Problem 20 - Calculus Early Transcendentals

Question

a. Find the first four nonzero terms of the Maclaurin series for the given function. b. Write the power series using summation notation. c. Determine the interval of convergence of the series. $$f(x)=\sinh 2 x$$

Step-by-Step Solution

Answer

Q: Determine the first four nonzero terms of the Maclaurin series for the function \(\sinh(2x)\) and identify the interval of convergence. A: The first four nonzero terms of the Maclaurin series for \(\sinh(2x)\) are given by: $$f(x) = 2x + \frac{4}{3}x^3$$ The interval of convergence for the series is \((-\infty, \infty)\).

Step 1: Find the first four nonzero terms of the Maclaurin series for the function \(\sinh 2x\).

Recall that the Maclaurin series for a function \(f(x)\) is given by: $$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!} x^3 + ... $$ First, find the first four derivatives of the function \(\sinh(2x)\): \(f(x) = \sinh(2x)\) \(f'(x) = 2\cosh(2x)\) \(f''(x) = 4\sinh(2x)\) \(f'''(x) = 8\cosh(2x)\) \(f^{(4)}(x) = 16\sinh(2x)\) Now, evaluate these derivatives at \(x = 0\): \(f(0) = \sinh(0) = 0\) \(f'(0) = 2\cosh(0) = 2\) \(f''(0) = 4\sinh(0) = 0\) \(f'''(0) = 8\cosh(0) = 8\) Therefore, the first four nonzero terms of the Maclaurin series are: $$f(x) = 2x + \frac{8}{3!}x^3 = 2x + \frac{4}{3}x^3$$

Step 2: Write the power series using summation notation.

To write the power series using summation notation, we need to find a pattern that generates the above series. The pattern is noticed that for the odd derivative evaluated at 0, we get nonzero terms: $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(2n+1)}(0)}{(2n+1)!}x^{2n+1}$$

Step 3: Determine the interval of convergence of the series.

To determine the interval of convergence of the series, we use the ratio test: $$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{2^{2n+3}x^{2n+3}}{(2n+3)!} \cdot \frac{(2n+1)!}{2^{2n+1}x^{2n+1}} = \lim_{n \to \infty} \frac{4x^2(2n+1)!}{(2n+3)!}$$ Now, notice that the denominator has an extra factor of \((2n+2)(2n+3)\) when compared to the numerator. So, the limit turns out to be: $$\lim_{n \to \infty} \frac{4x^2}{(2n+2)(2n+3)} = 0 $$ Since the limit is 0, independently of the value of \(x\), the series converges for all \(x\), so the interval of convergence is \((-\infty, \infty)\).