Convert to parfrac performs a partial fraction decomposition of the rational function f in the variable x.

If no x is provided, parfrac attempts to determine a suitable x, and proceeds if the operation is not ambiguous. For example, an expression that is a rational polynomial in both $a$ and $b$ requires that the variable be specified.

The optional argument K specifies how the denominator in f is to be factored.  If this argument is not specified, the denominator is factored by the factor command, which factors over the field implied by the coefficients present.

If the optional argument K is real (or complex), then a real (complex) floating-point factorization of the denominator is performed.

Note: This is implemented only for the univariate case.

If the argument K is RootOf or a radical or a list or set of RootOfs or radicals, then the denominators are factored over the algebraic number field implied by the field extensions K.

If the argument K is the name `sqrfree' then a square-free partial factorization is computed. A square-free factorization of the denominator in x is computed.

If the last argument is `true', this declares that the denominator of f is already in the desired factored form, and no factorization is required.

Note: Such a partial fraction decomposition can be done only if the factors in the denominator are relatively prime to each other.

If the programmer entry point form is used, then x must be a name, and the input flist must have the form:

 $\[n,\[\mathrm{f1},\mathrm{p1}\],\[\mathrm{f2},\mathrm{p2}\],...\]$ 

where $n$ is the numerator, and ${\mathrm{f1}}^{\mathrm{p1}},{\mathrm{f2}}^{\mathrm{p2}},...$ are the denominator factors. All $\mathrm{f1},\mathrm{f2},...$ must be relatively prime, and all $\mathrm{p1},\mathrm{p2},...$ must be positive integer values.

The programmer form also provides the output in a different form:

 $\[p,\[\mathrm{f1},\mathrm{n11},\mathrm{n12},...\],\[\mathrm{f2},\mathrm{n21},\mathrm{n22},...\],...\]$ 

where $p$ is the polynomial part, $\mathrm{f1},\mathrm{f2},...$ are as in the input, and the $\mathrm{n11},\mathrm{n12},...$ are the numerators of the partial fraction form such that the algebraic partial fraction form can be obtained as:

 $p+\frac{\mathrm{n11}}{\mathrm{f1}}+\frac{\mathrm{n12}}{{\mathrm{f1}}^2}+...+\frac{\mathrm{n21}}{\mathrm{f2}}+\frac{\mathrm{n22}}{{\mathrm{f2}}^2}+...$ 

Note: The programmer form output is dense, meaning all zero coefficients are included, and the polynomial part is always included (even if zero).

$f≔\frac{x^5+1}{x^4-x^2}$

$f≔\frac{x^5+1}{x^4-x^2}$

$\mathrm{convert}\left(f\,\mathrm{parfrac}\,x\right)$

$x+\frac{1}{x-1}-\frac{1}{x^2}$

$\mathrm{convert}\left(f\,\mathrm{parfrac}\right)$

$x+\frac{1}{x-1}-\frac{1}{x^2}$

$f≔\frac{x}{{\left(x-b\right)}^2}$

$f≔\frac{x}{{\left(x-b\right)}^2}$

$\mathrm{convert}\left(f\,\mathrm{parfrac}\,x\right)$

$\frac{b}{{\left(x-b\right)}^2}+\frac{1}{x-b}$

$\mathrm{convert}\left(f\,\mathrm{parfrac}\right)$

Error, (in `convert/parfrac`) the variable name (for conversion to partial fractions) must be provided

$f≔\frac{2.3x}{5.4x^3-2.3x+1}$

$f≔\frac{2.3x}{5.4x^3-2.3x+1}$

$\mathrm{convert}\left(f\,\mathrm{parfrac}\,x\right)$

$\frac{0.2240312285x+0.06336062145}{x^2-0.8091847442x+0.2288540244}-\frac{0.2240312285}{x+0.8091847442}$

$\mathrm{convert}\left(f\,\mathrm{parfrac}\,x\,\mathrm{complex}\right)$

$\frac{0.1120156142-0.3016538619\mathrm{I}}{x-0.4045923721-0.2552626820\mathrm{I}}+\frac{0.1120156144+0.3016538619\mathrm{I}}{x-0.4045923721+0.2552626820\mathrm{I}}-\frac{0.2240312285}{x+0.8091847442}$

$f≔\frac{4x^3-6x^2-2}{x^4-2x^3-2x+4}$

$f≔\frac{4x^3-6x^2-2}{x^4-2x^3-2x+4}$

$\mathrm{convert}\left(f\,\mathrm{parfrac}\,x\right)$

$\frac{3x^2}{x^3-2}+\frac{1}{x-2}$

$\mathrm{convert}\left(f\,\mathrm{parfrac}\,x\,2^{\frac{1}{3}}\right)$

$-\frac{6}{\left(-2+2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)\left(2^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+22^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+4\right)\left(x-2\right)}+\frac{\left(2^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-1\right)2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{\left(-2+2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)\left(-x+2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}+\frac{-2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-2x}{-2^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}x-x^2}$

$\mathrm{convert}\left(f\,\mathrm{parfrac}\,x\,\mathrm{real}\right)$

$\frac{1.000000000}{x-1.259921050}+\frac{1.000000000}{x-2.}+\frac{2.000000000x+1.259921050}{x^2+1.259921050x+1.587401052}$

$\mathrm{convert}\left(f\,\mathrm{parfrac}\,x\,\mathrm{complex}\right)$

$\frac{1.000000000+2.837357306\times {10}^{\mathrm{-10}}\mathrm{I}}{x-1.259921050}+\frac{1.000000000}{x+0.6299605249+1.091123636\mathrm{I}}+\frac{1.000000000-4.166666666\times {10}^{\mathrm{-10}}\mathrm{I}}{x-2.}+\frac{1.000000000}{x+0.6299605249-1.091123636\mathrm{I}}$

$p≔x^5-2x^4-2x^3+4x^2+x-2\:$

$f≔\frac{36}{p}$

$f≔\frac{36}{x^5-2x^4-2x^3+4x^2+x-2}$

$\mathrm{convert}\left(f\,\mathrm{parfrac}\,x\right)$

$-\frac{9}{{\left(x-1\right)}^2}+\frac{4}{x-2}-\frac{4}{x+1}-\frac{3}{{\left(x+1\right)}^2}$

$\mathrm{convert}\left(f\,\mathrm{parfrac}\,x\,\mathrm{sqrfree}\right)$

$\frac{4}{x-2}+\frac{-12x-24}{{\left(x^2-1\right)}^2}+\frac{-4x-8}{x^2-1}$

$f≔\frac{36}{\mathrm{convert}\left(p\,\mathrm{sqrfree}\,x\right)}$

$f≔\frac{36}{\left(x-2\right){\left(x^2-1\right)}^2}$

$\mathrm{convert}\left(f\,\mathrm{parfrac}\,x\,\mathrm{true}\right)$

$\frac{4}{x-2}+\frac{-12x-24}{{\left(x^2-1\right)}^2}+\frac{-4x-8}{x^2-1}$

$\mathrm{fl}≔\left[4x^3-6x^2-2\,\left[x-2\,2\right]\,\left[x^3-2\,1\right]\right]$

$\mathrm{fl}≔\left[4x^3-6x^2-2\,\left[x-2\,2\right]\,\left[x^3-2\,1\right]\right]$

$\mathrm{pl}≔\mathrm{convert}\left(\mathrm{fl}\,\mathrm{parfrac}\,x\right)$

$\mathrm{pl}≔\left[0\,\left[x-2\,2\,1\right]\,\left[x^3-2\,-2x^2-x-2\right]\right]$

$f≔\frac{\mathrm{fl}\left[1\right]}{\mathrm{mul}\left({i\left[1\right]}^{i\left[2\right]}\,i=\mathrm{fl}\left[2..-1\right]\right)}$

$f≔\frac{4x^3-6x^2-2}{{\left(x-2\right)}^2\left(x^3-2\right)}$

$p≔\mathrm{convert}\left(f\,\mathrm{parfrac}\,x\,\mathrm{true}\right)$

$p≔\frac{-2x^2-x-2}{x^3-2}+\frac{2}{x-2}+\frac{1}{{\left(x-2\right)}^2}$

$\mathrm{pl}\left[1\right]+\mathrm{add}\left(\mathrm{add}\left(\frac{i\left[j\right]}{{i\left[1\right]}^{j-1}}\,j=2..\mathrm{nops}\left(i\right)\right)\,i=\mathrm{pl}\left[2..-1\right]\right)$

$\frac{-2x^2-x-2}{x^3-2}+\frac{2}{x-2}+\frac{1}{{\left(x-2\right)}^2}$