Question

Find the Taylor Series for the function \(f(x)=e^{-6x}\) about x=-4.

(a) \(\sum_{n=0}^{\infty} \frac{(-6)^n}{n!} e^{12} (x+4)^n\)

(b) \(\sum_{n=0}^{\infty} \frac{(-6)^n}{n!} e^{24} (x-4)^n\)

(c) \(\sum_{n=0}^{\infty} \frac{(-6)^n}{n!} e^{24} (x+4)^n\)

(d) \(\sum_{n=0}^{\infty} \frac{(-4)^n}{n!} e^{24} (x+4)^n\)

I had been asked this question in exam.

The question is from Expansion of Trigonometric Functions in division Complex Numbers of Engineering Mathematics

Answer 1

The correct option is (c) \(\sum_{n=0}^{\infty} \frac{(-6)^n}{n!} e^{24} (x+4)^n\)

Best explanation: We start by finding the derivative of the given function,

\(f(x)=e^{-6x}\)

\(f'(x) = -6e^{-6x}\)

\(f”(x) = 36e^{-6x}\)

\(f”'(x) = -216e^{-6x}\)

\(f””(x) = 1296e^{-6x}\)

Thus we take derivative of maximum to the fourth order.

Thus according to formula of Taylor series about x=-4

\(e^{-6x}=\sum_{n=0}^{\infty} \frac{f^n (-4)}{n!} (x+4)^n\)

\(e^{-6x}=\sum_{n=0}^{\infty} \frac{(-6)^n}{n!} e^{24} (x+4)^n\)

Thus the Taylor Series is given by

\(e^{-6x}=\sum_{n=0}^{\infty} \frac{(-6)^n}{n!} e^{24} (x+4)^n\).