The math formula for expm1() function is e ** x -1
import math
print(math.e ** -34.11 -1 ) # -0.9999999999999984
e : is a numerical constant of value equal to 2.71828
This is the base of natural logarithm and known as Euler's number.
Additional Explanation and Use Cases for expm1()
The math.expm1(x) function calculates ( e^x - 1 ) accurately for small values of x, avoiding precision errors associated with directly computing `math.exp(x) - 1`.
Example 1: Why Use expm1() for Small Values of x
For very small values of x, using `expm1()` is more precise than `exp(x) - 1`.
import math
x = 1e-10
print(math.exp(x) - 1) # Less accurate due to floating-point rounding
print(math.expm1(x)) # More accurate for small x
Example 2: Comparing exp() and expm1()
Comparing results from `exp()` and `expm1()` for different values of x.
import math
x_values = [1, 0.1, 1e-10, -1]
for x in x_values:
print(f"x: {x}, exp(x) - 1: {math.exp(x) - 1}, expm1(x): {math.expm1(x)}")
Use `expm1()` for financial computations where precision is essential, like calculating continuous interest.
import math
principal = 1000 # Principal amount
rate = 0.05 # Annual interest rate
time = 1 # 1 year
# Continuous compound interest
interest = principal * math.expm1(rate * time)
print("Accrued interest:", interest)
Output
Accrued interest: 51.271096376024055
Applications of expm1()
Mathematical Computations: In scientific computing where small x values are common.
Continuous Compounding in Finance: Useful for precise financial calculations.
Physics and Engineering: Applicable in exponential decay/growth calculations.
These enhancements offer practical insights into `expm1()` and show when itโs preferable over `exp() - 1` for computational accuracy.
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