- #1

mathmari

Gold Member

MHB

- 5,049

- 7

Hey!

"Let the polynomials

$$f(x)=1+\sum_{k=1}^{8}{(2k)x^{2k}} \text{ and } g(x)=1+ \sum_{k=1}^{8}{(3k)x^{3k}}$$

of $\mathbb{Q}$. Without calculating $f(x)g(x)$, find the coefficient of $x^{21}$ at $f(x)g(x)$."

Let's consider $f(x)=\sum_{k=0}^{\infty}{a_kx^k}$ and $g(x)=\sum_{k=0}^{\infty}{b_kx^k}$

So the following coefficients exist:

\begin{matrix}

a_0 & b_0 \\

a_2 & b_3 \\

a_4 & b_6 \\

a_6 & b_9 \\

a_8 & b_{12} \\

a_{10} & b_{15} \\

a_{12} & b_{18} \\

a_{14} & b_{21} \\

a_{16} & b_{24} \\

\end{matrix}

Since $c_{21}=\sum_{i+j=21}{a_i b_j}$, we have to find each time two coefficients for which the sum of their indices is equal to $21$. So:

$a_0, b_{21}$

$a_6, b_{15}$

$a_{12}, b_9$

Therefore, $c_8=a_0 b_{21}+a_6 b_{15}+ a_{12} b_9=21+6 \cdot 15+ 12 \cdot 9=219$.

Is this correct?

Is the way I solved it ok, or is there a better way to write it?

"Let the polynomials

$$f(x)=1+\sum_{k=1}^{8}{(2k)x^{2k}} \text{ and } g(x)=1+ \sum_{k=1}^{8}{(3k)x^{3k}}$$

of $\mathbb{Q}$. Without calculating $f(x)g(x)$, find the coefficient of $x^{21}$ at $f(x)g(x)$."

Let's consider $f(x)=\sum_{k=0}^{\infty}{a_kx^k}$ and $g(x)=\sum_{k=0}^{\infty}{b_kx^k}$

So the following coefficients exist:

\begin{matrix}

a_0 & b_0 \\

a_2 & b_3 \\

a_4 & b_6 \\

a_6 & b_9 \\

a_8 & b_{12} \\

a_{10} & b_{15} \\

a_{12} & b_{18} \\

a_{14} & b_{21} \\

a_{16} & b_{24} \\

\end{matrix}

Since $c_{21}=\sum_{i+j=21}{a_i b_j}$, we have to find each time two coefficients for which the sum of their indices is equal to $21$. So:

$a_0, b_{21}$

$a_6, b_{15}$

$a_{12}, b_9$

Therefore, $c_8=a_0 b_{21}+a_6 b_{15}+ a_{12} b_9=21+6 \cdot 15+ 12 \cdot 9=219$.

Is this correct?

Is the way I solved it ok, or is there a better way to write it?

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