Chapter 1: Problem 15

Use the methods of this section to find the first few terms of the Maclaurin series for each of the following functions. $$ e^{\sin x} $$

Short Answer

Expert verified

1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots

Step by step solution

Understand the Maclaurin series

The Maclaurin series for a function f(x) is given by \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \]

Identify the function and its derivatives

The function here is $ e^{\sin x} $. We need to find the derivatives and evaluate them at x=0.

Compute the first few derivatives

First derivative: \[ f'(x) = e^{\sin x} \cos x \]At x=0, $ f'(0) = e^0 \cdot \cos 0 = 1 $Second derivative: \[ f''(x) = e^{\sin x} ( \cos^2 x - \sin x ) \]At x=0, $ f''(0) = e^{0} (1^2 - 0) = 1 $Third derivative: \[ f'''(x) = e^{\sin x} ( \cos^3 x - 3 \sin x \cos x ) - e^{\sin x} \sin x \cos x \]At x=0, $ f'''(0) = e^0 (1 - 0) - 0 = 1 $

Form the Maclaurin series

Using the values found, the first few terms of the Maclaurin series are: \[ e^{\sin x} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Calculus

Calculus is all about studying how things change. It uses concepts like derivatives and integrals to explore changes and areas under curves. The derivatives of a function help us understand how the function is changing at any given point. When we look at a function like $e^{\sin x}$, we can use calculus to find its Maclaurin series and approximate the function around x = 0.

Exponential Functions

Exponential functions are functions that involve constants raised to a variable's power, such as $e^{x}$. The number e is a unique and important constant in mathematics, approximately equal to 2.71828. In our problem, we deal with the function $e^{\sin x}$, which means we have an exponential function where the exponent is itself a function of x. This makes the problem interesting and a bit more complex.

Taylor Series Expansion

The Taylor series expansion is a way to represent functions as an infinite sum of terms calculated from the values of their derivatives at a single point. When this single point is x = 0, it's called a Maclaurin series. For a function f(x), the Maclaurin series looks like this: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \] This series allows us to approximate functions around the point x = 0. By finding the derivatives of our function $e^{\sin x}$ at x = 0, we can write its Maclaurin series.

Derivatives and Their Computation

Derivatives represent the rate at which a function is changing at any given point. To find the Maclaurin series of $e^{\sin x}$, we need to compute its derivatives at x = 0. Here’s how we do it step by step:

The first derivative of $e^{\sin x}$ is $ f'(x) = e^{\sin x} \cos x $. At x=0, this is 1.
The second derivative involves some chain rule and product rule: $ f''(x) = e^{\sin x} ( \cos^2 x - \sin x ) $. At x=0, this is also 1.
The third derivative becomes more complex: $ f'''(x) = e^{\sin x} ( \cos^3 x - 3 \sin x \cos x ) - e^{\sin x} \sin x \cos x $. At x=0, this too simplifies to 1.

With these derivatives, the terms of the Maclaurin series for $e^{\sin x}$ can be built up one by one.