TIMOTHY NASRALLA
11:36 AM (2 hours ago)
to me

For 1a
i - continuous
ii - x'(t) = -2x
iii f(z) = -2x
iv FPs: x= 0
Fixed points are found by f(x(t)) = 0
v no SFP's since the derivative of f(z) is greater than 0 


For 1b 
i- discrete time
ii- x(n) = 1/2 x(n-1)
iii - z = 1/2z
iv- FP's are setting the transformation equal to itself, i.e z = 1/2z, and solving for z. The FP of this transformation is 0 = z
v-  z=0 is also a stable fixed point Since the derivative of the transformation is 1/2, which is always less than 1.


For c
I - discrete time
ii - x(n) = 2*x(n-1) * (1-x(n-1)
iii -  z = 2*z*(1-z) or z = 2z- 2z^2
iv- FP: x= 0, x = 1/2
FP's are z - 2z^2 = 0, so z = 0 or z = 1/2
v- SFPs are x = 1/2
derivative of f(z) =  2-4z, at 0 it is not fixed since f(0) prime = 2. At 1/2 it is fixed since f(1/2) = 2

 For d
i- continuous time 
ii- x'(t) =