Chapter 1: Problem 21

Use the methods of this section to find the first few terms of the Maclaurin series for each of the following functions. $$ \text { cosh } x=\frac{e^{x}+e^{-x}}{2} $$

Short Answer

Expert verified

The first few terms of the Maclaurin series for $ \text{cosh}(x) $ are 1 + $ \frac{x^2}{2} $.

Step by step solution

Understand the Maclaurin series expansion

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

Find the zeroth derivative

Evaluate the function at x = 0. \[\text{cosh}(0) = \frac{e^0 + e^{-0}}{2} = 1 \]

Find the first derivative

Find the first derivative of cosh(x). The derivative of $ \text{cosh}(x) $ is $ \text{sinh}(x) $.So, \[\text{sinh}(x) = \frac{e^x - e^{-x}}{2} \]Then evaluate at x = 0. \[ \text{sinh}(0) = \frac{e^0 - e^{-0}}{2} = 0 \]

Find the second derivative

Find the second derivative of cosh(x). The derivative of $ \text{sinh}(x) $ is $ \text{cosh}(x) $.So, \[ \text{cosh}(x) = \frac{e^x + e^{-x}}{2} \] Then evaluate at x = 0. \[ \text{cosh}(0) = 1 \]

Find the third derivative

Find the third derivative of cosh(x). The third derivative of $ \text{cosh}(x) $ is $ \text{sinh}(x) $. Then evaluate at x = 0. \[ \text{sinh}(0) = 0 \]

Combine the terms

Combine the values into the Maclaurin series form: \[ \text{cosh}(x) \approx 1 + 0 \cdot x + \frac{1}{2!} x^2 + 0 \cdot \frac{x^3}{3!} \] Simplify the expression to get: \[ \text{cosh}(x) \approx 1 + \frac{x^2}{2} \]

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Key Concepts

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

cosh function

The hyperbolic cosine function, denoted as $\text{cosh}(x)$, is a fundamental part of hyperbolic functions. The formula for $\text{cosh}(x)$ is given by:
\[ \text{cosh}(x) = \frac{e^x + e^{-x}}{2} \]
It is similar to the trigonometric cosine function but is based on the exponential function instead. It is an even function, meaning it is symmetric around the y-axis. The significant aspect of $\text{cosh}(x)$ is its role in describing shapes like catenaries and its applications in various fields including engineering and physics.

hyperbolic functions

Hyperbolic functions include $\text{cosh}(x)$, $\text{sinh}(x)$, $\text{tanh}(x)$, and their inverses. These functions are analogs of the common trigonometric functions but for a hyperbola rather than a circle. They are defined as follows:

$\text{cosh}(x) = \frac{e^x + e^{-x}}{2} $
$\text{sinh}(x) = \frac{e^x - e^{-x}}{2} $
$\text{tanh}(x) = \frac{\text{sinh}(x)}{\text{cosh}(x)} $

Hyperbolic functions are useful in solving many real-world problems, such as calculating the length of a hanging cable or the growth rate of populations under certain conditions.

derivatives

Taking derivatives is a crucial skill for finding the Maclaurin Series and understanding the behavior of functions. For the function $\text{cosh}(x)$, let's recall the derivatives:

The first derivative of $\text{cosh}(x)$ is $\text{sinh}(x)$, thus $\frac{d}{dx} \text{cosh}(x) = \text{sinh}(x) $.
The second derivative of $\text{cosh}(x)$ is $\text{cosh}(x)$, thus $\frac{d^2}{dx^2} \text{cosh}(x) = \text{cosh}(x) $.
The third derivative returns to $\text{sinh}(x)$, thus $\frac{d^3}{dx^3} \text{cosh}(x) = \text{sinh}(x) $.

Each derivative cycles between $\text{cosh}(x)$ and $\text{sinh}(x)$, illustrating how the hyperbolic sine and cosine functions are interconnected. These derivatives help in constructing series approximations like the Maclaurin series to simplify complex functions into polynomial forms.