Question

Find the maclaurin series for the function. (use the table of power series for elementary functions.) f(x) = (cos(x 6))2

Answer 1

Answer:

$\displaystyle f(x) = \sum^{\infty}_{n = 0} \frac{x^{24n}}{[(2n)!]^2}$

General Formulas and Concepts:

Calculus

Sequences

Series

Power Series

• Power Series of Elementary Functions
• MacLaurin Series:                                                                                                $\displaystyle P(x) = \sum^{\infty}_{n = 0} \frac{f^n(0)}{n!}x^n$
• Taylor Series

Step-by-step explanation:

We are given the function:

$\displaystyle f(x) = [cos(x^6)]^2$

Recall that the power series for cos(x) is:

$\displaystyle cos(x) = \sum^{\infty}_{n = 0} \frac{(-1)^n x^{2n}}{(2n)!}$

To find the power series for cos(x⁶), substitute in x = x⁶:

$\displaystyle cos(x^6) = \sum^{\infty}_{n = 0} \frac{(-1)^n (x^6)^{2n}}{(2n)!}$

Simplifying it, we have:

$\displaystyle cos(x^6) = \sum^{\infty}_{n = 0} \frac{(-1)^n x^{12n}}{(2n)!}$

Rewrite the original function:

$\displaystyle f(x) = \bigg[ \sum^{\infty}_{n = 0} \frac{(-1)^n x^{12n}}{(2n)!} \bigg]^2$

Simplify:

$\displaystyle f(x) = \sum^{\infty}_{n = 0} \frac{(-1)^{2n} x^{24n}}{[(2n)!]^2}$

Simplify down further:

$\displaystyle f(x) = \sum^{\infty}_{n = 0} \frac{x^{24n}}{[(2n)!]^2}$

And we have our final answer.

Topic: AP Calculus BC (Calculus I + II)  

Unit: Power Series  

Book: College Calculus 10e