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Infinities 6
Iteration
Number, 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  1
x0  0
x1  2  0  1  1
x2  2  1  1  3
Iterative solution of equations
x2  x  1  0
x2  x  1
Graph
yx
1
x  1
x
1
y  1
x
and make a first
approximation of
the root.
x0  1
1
x1   1
x0
1
x1   1  2
1
1
x2   1
x1
1
x2   1  1.5
2
1
x3   1
x2
1
x3 
 1  1.67
1.5
x1=2
x3=1.67
x0=1
x2=1.5
Try solving:
x 2  6x  1
2x 2  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
A:
1
A
xn1   xn  
2
xn 
To find
40 :
x0  6
1
40 
x1   6  
2
6
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 selfsimilarity 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.