import numpy as np import matplotlib.pyplot as plt from mpl_toolkits.mplot3d import Axes3D

Question 4

def randomDirectionThree(): array=[0,0,0] number = np.random.randint(1, 7) if number == 1: array = [1,0,0] return array if number == 2: array = [-1,0,0] return array if number == 3: array = [0,1,0] return array if number == 4: array = [0,-1,0] return array if number == 5: array = [0,0,1] return array if number == 6: array = [0,0,-1] return array def polymer(n): array = np.zeros((n,3)) for i in range(n): array[i] = array[i-1] + randomDirectionThree() return array def coords(n): P = polymer(n) x = np.zeros(n) for i in range(n): x[i] = P[i][0] y = np.zeros(n) for i in range(n): y[i] = P[i][1] z = np.zeros(n) for i in range(n): z[i] = P[i][2] return x,y,z fig = plt.figure() ax = fig.gca(projection='3d') ax.plot(coords(100)[0], coords(100)[1], coords(100)[2]) ax.plot(coords(100)[0], coords(100)[1], coords(100)[2]) ax.plot(coords(100)[0], coords(100)[1], coords(100)[2]) ax.plot(coords(100)[0], coords(100)[1], coords(100)[2]) plt.show()

/shared-libs/python3.7/py-core/lib/python3.7/site-packages/ipykernel_launcher.py:47: MatplotlibDeprecationWarning: Calling gca() with keyword arguments was deprecated in Matplotlib 3.4. Starting two minor releases later, gca() will take no keyword arguments. The gca() function should only be used to get the current axes, or if no axes exist, create new axes with default keyword arguments. To create a new axes with non-default arguments, use plt.axes() or plt.subplot().

The polymers look pretty globular. They all have 100 monomers and we see that they are never really further than 10 monomer units from the origin.

def randomDirectionTwo(): array=[0,0] number = np.random.randint(1, 5) if number == 1: array = [1,0] return array if number == 2: array = [-1,0] return array if number == 3: array = [0,1] return array if number == 4: array = [0,-1] return array def polymer2(n): array = np.zeros((n,2)) for i in range(n): array[i] = array[i-1] + randomDirectionTwo() return array def endToEndVector2(n): vector = polymer2(n)[-1] return vector #gives the mean end to end vecotor of a sample of n end to end vectors def sampleMeanPolymer(n): meanArray = np.zeros((n,2)) for i in range(n): meanArray[i] = endToEndVector2(500) firstTerm = np.zeros(n) for i in range(n): firstTerm[i] = meanArray[i][0] secondTerm = np.zeros(n) for i in range(n): secondTerm[i] = meanArray[i][1] meanPolymer= [np.sum(firstTerm)/n , np.sum(secondTerm)/n] return meanPolymer def numberOfZeros(n): data = np.zeros((n,2)) for i in range(n): data[i] = endToEndVector2(500) number = 0 for i in range(n): if data[i][0] == 0 and data[i][1]==0: number = number+1 return number print('The mean total displacement of 10 polymers of 500 monomers is ', sampleMeanPolymer(10)) print('The mean total displacement of 100 polymers of 500 monomers is ', sampleMeanPolymer(100)) print('The mean total displacement of 1000 polymers of 500 monomers is ',sampleMeanPolymer(1000)) print('The number of polymers out of 1000 polymers of 500 monomers that are equal to the mean value, [0,0], are ', numberOfZeros(1000))

The mean total displacement of 10 polymers of 500 monomers is  [9.7, -4.7]
The mean total displacement of 100 polymers of 500 monomers  is  [-2.21, -1.85]
The mean total displacement of 1000 polymers of 500 monomers  is  [0.682, -0.264]
The number of polymers out of 1000 polymers of 500 monomers that are equal to the mean value, [0,0], are  3

I assumed here that when it asked how many simulations actually met the mean value you wanted us to use the mean value we calculated in 3 not the experimental ones we have above because the elements of the experimental ones are not even integers.

c) The root mean squared end to end distance is a better representation of length than the average end to end vector because of the symmetry of the end to end vectors. This causes the average to be zero even tho practically no polymer has end to end distance 0. Thus it is better to use the root mean squared end to end distance as a measure of the extent of a polymer because it will give us something similar to the length of a polymer with respect to distance from the origin. Then because the length is always positive we get a distribution of the extent of each polymer and dont get any cancelations from negative terms. We also saw above that practically no polymers actually meet the mean value so it is still quite unlikely.

def endToEndVector3(n): vector = polymer(n)[-1] return vector def dotProduct(n): R = endToEndVector3(n) return np.dot(R, R) #averageDot gives the average value of 1000 dot products between two random 3d polymers of length n def averageDot(n): average = np.zeros(1000) for i in range(1000): average[i] = dotProduct(n) return np.sum(average)/1000 def rootMean(n): return np.sqrt(averageDot(n)) print('the root mean squared end to end distance for 1000 polymers of length 100 is ', rootMean(100)) print('the root mean squared end to end distance for 1000 polymers of length 500 is ', rootMean(500)) print('the root mean squared end to end distance for 1000 polymers of length 1000 is ', rootMean(1000))

the root mean squared end to end distance for 1000 polymers of length 100 is  9.993097617856037
the root mean squared end to end distance for 1000 polymers of length 500 is  22.911656422004935
the root mean squared end to end distance for 1000 polymers of length 1000 is  31.909810403698735

This answer is consistent with the scaling we calculated in 3d! because all of our values are quite close to the square root of length. The square roots of the different lengths are , 10, ~22.36, ~31.62 which are all very similar to our values.