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I have the function $\frac{x^4}{(x^2+1)^5}$ but I can't remember how to calculate the fractions residues. I know that you can calculate the residues when the root is real and have multiplicity (r) >1 by using $b_{k}$ = $\frac{1}{(r-k)!}$ $\frac{d^{r-k}}{ds^{r-k}}$ ($\frac{N(s)}{D(s)}(s+a)^r)$ with 1<=k<=r

Is there a similar way as above for complex roots with r>1? For example, something like $\beta_{k}s+\beta_{l}$ = $\frac{1}{(r-k)!}$ $\frac{d^{r-k}}{ds^{r-k}}$ ($\frac{N(s)}{D(s)}((s+a)(s-a))^r)$

Nuno Silveira
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  • That formula is usually inconvenient for computing residues. You can rather expand in power series around the relevant point. – OR. Dec 19 '13 at 19:36
  • So, suppose you want to compute residue at $x=i$. We can write $x^4/(x^2+1)^5=(x-i+i)^4(x-i)^{-5}(x-i+2i)^{-5}$. Use the binomial formula to expand in series to the term you need. – OR. Dec 19 '13 at 19:38
  • Recall that when you are dividing series you can use long division to compute successively the term of the series of the quotient. – OR. Dec 19 '13 at 19:40
  • By the way, I think the formula you posted works for any complex numbers, not only for residues at real points. – OR. Dec 19 '13 at 19:41

1 Answers1

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$$x^4=((x^2+1)-1)^2=(x^2+1)^2-2(x^2+1)+1$$

should reduce your expression to friendlier looking fractions.

Lutz Lehmann
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