Program to find ex in an efficient way in Python

The mathematical constant e raised to the power x (e^x) can be calculated efficiently using the Taylor series expansion without library functions. The formula for e^x is:

e^x = 1 + x + x²/2! + x³/3! + x?/4! + ...

For example, if x = 5, then e⁵ ? 148.4131 because e⁵ = 1 + 5 + (5²/2!) + (5³/3!) + ... = 148.4131...

Algorithm

To solve this efficiently, we follow these steps ?

• Initialize factorial = 1, result = 1, and numerator = x
• Set n = 20 (number of terms for precision)
• For each term from 1 to n:
  • Add (numerator/factorial) to result
  • Update numerator = numerator × x
  • Update factorial = factorial × (i+1)
• Return the calculated result

Implementation

def calculate_exp(x):
    factorial = 1
    result = 1
    n = 20  # Number of terms for precision
    numerator = x
    
    for i in range(1, n):
        result += numerator / factorial
        numerator = numerator * x
        factorial = factorial * (i + 1)
    
    return result

# Test the function
x = 5
exp_value = calculate_exp(x)
print(f"e^{x} = {exp_value:.4f}")

# Compare with different values
test_values = [1, 2, 3, 5]
for val in test_values:
    result = calculate_exp(val)
    print(f"e^{val} = {result:.4f}")

e^5 = 148.4132
e^1 = 2.7183
e^2 = 7.3891
e^3 = 20.0855
e^5 = 148.4132

How It Works

The algorithm builds each term of the Taylor series incrementally. Instead of recalculating xⁱ and i! from scratch for each term, it updates the numerator and factorial efficiently ?

• Term 1: numerator = x, factorial = 1
• Term 2: numerator = x², factorial = 2
• Term 3: numerator = x³, factorial = 6
• And so on...

Precision Control

def calculate_exp_precision(x, terms=20):
    factorial = 1
    result = 1
    numerator = x
    
    for i in range(1, terms):
        result += numerator / factorial
        numerator = numerator * x
        factorial = factorial * (i + 1)
    
    return result

# Test with different precision levels
x = 2
for terms in [10, 15, 20, 25]:
    result = calculate_exp_precision(x, terms)
    print(f"e^{x} with {terms} terms = {result:.6f}")

e^2 with 10 terms = 7.389056
e^2 with 15 terms = 7.389056
e^2 with 20 terms = 7.389056
e^2 with 25 terms = 7.389056

Conclusion

The Taylor series method efficiently calculates e^x by incrementally building terms instead of recalculating powers and factorials. Using 20 terms provides sufficient precision for most applications, with a time complexity of O(n).