Problem 20 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The Maclaurin series for \(\sinh x\) is \(\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}\).

Step 1: Understand the Function

The function \(\sinh x\) defined as \(\frac{e^{x}-e^{-x}}{2}\) is hyperbolic sine. We need to express this function as a Maclaurin series, which is a Taylor series expansion around \(x=0\).

Step 2: Recall the Exponential Series

Recall the Maclaurin series for the exponential function: \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots\) and \(e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \ldots\)

Step 3: Expand Each Exponential

Write the expansions for \(e^{x}\) and \(e^{-x}\): \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots\) \(e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \cdots\)

Step 4: Find the Series for \(e^x - e^{-x}\)

Subtract the series for \(e^{-x}\) from \(e^{x}\): \((e^x - e^{-x}) = (1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots) - (1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots)\).The constants cancel: \((e^x - e^{-x}) = 2x + \frac{2x^3}{3!} + \frac{2x^5}{5!} + \ldots\)

Step 5: Divide by 2

The formula for \(\sinh x\) is \(\frac{e^{x} - e^{-x}}{2}\). Thus, divide each term of the previous step by 2:\[\sinh x = \frac{2x}{2} + \frac{2x^3}{3! \times 2} + \frac{2x^5}{5! \times 2} + \cdots = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots\]

Step 6: Write the General Term

The series for \(\sinh x\) can be written as:\[\sinh x = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}\]This implies the general term is \(\frac{x^{2n+1}}{(2n+1)!}\).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola rather than a circle. The hyperbolic sine, written as \(\sinh x\), which you encountered in the given problem, is one of these fascinating functions. Hyperbolic functions have applications in various fields such as physics, engineering, and complex analysis.
They are particularly useful in scenarios involving hyperbolas and exponential growth or decay. The hyperbolic sine \(\sinh x = \frac{e^x - e^{-x}}{2}\) behaves similarly to the sine function, but it scales differently over the coordinate space.
Some key properties of hyperbolic functions include:

• They are defined using exponential functions, as seen in \(\sinh x\) and \(\cosh x = \frac{e^x + e^{-x}}{2}\).
• They provide solutions to certain types of differential equations common in physics.
• They can model the shape of a hanging cable or chain, known as a catenary.

The understanding of hyperbolic functions is not only crucial for doing mathematics but also for applying mathematical concepts to real-world problems.

Exponential Function

The exponential function is one of the most important functions in mathematics. It forms the basis of hyperbolic functions and is defined as \(e^x\), where \(e\) is the base of the natural logarithm, approximately equal to 2.71828. The unique property of \(e^x\) is that its derivative is itself, making it highly significant in calculus and complex analysis.
The exponential function appears frequently in applications involving growth and decay, such as population dynamics, radioactive decay, and compound interest calculations. Its series expansion is given by the Maclaurin series:
\[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots\]For negative exponents, the series becomes:
\[e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \ldots\]
These expansions help us approximate the value of exponential functions and are foundational for understanding more complex series, like those involved with hyperbolic functions.

Series Expansion

Series expansion is a powerful tool in mathematics used to approximate functions by expressing them as an infinite sum of terms calculated from the values of their derivatives at a single point. The Maclaurin series is a specific type of series expansion that takes this point as \(x = 0\). For instance, each term in the series expansion of a function is derived from the function's derivatives at zero.
The purpose of series expansion is to simplify complex functions into manageable pieces that can be calculated more easily, especially when exact calculations are infeasible. In the problem, converting \(\sinh x\) into its series form allowed us to express it as:
\[\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots\]
Key benefits of using series expansions include:

• They provide an approximation of functions to a desired degree of accuracy.
• They are used in numerical analysis to facilitate computing the value of functions.
• They help in deriving solutions to differential equations and modeling continuous data.

Mastering the series expansion technique is crucial for tackling a wide range of mathematical and engineering problems.