Mudassir Lodi
Homework for Lecture 12 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. 18,, 2021. Subject: hw12 with an attachment hw12FirstLast.pdf and/or hw12FirstLast.txt
Also please indicate (EITHER way) whether it is OK to post

1. For each of the following two (single variable) functions find all the fixed points, and for each of them, decide whether they are stable fixed points.
(i)
x → x3− 6x2 + 12x – 6
x = 1, 2, 3 are the fixed points. x = 2 is the only stable fixed point.  
(ii)
x = 1/3, ½, -1/3, -1/2 are the fixed points. There are no stable fixed points. 

2.	Find the linearizations of the given functions at the designated points, and compare the exact value with the approximate values given by the linearization.
√ 
(i): f(x,y) =	x + 4y at (1,2). The values at (0.95,1.02).
(1.6683 + 1/2sqrt(1+4y), 1.8904)
(ii): f(x,y,z) = x3y4z5 at (1,1,1). The values at (1.01,1.02,0.99).
√ 
(3y^4z^5 + 4.12z^5 + 6.576, 7.3685z^5 + 6.576, 13.5835)

(iii): f(x1,x2,x3,x4) =	x1 + x2 + x3 + x4, at (1,1,1,1). The values at (1.01,1.01,0.99,0.99).
2.0000

3.	What is the Jacobian matrix (not to be confused with the Jacobian determinant) of the following transformation
	 	,
at the point (1,1).
{-x/(1+y)^2, 1/(1+x)}

4.	What is the Jacobian matrix (not to be confused with the Jacobian determinant) of the following transformation
	(x,y,z) → (x + y + z , x2 + y2 + z2 , x3 + y3 + z3 )	,
at the point (1,1,1).

{{1,0,0}->{1,2x,3x^2}, {0,1,0}->{1,2y,3y^2}, {0,0,1}->{1,2z,3z^2}}

5.	In your words explain why it makes sense that a fixed point (x0,y0) of a transformation (x,y) → (f(x,y),g(x,y)), i.e. a point in R2 such that x0 = f(x0,y0) and y0 = g(x0,y0) is a stable fixed point if its Jacobian matrix , at that point, i.e.
the 2 × 2 numerical matrix
	 	,
has all its eigenvalues with absolute value less than 1.

Eigenvalues represent a linear scalar of values associated with a particular system. Because a Jacobian matrix also represents vectors, it can be assumed that eigenvalues with an absolute value of less than 1 in these matrices are stable fixed points.