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