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Find the terms through \(x^{5}\) in the Maclaurin series for \(f(x) .\) Hint: It may be easiest to use known Maclaurin series and then perform multiplications, divisions, and so on. For example, \(\tan x=(\sin x) /(\cos x) .\) \(f(x)=\tanh x\)

Short Answer

Expert verified
The terms through \(x^5\) in the Maclaurin series for \(f(x) = \tanh x\) are \(x - \frac{x^3}{3} + \frac{x^5}{5}\).

Step by step solution

01

Recall the definition of Maclaurin series

The Maclaurin series for a function \(f(x)\) is given by \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \). This involves evaluating the function and its derivatives at \(x=0\).
02

Express \(\tanh x\) in terms of known series

\(\tanh x = \frac{\sinh x}{\cosh x}\). We can use the known series expansions for \(\sinh x\) and \(\cosh x\): - \(\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots\)- \(\cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots\)
03

Perform division of the series

To find \(\tanh x\), perform long division of the \(\sinh x\) series by the \(\cosh x\) series, maintaining terms up to \(x^5\). Start dividing:- The leading term \(x\) divided by 1 gives \(x\).- Multiply this by the \(\cosh x\) series to subtract: \((x)(1 + \frac{x^2}{2} + \frac{x^4}{24}) = x + \frac{x^3}{2} + \frac{x^5}{24}\).- Subtract from \(\sinh x\): \((x + \frac{x^3}{6} + \frac{x^5}{120}) - (x + \frac{x^3}{2} + \frac{x^5}{24}) = -\frac{x^3}{3} - \frac{x^5}{40}\).- Divide \(-\frac{x^3}{3}\) by 1 to add \(-\frac{x^3}{3}\) to the quotient. Multiply and subtract similarly for resulting terms.
04

Simplify to obtain the series terms

Continuing the division process outlined, we obtain the Maclaurin series up to \(x^5\):\[ \tanh x = x - \frac{x^3}{3} + \frac{x^5}{5} + \cdots \]

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Key Concepts

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

Hyperbolic Functions
Hyperbolic functions, such as \ \( \tanh x \ \), offer intriguing connections between trigonometry and exponential functions. They behave similarly to trigonometric functions, but instead, they relate to hyperbolae rather than circles. The important hyperbolic functions include \( \sinh x \) and \( \cosh x \), which are defined as follows:
  • \( \sinh x = \frac{e^x - e^{-x}}{2} \)
  • \( \cosh x = \frac{e^x + e^{-x}}{2} \)
Using these definitions, \ \( \tanh x = \frac{\sinh x}{\cosh x} \ \) acts as the hyperbolic equivalent of \( \tan x \).
These functions have unique properties and applications, especially in solving certain differential equations and in the field of complex analysis. Understanding their series expansions helps us in approximations and computations, making them valuable in various mathematical contexts.
The \ \( \tanh x \ \) function, in particular, is commonly used in statistical and neural network applications because of its ability to provide a smooth output between -1 and 1.
Series Expansion
Series expansion is a powerful mathematical tool that allows functions to be expressed as infinite sums of terms. Among the various series types, the Maclaurin series is heavily used in calculus and involves expanding a function at x = 0. The general form of a Maclaurin series is given by:\[f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots\]Where:
  • \( f(n)(0) \) denotes the nth derivative evaluated at \( x = 0 \)

Using this formula, we can approximate functions around a certain point, providing a straightforward method for evaluating complex functions.
In our case, by expressing \( \tanh x \) as the ratio of \( \sinh x \) and \( \cosh x \), we utilized known series of these hyperbolic functions to construct the Maclaurin series of \( \tanh x \) up to a desired degree.
This allows us to explore the local behavior of the function and simplifies calculations for practical uses, such as computations in physics and engineering.
Derivative Evaluation
Evaluating the derivatives of a function at a specific point is crucial for constructing its series expansion. For \( \tanh x \), we consider derivatives of both \( \sinh x \) and \( \cosh x \) to determine the series expansion accurately. The derivatives of these basic hyperbolic functions at any point are:
  • \( \frac{d}{dx}[\sinh x] = \cosh x \)
  • \( \frac{d}{dx}[\cosh x] = \sinh x \)
Evaluating at \( x = 0 \, \) we get:
  • For \( \sinh x \), all odd derivatives are zero and even order derivatives will evaluate to powers of one.
  • For \( \cosh x \), all odd derivatives are zero at \( x = 0 \) while even derivatives translate directly to factorial terms, such as \( 1, \frac{x^2}{2!}, \frac{x^4}{4!} \) for \( x^2, x^4 \) respectively.
We use these evaluations to form the Maclaurin series by integrating them into the power series formula for \( \tanh x \,\) leading to the accurate series representation up to \( x^5 \).
Long Division Method
Long division of series allows us to find a more exact function series by dividing one series by another, akin to numerical long division.In the case of \( \tanh x = \frac{\sinh x}{\cosh x} \,\) we must perform a careful step-by-step division.
This method starts by:
  • Taking the leading term of \( \sinh x \), which is \( x \), and dividing it by \( 1 \), the leading term of \( \cosh x \).
  • Using this result, multiply the \( \cosh x \) series to subtract from \( \sinh x \), and obtain the next terms.
  • Continuing until reaching the desired power degree.
With each division step, more terms are added to the quotient, gradually building the \( \tanh x \) series.
This allows us to break complex functions into simpler components, aiding in analysis and applications where precise functional forms are needed.
The result offers a glimpse into the structural behavior of functions, particularly useful in scenarios requiring high precision in mathematical and scientific calculations.

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Most popular questions from this chapter

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