| Version 24 |
Version 23 |
| \PMlinkescapeword{decomposition} |
|
|
|
| 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\, $b_1,\,b_2,\,\ldots,\,b_t$\, with the multiplicities\, $\tau_1,\,\tau_2,\,\ldots,\,\tau_t$ ($t \ge 1$), 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\, $b_1,\,b_2,\,\ldots,\,b_t$\, with the multiplicities\, $\tau_1,\,\tau_2,\,\ldots,\,\tau_t$ ($t \ge 1$), 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}+\cdots |
\sum_{j=1}^t\left(\frac{A_{j1}}{z-b_j}+\frac{A_{j2}}{(z-b_j)^2}+\cdots |
| +\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 \PMlinkescapetext{{\em imaginary}} (i.e. non-real) zeros of $Q(z)$ are pairwise complex conjugates, with same multiplicities, and the corresponding linear \PMlinkname{factors}{Product} 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 \PMlinkescapetext{{\em imaginary}} (i.e. non-real) 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}+\cdots |
\sum_{i=1}^m\left(\frac{A_{i1}}{x-b_i}+\frac{A_{i2}}{(x-b_i)^2}+\cdots |
| +\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}+\cdots |
\frac{B_{j2}x+C_{j2}}{( x^2+p_jx+q_j)^2}+\cdots |
| +\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 \PMlinkescapetext{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 \PMlinkescapetext{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.} |
| $$\frac{-x^5\!+\!6x^4\!-\!7x^3\!+\!15x^2\!-\!4x\!+\!3} |
$$\frac{-x^5\!+\!6x^4\!-\!7x^3\!+\!15x^2\!-\!4x\!+\!3} |
| {(x\!-\!1)^3(x^2\!+\!1)^2} \,=\, |
{(x\!-\!1)^3(x^2\!+\!1)^2} \,=\, |
| -\frac{1}{x\!-\!1}\!+\!\frac{3}{(x\!-\!1)^3}\!+ |
-\frac{1}{x\!-\!1}\!+\!\frac{3}{(x\!-\!1)^3}\!+ |
| \!\frac{x}{x^2\!+\!1}\!+\!\frac{2x\!-\!1}{(x^2\!+\!1)^2}$$ |
\!\frac{x}{x^2\!+\!1}\!+\!\frac{2x\!-\!1}{(x^2\!+\!1)^2}$$ |
