z-b_j}+\frac{A_{j2}}{(z-b_j)^2}+... \sum_{j=1}^t\left(\frac{A_{j1}}{z-b_j}+\frac{A_{j2}}{(z-b_j)^2}+...
+\frac{A_{j\tau_j}}{(z-b_j)^{\tau_j}}\right),$$ +\frac{A_{j\tau_j}}{(z-b_j)^{\tau_j}}\right),$$
where $H(z)$ is a polynomial and the $A_{jk}$'s are certain complex numbers. where $H(z)$ is a polynomial and the $A_{jk}$'s are certain complex numbers.
Let us now take the special case that all coefficients of $P(z)$ and $Q(z)$ are real. \,Then the imaginary zeros of $Q(z)$ are pairwise complex conjugates, with same multiplicities, and the corresponding linear factors of $Q(z)$ may be pairwise multiplied to quadratic polynomials of the form \,$z^2+pz+q$\, with real $p$'s and $q$'s and \,$p^2 < 4q$. \,Hence the above decomposition leads to the unique decomposition of the form Let us now take the special case that all coefficients of $P(z)$ and $Q(z)$ are real. \,Then the imaginary zeros of $Q(z)$ are pairwise complex conjugates, with same multiplicities, and the corresponding linear factors of $Q(z)$ may be pairwise multiplied to quadratic polynomials of the form $z^2+pz+q$ with real $p$'s and $q$'s and \,$p^2 < 4q$. \,Hence the above decomposition leads to the unique decomposition of the form
\begin{align*}R(x) = \quad & H(x)+ \begin{align*}R(x) = \quad&H(x)+
\sum_{i=1}^m\left(\frac{A_{i1}}{x-b_i}+\frac{A_{i2}}{(x-b_i)^2}+... \sum_{i=1}^m\left(\frac{A_{i1}}{x-b_i}+\frac{A_{i2}}{(x-b_i)^2}+...
+\frac{A_{i\mu_i}}{(x-b_i)^{\mu_i}}\right)\\ +\frac{A_{i\mu_i}}{(x-b_i)^{\mu_i}}\right)\\
&+\sum_{j=1}^n\left(\frac{B_{j1}x+C_{j1}}{x^2+p_jx+q_j}+ &+\sum_{j=1}^n\left(\frac{B_{j1}x+C_{j1}}{x^2+p_jx+q_j}+
\frac{B_{j2}x+C_{j2}}{( x^2+p_jx+q_j)^2}+... \frac{B_{j2}x+C_{j2}}{( x^2+p_jx+q_j)^2}+...
+\frac{B_{j\nu_j}x+C_{j\nu_j}}{( x^2+p_jx+q_j)^{\nu_j}}\right), +\frac{B_{j\nu_j}x+C_{j\nu_j}}{( x^2+p_jx+q_j)^{\nu_j}}\right),\\
\end{align*} \end{align*}
where $m$ is the number of the distinct real zeros and $2n$ the number of the distinct imaginary zeros of the denominator $Q(x)$ of the fractional expression \,$R(x) = \frac{P(x)}{Q(x)}$. \,The coefficients $A_{ik}$, $B_{jk}$ and $C_{jk}$ are uniquely determined real numbers. where $m$ is the number of the distinct real zeros and $2n$ the number of the distinct imaginary zeros of the denominator $Q(x)$ of the fractional expression $R(x) = \frac{P(x)}{Q(x)}$. \,The coefficients $A_{ik}$, $B_{jk}$ and $C_{jk}$ are uniquely determined real numbers.
Cf. the partial fractions of {\em fractional numbers}. Cf. the partial fractions of {\em fractional numbers}.
\textbf{Example.} \textbf{Example.}
$$\frac{-x^5+6x^4-7x^3+15x^2-4x+3}{(x-1)^3(x^2+1)^2} = $$\frac{-x^5+6x^4-7x^3+15x^2-4x+3}{(x-1)^3(x^2+1)^2} =
-\frac{1}{x-1}+\frac{3}{(x-1)^3}+\frac{x}{x^2+1}+\frac{2x-1}{(x^2+1)^2}$$ -\frac{1}{x-1}+\frac{3}{(x-1)^3}+\frac{x}{x^2+1}+\frac{2x-1}{(x^2+1)^2}$$