Question

For the following indefinite integral, find the full power series centered at x=0 and then give the first 5 nonzero terms of the power series.

f(x)=∫x9cos(x3) dx

Answer 1

Answer:

a.  $\displaystyle f(x) = \sum^{\infty}_{n = 0} \frac{(-1)^n x^{6n + 10}}{(2n)!} + C_1$

b.  $\displaystyle P_5(x) = \frac{x^{10}}{10} - \frac{x^{16}}{16 \cdot 2!} + \frac{x^{22}}{22 \cdot 4!} - \frac{x^{28}}{28 \cdot 6!} + \frac{x^{34}}{34 \cdot 8!} + C_1$

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Integration

• Integrals
• Indefinite Integrals
• Integration Constant C

Integration Rule [Reverse Power Rule]:                                                               $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Sequences

Series

MacLaurin Polynomials

• MacLaurin Polynomial:                                                                                 $\displaystyle P_n(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^n(0)}{n!}x^n + \cdots$
• MacLaurin Series:                                                                                         $\displaystyle P(x) = \sum^{\infty}_{n = 0} \frac{f^n(0)}{n!}x^n$

Power Series

• Power Series of Elementary Functions
• MacLaurin Polynomials
• Taylor Polynomials

Integration of Power Series:

1.  $\displaystyle f(x) = \sum^{\infty}_{n = 0} a_n(x - c)^n$
2.  $\displaystyle \int {f(x)} \, dx = \sum^{\infty}_{n = 0} \frac{a_n(x - c)^{n + 1}}{n + 1} + C_1$

Step-by-step explanation:

*Note:

You could derive the Taylor Series for cos(x) using Taylor polynomials differentiation but usually you have to memorize it.

We are given and are trying to find the power series:

$\displaystyle f(x) = \int {x^9cos(x^3)} \, dx$

We know 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(x³), substitute in x = x³:

$\displaystyle cos(x^3) = \sum^{\infty}_{n = 0} \frac{(-1)^n (x^3)^{2n}}{(2n)!}$

Simplifying it, we have:

$\displaystyle cos(x^3) = \sum^{\infty}_{n = 0} \frac{(-1)^n x^{6n}}{(2n)!}$

Rewrite the original function:

$\displaystyle f(x) = \int {x^9 \sum^{\infty}_{n = 0} \frac{(-1)^n x^{6n}}{(2n)!}} \, dx$

Rewrite the integrand by including the x⁹ in the power series:

$\displaystyle f(x) = \int {\sum^{\infty}_{n = 0} \frac{(-1)^n x^{6n + 9}}{(2n)!}} \, dx$

Integrating the power series, we have:

$\displaystyle f(x) = {\sum^{\infty}_{n = 0} \frac{(-1)^n x^{6n + 10}}{(6n + 10)(2n)!}} + C_1$

To find the first 5 nonzero terms of the power series, we simply expand it as a MacLaurin polynomial:

$\displaystyle P_5(x) = {\sum^4_{n = 0} \frac{(-1)^n x^{6n + 10}}{(6n + 10)(2n)!}} + C_1 = \frac{x^{10}}{10} - \frac{x^{16}}{16 \cdot 2!} + \frac{x^{22}}{22 \cdot 4!} - \frac{x^{28}}{28 \cdot 6!} + \frac{x^{34}}{34 \cdot 8!} + C_1$

Topic: AP Calculus BC (Calculus I + II)

Unit: Power Series

Book: College Calculus 10e