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Maclaurin expansion of e^x log(1+x)

Short answer

Multiply the Maclaurin series of e^x and log(1+x): e^x = 1 + x + x^2/2 + x^3/6 + x^4/24 + ... and log(1+x) = x - x^2/2 + x^3/3 - x^4/4 + ... Collecting terms up to x^4 gives e^x log(1+x) = x + x^2/2 + x^3/3 + 0·x^4 + ... The x^4 term vanishes.

Expand eˣ log_e(1+x) in ascending powers of x up to the term containing x⁴ in Maclaurin's series.

Maclaurin's series: f(x) = f(0) + x f′(0) + (x²/2!) f″(0) + ... For products of standard functions we multiply their Maclaurin series (valid by uniqueness of the expansion).

eˣ        = 1 + x + x²/2 + x³/6 + x⁴/24 + ...
log(1+x)  = x − x²/2 + x³/3 − x⁴/4 + ...

Multiply and collect terms up to x⁴:

x¹ :  1                                   = 1
x² :  −1/2 + 1                            = 1/2
x³ :  1/3 − 1/2 + 1/2                     = 1/3
x⁴ :  −1/4 + 1/3 − 1/4 + 1/6              = 0

(x³ term: from 1·(x³/3), x·(−x²/2) and (x²/2)·x. x⁴ term: from 1·(−x⁴/4), x·(x³/3), (x²/2)(−x²/2) and (x³/6)·x.)

Answer: eˣ log(1+x) = x + x²/2 + x³/3 + 0·x⁴ + ..., so the x⁴ coefficient is zero.

✓ Answer checked by Claude Opus 5.5 (high effort)

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