# Double Summation question from Hoffman/Kunze

I've been self-studying Hoffman/Kunze and I'm puzzled by this seemingly easy double summation. I still don't see the jump in logic from the 3rd to the 4th step on pg 118 in Ch4 polynomials.

$[(fg)h]_n=\displaystyle\sum_{i=0}^n(fg)_ih_{n-i}$

$=\displaystyle\sum_{i=0}^n\left(\sum_{j=0}^if_ig_{i-j}\right)h_{n-i}$

$=\displaystyle\sum_{i=0}^n\sum_{j=0}^if_ig_{i-j}h_{n-i}$

$=\displaystyle\sum_{j=0}^nf_j\sum_{i=0}^{n-j}g_ih_{n-i-j}$

I reached the 4th step but got $f_{i+j}$ instead of $f_j$. Have I made a mistake?

P.S : I can't read math format well on chrome browser. Is there an issue?

## 1 Answer

Your calculation is correct.

We obtain \begin{align*} \sum_{i=0}^n\sum_{j=0}^if_ig_{i-j}h_{n-j} &=\sum_{0\leq j\leq i\leq n}f_ig_{i-j}h_{n-i}\\ &=\sum_{j=0}^n\sum_{i=j}^nf_ig_{i-j}h_{n-i}\tag{1}\\ &=\sum_{j=0}^n\sum_{i=0}^{n-j}f_{i+j}g_ih_{n-i-j}\tag{2} \end{align*}

Comment:

• In (1) we exchange the sums.

• In (2) we shift the index $i$ to start from zero.