High School Mathematics Extensions/Supplementary/Summation Sign

Summation Notation
We normally use the "+" sign to represent a sum, but if the sum expression involved is complex and long, it can be confusing.

For example:$$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} ...... + \frac{1}{100 \times 101}$$

Writing the above would be a tedious and messy task!

To represent expression of this kind more compactly and nicely, people use the summation notation, a capital Greek letter "Sigma". On the right of the sigma sign people write the expression of each term to sum, and write the upper and lower limit of the variable on top and under the sigma sign.

Example 1: $$\sum_{k=3}^{10} 2k+1$$ $$=\ (2(3)+1)\ +\ (2(4)+1)\ +\ (2(5)+1)\ +......+\ (2(10)+1)$$ $$=\ 7\ +\ 9\ +\ 11\ +......+\ 21$$

 Misconception: From the above there is a common misconception that the number on top of the Sigma sign is the number of terms. This is wrong. The number on top is the number to substitute back in the last term.

 It would be useful here to indicate what values the lower limit of summation may take on.

Example 2: $$\frac{1}{4}-\frac{1}{9}+\frac{1}{16}-\frac{1}{25}......-\frac{1}{9801}+\frac{1}{10000}$$ $$=\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{4^2}-\frac{1}{5^2}......-\frac{1}{99^2}+\frac{1}{100^2}$$ $$=\sum_{k=2}^{100} (-1)^k \frac{1}{k^2}$$

 Tip:If the terms alternate between plus and minus, we can use the sequence $$(-1)^k=-1,\ 1,\ -1,\ 1...$$

Exercise
Change the following into sum notation: Change the following sum notation into the normal representation: (Need more exercise,especially "reading" sigma notation and change back into the old form)
 * 1) Use the summation notation to represent the expression in the first example.
 * 1) $$23\ +\ 24\ +\ 25\ +\ 26\ +......+\ 1927$$
 * 2) $$13\ +\ 16\ +\ 19\ +\ 22\ +......+\ 301$$
 * 3) *$$1\ -\ 2\ -\ 3\ +\ 4\ +\ 5\ -\ 6\ -\ 7\ +\ 8......+\ 400$$(Hint:reorder the terms, or get more than one term in the expression)
 * 4) *$$1000\ -\ \frac{3}{1\times(1+3+5)}\ -\ \frac{5}{(1+3)\times(1+3+5+7)}\ -\ \frac{7}{(1+3+5)\times(1+3+5+7+9)}......$$(Hint:You need to use more than one sigma sign)
 * 1) $$\pi = 4\sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1}$$
 * 2) $$\sin x = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}x^{2k+1}$$

Operations of sum notation
Although most rules related to sum makes sense in the ordinary system, in this new system of sum notation, things may not be as clear as before and therefore people summarize some rules related to sum notation (see if you can identify what they correspond to!)

(Note:I suggest getting a visual aid on this one:showing that you can sum a two dimensional array in either direction)
 * $$\sum_{i=p}^{q} A_i\pm c\ =\ \pm(q-p+1)c\ +\ \sum_{i=p}^{q} A_i$$
 * $$\sum_{i=p}^{q} A_i \pm B_i \ =\ \sum_{i=p}^{q} A_i \pm \sum_{i=p}^{q} B_i$$
 * $$\sum_{i=p}^{q} c A_i\ =\ c\sum_{i=p}^{q} A_i$$
 * $$\sum_{i=p}^{q} \left [ \sum_{j=r}^{s} A_{ij} \right ] \ =\ \sum_{j=r}^{s} \left [ \sum_{i=p}^{q} A_{ij} \right ]$$
 * $$\sum_{i=p}^{q} A_i \ =\ \sum_{i=p-k}^{q-k} A_{i+k} $$(Index substutition)
 * $$\sum_{i=p}^{q} A_i \ =\ \sum_{i=p}^{r} A_i + \sum_{i=r+1}^{q} A_i \;, where\; p \le r < q $$(Decomposition)
 * $$\left ( \sum_{i=p}^{q} a_i \right ) \times \left ( \sum_{j=r}^{s} b_j \right )\ =\ \sum_{i=p}^{q} \sum_{j=r}^{s} a_i b_j$$(Factorization/Expansion)

Exercise
(put up something here please)

Beyond
"To iterate is human; to recurse, divine."

When human repeated summing, they have decided to use a more advanced concept, the concept of product. And of course everyone knows we use $$\times$$. And when we repeat product, we use exponential. Back to topic, we now have a notation for complex sum. What about complex product? In fact, there is a notation for product also. We use the capital Greek letter "pi" to denote product, and basically everything else is exactly the same as sum notation, except that the terms are not summed, but multiplied. Example: $$\prod_{h=2}^{5} 2h-3$$ = $$[(2\times2)-3]\times[(2\times3)-3]\times[(2\times4)-3]\times[(2\times5)-3]$$

Exercise
1. It has been known that the factorial is defined inductively by: $$0!\ =\ 1$$ $$n!\ \times\ (n+1)\ =\ (n+1)!$$

Now try to define it by product notation.