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It's all Greek to me! Sigma notation
Lecture Level: Easy
Lecture Length:9:50 minutes

Lecture Index

In the last lecture. We talked about generalizations as a way to discover new results in mathematics.

Here we are presenting a few more.

The sum $1+2+3+ \cdots +n$ could be seen as $1^1+2^1+3^1 + \cdots + n^1$ and that immediately suggests

the sum $1^2+2^2+3^2+3^2 + \cdots + n^2$ and from there to $1^k+2^k+3^k+ \cdots + n^k$ for any $k$ natural.

That may also lead us to $1^{-1}+2^{-1}+3^{-1}+ \cdots + n^{-1}$ and that sum is equivalent to $1 + 1/2 + 1/3 + \cdots + 1/n$.

We can create new mathematics, when we create generalizations of existing results or new ways to approach the solution of old problems.These generalized results give us new understanding to a particular topic.As you can see there are many ways to generate new questions, similar in principle to our original question, but each one of them sufficiently different to be considered a new question. There is something that is common to all of them. And that is that in all that we have seen so far. We have been adding multiple terms of a progression of some sort. Since this addition of multiple elements happen often in mathematics it is convenient to introduce a notation created by the french mathematician Joseph Fourier. Using Fourie's idea we will use the upper case sigma letter from the Greek alphabet to denoted summation. So we write.$$ \sum\limits_{k=1}^n a_k = a_1+a_2+a_3+ \cdots + a_n $$
The numbers on the top and bottom define as in this case that k goes from $1$ to $n$.
How will you represent the sum $1+ 2 + 3 + \cdots +n$ with the summation symbol?$$ \sum\limits_{k=1}^n k$$
and how will you represent $1^2 + 2^2 + 3^2 + \cdots + n^2$?$$ \sum\limits_{k=1}^n k^2$$
and for $3^3 +4^3+ \cdots +(n-3)^3$?$$ \sum\limits_{k=3}^{n-3} k^3 $$
There are some very immediate properties of summations and we will not demonstrate them but if you feel the urge to place them in proper ground you may or you may find the proof in some book.
The first one is $$\sum\limits_{k} (a_k + b_k)= \sum\limits_{k} a_k +\sum\limits_{k} b_k$$
this will allow us to divide a summation into two summations and the other way around.
The next property is $$\sum\limits_{k} c a_k = c \sum\limits_{k} a_k$$
This means that we can factor any term that does not involve the summation index $k$.
Let us try now to find the sum $$\sum\limits_{k=1}^n k^2$$
We are tempted to use the formula we have found before. That is $T_n=n(n+1)/2$ which is equal to $(n^2+n)/2$.
Using the sigma notation one could write.$$ \sum\limits_{k=1}^n T_k=1/2 \sum\limits_{k=1}^n k(k+1)$$
Thinking of these relations from a geometric viewpoint one is immediately taken into three dimensional space and if one places balls forming a tetrahedron with $T_n$ balls on the base and then $T_{(n-1)}$ on the top until we get to the apex with just $1$.
These numbers are called tetrahedral numbers. In similar fashion one could could place balls in a pyramidal shape with square base. $n^2$ and $(n-1)^2$ balls on the top and so on until we get to the apex. Again with just one ball. These numbers are called pyramidal numbers.Let us denoted the $n$ tetrahedral number by $Tetr (n)$ and that is $\sum\limits_{k=1}^{n} T_k$ and the $n$ Pyramidal number by $Pyr (n)$ and that is equal to $\sum\limits_{k=1}^{n} k^2$.
Then we can translate our prior relation into $$2 Tetr (n)=T_n+Pyr(n)$$
That is a very nice relation that links triangular numbers, tetrahedral numbers and pyramidal numbers.
We set out to find a formula for the pyramidal $n$ and now we have in hand a relation that involve the tetrahedral numbers so let us find a formula for them now. After computing a few tetrahedral numbers$1,4,10,20,...$ one is lead to suspect that the following relation is true. $3$ times tetrahedral $n$ equals $(n+2)T_n$also there is a web site at where one is able to place a few elements of a sequence and one can get what known sequences contain those elements. The more elements one place the more tuned the answer will be. So we are conjecturing that this relation is true.$$3 Tetr (n) = (n+2) T_n$$
There is a very beautiful demonstration of this fact in the book of numbers by Conway and Guy but I will not reproduce that here instead I will show that this relation is true thinking $3$ dimensionally.
It is immediate to see this geometrically as $3$ tetrahedral and the $(n+2) T_n$ as a prism or wedge with triangular base. We have seen the prior relation relating two tetrahedral to a triangular and a pyramidal so is natural to think on applying that relation here so from the $3$ tetrahedral one is lead to one tetrahedral plus one pyramidal plus one triangular.

Can you see now $3$ dimensionally how to get from there to the prism $(n+2) T_n$ ?

It's easier to visualize this if you think of these figures as continuous figures as a tetrahedra a pyramid and the wage and we are able to transform them using a principle similar to the cavalieri principle for volume of figures. But in this case deformations that keep the numbers of balls by layers the same. In the sense that they still bear the same discrete numbers. This is a $3$ dimensional analogous to Gauss's solution.We have in fact conveniently rearrange the numbers this time with the help of a $3$ dimensional space visualization to get a sum where all the numbers are equal to in this case $T_n$. Therefore equals to $T_n (2n+1) /3$.

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