Maclaurin expansion of eˣ/(1+eˣ)

Short answer

Using Maclaurin's series, f(x)=e^x/(1+e^x) gives f(0)=1/2, f'(0)=1/4, f''(0)=0, f'''(0)=-1/8, so the expansion up to x³ is 1/2 + x/4 − x³/48 + ...

Assuming the validity of expansion, find the expansion of the function eˣ/(1+eˣ) by Maclaurin's theorem (as far as the term in x³).

Maclaurin's series: f(x) = f(0) + x f′(0) + (x²/2!) f″(0) + (x³/3!) f‴(0) + ...

Let f(x) = eˣ/(1+eˣ) = 1 − 1/(1+eˣ). Differentiate:

f′ = eˣ/(1+eˣ)² = f(1 − f)

Since f′ = f − f², we get

f″  = f′ − 2f f′ = f′(1 − 2f)
f‴  = f″(1 − 2f) − 2(f′)²

At x = 0: f(0) = 1/2.

f′(0)  = (1/2)(1/2) = 1/4
f″(0)  = (1/4)(1 − 1) = 0
f‴(0)  = 0 − 2(1/16) = −1/8

Substitute:

f(x) = 1/2 + x(1/4) + (x²/2)(0) + (x³/6)(−1/8) + ...

Answer: eˣ/(1+eˣ) = 1/2 + x/4 − x³/48 + ...

✓ Answer checked by Claude Opus 5.5 (high effort)

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