Answered Essay: Exponential function. The exponential function may be approximated by Taylor polynomials e

In PProblem 1: Exponential function. The exponential function may be approximated by Taylor polyno- mials 1n n=0 The sum may be computed directly term by term using AlgorithmI input: x t=x for n-1 to N{ S-s+t t-t*x/n return S r using the identity, Algorithm II input: x N-1N for n-N to 1 { s-s*x/n+1 return S Use these algorithms to compute e50 and e-50. Assume that the built-in function in your environment (likely called exp()) provides the exact value and plot the the error of your numerical approxima tions, i.e., |PN(50) et, as the function of N. Values of N a 200 are probably as far as you need to go. (You should use the logarithmic plot to be able to display the entire range of error values in one figure, i.e., plot the logarithm of the error rather than the error itself.) If your environment uses very high precision arithmetic, you may not notice any significant difference between the algorithms, the error should become more noticeable if you use just single precision (32 bit) arithmetic. Discuss your findingsython

Exponential function. The exponential function may be approximated by Taylor polynomials e^x almostequalto P_N(x) = sigma^N_n = 0 x^n/n!. The sum may be computed directly term by term using Algorithm I input: x s = 1.0 t = x for n = 1 to N{ s = s + t t = t * x/n } return s Or using the identity, P_N(x) = 1 + x(x/2(1 + x/2(1 + x/3(1 + x/N – 1(1 + x/N)…))): Algorithm II input: x s = 1.0 for n = N to 1 { s = s*x/n+1 } return s Use these algorithms to compute e^50 and e^-50. Assume that the built-in function in your environment (likely called “exp()”) provides the exact value and plot the error of your numerical approximations, i.e., |P_N(50) – e^plusminus 50|, as the function of N. Values of N almostequalto 200 are probably as far as you need to go. (You should use the logarithmic plot to be able to display the entire range of error values in one figure, i.e., plot the logarithm of the error rather than the error itself.) If your environment uses very high precision arithmetic, you may not notice any significant difference between the algorithms, the error should become more noticeable if you use just single precision (32 bit) arithmetic. Discuss your findings.

Expert Answer

 

import math

def myexp(x,n):
e=0
for i in range(0,n):
e=e+(x**i)/math.factorial(i)
return e
lst=[]
for i in range(1,200):
error=abs(myexp(50,i)-math.exp(50))
lst.append(math.log(error))

import matplotlib.pyplot as plt
plt.plot(lst)
plt.ylabel(‘errors’)
plt.show()

plot:

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