Problem 46 - Calculus Early Transcendentals

Question

Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to the following numbers. $$\sqrt{e}$$

Step-by-Step Solution

Answer

Answer: The first four nonzero terms of the Taylor series for the given function are $$1 + \frac{1}{2}x + \frac{1}{8}x^2 + \frac{1}{48}x^3$$.

Step 1: Define the Function

Let's start by defining the function we want to approximate with a Taylor series: $$f(x) = \sqrt{e^{x}}$$

Step 2: Find the first 4 derivatives of the function f(x)

Now, find the first four derivatives of the function f(x): 1. First Derivative: $$f'(x) = \frac{d}{dx}\sqrt{e^{x}} = \frac{1}{2}(e^{x})^{-\frac{1}{2}}(e^{x}) = \frac{1}{2}e^{\frac{x}{2}}$$ 2. Second Derivative: $$f''(x) = \frac{d}{dx}\left(\frac{1}{2}e^{\frac{x}{2}}\right) = \frac{1}{4}e^{\frac{x}{2}}$$ 3. Third Derivative: $$f^{(3)}(x) = \frac{d}{dx}\left(\frac{1}{4}e^{\frac{x}{2}}\right) = \frac{1}{8}e^{\frac{x}{2}}$$ 4. Fourth Derivative: $$f^{(4)}(x) = \frac{d}{dx}\left(\frac{1}{8}e^{\frac{x}{2}}\right) = \frac{1}{16}e^{\frac{x}{2}}$$

Step 3: Evaluate the derivatives at x=0

Now, evaluate all the derivatives obtained in step 2 at x=0. 1. First Derivative: $$f'(0) = \frac{1}{2}e^{\frac{0}{2}} = \frac{1}{2}$$ 2. Second Derivative: $$f''(0) = \frac{1}{4}e^{\frac{0}{2}} = \frac{1}{4}$$ 3. Third Derivative: $$f^{(3)}(0) = \frac{1}{8}e^{\frac{0}{2}} = \frac{1}{8}$$ 4. Fourth Derivative: $$f^{(4)}(0) = \frac{1}{16}e^{\frac{0}{2}} = \frac{1}{16}$$

Step 4: Construct the Taylor series

Now, let's write the Taylor series using the values obtained in step 3. $$\begin{aligned} f(x) &= f(0) + \frac{f'(0)}{1!}(x-0)^1 + \frac{f''(0)}{2!}(x-0)^2 + \frac{f^{(3)}(0)}{3!}(x-0)^3 + \cdots \\ &= e^0 + \frac{1}{2}x + \frac{1}{4}\frac{1}{2}x^2 + \frac{1}{8}\frac{1}{6}x^3 + \cdots \\ &= 1 + \frac{1}{2}x + \frac{1}{8}x^2 + \frac{1}{48}x^3 + \cdots \end{aligned}$$ The first four nonzero terms of the Taylor series of $$f(x) = \sqrt{e^x}$$ are: $$1 + \frac{1}{2}x + \frac{1}{8}x^2 + \frac{1}{48}x^3$$