Problem 13 - 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. $$f(x)=e^{2 x}$$

Step-by-Step Solution

Answer

Question: Express the Maclaurin series for the function \(f(x) = e^{2x}\) up to the fourth nonzero term, and determine its interval of convergence. Answer: The Maclaurin series for the function \(f(x) = e^{2x}\) up to the fourth nonzero term is: $$1 + 2x + 2x^2 + \frac{4}{3}x^3$$. The interval of convergence for this series is \(-\infty < x < \infty\), as the exponential function converges for all real values of x.

Step 1: Find the derivatives of the function

To find the first four nonzero terms of the Maclaurin series, we need to find the first four derivatives of the function \(f(x) = e^{2x}\). First derivative: $$f'(x) = \frac{d}{dx}e^{2x} = 2e^{2x}$$ Second derivative: $$f''(x) = \frac{d^2}{dx^2}e^{2x} = 4e^{2x}$$ Third derivative: $$f^{(3)}(x) = \frac{d^3}{dx^3}e^{2x} = 8e^{2x}$$ Fourth derivative: $$f^{(4)}(x) = \frac{d^4}{dx^4}e^{2x} = 16e^{2x}$$

Step 2: Evaluate the derivatives at x = 0

Now that we have the derivatives, we need to evaluate them at \(x = 0\). $$f(0) = e^{2(0)} = 1$$ $$f'(0) = 2e^{2(0)} = 2$$ $$f''(0) = 4e^{2(0)} = 4$$ $$f^{(3)}(0) = 8e^{2(0)} = 8$$ $$f^{(4)}(0) = 16e^{2(0)} = 16$$

Step 3: Write the first four nonzero terms of the Maclaurin series

Now, we can write the Maclaurin series up to the fourth nonzero term by replacing \(f^{(n)}(0)\) with the values we found in step 2 and plugging them into the Taylor series formula. $$f(x) \approx 1 + 2x + \frac{4}{2!}x^2 + \frac{8}{3!}x^3 = 1 + 2x + 2x^2 + \frac{4}{3}x^3$$

Step 4: Write the power series using summation notation

We can express the general term of the Maclaurin series as: $$a_n = \frac{2^n}{n!}x^n$$ Now, we can write the power series using summation notation: $$f(x) = \sum_{n=0}^{\infty}\frac{2^n}{n!}x^n$$

Step 5: Determine the interval of convergence

As the function is \(f(x) = e^{2x}\), and the exponential function converges for all real values of x, the interval of convergence for this Maclaurin series is: $$\text{Interval of convergence:} \, -\infty < x < \infty$$