Maclaurin Series for e^x
Taylor series of the exponential at 0.
Formula
Variables
Example
e^1 = 1 + 1 + 1/2 + 1/6 + 1/24 + … ≈ 2.718.
How it works
This identity says that the exponential function can be rebuilt exactly from an infinite polynomial-like sum centered at x = 0: 1 + x + x^2/2! + x^3/3! + .... It comes from the Taylor series formula, because every derivative of e^x is again e^x, and at x = 0 all derivatives equal 1, so the coefficients become 1/n!. The factorials grow quickly, which makes the series converge for every real or complex x, not just near 0.
When to use it
Use this when you need to approximate e^x numerically, especially for small or moderate x, or when you want to analyze exponential behavior with algebraic tools. It is also essential in calculus, differential equations, and any setting where exponentials are expanded, linearized, or compared term-by-term.
Worked example, step by step
Approximate e^0.5 using the first five terms of the Maclaurin series. Start with e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + ... Substitute x = 0.5: e^0.5 ≈ 1 + 0.5 + (0.5^2)/2 + (0.5^3)/6 + (0.5^4)/24 Compute each term: 1 = 1 0.5 = 0.5 0.5^2/2 = 0.25/2 = 0.125 0.5^3/6 = 0.125/6 ≈ 0.0208333 0.5^4/24 = 0.0625/24 ≈ 0.00260417 Add them: 1 + 0.5 + 0.125 + 0.0208333 + 0.00260417 ≈ 1.64843747 So e^0.5 ≈ 1.64844.
Common mistakes to avoid
- Forgetting that 0! = 1, so the constant term is 1
- Using x^n/n instead of x^n/n!, which changes the series completely
- Assuming the series is only an approximation for small x, when in fact it converges for all x
Did You Know?
Setting x=1 in the series for e^x gives the famous sum 1 + 1 + 1/2 + 1/6 + ··· = e.
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