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

Answer:[tex]\displaystyle f(x) = \sum^{\infty}_{n = 0} \frac{x^{24n}}{[(2n)!]^2}[/tex]General Formulas and Concepts:CalculusSequencesSeriesPower SeriesPower Series of Elementary FunctionsMacLaurin Series:                                                                                                [tex]\displaystyle P(x) = \sum^{\infty}_{n = 0} \frac{f^n(0)}{n!}x^n[/tex]Taylor SeriesStep-by-step explanation:We are given the function:[tex]\displaystyle f(x) = [cos(x^6)]^2[/tex]Recall that the power series for cos(x) is:[tex]\displaystyle cos(x) = \sum^{\infty}_{n = 0} \frac{(-1)^n x^{2n}}{(2n)!}[/tex]To find the power series for cos(x⁶), substitute in x = x⁶:[tex]\displaystyle cos(x^6) = \sum^{\infty}_{n = 0} \frac{(-1)^n (x^6)^{2n}}{(2n)!}[/tex]Simplifying it, we have:[tex]\displaystyle cos(x^6) = \sum^{\infty}_{n = 0} \frac{(-1)^n x^{12n}}{(2n)!}[/tex]Rewrite the original function:[tex]\displaystyle f(x) = \bigg[ \sum^{\infty}_{n = 0} \frac{(-1)^n x^{12n}}{(2n)!} \bigg]^2[/tex]Simplify:[tex]\displaystyle f(x) = \sum^{\infty}_{n = 0} \frac{(-1)^{2n} x^{24n}}{[(2n)!]^2}[/tex]Simplify down further:[tex]\displaystyle f(x) = \sum^{\infty}_{n = 0} \frac{x^{24n}}{[(2n)!]^2}[/tex]And we have our final answer.Topic: AP Calculus BC (Calculus I + II)  Unit: Power Series  Book: College Calculus 10e