Numerical test of \(e^{-x} \approx (1 -\frac{x}{N})^N\) for large \(N\)#

In this notebook we test:

  1. For what \(x\) does this work?

  2. How big does \(N\) have to be?

Plan: calculate the error for a range of \(x\) and \(N\) and make a table.

Standard Python imports plus seaborn (to make plots looks nicer).

import numpy as np
import scipy.linalg as la

import matplotlib.pyplot as plt
import seaborn as sns; sns.set_style("darkgrid"); sns.set_context("talk")

Set up the \(x\) and \(N\) arrays.

delta_x = 0.1  # spacing of x points
x_values = np.arange(0, 2+delta_x, delta_x)  # Mesh points rom 0 to 2 inclusive.
N_values = [10, 100]  # You can add more to this list

print(x_values)  # just to check they are what we want

print(N_values)

Write functions to evaluate the approximation and for relative errors

def rel_error(x1, x2):
    """
    Calculate the (absolute value of the) relative error between x1 and x2. 
    Notice that we use the average for the denominator.
    """
    return np.abs( (x1 - x2) / ((x1 + x2)/2) )

def exp_approx(z, N):
    """
    Calculate (1 + z/N)^N
    """
    return (1 + z/N)**N

Step through \(x\) array and for each \(x\) step through \(N\), making a table of results

print(' x   exp(-x)  N: ', end=" ")   # The end=" " option suppresses a return.
for N in N_values:
   print(f'  {N}     ', end=" ")
print('\n')

for x in x_values:
    f_of_x = np.exp(-x)
    print(f"{x:.1f}   {f_of_x:.3f}   ", end=" ")
    for N in N_values:
        approx = exp_approx(-x, N)
        print(f"  {rel_error(f_of_x, approx):.2e}", end=" ")
    print(" ")

Things to do#

  1. How does the relative error scales with \(N\)? (To be sure, add additional \(N\) values to N_values.) Explain the observed scaling.

  2. Investigate how well the approximation works for \(x > 2\) by increasing the range of \(x\). Describe the behavior.