I'm trying to figure out double summations.
I was wondering, when trying to simplify them. Can I just follow the same rules as with a standard summation?
I know that I'm allowed to split a summation, like this:
$$\sum_{i=1}^{n}(x_{i}+y_{j})=\sum_{i=1}^{n}x_{i}+\sum_{i=1}^{n}y_{j}$$
But am I allowed to the same with double summations like so?
$$\sum_{i=1}^{n}\sum_{j=1}^{n}(i-j)=\sum_{i=1}^{n}\sum_{j=1}^{n}(i)-\sum_{i=1}^{n}\sum_{j=1}^{n}(j)$$
This would allow me to remove the double summation and turn it into a single one, I believe.
I couldn't find any clear answers anywhere else, so I'm hoping you guys would be able to help me.
Thanks in advance!
Best Answer
First, let's prove that for any sequences $(a_{k})$ and $(b_{k})$ and a positive integer $n$, $$ \sum_{k=1}^{n}\left[a_{k}+b_{k}\right]=\sum_{k=1}^{n}a_{k}+\sum_{k=1}^{n}b_{k}. $$ Under the convention $\sum_{k=1}^{0}f_{k}=0$, the statement above is trivially true for $n=0$. Now, suppose the statement holds for some particular $n=n_{0}-1$ with $n_{0}\geq1$. Then, $$ \sum_{k=1}^{n_{0}}\left[a_{k}+b_{k}\right]=a_{n_{0}}+b_{n_{0}}+\sum_{k=1}^{n_{0}-1}\left[a_{k}+b_{k}\right]=a_{n_{0}}+b_{n_{0}}+\sum_{k=1}^{n_{0}-1}a_{k}+\sum_{k=1}^{n_{0}-1}b_{k}=\sum_{k=1}^{n_{0}}a_{k}+\sum_{k=1}^{n_{0}}b_{k}. $$ The desired result follows by induction.
Now, you should be able to use the above result to prove that for any sequences $(a_{k\ell})$ and $(b_{k\ell})$ and positive integers $n$ and $m$, $$ \sum_{k=1}^{n}\sum_{\ell=1}^{m}\left[a_{k\ell}+b_{k\ell}\right]=\sum_{k=1}^{n}\sum_{\ell=1}^{m}a_{k\ell}+\sum_{k=1}^{n}\sum_{\ell=1}^{m}b_{k\ell}. $$