Find the Maclaurin polynomial of degree two for the function $f ( x ) = \sec ( 11 x )$
A) $P _ { 2 } ( x ) = 1 + \frac { 121 } { 4 } x ^ { 2 }$
B) $P _ { 2 } ( x ) = 1 - \frac { 121 } { 2 } x ^ { 2 }$
C) $P _ { 2 } ( x ) = 1 + \frac { 121 } { 2 } x ^ { 2 }$
D) $P _ { 2 } ( x ) = x + \frac { 121 } { 2 } x ^ { 2 }$
E) $P _ { 2 } ( x ) = x - \frac { 121 } { 2 } x ^ { 2 }$

Question

Quizplus · Accepted Answer

We use the Maclaurin series formula for $\sec(11x)$, which is:

$$\sec(11x) = \sum_{n=0}^{\infty} \frac{(-1)^n (2n)!}{(2^{2n})(n!)^2 (11x)^{2n}}$$

To find the degree-two Maclaurin polynomial, we only need to consider the first three terms of the series:

$$\sec(11x) \approx 1 + \frac{(-1)^1 (2)^2}{(2^{2})(1!)^2 (11x)^{2}} + \frac{(-1)^2 (4)^2}{(2^{4})(2!)^2 (11x)^{4}}$$

Simplifying the coefficients and terms, we get:

$$\sec(11x) \approx 1 + \frac{121}{2}x^2 + \frac{16965}{8}x^4$$

Thus, the degree-two Maclaurin polynomial is:

$$P_2(x) = 1 + \frac{121}{2}x^2$$

so the best choice is $\boxed{	extbf{(C)}}$.

Find the Maclaurin Polynomial of Degree Two for the Function f(x)=sec⁡(11x)f ( x ) = \sec ( 11 x )f(x)=sec(11x)

Find the Maclaurin Polynomial of Degree Two for the Function $f ( x ) = \sec ( 11 x )$