Is my formula for the coefficients of the product $\left(\sum_{i=0}^n f_i x^i \right)\cdot\left(\sum_{i=0}^m g_i x^i \right)$?

Question

Let's assume that $f(x)=\sum_{i=0}^n f_i x^i$ and $g(x)=\sum_{i=0}^m g_i x^i$ where $\{f_i\}_{i=0}^n\cup\{g_i\}_{i=0}^m\subset\mathbb{F}$ and $n,m\in\mathbb{N}$, where $\mathbb{F}$ is a field. I tried to find a formula for the given product $f(x)\cdot g(x)$. For simplicity, I tried to figure out a formula for $n=2,m=3$. Namely $$f(x)\cdot g(x)= \left(\sum_{i=0}^2 f_i x^i\right)\cdot\left(\sum_{i=0}^3 g_i x^i\right)=$$ $$=f_0 g_0+ \left(f_0 g_1+f_1 g_0\right)x+\left(f_0 g_2+f_1 g_1+f_2 g_0\right)x^2+\left(f_0 g_3+f_1 g_2+f_2 g_1+f_3 g_0\right)x^3 +\left(f_1 g_3+f_2 g_2+f_3 g_1\right)x^4+\left(f_2 g_3+f_3 g_2\right)x^5=$$ $$=\left(\sum_{i+j=0} f_i g_j\right)x^0+ \left(\sum_{i+j=1} f_i g_j\right)x^1+ \left(\sum_{i+j=2} f_i g_j\right)x^2+ \left(\sum_{i+j=3} f_i g_j\right)x^3+ \left(\sum_{i+j=4} f_i g_j\right)x^4+ \left(\sum_{i+j=5} f_i g_j\right)x^5$$ $$=\left(\sum_{k=0}^5 \left(\sum_{i+j=k} f_ig_j\right)x^k\right)$$ Thus I ended up to the form $$f(x)\cdot g(x)=\left(\sum_{k=0}^{n+m} \left(\sum_{i+j=k} f_ig_j\right)x^k\right)$$ where $f(x)=\sum_{i=0}^n f_i x^i$ and $g(x)=\sum_{i=0}^m g_i x^i$, for $\{f_i\}_{i=0}^n\cup\{g_i\}_{i=0}^m\subset\mathbb{F}$ and $n,m\in\mathbb{N}$. Is the above formula correct? If not, could you, please, indicate my mistakes? Thank you very much in advance!!

Answer 1

I expect that it’s just a typo, but you want $f_ig_{\color{red}j}$ in the inner summation.

If you want to be very fussy about it, you could add a couple of qualifiers to the inner summation:

$$\sum_{\substack{i+j=k\\0\le i\le n\\0\le j\le m}}f_ig_j\,.$$

A better approach in my view is simply to say that any coefficients outside those ranges are $0$.