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I am trying to get from

$$\frac{z^7 + 1}{z^2(z^4+1)}$$

to

$$\frac{1}{z^2} + z - \frac{z+z^2}{1+z^4}.$$

The author did this by doing a partial fractions decomposition. I don't see how, however.. If I compute the partial fractions decomposition, I first find the roots of the denominator, but that's not what's done here. What he does is something that he calls "partial" partial fractions decomposition. Any help?

Latin Wolf
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Pascal Engeler
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  • First decompose it into normal partial fractions. Then add certain terms together. – Hui Yu Jul 17 '13 at 13:44
  • I don't think that's what's done. The context is integrating the initial term over a closed curve. Thus the purpose of doing this strange decomposition is to say "aha, the terms $z$ and $\frac{1}{z^2}$ both have anti-derivatives, so their integral is zero. Then the next step is to decompose the remaining term in order to integrate it. I know how a PFD is done, but I'm asking myself how the above is done directly. – Pascal Engeler Jul 17 '13 at 13:49
  • Yes. That is how I would do it. Do you know how to calculate the coefficients of a Taylor series expansion? Or calculate the residue? This is just a higher order residue. – Hui Yu Jul 17 '13 at 13:51
  • Yes and yes, but I don't see the context... – Pascal Engeler Jul 17 '13 at 13:59

3 Answers3

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Indeed, it is possible to do a version of partial fractions without using linear factors: whenever we have _relatively_prime_ polynomials $f,g$, we can use the Euclidean algorithm for polynomials (with coefficients in a field such as $\mathbb R$) to find two other polynomials $a,b$ such that $af+bg=1$. Then for any polynomial $h$ $$ {h\over fg} \;=\; {h\cdot 1\over fg} \;=\; {h(af+bg)\over fg} \;=\; {ha\over g}+{hb\over f} $$ And of course one can divide-with-remainder to make the numerators of smaller degree than the denominators.

paul garrett
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  • Is this guaranteed to "work", as in $ha$ and $hb$ having lower degree than $h$? – ShreevatsaR Jul 17 '13 at 13:53
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    This is how partial fractions should be taught in school. – OR. Jul 17 '13 at 13:53
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    @ShreevatsaR Read the last sentence in the answer. – OR. Jul 17 '13 at 13:54
  • @RGB: Read the last word in my comment. :-) I read the answer to the end, which is why I asked: note that I asked for lower degree than $h$, not lower degree than $f$ and $g$. To answer my own question: I guess the answer is no, it is not guaranteed (e.g. start with $h = 1$), but that doesn't matter. – ShreevatsaR Jul 17 '13 at 14:07
  • I assumed you were asking the wrong question since it is clear you can't expect $ha$ having lower degree than $h$ unless $a=0$. – OR. Jul 17 '13 at 14:10
  • @RGB: Good point. I was unclear; what I meant by $ha$ and $hb$ was actually the numerators of $ha/g$ and $hb/f$ after dividing them, as the answer suggests in the last sentence (which is what I meant by "which is why I asked" in the comment above). Anyway, now I realize that this is arguably the "wrong question" nevertheless, for as far as the goals of partial fraction decomposition go, it really doesn't matter whether the numerators of (what's left of) $ha/g$ and $hb/f$, after simplification, have lower degree than $h$ or not. But that is what I was thinking of, for whatever reason. – ShreevatsaR Jul 17 '13 at 17:29
  • You can divide out the $z$, after that you have a denominator in $z^2$. – vonbrand Jan 19 '21 at 15:20
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This is another way of doing what is in Paul Garret answer, avoiding to compute more than we are looking for.

You get the polynomial part $z$ by doing long division of $z^7+1$ by $z^2(z^4+1)$.

You get the principal part corresponding to the factor $z^2$ in the denominator by computing the first two steps of long division of $z^7+1$ by $(z^4+1)$ but organizing the terms in decreasing degrees before dividing. The first two coefficients you get are the coefficients of $\frac{A}{z^2}+\frac{B}{z}$ in that order. The other fraction you can get subtracting $\frac{1}{z^2}+z$ from $\frac{z^7+1}{z^2(z^4+1)}$.

OR.
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Hint

First by Euclidean divsion we have

$$\frac{z^7 + 1}{z^2(z^4+1)}=z-\frac{z^3-1}{z^2(z^4+1)}$$

Now we know that the partial fractional decompostion takes the form

$$\frac{z^3-1}{z^2(z^4+1)}=\frac{a}{z^2}+\frac{b}{z}+F(z)$$ where $0$ isn't a pole for the fraction $F$ and with routine calculus we find $a=-1$ and $b=0$.

Finally to find $F$ just do $$F(z)=\frac{z^3-1}{z^2(z^4+1)}+\frac{1}{z^2}=\frac{z+z^2}{1+z^4}$$