Computing the coefficient of the term of a certain degree in a polynomial

Question

Given the polynomial

${1\over8}((1+z)^9 + 3(1-z)^4(1+z)^5 + (1-z)^6(1+z)^3)$

(which is the weight enumerator of a code)

how do I find out the coefficient of $z^2$?

The solution given is ${1 \over 8}(36-12+0) = 3$.

I got $36$ for the $z^2$ coefficient of $(1+z)^9$ using the Binomial Theorem, but I don't know how to get $-12$ for the $z^2$ coefficient of $3(1-z)^4(1+z)^5$. By using the Binomial Theorem separately on $(1-z)^4$ and $(1+z)^5$ I get the following two polynomials, repsectively:

$z^4-4z^3+6z^2-4z+1$

$z^5+5z^4+10z^3+10z^2+5z+1$

I am unsure what to do next, or even if this is going in the right direction.

Answer 1

A short cut or two: $$(1-z)^4(1+z)^5=(1-z^2)^4(1+z)=(1-4z^2+\cdots)(1+z)=1+z-4z^2+\cdots$$ and $$(1-z)^6(1+z)^3=(1-z^2)^3(1-z)^3=(1-3z^2+\cdots)(1-3z+3z^2-z^3) =1-3z+0z^2+\cdots$$ etc.

Answer 2

Notice that the coefficient of $z^2$ in $(a+bz+cz^2+\cdots)(d+ez+fz^2+\cdots)$ is $af+be+cd$.

After expansion of the powers,

$$1+9z+36z^2+\cdots+\\ 3(1-4z+6z^2-\cdots)(1+5z+10z^2+\cdots)+\\ (1-6z+15z^2-\cdots)(1+3z+3z^2+\cdots) $$

Which gives

$$\frac{36+3(10-20+6)+(3-18+15)}8=\frac{24}8.$$

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Using Lord Shark the Unknown's shortcuts, and the rule $(a+bz+cz^2+\cdots)(d+ez^2+\cdots)$ that gives $ae+cd$,

$$1+9z+36z^2+\cdots+\\ 3(1-4z^2+\cdots)(1+z)+\\ (1-3z^2+\cdots)(1-3z+3z^2+\cdots) $$

Now

$$\frac{36+3(-4)+(3-3)}8=\frac{24}8.$$

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Using a CAS,

$$\frac{5z^9+9z^8+24z^7+80z^6+138z^5+138z^4+80z^3+24z^2+9z+5}8.$$

Answer 3

We can work this out by establishing a general formula for

$$(1-z)^m(1+z)^n=(1-mz+\frac{m(m-1)}2-\cdots)(1+nz+\frac{n(n-1)}2z^2+\cdots).$$

The $z^2$ term has the coefficient $$\frac{m(m-1)}2-mn+\frac{n(n-1)}2=\frac{(m-n)^2-(m+n)}2.$$

Hence, $$9,0\to 36\\4,5\to-4,\\6,3\to0$$ and the solution is

$$\frac{36-3\cdot4}8.$$

Answer 4

It is convenient to use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$. This way we can write for instance \begin{align*} [z^k](1+z)^n=\binom{n}{k}\tag{1} \end{align*}

We obtain \begin{align*} \color{blue}{[z^2]}&\color{blue}{\frac{1}{8}\left((1+z)^9+3(1-z)^4(1+z)^5+(1-z)^6(1+z)^3\right)}\\ &=\frac{1}{8}\left([z^2](1+z)^9+3[z^2](1-z)^4(1+z)^5+[z^2](1-z)^6(1+z)^3\right)\tag{2}\\ &=\frac{1}{8}\left\{\binom{9}{2}+3\left(\binom{4}{0}[z^2]-\binom{4}{1}[z^1]+\binom{4}{2}[z^0]\right)(1+z)^5\right.\\ &\qquad \qquad\qquad \left.+\left(\binom{6}{0}[z^2]-\binom{6}{1}[z^1]+\binom{6}{2}[z^0]\right)(1+z)^3\right\}\tag{3}\\ &=\frac{1}{8}\left\{\binom{9}{2}+3\left(\binom{4}{0}\binom{5}{2}-\binom{4}{1}\binom{5}{1}+\binom{4}{2}\binom{5}{0}\right)\right.\\ &\qquad\qquad\qquad\left.+\left(\binom{6}{0}\binom{3}{2}-\binom{6}{1}\binom{3}{1}+\binom{6}{2}\binom{3}{0}\right)\right\}\tag{4}\\ &=\frac{1}{8}\left(36+3\left(10-20+6\right)+\left(3-18+15\right)\right)\\ &\,\,\color{blue}{=\frac{1}{8}\left(36-12+0\right)}\\ &\,\,\color{blue}{=3} \end{align*} and the claim follows.

Comment:

• In (2) we use the linearity of the coefficient of operator.

• In (3) we select the coefficient of $z^2$ in $(1+z)^9$ and we again use the linearity of the coefficient of operator together with the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

• In (4) we select the coefficient of $z^k, k=0,1,2$ accordingly.