I got the question on a midterm and got it wrong. I'd like to know where I went wrong. We were supposed to find the coefficient of $x^{15}$ of$$(1-x^2)^{-10}(1-2x^9)^{-1}$$

My answer

The only way to get $x^{15}$ is to select three $-x^2$'s and one $-2x^9$

There are $-10 \choose 3$ ways to choose three $-x^2$ with $${-10 \choose 3} = {10 + 3 - 1 \choose 3} = {12 \choose 3}$$ For each of the $12 \choose 3$ ways to get three $-x^2$'s, there's only one way to get a $-2x^9$.

The -1 from the $-x^2$ and the -2 from the $-2x^9$ are part of the coefficients, and so the coefficient of $x^{15}$ is $$2{12 \choose 3}$$

Is this the correct answer? If not, where did I go wrong? Thanks

  • $\begingroup$ U forgot to say what question you are trying answer? $\endgroup$ – Piotr Benedysiuk Jul 8 '16 at 21:47
  • $\begingroup$ Sorry, Just fixed it. $\endgroup$ – The_Questioner Jul 8 '16 at 21:50
  • $\begingroup$ @AlexProvost You can expand binomials to negative powers using an extension of the binomial theorem called the Newton binomial series. In generating function theory, writing $[x^n]$ in front of a power series (or a formula equal to a power series) means extracting the coefficient of $x^n$. $\endgroup$ – arctic tern Jul 8 '16 at 21:59
  • $\begingroup$ x^15. Sorry, I thought I was being clear. $\endgroup$ – The_Questioner Jul 8 '16 at 21:59
  • $\begingroup$ @arctictern The original formatting made me think he was looking at $x^{15}(1-x^2)^{-10}(1-2x^9)^{-1}$. Interesting, I didn't know about the $[x^n]$ notation! $\endgroup$ – Alex Provost Jul 8 '16 at 22:01

You're correct the only way to make $x^{15}$ out of $-x^2$ and $-2x^9$ is to use three of the former and one of the latter. Therefore, using the binomial series, the coefficient should be

$$\binom{-10}{3}(-x^2)^3\,\cdot\,\binom{-1}{1}(-2x^9)^1. $$


$$\binom{-10}{3}=\frac{-10(-10-1)(-10-2)}{3\cdot2\cdot1}=-220, \qquad \binom{-1}{1}=-1 $$

(since $\binom{n}{1}=n$ identically), so the term is $440x^{15}$.

In your calculations, you seem to have missed the negative sign in $\binom{-10}{3}=-220$, and also didn't even reference $\binom{-1}{1}$ at all, which makes it seem like you've got two sign errors that cancelled each other out to give you the correct answer.

  • $\begingroup$ I mentioned the -1 and -2 which are included in the coefficient. Also, I thought the (-1 choose 1) was unnecessary. I guess I should've elaborated more. Thanks. $\endgroup$ – The_Questioner Jul 8 '16 at 22:09
  • $\begingroup$ @The_Questioner You wrote $\binom{-10}{3}=\binom{12}{3}$, which is false. Don't write false things. There is a total of four different negative signs at play. One comes from $\binom{-10}{3}$, one comes from $(-x^2)^3$, one comes from $\binom{-1}{1}$, and one comes from $(-2x^9)^1$. $\endgroup$ – arctic tern Jul 8 '16 at 22:14

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