Question

Product of two summations?

Say I am multiplying two polynomials:
\[ \left( \sum_{k=0}^na_kx^k\right) \left( \sum_{j=0}^mb_jx^j\right)\] apparently this is equal to \[ \sum_{r=0}^{n+m}\left(\sum_{k=0}^ra_kb_{r-k}\right)x^r\]

Answer 1

What I did:
\[ \begin{align}
\left( \sum_{k=0}^na_kx^k\right) \left( \sum_{j=0}^mb_jx^j\right)
&= \sum_{k=0}^n\sum_{j=0}^ma_kb_jx^{k+j}\\

&=\sum_{k=0}^n\sum_{r-k=0}^ma_kb_{r-k}x^r \\
&=\sum_{k=0}^n\sum_{r=k}^ma_kb_{r-k}x^r
\end{align}\]
Obviously I'm having trouble transitioning to the required double sum with the required indices. I tried drawing a Cartesian diagram in both cases using \(m=5\) and \(n=2\) as an example to see what points are covered by the sum, but they don't seem to be the same? Any help would be appreciated.

Answer 2

Oh yeah I let \(r=j+k\)

Answer 3

Do you still need help?

Answer 4

why u let r=k-j ?

Answer 5

that change second summation , each loop the second summation need to start from zero
so i guess to the second step is enough unless ur trying to prove something else ,
whats ur whole question anyway ?

Answer 6

@wio @ikram002p

I let \(r=j+k\) since the formula to prove was to get:
\[\sum_{r=0}^{n+m}\left(\sum_{k=0}^ra_kb_{r-k}\right)x^r\] . I assumed this is how they got their \(r\) variable? So my problem was trying to go from where I left off in my last post. I tried drawing the points covered by the double summation here using \(m=5,n=2\) as an example (attached)

Where the left drawing was for \( =\sum\limits_{k=0}^n\sum\limits_{r=k}^ma_kb_{r-k}x^r\)

And the right drawing was for \(\sum\limits_{r=0}^{n+m}\left(\sum\limits_{k=0}^ra_kb_{r-k}\right)x^r\)

I guess my understanding of the double summation where an index depends on the other is what is making me not understand ?