Mathematics, 26.10.2019 06:43 solomander4

Evaluate the indefinite integral as an infinite series
∫sinx /3x dx.
find the first five non-zero terms of series representation centered at x=0.
f(x)=
what is the radius of convergence?

Answers: 3

Show answers

Answers

Answer from: Suphat

$\displaystyle f(x) = \frac{x}{3} - \frac{x^3}{9 \cdot 3!} + \frac{x^5}{15 \cdot 5!} - \frac{x^7}{21 \cdot 7!} + \frac{x^9}{27 \cdot 9!}$

Radius of Convergence: $\displaystyle \mathbb{R}$ (all real numbers)

General Formulas and Concepts:

Pre-Calculus

Summation Properties

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 FunctionsMacLaurin and Taylor PolynomialsTaylor 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:

$\displaystyle \lim_{n \to \infty} \bigg| \frac{(-1)^\big{n + 1}x^\big{2(n + 1) + 1}}{[2(n + 1) + 1]^\big2[2(n + 1)]!} \cdot \frac{(2n + 1)^\big2(2n)!}{(-1)^\big{n}x^\big{2n + 1}} \bigg|$