Sigma Sum Manipulation: How does the second equality hold? I'm only familiar with multiplying sigma sums by integers and adding them together, and so the second equality is quite confusing for me.
$f_i$ is a polynomial of degree $i$.
How does the solution below isolate $x^j$ like that?

I got up to here
$$\sum_{i=0}^{n}c_i(\sum_{j=0}^{i}a_{ij}x^j)=\sum_{i=0}^{n}(\sum_{j=0}^{i}c_ia_{ij}x^j) = ?$$
 A: Here we use an intermediate representation which might help to better see the  index region. We obtain

\begin{align*}
\sum_{i=0}^nc_if_i&=\sum_{i=0}^n c_i\left(\sum_{j=0}^i a_{ij}x^j\right)\\
&=\sum_{\color{blue}{0\leq j\leq i\leq n}}c_ia_{ij}x^j\\
&=\sum_{j=0}^n\left(\sum_{i=j}^nc_ia_{ij}\right)x^j
\end{align*}

A: Notice that as we proceed through the second equality, the order of the sums is reversed (from "for each $i$ sum over $j$" to "for each $j$ sum over $i$").  It can be helpful to draw a diagram of which pairs of indices, $(i,j)$ actually appear in the sum.
On the left, we have $i \in [0, n]$ and for each of those $j \in [0,i]$  Let's graph those pairs.

To reverse the order of summation, we need to study the range of $i$s for each $j$.  When $j = 0$, we have terms for $i \geq 0$.  When $j = 1$, for $i \geq 1$.  In fact, for each $j \geq 0$, we have $i \geq j$.  So our sum should be of the shape
$$  \sum_{j=0}^n \sum_{i=j}^n \dots $$
So lets' show a few more steps in the derivation of the equality you ask about:
\begin{align*}
\sum_{i=0}^n c_i \left( \sum_{j=0}^i a_{ij} x^j \right) &= \sum_{i=0}^n \left( \sum_{j=0}^i \left( a_{ij} c_i x^j \right) \right)  &  &\text{distributivity of multiplication over addition}  \\
    &= \sum_{j=0}^n \left( \sum_{i=j}^n \left( a_{ij} c_i x^j \right) \right)  &  &\text{exchange order of sums}  \\
    &= \sum_{j=0}^n \left( \left( \sum_{i=j}^n a_{ij} c_i\right) x^j \right)  &  &\text{(un-)distribute}  \\
\end{align*}
We are able to undistribute $x^j$ out of the inner summand because "$x^j$" does not depend on the inner index, $i$.  We cannot un-distribute $c_i$ or $a_{ij}$ from the final inner sum because these depend on $i$, so are not (necessarily) the same for each term in the sum over $i$.
A: If you examine carefully just which combinations of values of $i$ and $j$ occur in it, you’ll see that the first double sum can be rewritten like this:
$$\sum_{i=0}^nc_i\left(\sum_{j=0}^ia_{ij}x^j\right)=\sum_{0\le j\le i\le n}c_ia_{ij}x^j\,.$$
There is actually one term in the double sum for each pair $\langle i,j\rangle$ with $0\le j\le i\le n$. It may help to look at a small example, say with $n=3$; the following table shows exactly which combinations occur, and each of them occurs only once. If you move the $c_i$ inside the inner sum, the terms are:
$$\begin{array}{c|cc}
i\backslash j&0&1&2&3\\\hline
0&c_0a_{00}x^0\\
1&c_1a_{10}x^0&c_1a_{11}x^1\\
2&c_2a_{20}x^0&c_2a_{21}x^1&c_2a_{22}x^2\\
3&c_3a_{30}x^0&c_3a_{31}x^1&c_3a_{32}x^2&c_3a_{33}x^3
\end{array}$$
Once the $c_i$ multipliers have been brought inside the inner summation, it is adding across the rows of this table; the outer one is then adding up these sums.
Now change the order of summation: add up the columns, and then add the column sums. The sum of column $j$ is
$$\sum_{i=j}^nc_ia_{ij}x^j\,,$$
so the sum of the whole array is
$$\sum_{j=0}^n\sum_{i=j}^nc_ia_{ij}x^j\,.$$
Now the index $j$ is constant within the inner summation, so we can factor out $x^j$, just as $c_i$ was outside the original inner summation:
$$\sum_{j=0}^n\sum_{i=j}^nc_ia_{ij}x^j=\sum_{j=0}^nx^j\sum_{i=j}^nc_ia_{ij}\,.$$
Clearly the inner summation is just the coefficient of $x^j$, so we could just as well write this as
$$\sum_{j=0}^n\left(\sum_{i=j}^nc_ia_{ij}\right)x^j\,.$$
A: Compare coefficients: $$c_0(a_{00}x^0)+c_1(a_{10}x^0+a_{11}x^1)+\dots+c_n(a_{n0}x^0+a_{n1}x^1+\dots a_{nn}x^n)$$
can be rewritten as,
$$(c_0a_{00}+c_1a_{10}+\dots+c_na_{n0})x^0+(c_1a_{11}+c_2a_{21}+\dots+c_na_{n1})x^1+\dots+c_na_{nn}x^n$$
