Question Bank • QB_EM_145 NAT • 2 Marks

Section 1: Engineering Mathematics

Calculus

Subtopic: Fourier, Taylor and MacLaurin series

Question Statement:

In the Taylor series expansion of $f(x) = e^{2x}$ about $x = 0$ (Maclaurin series), what is the coefficient of $x^3$ (round off to 3 decimal places)?

Answer Key:

1.333

Step-by-Step Solution & Explanation

Maclaurin series of $e^{2x}$ is: $e^{2x} = \sum_{n=0}^\infty \frac{(2x)^n}{n!} = 1 + 2x + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \dots$ The coefficient of $x^3$ is: $\frac{8}{3!} = \frac{8}{6} = \frac{4}{3} \approx 1.333$
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