Complex Analysis Multiple Choice Questions 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
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.
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)
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).
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