# 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

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

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

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. 