Using MimeTeX 3

Brackets

As students advance through Maths, they come into contact with brackets. Algebraic notation depends heavily on brackets. The usual keyboard values of ( and ) are useful, for example:

  $d=2\ \times \ (4\ -\ j)$

This is written as:

 $$ d = 2 \ \times \ (4 \ - \ j) $$

Usually, these brackets are enough for most formulae but they will not be in some circumstances. Consider this:

 $4x^{3}\ +\ (x\ +\ {\frac {42}{1+x^{4}}})$ 

Is OK, but try it this way:

 $4x^{3}\ +\ \left(x\ +\ {\frac {42}{1+x^{4}}}\right)$ 

This can be achieved by:

 $$ 4x^3 \ + \ \left(x \ + \ \frac{42}{1 + x^4}\right) $$

A simple change using the \left( and \right) symbols instead. Note the actual bracket is both named and presented.

Ellipsis

The Ellipsis is a simple code:

 $x_{1},\ x_{2},\ \ldots ,\ x_{n}$ 

Written like:

 $$ x_1, \ x_2, \ \ldots, \ x_n  $$

A more practical application could be:

Question:

 "Add together all the numbers from 1 $\ldots$ 38.
 What is an elegant and simple solution to this problem?
 Can you create an algebraic function to explain this solution?
 Will your solution work for all numbers?"

Answer: The question uses an even number to demonstrate a mathematical process and generate an algebraic formula.

Part 1:	Part 2.	Part 3.
$1.\ 1\ +\ 38\ =\ 39$ $2.\ 2\ +\ 37\ =\ 39$ $3.\ 3\ +\ 36\ =\ 39$ $\ldots$ $19.19\ +\ 20\ =\ 39$ $\therefore x\ =\ 39\ \times \ 19$ $\therefore x\ =\ 741$ </math>	The solution is that, beginning form each end, the numbers are added and then multiplied by the number of different combinations. The answer must depend on the number, ${\frac {n}{2}}$ being a whole number. Therefore, the solution will not work for an odd range of numbers, only an even range.	An algebraic solution might read something like: $t=(1+n)\times n/2$ Where t = total and n = the last even number.