pptx - Williams College

```Using Origami to Find
Rational Approximations of Irrational Roots
Jeremy Lee
Amherst Regional High School
Williams College
April 6th, 2013
Origami Instructions
• Folding paper allows us to find rational
approximations of the square root of
two.
Side Unit fits
Diagonal Once with
remainder.
Origami Instructions
Side Unit fits
Diagonal Once with
remainder.
Remainder fits side
unit twice with
remainder2 space
Origami Instructions
Side Unit fits
Diagonal Once with
remainder.
Remainder fits side
unit twice with
remainder2 space
Remainder2 fits
Remainder unit twice
with Remainder3 left
over.
Origami Instructions
Summary of Origami Exercise
• As we keep doing the exercise, you
always get the number two after the
first step. The next remainder will fit
into previous remainder twice again.
Summary of Origami Exercise
• All these approximations of the
square root of two come from the
continued fractions representation
of the square root of two.
Continued Fractions Method
1
2  a0 
1
a1 
a2 
1
a3 
1

Continued Fractions Method
1
2  1
1
a1 
a2 
1
a3 
1

Continued Fractions Method
1
2  1
1
2
a2 
1
a3 
1

Continued Fractions Method
1
2  1
1
2
2
1
a3 
1

Continued Fractions Method
1
2  1
1
2
2
1
2
1

There will be twos continuing throughout the fraction after one.
How to Use Continued Fractions to
Build Approximations
hi
2 1
2  1
1

2
2  1
2
1
2
3
1

7
5
2
1
2  1
2

1
2
17
12
1
2
ki
Rational Approximations Of the Square Root of Two
hi
ki
Denoting Parts of the Square Root of Two
Continued Fraction
1
2  1
1
2
2
1
2
1

Let Big Box equal x
What does the big box converge to?
Algebra that Continued Fractions
Converges to the Square Root of Two
x
1
2 x
2  1  (1 
2x  x  1
2
2)
x  2x 1  0
2
2
8
2

22 2
 1 
2
2
Why does the answer have to be  1 
2?
Assuming that this continued fraction converges, that is what x converges to.
Therefore the continued fraction equals the square root of 2.
The Babylonian Method
Let’s try it with the square root of 2
Have x0 equal 1, our first approximate.
x0  1
S  x0
x n 1
1
S 

  x n 

2
xn 
x1 
1
2 3
1   
2
1 2




13
2  17
x2 




3
2 2  
12
 

2





1  17
2  577
x3 




2 12  17 
408




12




Algebra that Babylonian Method
Converges to the Square Root of Two
1 
2
x 
x 

2
x
x
x 

2
x

2
1
x
x
2
1
2
2x
x
2
2
x 
 x
 2
2
2
 2
The reason why x is positive is
because if the Babylonian
Method converges , the result
is what it will converge to.
If x is negative then the result
is the negative square root of
two.
There is a wonderful blog entry that is focused
on the Babylonian Method and how it works
with the square root of two.
http://johncarlosbaez.wordpress.com/2011/12
/02/babylon-and-the-square-root-of-2/
hi
ki
hi+1/hi
ki+1/ki
What are the explicit formulas for each
the numerator and the denominator of
all rational approximations of the square
root of two?
There is almost a common ratio between
consecutive numerators of the rational
approximations. There is almost a common
ratio between consecutive denominators of
the rational approximations. The ratio is
around 1 + sqrt(2) for both of these
relationships.
Maybe, hi and ki are the sums of two
geometric sequences with a common ratio of 1
+ sqrt(2)
Rational Approximations Numerator
 2   c r   1
h  c 3  2 2   c  r   3
2 h  c 2  2 2   c  2 r   2
h1  c1 1 
2

2
2
1
1
1

c1  1  c 2 r  2 r
2


5 h1  1  c 2 r  2 r

2
2
4
4

2  6 c 2 r  18
2
1
c1  c 2  1
3
2
3

c1  c 2 6 r  r
2
2
3
2  c 2 r  17
6 h 2  c1 18  12
2
5  5 2   c 5 r   5
h  1  c r  2 r 7  5 2   c  r   7
0  c r  2 r  r 


h 4  c1 17  12
2
2
2
2
0  r  2r  1
2
r  1
2

h1  c1 1 
c1  c 2 
1
2 c1 
c1  c 2  0


2  c2 1 
2 c1 
2c2  1
2c2  1

2 1
hi 
1
2
1 
2

i

1
2
1 
2

i
i
hi
Rational Approximations Denominator
k 1  c 1 (1 
2 )  c2r  1
k 2  c1 (3  2 2 )  c 2 r
2
 2
c1 (1 

2 )  c 2 (1 
c1   c 2
c1  c 2 r  2 c 2 r  0
 c 2 (1 
k 3  c1 ( 7  5 2 )  c 2 r
3
5
5 k 1  c1 (5  5 2 )  5 c 2 r  5
2 c1  2 c 2 r  5 c 2 r  0
3
r 1
2
2) 1
2
c2  
c1 
2 c1  2 c 2 r  4 c 2 r  0
2
2 )  c 2 (1 

2 0
 2 2c2  1
2
r  2r  1  0

c1  c 2 3  2 2  2 c 2 1 
2 k 1  c1 ( 2  2 2 )  2 c 2 r  2
2

2) 1
ki 
4
2
4
2
4
1 
2

i

2
4
1 
2

i
i
hi
ki
Connection between Continued Fractions
and Babylonian Method
What happens when the continued fraction approximation is put into the Babylonian Method?
 hi 2

 k i 2

1
4
1
8
1 
2
1 
2
2 k i  
1
hi k i 
2
2
2 hi k i 
hi 
2
4
8



2i
2i
2
1 
2
 2 k i  
2
1
4

2i

1
2
2
2
1  ( hi )  2 ( k i )

2 
hi k i
1
  1 i 
1
1
 1
8

1 
2

4

2i
2
 h2 i

 k
2i

i
4
8

1 
2

1 
2

1
4
1 
2

2i
2
  1 i 
2
2i
1 
1
2
1 
2
4

2i
1 
2
2
1  hi 2 k i  1  ( hi )  2 ( k i )

 


2  ki
h i  2 
hi k i

2i
2
2




2i
1 
2
1
1 

2i
2

2i
This is the Babylonian Method by taking a
continued fraction approximation and
average it with twice its reciprocal.
 k 2i

2i
2
 h2 i
h2 i
k 2i
The result is the continued fraction approximation
of the square root of two, it’s just the number of
steps of the continued fractions approximation is
doubled for the Babylonian Method. You only did
one step of the Babylonian Method.
Conclusion
•We did an origami experiment that finds the same rational
approximations as the continued fractions method.
•We found that the continued fraction and Babylonian Methods
produced the same rational approximations for the square root of
two.
The Blog Entry
http://johncarlosbaez.wordpress.com/2011/12/02/babylon-and-the-square-root-of-2/
Conclusion
•We did an origami experiment that finds the same rational
approximations as the continued fractions method.
•We found that the continued fraction and Babylonian Methods
produced the same rational approximations for the square root of
two.
•Could we do something similar with Fibonacci Numbers and the
Golden Ratio which has the square root of five in it?
The Blog Entry
http://johncarlosbaez.wordpress.com/2011/12/02/babylon-and-the-square-root-of-2/
```

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