Question

I have to use generating function to figure out the coefficient of terms, I want to know why the formula turn different, and which one is right or this is another form of that?

Answer 1

the first one is the sum of geometric series
\[1+x^2+x^4+\ldots=\sum_{k=0}^\infty a_k\qquad a_k=(x^2)^k\\
\qquad={1\over 1-x^2}
\]

Answer 2

so, for (1),
\[
\frac{1}{(1-x^2)^3}=(1-x)^{-3}=\sum_{k=0}^\infty{k-(-3)-1\choose-(-3)-1}(x^2)^k\\
\qquad=\sum_{k=0}^\infty{k+2\choose2}x^{2k}\\
\Large{x^8=x^{2k}\implies2k=8\implies k=4}\\
{\rm coefficient:}\quad{4+2\choose2}={6\choose2}={6\times5\times4!\over2!\times4!}=\color{red}{\mathbf 15}
\]

Answer 3

\[\Large{{x^7\over1-x}+{2x^8\over1-x}=x^7\sum_{k=0}^\infty x^k+2x^8\sum_{k=0}^\infty x^k\\
}\]
coeff of each term on the left is "1"
coeff of each term on the right is 2(1)="2"

coeff of x^8 = 1+2=3

Answer 4

Kid, that what I get, but on the second paper, my prof correct something like \[\sum_{k=0}^{\infty}\left(\begin{matrix}k+1 \\ 1\end{matrix}\right) x^(k+7) +2 \sum_{k=0}^{\infty}\left(\begin{matrix}k+1 \\ 1\end{matrix}\right) x^(k+8)\]
and then \[\left(\begin{matrix}2 \\ 1\end{matrix}\right)+ 2 \left(\begin{matrix}0 \\ 1\end{matrix}\right)=3\]
I have no idea about that stuff

Answer 5

for the part I messed up, I know my mistake and know how to fix it. For this part I don't know because I m not wrong.

Answer 6

it'd be
\[\Large{
\sum_{k=0}^\infty x^{k+7}+2\sum_{k=0}^\infty x^{k+8}
}\]

Answer 7

yeap, the net let it as it is. not me

Answer 8

the coeffs are just ones...
dont worry about the other.
\[{k\choose1}=1\]ALWAYS

Answer 9

hey, did you see the next page I post with mine. it has 2 pages

Answer 10

yes I did

Answer 11

you saw the differences?

Answer 12

yes. but the second page is not easy to understand

Answer 13

ok, if you said so. I'm ok, now I study for tomorrow

Answer 14

that's my prof's stuff