Problem 82 - Calculus Early Transcendentals

Question

Use composition of series to find the first three terms of the Maclaurin series for the following functions. a. \(e^{\sin x}\) b. \(e^{\tan x}\) c. \(\sqrt{1+\sin ^{2} x}\)

Step-by-Step Solution

Answer

Question: Find the first three terms of the Maclaurin series for the following functions: a. \(e^{\sin x}\) b. \(e^{\tan x}\) c. \(\sqrt{1+\sin^2 x}\) Answer: a. The first three terms of the Maclaurin series for \(e^{\sin x}\) are: \(1 + x + \frac{1}{2}x^2\) b. The first three terms of the Maclaurin series for \(e^{\tan x}\) are: \(1 + x + \frac{1}{2}x^2\) c. The first three terms of the Maclaurin series for \(\sqrt{1+\sin^2 x}\) are: \(1 + \frac{1}{2}x^2\)

Step 1: a. Finding the first three terms of the Maclaurin series for \(e^{\sin x}\)#

Step 1: Calculate the derivatives Calculate the first two derivatives of \(f(x)=e^{\sin x}\) with respect to x: \(f'(x) = \frac{d}{dx}(e^{\sin x})=e^{\sin x}\cos x\) \(f''(x) = \frac{d^2}{dx^2}(e^{\sin x})=-e^{\sin x}\sin x + e^{\sin x}\cos^2 x\) Step 2: Evaluate the derivatives at x=0 Evaluate the first two derivatives and the function itself at x=0: \(f(0) = e^{\sin(0)} = e^0 = 1\) \(f'(0) = e^{\sin(0)}\cos(0) = e^0\cdot 1 = 1\) \(f''(0) = -e^{\sin(0)}\sin(0) + e^{\sin(0)}\cos^2(0) = -e^0\cdot 0 + e^0\cdot 1^2 = 1\) Step 3: Write the first three terms of the Maclaurin series Using the results from Steps 1 and 2, we can write the Maclaurin series for \(e^{\sin x}\) as: \(e^{\sin x} \approx 1 + x + \frac{x^2}{2!} = 1 + x + \frac{1}{2}x^2\)

Step 2: b. Finding the first three terms of the Maclaurin series for \(e^{\tan x}\)#

Step 1: Calculate the derivatives Calculate the first two derivatives of \(f(x) = e^{\tan x}\) with respect to x: \(f'(x) = \frac{d}{dx}(e^{\tan x})=e^{\tan x}\sec^2 x\) \(f''(x) = \frac{d^2}{dx^2}(e^{\tan x})= 2e^{\tan x}\sec^2 x\tan x + e^{\tan x}\sec^4 x\) Step 2: Evaluate the derivatives at x=0 Evaluate the first two derivatives and the function itself at x=0: \(f(0) = e^{\tan(0)} = e^0 = 1\) \(f'(0) = e^{\tan(0)}\sec^2(0) = e^0\cdot 1 = 1\) \(f''(0) = 2e^{\tan(0)}\sec^2(0)\tan(0) + e^{\tan(0)}\sec^4(0) = 2e^0\cdot 1\cdot 0 + e^0\cdot 1^4 = 1\) Step 3: Write the first three terms of the Maclaurin series Using the results from Steps 1 and 2, we can write the Maclaurin series for \(e^{\tan x}\) as: \(e^{\tan x} \approx 1 + x + \frac{x^2}{2!} = 1 + x + \frac{1}{2}x^2\)

Step 3: c. Finding the first three terms of the Maclaurin series for \(\sqrt{1+\sin^2 x}\)#

Step 1: Calculate the derivatives Calculate the first two derivatives of \(f(x) =\sqrt{1+\sin^2 x}\) with respect to x: \(f'(x) = \frac{d}{dx}(\sqrt{1+\sin^2 x})= \frac{\sin x\cos x}{\sqrt{1+\sin^2 x}}\) \(f''(x) = \frac{d^2}{dx^2}(\sqrt{1+\sin^2 x}) = -\frac{\sin^2 x\cos^2 x}{(1+\sin^2 x)^{3/2}} + \frac{\cos^2 x}{(1+\sin^2 x)^{3/2}}\) Step 2: Evaluate the derivatives at x=0 Evaluate the first two derivatives and the function itself at x=0: \(f(0) = \sqrt{1+\sin^2(0)} = \sqrt{1+0} = 1\) \(f'(0) = \frac{\sin(0)\cos(0)}{\sqrt{1+\sin^2(0)}} = \frac{0}{1} = 0\) \(f''(0) = -\frac{\sin^2(0)\cos^2(0)}{(1+\sin^2(0))^{3/2}} + \frac{\cos^2(0)}{(1+\sin^2(0))^{3/2}} = -\frac{0}{1} + \frac{1}{1} = 1\) Step 3: Write the first three terms of the Maclaurin series Using the results from Steps 1 and 2, we can write the Maclaurin series for \(\sqrt{1+\sin^2 x}\) as: \(\sqrt{1+\sin^2 x} \approx 1 + \frac{1}{2}x^2\)