Question

Show that a double summation can be interchanged

Answer 1

what does this mean?

Answer 2

\[\sum_{i=a}^{n} \sum_{k=b}^m f(i,k) = ? \]
what do we want to interchange here?

Answer 3

\[\sum_{i=1}^{m}\sum_{j=1}^{n}aij =\sum_{j=1}^{n}\sum_{i=1}^{m}aij\]

Answer 4

I've tried this with infinite limits and other stuff, but I can't get a formal proof

Answer 5

I really understand why they are the same, but I need to show that.

Answer 6

\[\text{ \let's look at this for some } m=3 \text{ and } n=4 \\ \sum_{i=1}^{3} \sum_{k=1}^4 f(i,k)=\sum_{i=1}^{3}(f(i,1)+f(i,2)+f(i,3)+f(i,4)) \\ = f(1,1)+f(1,2)+f(1,3)+f(1,4) \\ + f(2,1)+f(2,2)+f(2,3)+f(2,4) \\ + f(3,1)+f(3,2)+f(3,3)+f(3,4) \\ \\ \text{ whereas } \\ \sum_{k=1}^4 \sum_{i=1}^3 f(i,k)=\sum_{k=1}^{4}(f(1,k)+f(2,k)+f(3,k)) \\ =f(1,1)+f(2,1)+f(3,1) \\ +f(1,2)+f(2,2)+f(3,2) \\ +f(1,3)+f(1,3)+f(3,3) \\ +f(1,4)+f(2,4)+f(3,4) \\ \]

so the rows became the columns in the next sum...
but we do have the same sum either way

Answer 7

so do just do similar thing here
compare rows to columns in the expansion
\[\sum_{i=1}^{n} \sum_{k=1}^{m} f(i,k)=\sum_{i=1}^{n}(f(i,1)+f(i,2)+f(i,3)+ \cdots + f(i,m)) \\ =f(1,1)+f(1,2)+f(1,3)+ \cdots +f(1,m) \\ +f(2,1)+f(2,2)+f(2,3)+ \cdots + f(2,m) \\ + \cdots + \\ +f(n,1)+f(n,2)+f(n,3)+ \cdots +f(n,m)\]

Answer 8

so expand
\[\sum_{k=1}^{m} \sum_{i=1}^{n} f(i,k)\]
and you should see the columns of the previous sum appear as the rows in this sum

Answer 9

Thanks a lot, much better now. Light went off yesterday. Thanks!