Respuesta :
Answer:
[tex]\displaystyle f(x) = \frac{x}{3} - \frac{x^3}{9 \cdot 3!} + \frac{x^5}{15 \cdot 5!} - \frac{x^7}{21 \cdot 7!} + \frac{x^9}{27 \cdot 9!}[/tex]
Radius of Convergence: [tex]\displaystyle \mathbb{R}[/tex] (all real numbers)
General Formulas and Concepts:
Pre-Calculus
- Summation Properties
Calculus
Integration
- Integrals
- [Indefinite Integrals] Integration Constant C
Integration Property [Multiplied Constant]: [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
Sequences
Series Convergence Tests
- Ratio Test: [tex]\displaystyle \lim_{n \to \infty} \bigg| \frac{a_{n + 1}}{a_n} \bigg|[/tex]
Radius of Convergence
Series
Taylor Polynomials and Approximations
- MacLaurin Polynomial: [tex]\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[/tex]
Power Series
- Power Series of Elementary Functions
- MacLaurin and Taylor Polynomials
- Taylor Series: [tex]\displaystyle P(x) = \sum^{\infty}_{n = 0} \frac{f^n(c)}{n!}(x - c)^n[/tex]
Integration of Power Series:
- [tex]\displaystyle f(x) = \sum^{\infty}_{n = 0} a_n(x - c)^n[/tex]
- [tex]\displaystyle \int {f(x)} \, dx = \sum^{\infty}_{n = 0} \frac{a_n(x - c)^{n + 1}}{n + 1} + C_1[/tex]
Step-by-step explanation:
*Note:
You can derive the power series of sin(x) using Taylor polynomials (not shown here).
We are given the integral:
[tex]\displaystyle \int {\frac{\sin x}{3x}} \, dx[/tex]
Recall the power series for sin(x):
[tex]\displaystyle \sin x = \sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n + 1}}{(2n + 1)!}[/tex]
To find power series for [tex]\displaystyle \frac{\sin x}{3x}[/tex] , divide the power series by 3x:
[tex]\displaystyle \frac{\sin x}{3x} = \sum^{\infty}_{n = 0} \bigg[ \frac{(-1)^nx^{2n + 1}}{(2n + 1)!} \cdot \frac{1}{3x} \bigg][/tex]
Simplify the power series:
[tex]\displaystyle \frac{\sin x}{3x} = \frac{1}{3} \sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n}}{(2n + 1)!}[/tex]
Rewrite the original integral:
[tex]\displaystyle \int {\frac{\sin x}{3x}} \, dx = \int {\frac{1}{3} \sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n}}{(2n + 1)!}} \, dx[/tex]
Rewrite it once more:
[tex]\displaystyle \int {\frac{\sin x}{3x}} \, dx = \frac{1}{3} \int {\sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n}}{(2n + 1)!}} \, dx[/tex]
Integrate the power series:
[tex]\displaystyle \int {\frac{\sin x}{3x}} \, dx = \frac{1}{3} \sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n + 1}}{(2n + 1)(2n + 1)!}} + C[/tex]
Simplify the integrated power series:
[tex]\displaystyle \int {\frac{\sin x}{3x}} \, dx = \frac{1}{3} \sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n + 1}}{(2n + 1)^2(2n)!}} + C[/tex]
To find the first 5 nonzero terms of the power series, we simply expand it as a MacLaurin polynomial:
[tex]\displaystyle f(x) = \frac{1}{3} \sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n + 1}}{(2n + 1)^2(2n)!}} = \frac{x}{3} - \frac{x^3}{9 \cdot 3!} + \frac{x^5}{15 \cdot 5!} - \frac{x^7}{21 \cdot 7!} + \frac{x^9}{27 \cdot 9!}[/tex]
To find the radius of convergence, we apply the Ratio Test to the power series:
[tex]\displaystyle \lim_{n \to \infty} \bigg| \frac{(-1)^\big{n + 1}x^\big{2(n + 1) + 1}}{[2(n + 1) + 1]^\big2[2(n + 1)]!} \cdot \frac{(2n + 1)^\big2(2n)!}{(-1)^\big{n}x^\big{2n + 1}} \bigg|[/tex]
Simplify the limit:
[tex]\displaystyle \lim_{n \to \infty} \bigg| \frac{x^\big{2n + 3}}{(2n + 3)^\big2(2n + 2)!} \cdot \frac{(2n + 1)^\big2(2n)!}{x^\big{2n + 1}} \bigg|[/tex]
Simplify the limit further:
[tex]\displaystyle \lim_{n \to \infty} \bigg| \frac{x(2n + 1)}{(2n + 3)(2n + 2)} \bigg|[/tex]
Evaluate the limit:
[tex]\displaystyle \lim_{n \to \infty} \bigg| \frac{x(2n + 1)}{(2n + 3)(2n + 2)} \bigg| = 0[/tex]
∴ since the limit equals 0, the series converges absolutely and the radius of convergence is all real numbers.
Topic: AP Calculus BC (Calculus I + II)
Unit: Power Series