# Coefficient of specific term in a Series Expansion

Given series expansion: $(z^5 + z^6 + z^7 + ...)^8$ Question: Find the coefficient of $z^{10}$.

Given series expansion: $1/(1+z^3)^2$ Question: Find coefficient of $z^{12}$

Can someone help me solve these types of questions? Where can I find more information on how to find the coefficients of the $z$ term in a series expansion?

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HINTS:

• $(z^5+z^6+z^7+\ldots)^8=\left(z^5(1+z+z^2+\ldots)\right)^8=z^{40}(1+z+z^2+\ldots)^8$, so the smallest power of $z$ with a non-zero coefficient is ... ?

• $\displaystyle\left(\frac1{1+u}\right)^2=\frac1{(1+u)^2}=\frac{d}{du}\left(\frac{-1}{1+u}\right)$, and you know the series expansion of $\displaystyle\frac1{1+u}$. Start with that, multiply by $-1$, and differentiate to get the series expansion of $\dfrac1{(1+u)^2}$. To get the series expansion of $\dfrac1{(1+z^3)^2}$, just replace $u$ by ... ?

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I don't know the steps so why are you taking the derivative? – Sol Bethany Reaves Apr 15 '13 at 13:21
@SolBethanyReaves: You know the series expansion of $\frac1{1+z}$. You can easily differentiate it and use the result to find the series expansion of $\frac1{(1+z)^2}$. Then make a pretty obvious substitution of something else for $z$, and you’ll be able to read off the coefficient of $z^{12}$ very easily. – Brian M. Scott Apr 15 '13 at 13:24
Brian, Sol has $z^3$ where you have $z$, and (judging from the comment on my answer) that seems to be a stumbling block for Sol. – Gerry Myerson Apr 16 '13 at 0:06
@Gerry: I was hoping that the comment about making a substitution would suffice, but it appears that it won’t. – Brian M. Scott Apr 16 '13 at 2:50
@SolBethanyReaves: I’ve expanded both hints a bit. – Brian M. Scott Apr 16 '13 at 2:55

For the first one, use what you know about geometric series to get an expression involving $(1-z)^{-8}$, then use the binomial theorem to expand that expression into a series in which you know the coefficients. The second is also an exercise in the use of the binomial theorem.

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For the latter one, I don't know how to transform 1/(1+z^3)^2 to 1/(1-z)^n, which is the binomial theorem. – Sol Bethany Reaves Apr 15 '13 at 19:57
@Gerry : For the first one, isn't the lowest power of $z$ $z^{40}$, making the answer $0$? What am I missing? – Stefan Smith Apr 15 '13 at 20:53
@Stefan, you're not missing anything. Of course, if the problem had been the coefficient of $z^{100}$, then my suggestion would be more useful. – Gerry Myerson Apr 16 '13 at 0:02
Sol, if you introduce a new variable $Q$, defined by $Q=-z^3$, then your function becomes $(1-Q)^{-n}$, and the binomial theorem applies; after applying it, you can then replace $Q$ everywhere with $-z^3$ to get an answer involving $z$. – Gerry Myerson Apr 16 '13 at 0:04