Complex Numbers Questions and Answers – Expansion of Trigonometric Functions

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This set of Complex Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Expansion of Trigonometric Functions”.

1. The Taylor series for f(x)=7x2-6x+1 at x=2 is given by a+b(x-2)+c(x-2)2. Find the value of a+b+c.
a) -1
b) 0
c) 17
d) 46
View Answer

Answer: d
Explanation: We know
\(f(x)=7x^2-6x+1\)
\(f'(x)=14x-6\)
\(f”(x)=14\)
\(f”'(x)=0\)
Thus for n>=3, the derivative of the function is 0.
As per the Taylor Series,
\(7x^2-6x+1=\sum_{n=0}^{\infty} \frac{f^n (2)(x-2)^n}{n!}\)
\(7x^2-6x+1=f(2)+f'(2)(x-2)+\frac{1}{2} f”(2) (x-2)^2+0\)
\(7x^2-6x+1=17+22(x-2)+7(x-2)^2\)
Thus, a=17, b=22, c=7
a+b+c=46
Thus the answer is 46.
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2. 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) \(\sum_{n=0}^{\infty} \frac{(-4)^n}{n!} e^{24} (x+4)^n\)
View Answer

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

Sanfoundry Global Education & Learning Series – Complex Analysis.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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