Algebraic manipulation of product of two summations

Feb 2015
16
1
United Kingdom
If

$F_1(x)=\sum_{n\ge 0}a_nx^n$, $F_2(x)=\sum_{n\ge 0}b_nx^n$,

why does the following hold?

$$\begin{align*}
F_1(x)F_2'(x)&=\sum_{n\ge 0}a_nx^n\sum_{n\ge 0}nb_nx^{n-1}\\
&=\sum_{n\ge 0}a_nx^n\sum_{n\ge 0}(n+1)b_{n+1}x^n\\
&=\sum_{n\ge 0}\sum_{k=0}^na_k(n-k+1)b_{n-k+1}x^n
\end{align*}$$ More specifically, how does one move from the second to the third line of the derivative-related equation?
 

v8archie

Math Team
Dec 2013
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2,682
Colombia
It's just a term by term treatment of the product, looking at increasing powers of $x$.
 
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mathman

Forum Staff
May 2007
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To save writing, let \(\displaystyle c_n=(n+1)b_{n+1}\).

The second line is then \(\displaystyle \sum_{n\ge 0}\sum_{k\ge 0}a_nc_kx^{n+k}\).
Now let j=n+k so the double sum becomes \(\displaystyle \sum_{n\ge 0}\sum_{j\ge n}a_nc_{j-n}x^j\).

Finally switch the order of summations. \(\displaystyle \sum_{j\ge 0}x^j\sum_{n=0}^j a_nc_{j-n}\).
 
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