Euler's Number Calculator
Explore Euler's number e with Taylor series approximation, continuous compounding, and normal distribution.
Euler's Number (e)
e = 2.718281828459045
常见问题
代码实现
import math
from decimal import Decimal, getcontext
# 1. Taylor series approximation of e
def euler_taylor(n_terms: int) -> float:
"""Approximate e using n terms of Taylor series: sum(1/k!) for k=0..n"""
total = 0.0
factorial = 1
for k in range(n_terms):
if k > 0:
factorial *= k
total += 1 / factorial
return total
# 2. High-precision e using Python's decimal module
def euler_high_precision(decimal_places: int) -> Decimal:
getcontext().prec = decimal_places + 10 # extra guard digits
e = Decimal(0)
factorial = Decimal(1)
for k in range(200): # 200 terms is enough for 100+ decimal places
if k > 0:
factorial *= k
term = Decimal(1) / factorial
e += term
if term < Decimal(10) ** -(decimal_places + 5):
break
return +e # re-apply precision
# 3. Common formulas involving e
def compound_continuous(principal: float, rate: float, years: float) -> float:
"""A = P × e^(r×t) — continuous compounding formula"""
return principal * math.exp(rate * years)
def normal_pdf(x: float, mu: float = 0, sigma: float = 1) -> float:
"""Normal distribution PDF: f(x) = (1/σ√2π) × e^(-(x-µ)²/2σ²)"""
return (1 / (sigma * math.sqrt(2 * math.pi))) * math.exp(-((x - mu)**2) / (2 * sigma**2))
# Examples
print(f"e (Python math): {math.e}")
print(f"e (5 terms): {euler_taylor(5):.8f}")
print(f"e (20 terms): {euler_taylor(20):.15f}")
# Euler's identity: e^(iπ) + 1 = 0
e_to_ipi = math.exp(complex(0, math.pi))
print(f"\ne^(iπ) = {e_to_ipi.real:.10f} + {e_to_ipi.imag:.0f}i")
print(f"e^(iπ) + 1 = {e_to_ipi.real + 1:.2e} (≈ 0, Euler's identity)")
# Continuous compounding
amount = compound_continuous(1000, 0.05, 10)
print(f"\n$1000 at 5% for 10 years (continuous): ${amount:.2f}")
# Normal distribution at x=0
print(f"Normal PDF at x=0: {normal_pdf(0):.6f}")Comments & Feedback
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