Problem 19 - Thomas Calculus

Question

Find the Maclaurin series for the functions. \(\cosh x=\frac{e^{x}+e^{-x}}{2}\)

Step-by-Step Solution

Answer

\( \cosh x = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \) is the Maclaurin series for \( \cosh x \).

Step 1: Recall the Definition of Maclaurin Series

The Maclaurin series is a Taylor series expansion around the point zero. The general formula for the Maclaurin series of a function \( f(x) \) is given by:\[f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots \]

Step 2: Determine the Derivatives of \( \cosh x \)

Calculate the successive derivatives of the function \( \cosh x = \frac{e^x + e^{-x}}{2} \). - \( f(x) = \cosh x = \frac{e^x + e^{-x}}{2} \)- \( f'(x) = \sinh x = \frac{e^x - e^{-x}}{2} \)- \( f''(x) = \cosh x = \frac{e^x + e^{-x}}{2} \)- \( f'''(x) = \sinh x = \frac{e^x - e^{-x}}{2} \)Notice the alternating pattern of derivatives.

Step 3: Evaluate the Derivatives at \( x = 0 \)

Substitute \( x = 0 \) into the derivatives:- \( f(0) = \cosh 0 = 1 \)- \( f'(0) = \sinh 0 = 0 \)- \( f''(0) = \cosh 0 = 1 \)- \( f'''(0) = \sinh 0 = 0 \)This pattern continues for higher derivatives, with even derivatives evaluated at zero yielding 1, and odd derivatives yielding 0.

Step 4: Construct the Series

Using the values obtained, plug them into the Maclaurin series formula:\[\cosh x = 1 + 0 \cdot x + \frac{1 \cdot x^2}{2!} + 0 \cdot x^3 + \frac{1 \cdot x^4}{4!} + \cdots\]This simplifies to:\[\cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots\]

Step 5: Write the General Formula

Recognize the pattern where the non-zero terms occur only at even powers of \( x \):\[\cosh x = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}\]This is the Maclaurin series for \( \cosh x \).

Key Concepts

Taylor series expansion

The Taylor series is a powerful tool used to express a function as an infinite sum of terms. Each term is derived from the values of the function's derivatives at a single point. A Maclaurin series is a type of Taylor series expansion centered at zero, which makes calculating the series terms easier for many functions. The general formula for a Maclaurin series of a function \( f(x) \) is:
\[ f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots \]
This formula highlights how each term in the series uses the derivatives evaluated at zero, multiplied by a power of \( x \) and divided by a factorial. Factoring in the derivatives and their corresponding factorial helps ensure the approximation works well, even for complex functions.

hyperbolic cosine function

The hyperbolic cosine function, \( \cosh x \), is defined as:
\[ \cosh x = \frac{e^x + e^{-x}}{2} \]
This function is similar in form to the trigonometric cosine function but uses exponential functions with base \( e \) instead. The hyperbolic functions, including \( \cosh x \), have interesting identities and properties that make them useful in various mathematical and physics applications.

• Unlike trigonometric functions, hyperbolic functions are defined for all real numbers.
• They have properties akin to the trigonometric functions but differ in their nature and periodicity.
• The graph of \( \cosh x \) shows symmetry about the y-axis, making it an even function.

Understanding \( \cosh x \) and its derivatives is crucial for evaluating its series expansion accurately.

derivatives

Derivatives play a vital role in constructing the Taylor series for any function. They help us approximate the function around a specific point. For the hyperbolic cosine function, \( \cosh x \), calculating derivatives is made easier due to the repeating nature of its terms:

• \( f(x) = \cosh x = \frac{e^x + e^{-x}}{2} \)
• \( f'(x) = \sinh x = \frac{e^x - e^{-x}}{2} \)
• \( f''(x) = \cosh x = \frac{e^x + e^{-x}}{2} \)
• \( f'''(x) = \sinh x = \frac{e^x - e^{-x}}{2} \)

The derivatives alternate between \( \cosh x \) and \( \sinh x \), with the even derivatives representing \( \cosh x \) and odd ones \( \sinh x \). Evaluating these derivatives at \( x = 0 \) shows a pattern where even derivatives yield 1, and odd derivatives yield 0, simplifying the calculation of series coefficients.

series evaluation

Evaluating the series involves substituting the derivatives' values into the Maclaurin series formula. For the hyperbolic cosine function, this means recognizing the terms in the formula:

• \( f(0) = \cosh 0 = 1 \)
• \( f'(0) = \sinh 0 = 0 \)
• \( f''(0) = \cosh 0 = 1 \)
• \( f'''(0) = \sinh 0 = 0 \)

Using these evaluated derivatives, the Maclaurin series for \( \cosh x \) becomes:
\[ \cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots \]
These series terms capture the behavior of \( \cosh x \), especially near \( x = 0 \). The even powers reflect the even function nature of \( \cosh x \), leading to the concise format:
\[ \cosh x = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \]
This formula not only presents \( \cosh x \) as an infinite sum but also provides a useful approximation tool for practical calculations.