Question

Please write python code and please answer all parts and separate them into their parts, thank you.​​​​​​

Given the sequence:

with the initial condition Xo chosen in the interval [0,1], and "a" is a given parameter between 0 and4 This sequence is a simple model to describe how a population (or civilization) evolves in a closed system with a finite amount of resources. The parameter "a" describes the rate of development and expansion of the civilization. The population is described by a simple variable x, which quantifies the total population (when x=0 the population is naught, when x=1 the population has reached its maximu m sustainable size) The civilization grows with the combined rate of reproduction (proportional factor x in the sequence) and starvation (term 1-x). Indeed, since the resources are limited the expansion is also proportional to "(1-X)" and if the civilization reaches the maximum population x-1 it will have exhausted all its resources and then collapses (x-0) 1. Write a program by using a "for loop" to compute the convergence point of the sequence for a-1 and a random number between 0 and 1 for Xo. We assume in this problem that the sequence reaches the converge point with n-100 sequence elements, so the convergence point x* can be approximated by the n-100 element of the sequence (xio0) 2. Store in an array the value of "a" and the obtained convergence point x* 3. Add a for loop to repeat the same operation for 50 different random initial conditions. All obtained pairs of data points (a, x*) should be stored in a NumPy array 4. Add one more for loop to repeat the same procedure (1-3) for N-2'000 different rate parameters"a spanning the interval [1.5,4]. This means that "a" needs to be equal to a-1.5 for the first cycle of your additional "for loop", and be equal to"a - 4" for the last cycle of this additional "for loop Plot the result (with the matploblib module) obtained for the convergence points x* (or X100) as a function of the parameter a. Note: if you are out of memory, or Python struggles with the amount of requested information, decrease N-2000 to a smaller number, N-100 or N-1000 5. 6. You should have obtained a graphic of the fix points x of the sequence xn obtained for different growing rate a. Repeat the same procedure (4-5) with a range of a in [3,4] and then [3.5,3.8]

Answer 1

Solution:-

Screenshot of the code for part 1-4

import numpy as np import matplotlib.pyplot as plt ##1 n 100 x0 np. random. uniform() for i in range(n): x a x0(1-x0) print(x) ##2 arr np. zeros (2) arr [0] a arr [1]-x ##3 intial-points 50 n 100 arr1 - np.zeros (shape (intial,2)) for i in range(intial_points): x8 np. random . uniform() for j in range(n): arr1 [i] [0] arr! [1] [1] -a print(arr1) ##4 intial-points 50 alower 1.5 a-upper = 4 n 100 N 2000 a -np.linspace(a_lower,a_upper,N) arr1 - np.zeros (shape- (N intial_points,2)) count 0 for t in range(N): - for i in range(intial_points): x8 np. random . uniform() for j in range(n): arr1 [count] [0] arr1 [count] [1] count + 1 a [t] = x print(arr1)

Screenshot of the code for part 5-6

##5 plt.plot(arr1:,0],arr1[:,1]) plt.show() ##6: intial-points 50 a-upper = 4 n二100 N二2000 a -np.linspace(a_lower,a_upper,N) arr1二np. zeros (shape-(N-intia _points, 2)) count 0 for t in range (N): for i in range(intial_points): x0 np.random.uniform() for j in range(n): arr1 [count] [0] arr1 [count] [1] count += 1 a [t] = x plt.plot(arr1:,0],arr1[:,1]) plt.show) intial-points 50 a-lower 3.5 a-upper 3.8 n二100 N二2000 a -np.linspace(a_lower,a_upper,N) arr1二np. zeros (shape-(N-intia _points, 2)) count 0 for t in range (N): for i in range(intial_points): x0 np.random.uniform() for j in range(n): arr1 [count] [0] arr1 [count] [1] count += 1 a [t] = x plt.plot(arr1:,0],arr1[:,1]) plt.show)

Output of part 5

10 0.8 0.6 0.4 0.2 0.0 15 2.0 2.5 3.0 3.5 4.0

Output of part 6

ї0 0.8 0.6 0.4 0.2 0.0 3.0 3.2 3.4 3.6 3.8 4.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 3.50 3.55 360 365 3.70 375 3.80

Code in text format...!!

import numpy as np
import matplotlib.pyplot as plt

# # 1

a = 1
n = 100
x0 = np.random.uniform()
for i in range(n):
x = a * x0 * (1-x0)
x0 = x
print(x)

# # 2
arr = np.zeros(2)
arr[0] = a
arr[1] = x

# # 3
intial_points = 50
a = 1
n = 100
arr1 = np.zeros(shape=(intial,2))
for i in range(intial_points):
x0 = np.random.uniform()
for j in range(n):
x = a * x0 * (1-x0)
x0 = x
arr1[i][0] = a
arr1[i][1] = x
print(arr1)

# # 4
intial_points = 50
a_lower = 1.5
a_upper = 4
n = 100
N = 2000
a = np.linspace(a_lower,a_upper,N)
arr1 = np.zeros(shape=(N*intial_points,2))
count = 0
for t in range(N):
for i in range(intial_points):
x0 = np.random.uniform()
for j in range(n):
x = a[t] * x0 * (1-x0)
x0 = x
arr1[count][0] = a[t]
arr1[count][1] = x
count += 1
print(arr1)

# # 5

plt.plot(arr1[:,0],arr1[:,1])
plt.show()

# # 6:

intial_points = 50
a_lower = 3
a_upper = 4
n = 100
N = 2000
a = np.linspace(a_lower,a_upper,N)
arr1 = np.zeros(shape=(N*intial_points,2))
count = 0
for t in range(N):
for i in range(intial_points):
x0 = np.random.uniform()
for j in range(n):
x = a[t] * x0 * (1-x0)
x0 = x
arr1[count][0] = a[t]
arr1[count][1] = x
count += 1
plt.plot(arr1[:,0],arr1[:,1])
plt.show()
  
intial_points = 50
a_lower = 3.5
a_upper = 3.8
n = 100
N = 2000
a = np.linspace(a_lower,a_upper,N)
arr1 = np.zeros(shape=(N*intial_points,2))
count = 0
for t in range(N):
for i in range(intial_points):
x0 = np.random.uniform()
for j in range(n):
x = a[t] * x0 * (1-x0)
x0 = x
arr1[count][0] = a[t]
arr1[count][1] = x
count += 1
plt.plot(arr1[:,0],arr1[:,1])
plt.show()