The number *e* is a famous irrational number called Euler’s number after Leonhard Euler a Swiss Mathematician (1707 – 1783). Number *e* is considered to be one of the most important numbers in mathematics.

The first few digits are: 2.7182818284590452353602874713527… It has an infinite number of digits with no recurring pattern. It cannot be written as a simple fraction.

Number *e* is the limit of (1 + ^{1}/_{n})^{n} as n approaches infinity:

Number *e* is a mathematical constant that is the base of the natural logarithm: the unique number whose natural logarithm is equal to one. Find out more on https://en.wikipedia.org/wiki/E_(mathematical_constant).

In this challenge we will use a Python script to calculate an approximation of *e* using three different approaches. The first approach as described above as already been implemented in the above Python trinket.

#### Calculating *e* using an iterative approach

Euler’s number,

*e*, can also be calculated as the sum of the infinite series:

Complete your Python script to implement this infinite series using an iterative approach. (Method 2)

#### Calculating *e* using a continuous fraction (iterative approach)

A less common approach to calculate number

*e*is to use a continued fraction based on the following sequence:

Continued Fraction:

Complete your Python script to implement this continued fraction. (Method 3)

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#### Solution...

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#### Solution...

You are viewing this solution as part of your full membership subscription!#### Python Code

#Calculating an approximation of Euler's Number #Method 1: e = limit of (1 + 1/n)^n as n approaches infinity print("Method 1: e = limit of (1 + 1/n)^n as n approaches infinity") n=10000000 e = (1+1/n)**n print(e) #Method 2: Using an infinite series (Iterative approach) print("\nMethod 2: Using an infinite series (Iterative approach)") e = 1 quotient = 1 for i in range(1,100): quotient = quotient * i e = e + 1/quotient print(e) #Method 3: Using a continued fraction (Iterative approach) print("\nMethod 3: Using a continued fraction (Iterative approach)") n = 100 # number of iterations e = 0 for i in range(n, 0, -1): if i % 3 == 1: j = 2 * int(i / 3) else: j = 1 e = 1.0 / (e + j) e = e+1 print(e)