Question

Example of double summation sigma operation

Answer 1

In short, a double summation is a summation inside of another summation.

It looks like this:

∑nj=1∑nk=1f(j,k)∑j=1n∑k=1nf(j,k)

The idea behind this is you’re doing a sum within a sum, and both indices will be inside the inner sum. For example, let’s consider

∑4k=1∑5j=1k+j∑k=14∑j=15k+j

and its value.

You would first evaluate the sum inside:

∑4k=1(k+1)+(k+2)+(k+3)+(k+4)+(k+5)=∑4k=15k+15∑k=14(k+1)+(k+2)+(k+3)+(k+4)+(k+5)=∑k=145k+15

And then evaluate the outer sum given the value of the inner sum:

∑4k=15k+15=(5(1)+15)+(5(2)+15)+(5(3)+15)+(5(4)+15)∑k=145k+15=(5(1)+15)+(5(2)+15)+(5(3)+15)+(5(4)+15)

=(5+15)+(10+15)+(15+15)+(20+15)=(5+15)+(10+15)+(15+15)+(20+15)

=20+25+30+35=20+25+30+35

=110=110

So, ∑4k=1∑5j=1k+j=110∑k=14∑j=15k+j=110

Summations like these are often used to set up arrays in programming or define certain mathematical concepts such as double or triple integrals.

So,

limn,m→∞∑ni=1∑mj=1f(x∗i,y∗j)ΔxΔylimn,m→∞∑i=1n∑j=1mf(xi∗,yj∗)ΔxΔy

is an example of the application usage of a double summation.

Let’s try another one:

∑5j=1∑5i=12ij∑j=15∑i=152ij

Same concept:

=∑5j=1(2j)+(4j)+(6j)+(8j)+(10j)=∑5j=1(30j)=∑j=15(2j)+(4j)+(6j)+(8j)+(10j)=∑j=15(30j)

Now evaluate what’s left:

∑5j=1(30j)=30(1)+30(2)+30(3)+30(4)+30(5)=30+60+90+120+150=450.∑j=15(30j)=30(1)+30(2)+30(3)+30(4)+30(5)=30+60+90+120+150=450.

So, ∑5j=1∑5i=12ij=450∑j=15∑i=152ij=450.

Remember that each summation must have its own variable to go inside the inner function, because otherwise it wouldn’t make sense.