History
Direct translation version:
How Many Pairs of Rabbits Are Created by One Pair in One Year?
A certain man had one pair of rabbits together in a certain enclosed place, and one wishes to know how many are created from the pair in one year when it is the nature of them in a single month to bear another pair, and in the second month those born to bear also.
Solution:
Because the above written pair in the first month bore, you will double it; there will be two pairs in one month. One of these, namely the first, bears in the second month, and thus there are in the second month 3 pairs; of these in one month two are pregnant, and in the third month 2 pairs of rabbits are born, and thus there are 5 pairs in the month; in this month 3 pairs are pregnant, and in the fourth month there are 8 pairs, of which 5 pairs bear another 5 pairs; these are added to the 8 pairs making 13 pairs in the fifth month; these 5 pairs that are born in this month do not mate in this month, but another 8 pairs are pregnant, and thus there are in the sixth month 21 pairs; [p284] to these are added the 13 pairs that are born in the seventh month; there will be 34 pairs in this month; to this are added the 21 pairs that are born in the eighth month; there will be 55 pairs in this month; to these are added the 34 pairs that are born in the ninth month; there will be 89 pairs in this month; to these are added again the 55 pairs that are born in the tenth month; there will be 144 pairs in this month; to these are added again the 89 pairs that are born in the eleventh month; there will be 233 pairs in this month.
Month  State  Count 

0  j  1 
1  a  1 
2  aj  2 
3  aj a  3 
4  aj a aj  5 
5  aj a aj aj a  8 
6  aj a aj aj a aj a aj  13 
7  aj a aj aj a aj a aj aj a aj aj a  21 
These are getting very long, so let's introduce a shorter notation using exponents to do the counting.
Polynomial rewrite system
j > a, a > aj, n > n+1
/ y x \ x + y y
G\a j + n/ = a j + n + 1
x(n+1) = y(n)
y(n+1) = y(n)+x(n)
F(n) = y(n) + x(n)
F(n+1) = F(n) + y(n)
F(n+2) = F(n+1) + y(n+1) = F(n+1) + y(n) + x(n) = F(n+1) + F(n)
++
 Governing equation 
++
 
 F(n+2) = F(n+1) + F(n) 
 
++
Solution....
Depending on whether F_1 = 1 or F_1 = 2, ...
t t
/ ___\ / ___ \ / ___ \ / ___ \
1 \/ 5  \/ 5 1  \/ 5 1  \/ 5 1
 +   +  +   +    + 
\2 2 / \ 10 2/ \ 2 2/ \ 10 2/
or
t t
/ ___\ / ___\ / ___ \ / ___ \
1 3 \/ 5  1 \/ 5   \/ 5 1  3 \/ 5 1
 +   +  +   +    + 
\2 10 / \2 2 / \ 2 2/ \ 10 2/
Note: Our sequence is integers, but the solution involves the \(\sqrt{5}\)!
GOLDEN RATIO: \(\phi:= (1+\sqrt{5})/2 \approx 1.618033\) \[1/\phi = ( \sqrt{5}1 )/2\] \[a \phi^t + b (\phi)^{t}\] \[F(t+1) / F(t) \rightarrow \phi\]
Classic case  selfsimilarity of a divided line segment
1 x
+++ (1/x) = x/(1+x)
Inscribed a Pentagon in the diagonals of a larger pentagon. Construct a rhombus using diagonals. Use to show similar triangles. Let z be the edge of the pentagon, y be the inner pentagon edge \(x/z = y/x\) (similar triangles) , \(z = x+y\) (rhombus)
Can be used to calculate \(2 cos(\pi/5) = \phi\).
Golden Rectangle
1 = a + b, a = b + c, (1+a)/1 = (b+c)/b
(1+a)*(1a) = a
a = 1/PHI
Difference equations can be much more complicated that just these matrix equations. For instance, ...
Mendel's law for 2allelle diploid inheritance
2
⎛ y(t)⎞
x(t + 1) = ⎜x(t) + ────⎟
⎝ 2 ⎠
⎛y(t) ⎞
y(t + 1) = (2⋅x(t) + y(t))⋅⎜──── + z(t)⎟
⎝ 2 ⎠
2
⎛y(t) ⎞
z(t + 1) = ⎜──── + z(t)⎟
⎝ 2 ⎠
Mandelbrot set equation
2
z(t+1) = z(t) + c
One of the useful things we would like to know is what can happen to the solutions as we iterate equations over time...
Linear equation x(t+1) = b x(t) + a
x = a / ( 1  b ) { Steadystate solution }
x(t+1) = b x(t) + a
x(t) = a/(1b) + ( x(0)  a/(1b) ) b**t
Example 1
x(t+1) = 2 + x(t)/2
x(t) = 4 + C 2**t
Example 2
x(t+1) = 3 + 2 x(t)
x(t) = 3 + C*2**t
Cobwebbing  example 1 converges  example 2 diverges
\[x(t+1) = r x(t) ( 1  x(t) )\]
Steadystate condition and local stability: \(x \in 0, 1  1/r\) solve the steadystate equation \(x = f(x)\). In general, a solution of this equation will be locally stable if \(1 < f'(x) < 1\).
Basin of atraction and global stability ...
a highquality bifurcation plot for the discrete logistic equation.
Discuss Smale's horseshoe, as exemplified by cresant dough folding and rolling. Shown iterates of the function to exemplify folding process. Talk about fixed points of period 1, 2, 3, 4 and their local stability from figure.
Also discuss supertracks and windowing within the bifurcation diagram.
Local Stability: If we perturb things off the equilibrium a little, do we stay there or go back to were we started?
+++
  
 b < 1  Oscillatory instability and divergence 
  
+++
  
 b = 1  Oscillatory stability 
  
+++
  
 1 < b < 0  Oscillatory convergence 
  
+++
  
 b = 0  Instant convergence 
  
+++
  
 0 < b < 1  Monotone convergence 
  
+++
  
 1 = b, a < 0  No fixed points, incremental decrease 
  
+++
  
 1 = b, a = 0  All points stationary 
  
+++
  
 1 = b, a > 0  No fixed points, incremental increase 
  
+++
  
 1 < b  Monotone divergence 
  
+++
x(t+1) = r x(t) ( 1  x(t) )
x in 0, 1  1/r
x(t+1) = f(x_t)
1 < f'(x) < 1
x = 0 is stable for r < 1
x = 1  1/r is stable for 1 < r < 3
r = 3 introduces oscillator instability
implement in python and show numerical experiments
iterated function maps ... period2 solutions period3 solutions period4 solutions
Smale's model of folding dough, and sensitivity to initial conditions while still remaining finite.
Birfurcation diagram
Mandelbrot = \({ c : z(0) = 0, z(t+1) = z(t)^2 + c, z(\infty) < \infty }\)
The Kolmogorov complexity of a string (or object) in a language is equal to the length of the shortest representation for the object in the language.
Suppose our language represents numbers in terms of decimals.
* one = 1
* half = 1/2 = 0.5
* 1/3 = 0.3333....
So if we are confined to decimal representations, 1/3 has infinite complexity! But we can extend the language to identify repeating components... Then 1/3 = 0.3, and again the represention is short and quick.
* sqrt(2) = 1.4142135623730951....
Now, we have infinite, nonrepeating decimals ... But suppose we use geometry or algebra?
* sqrt(2) = length of the diagonal of a unit square
* sqrt(2) = x : x**2  2 = 0
* sqrt(2) : Iterate y/2 + 1/y until convergence
Transcendental numbers = not algebraic, no polynomial...
* pi = limit( { (x,y) in Z2 : x**2 + y**2 < n**2}/n**2, n > oo)
Most objects have complexity similar to their length.
One interpretation: modelling is working to find simple ways to express the important features of complicated things.
Use a rewrite system to efficiently express the Kochy curve
Use a rewrite system to efficiently express the tree
of the previous example
Use a rewrite system to represent a plant's geometry
(Example A from ABOPs)

a ba a ba ba a ba a ba ba a ba ba a ba a ba ba a ba a b

abaababaabaab

abaab

ab

( > ,  > )**8 @ 

ba ba ba a bbbbba bba a a bbbba bba bba bbb
39 
24 
15 
9 
6 
(f++f++f++) @subs ( f > ff++ff )**oo
Sierpinski's Triangle gasket ... a rewrite rule in two dimensions, or one dimension
A @subs (A > BAB, B > A+B+A)**n
Dragon curve
fx @subs (x > x+yf+, y > fxy)**20
The fern  make the whole from parts similar to the pieces... Rewrite rules for constructing selfsimilar fractals... Fibonacci without rearrangement
(language depends on the machine which will interpret it)
 Does the language need a stack?
Turtle language:
A square
A 5pointed star
Simple triangle Kochy curve, 2 iterations
Introduce stack notation "[" = push, "]" = pop
A simple tree
\  /
\ \/ /
\  /
\  /
\/


One interpretation: modelling is working to find simple ways to express the important features of complicated things.