Mudassir Lodi
Homework for Lecture 9 of Dr. Z.’s Dynamical Models in Biology class
Email the answers (either as .pdf file or .txt file) to ShaloshBEkhad@gmail.com by 8:00pm Monday, Oct. 4,, 2021. Subject: hw9 with an attachment hw9FirstLast.pdf and/or hw9FirstLast.txt
Also please indicate (EITHER way) whether it is OK to post

1. Use BOTH procedure Orb and SPF
https://sites.math.rutgers.edu/~zeilberg/Bio21/M9.txt
To find the stable fixed points of the following non-linear recurrencnes
xn = f(xn−1)
for the following f(x)
(i) 2x(1−x)	,	(ii) (2.5)x(1−x) ,	(iii) (3,1)x(1−x) ,	( iv)	,	( v)	,	( vi)
.
Orb(f,x,
SPF(f,x)
i)	f’(x) = 2 – 4x
f’(0) = -2 (unstable)
ii)	f’(x) = 2.5 – 5x
f’(0) = -2.5 (unstable)
iii)	f’(x) = 3.1 – 6.2x
f’(0) = -3.1 (unstable)
iv)	f’(x) = -1/(x+3)^2
f’(0) = -1/9 (stable)
v)	f’(x) = 1/(x+4)^2
f’(0) = 1/16 (stable)
vi)	f’(x) = -(x^2 + 4x -1)/(2x^2 + x + 4)^2
f’(0) = 1/16 (stable)

2.	Consider the discrete dynamical system , where a and b are positive numbers.
Note that FP and SFP will not work if a and b are left as symbols. You are welcome to use the solve command to find explicit expressions, in terms of a and b of the two fixed points.
Then you are welcome to use diff command to find an expression C(a,b), such that the following is true
 , where a and b are positive numbers, has a stable fixed point if and only if
	−1 < C(a,b) < 1	.
Then experiment with a = 1,b = 2, a = 2,b = 3, and a = 12,b = 17 to see whether the condition is met, and check your answers against the outputs of Orb, FP and SFP for these numerical values of a and b.
i)	f(x) = x+1/x+2
f’(x) = 1/(x+2)^2
f’(0) = 1/4
ii)	x+2/x+3
f’(x) = 1/(x+3)^2
f’(0) = 1/9
iii)	x+12/x+17
f’(x) = 5/(x+17)^2
f’(0) = 5/289
The condition is met for x = 0 for all three of the a,b functions. 

3.	For an arbitrary k (between 1 and 4) find the equilibrium points of
	x → k x(1 − x)	.
Prove that x = 0 is never stable, but that the other one is sometimes stable. What values of k is that other point fixed point stable? Use this to find the first bifurcation point when it switches from very one ultimate population value to going back-and-forth between two population values. 
This function can be simplified to f(x) = kx – kx^2. The derivative f’(x) would be k – 2kx. Calculating x = 0 in this function would always result in k, which therefore would be unstable (if k is greater than 1). Any value of k between -1 and 1, therefore, would result in a stable fixed point. 

4. For an arbitrary k (between 1 and 4) find the equilibrium points of x → f(f(x))
	f(x) = kx(1 − x)	.
you are welcome to use the Maple solve command to get explicit expressions in k for these four fixed points. Two of them are obviously the same as the fixed points of x → kx(1 − x), and you already know that they are not stable.
Find a condition in k for the two new fixed points (of x → f(f(x)) to be stable fixed points.
Use it to find exactly the second bifurcation points, when the population stops converging to a period 2 orbit, and starts going into a period 4 orbit.
Verify this theoretical prediction using Orb with x0 = 0.5.
f(f(x)) = k(kx - kx^2))(1-(kx – kx^2)) = (k^2x – k^2x^2)(1 – kx - kx^2) 
f(f(x)) = (k^2x – k^2x^2) + (-k^2x^2 + k^3x^3) – (k^3x^3 + k^3x^4)
f(f(x)) = (k^2x – 2k^2x^2 + k^3x^4)
f’(f(x)) = (k^2 – 4k^2x + 4k^3x^3)
f’(f(x)) = k – 4kx + 4kx^3
k would need to be between 1 and -1 in order for x = 0 to be a stable fixed point. 

5. Read and understand (as much as possible) Mitchell Feigenbaum’s seminal article:
https://sites.math.rutgers.edu/~zeilberg/Bio21/MF78.pdf .