Question Bank • QB_EM_089
NAT • 2 Marks
Section 1: Engineering Mathematics
Calculus
Subtopic: Fourier, Taylor and MacLaurin series
Question Statement:
The coefficient of $(x - 2)^3$ in the Taylor series expansion of $f(x) = e^{2x}$ about $x = 2$ is $C e^4$. Calculate the numerical value of $C$.
Step-by-Step Solution & Explanation
Taylor series coefficient for $(x - a)^n$ is $\frac{f^{(n)}(a)}{n!}$.
Here $f(x) = e^{2x}$, $a = 2$, $n = 3$.
$f'(x) = 2 e^{2x}, \quad f''(x) = 4 e^{2x}, \quad f'''(x) = 8 e^{2x}$
$f'''(2) = 8 e^4$
$\text{Coefficient} = \frac{8 e^4}{3!} = \frac{8 e^4}{6} = \frac{4}{3} e^4 \approx 1.333 e^4$
So $C = \frac{4}{3} \approx 1.333$.
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