aanpalendmw

2022-07-22

partial fraction decomposition of $\frac{{k}^{4}}{(a\phantom{\rule{thinmathspace}{0ex}}{k}^{3}-1{)}^{2}}$

I have to perform complex partial fraction decomposition of the following term:

$\frac{{k}^{4}}{(a\phantom{\rule{thinmathspace}{0ex}}{k}^{3}-1{)}^{2}}$

where $a$ is a real positive number.

and I would like to know if it is possible to reduce it to a sum of fractions of the type $\frac{A}{k\pm z}$,$\frac{B\phantom{\rule{thinmathspace}{0ex}}k}{{k}^{2}\pm z}$,$\frac{C\phantom{\rule{thinmathspace}{0ex}}k}{{k}^{2}-y}$ or similar. Where $z$ is a complex number and $y$ is a real number.

If it is not possible to reduce it to the kind of fraction I listed above other type of decomposition might work too.

Any hint on the process, or any reference would be nice.

Thanks in advance

I have to perform complex partial fraction decomposition of the following term:

$\frac{{k}^{4}}{(a\phantom{\rule{thinmathspace}{0ex}}{k}^{3}-1{)}^{2}}$

where $a$ is a real positive number.

and I would like to know if it is possible to reduce it to a sum of fractions of the type $\frac{A}{k\pm z}$,$\frac{B\phantom{\rule{thinmathspace}{0ex}}k}{{k}^{2}\pm z}$,$\frac{C\phantom{\rule{thinmathspace}{0ex}}k}{{k}^{2}-y}$ or similar. Where $z$ is a complex number and $y$ is a real number.

If it is not possible to reduce it to the kind of fraction I listed above other type of decomposition might work too.

Any hint on the process, or any reference would be nice.

Thanks in advance

yelashwag8

Beginner2022-07-23Added 17 answers

I would start by making the change of variable $a{k}^{3}={x}^{3}$, which transforms the expression into

${a}^{3/4}\frac{{x}^{4}}{({x}^{3}-1{)}^{2}}.$

The denominator is now factored. Given that there is only a two-cube difference, this is simple.

${x}^{3}-1=(x-1)({x}^{2}+x+1).$

The quadratic is irreducible over the reals, as you can see. Then (ignoring the constant ${a}^{3/4}$) we have

$\frac{{x}^{4}}{(x-1{)}^{2}({x}^{2}+x+1)}=\frac{A}{x-1}+\frac{B}{(x-1{)}^{2}}+\frac{Cx+D}{{x}^{2}+x+1}+\frac{Ex+F}{({x}^{2}+x+1{)}^{2}}.$

Glenn Hopkins

Beginner2022-07-24Added 4 answers

Set $a={b}^{3}$and denominator will be factored as

${({b}^{3}{k}^{3}-1)}^{2}=(bk-1{)}^{2}{({b}^{2}{k}^{2}+bk+1)}^{2}$

After that, you split into partial fractions.

$\frac{e+fk}{{b}^{2}{k}^{2}+bk+1}+\frac{g+hk}{{({b}^{2}{k}^{2}+bk+1)}^{2}}+\frac{c}{bk-1}+\frac{d}{(bk-1{)}^{2}}$

where $c,d,e,f,g,h$ are unknowns and must be obtained solving a six equations linear system. Prior to considering the numerator of the result, you first sum all the fractions.

${b}^{5}c{k}^{5}+{b}^{4}c{k}^{4}+{b}^{4}d{k}^{4}+{b}^{4}e{k}^{4}+{b}^{4}f{k}^{5}+{b}^{3}c{k}^{3}+2{b}^{3}d{k}^{3}-{b}^{3}e{k}^{3}-{b}^{3}f{k}^{4}-{b}^{2}c{k}^{2}+3{b}^{2}d{k}^{2}+{b}^{2}g{k}^{2}+{b}^{2}h{k}^{3}-bck+2bdk-bek-bf{k}^{2}-2bgk-2bh{k}^{2}-c+d+e+fk+g+hk$

Numerator must be identical to ${k}^{4}$so coefficients (from degree 0 up to 5 must be $\{0,0,0,0,1,0\}$

The system offers several lovely solutions.

$c=\frac{2}{9{b}^{4}},d=\frac{1}{9{b}^{4}},e=\frac{4}{9{b}^{4}},f=-\frac{2}{9{b}^{3}},g=-\frac{1}{3{b}^{4}},h=0$

Hence, the initial fraction can be expressed as

$\frac{1}{9{b}^{4}}(-\frac{2(bk-2)}{{b}^{2}{k}^{2}+bk+1}-\frac{3}{{({b}^{2}{k}^{2}+bk+1)}^{2}}+\frac{2}{bk-1}+\frac{1}{(bk-1{)}^{2}})$

with $b={a}^{\frac{1}{3}}$

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