Question

Using a python program find the numerical derivative at the indicated point, using the backward, forward, and centered formula. Use ℎ=0.05h=0.05, then each case compute the error. Which one is more accurate?

1. ??ex at ?=0x=0.
2. tan−1(?2−?+1)tan−1⁡(x2−x+1) at ?=1x=1.
3. tan−1(100?2−199?+100)tan−1⁡(100x2−199x+100) at ?=1x=1.

Using a python program find the Monte Carlo integration, compute the value of π.

Answer 1

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# this is the template of our weight function g(x)

def g_of_x(x, A, lamda):

e = 2.71828

return A*math.pow(e, -1*lamda*x)

def inverse_G_of_r(r, lamda):

return (-1 * math.log(float(r)))/lamda

def get_IS_variance(lamda, num_samples):

"""

This function calculates the variance if a Monte Carlo

using importance sampling.

Args:

- lamda (float) : lamdba value of g(x) being tested

Return:

- Variance

"""

A = lamda

int_max = 5

  

# get sum of squares

running_total = 0

for i in range(num_samples):

x = get_rand_number(0, int_max)

running_total += (f_of_x(x)/g_of_x(x, A, lamda))**2

  

sum_of_sqs = running_total / num_samples

  

# get squared average

running_total = 0

for i in range(num_samples):

x = get_rand_number(0, int_max)

running_total += f_of_x(x)/g_of_x(x, A, lamda)

sq_ave = (running_total/num_samples)**2

  

  

return sum_of_sqs - sq_ave

# get variance as a function of lambda by testing many

# different lambdas

test_lamdas = [i*0.05 for i in range(1, 61)]

variances = []

for i, lamda in enumerate(test_lamdas):

print(f"lambda {i+1}/{len(test_lamdas)}: {lamda}")

A = lamda

variances.append(get_IS_variance(lamda, 10000))

clear_output(wait=True)

  

optimal_lamda = test_lamdas[np.argmin(np.asarray(variances))]

IS_variance = variances[np.argmin(np.asarray(variances))]

print(f"Optimal Lambda: {optimal_lamda}")

print(f"Optimal Variance: {IS_variance}")

print(f"Error: {(IS_variance/10000)**0.5}")