0

I have come across some Laurent series in which the denominator of a fraction contains a power series. Looking around I came across this Calculate Laurent series for $1/ \sin(z)$ which suggests that it is possible to calculate the reciprocal of a series up to a certain number of term. I have looked at the formula and because of the abstract notation, it is not clear to me how this formula is applied. So for example if I wanted to find the reciprocal of the following polynomial: $$x - 2x^2 + 3x^3 - 4x^4$$ How would I compute this?

  • Have you tried long division? – saulspatz Jan 14 '19 at 20:36
  • 1
    Is the Cauchy product formula simply long division? –  Jan 14 '19 at 20:40
  • I would say yes, essentially. If we write$$1=(x-2x^2+3x^2-4x^4)({1\over x}+a_0+a_1x+\dots)$$ then the calculations to get the $a_k$ are the same ones we'd make doing a long division, or so it seems to me. – saulspatz Jan 14 '19 at 20:48
  • So from there I just equate the coefficients right? Like equating powers of $x^1$ I would get $a_0 - 2 = 0 ==> a_0 = 2$? –  Jan 14 '19 at 21:11
  • Yes, that's right. – saulspatz Jan 14 '19 at 21:50
  • How did you know that reciprocal would start with $x^{-1}$? –  Jan 14 '19 at 23:23
  • 1
    To do long division in this context is simplicity itself. You have to write divisor and dividend in increasing degree, as you have already done with your polynomial $f(x)$. Write the algorithm down and you see immediately that the first monomial in the series for $1/f$ is certainly $x^{-1}$. – Lubin Jan 14 '19 at 23:48
  • Alternatively, $${1\over x-2x^2+3x^3-4x^4}={1\over x}{1\over 1-2x+3x^2-4x^3}$$ and the second factor is analytic at $x=0$ with value $1$. – saulspatz Jan 15 '19 at 01:27
  • Looking around I have found this https://mathoverflow.net/questions/53384/power-series-of-the-reciprocal-does-a-recursive-formula-exist-for-the-coeffic#. Are those formula based on a derivation using long division? –  Jan 15 '19 at 13:33

0 Answers0