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

`\displaystyle f(x) = \sum^(\infty)_(n = 0) (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) (f^n(0))/(n!)x^n`
• Taylor Series

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) ((-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) ((-1)^n (x^6)^(2n))/((2n)!)`

Simplifying it, we have:

`\displaystyle cos(x^6) = \sum^(\infty)_(n = 0) ((-1)^n x^(12n))/((2n)!)`

Rewrite the original function:

`\displaystyle f(x) = \bigg[ \sum^(\infty)_(n = 0) ((-1)^n x^(12n))/((2n)!) \bigg]^2`

Simplify:

`\displaystyle f(x) = \sum^(\infty)_(n = 0) ((-1)^(2n) x^(24n))/([(2n)!]^2)`

Simplify down further:

`\displaystyle f(x) = \sum^(\infty)_(n = 0) (x^(24n))/([(2n)!]^2)`

And we have our final answer.

Topic: AP Calculus BC (Calculus I + II)

Unit: Power Series

Book: College Calculus 10e