Infinities 6

IterationNumber, Algebra and Geometry

Iteration = Repetition of a procedure applied to the result of a previous application.

• sequences• approximation• iterative reasoning• generation of fractals

Iteratively defined sequences

xn+1 =f xn( )

e.g. xn+1 =2xn +1x0 =0x1 =2 × 0 +1=1x2 =2 ×1+1=3

Iterative solution of equations

x2 −x−1=0

x2 =x+1

x =1x+1

Graph

and make a first approximation of

the root.

y =x

y=1x+1

x1 =1x0

+1

x1 =11+1=2

x2 =1x1

+1

x2 =12

+1=1.5

x3 =1x2

+1

x3 =1

1.5+1=1.67

x0 =1

x0=1

x1=2

x2=1.5

x3=1.67

Try solving:

x2 −6x =1

2x2 −x =2

Calculating square roots without a calculator (Babylonian method)

• Make a first approximation to the root.

• Divide this into the number.

• Find the mean of the quotient and your first approximation; use this as the second approximation.

To find 40 :x0 =6

x1 =12

6 +406

⎛⎝⎜

⎞⎠⎟

To find A :

xn+1 =12

xn +Axn

⎛⎝⎜

⎞⎠⎟

Towers of Hanoi

• Move the tower from one peg to another.

• You may move only one ring at a time.

• A ring may not be placed on top of a ring of a smaller size.

• What is the smallest number of moves?

Fractal = A geometric pattern exhibiting self-similarity in that small details of its structure viewed at any scale repeat elements of the overall pattern.

- often generated by iterative processes

http://www.ph.biu.ac.il/~rapaport/java-apps/lsys.html

What is the Mandelbrot set? The locus of points, C, for which the series Zn+1 = Zn * Zn + C, Z0 = (0,0) is bounded by a circle of radius two, centered on the origin.

Multiply Z by itself. Add C. The answer is the new value for Z.

Repeat until the absolute value of Z is greater than two.

If abs(Z) ever exceeds two, then it will head off towards infinity which means that the point C is not in the Mandelbrot set. These points are typically assigned a colour based on how many iterations were done before abs(Z) exceeded two.

If abs(Z) doesn't exceed two after a large number of iterations, then we assume that C is in the Mandelbrot set. These points are typically coloured black.

The Koch snowflake is constructed as follows. Start with a line segment.

Divide it into 3 equal parts.

Erase the middle part and substitute it by the top part of an equilateral triangle.

Now, repeat this procedure for each of the 4 segments of this second stage.

Sierpinska’s Carpet

Start with a square of carpet of side 3 units long. (Make your units quite large.)It can be divided into 9 1-unit squares.Remove the middle square.

What is the area of the hole?What is the area of the carpet?What is the perimeter of the hole?What is the perimeter of the carpet?

Repeat the process for each of the smaller squares left in the carpet. (i.e. remove the middle) Consider the same questions.

Repeat . . . .

Chaos Game

• Randomly choose a corner of the triangle by throwing the die.

• Starting from the last point marked, mark the next point 1/2 of the way towards the corner you have selected.

• Continue the process.