Complex Numbers Questions and Answers – Expansion of Trigonometric Functions


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

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,
\(f'(x) = -6e^{-6x}\)
\(f”(x) = 36e^{-6x}\)
\(f”'(x) = -216e^{-6x}\)
\(f””(x) = 1296e^{-6x}\)
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
\(e^{-6x}=\sum_{n=0}^{\infty} \frac{(-6)^n}{n!} e^{24} (x+4)^n\).

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter