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In each part, rewrite the expression as a ratio of polynomials. $$ \begin{array}{ll}{\text { (a) } \cosh (\ln x)} & {\text { (b) } \sinh (\ln x)} \\\ {\text { (c) } \tanh (2 \ln x)} & {\text { (d) } \cosh (-\ln x)}\end{array} $$

Short Answer

Expert verified
(a) \( \frac{x^2 + 1}{2x} \), (b) \( \frac{x^2 - 1}{2x} \), (c) \( \frac{x^4 - 1}{x^4 + 6x^2 + 1} \), (d) \( \frac{x^2 + 1}{2x} \).

Step by step solution

01

Rewrite \( \cosh(\ln x) \) using exponential form

The hyperbolic cosine function \( \cosh(y) \) can be expressed as \( \frac{e^y + e^{-y}}{2} \). Substituting \( y = \ln x \), the expression becomes \( \cosh(\ln x) = \frac{e^{\ln x} + e^{-\ln x}}{2} \). Since \( e^{\ln x} = x \) and \( e^{-\ln x} = \frac{1}{x} \), this simplifies to \( \frac{x + \frac{1}{x}}{2} = \frac{x^2 + 1}{2x} \).
02

Rewrite \( \sinh(\ln x) \) using exponential form

The hyperbolic sine function \( \sinh(y) \) can be expressed as \( \frac{e^y - e^{-y}}{2} \). Substituting \( y = \ln x \), the expression becomes \( \sinh(\ln x) = \frac{e^{\ln x} - e^{-\ln x}}{2} \). Simplifying \( e^{\ln x} = x \) and \( e^{-\ln x} = \frac{1}{x} \), we get \( \frac{x - \frac{1}{x}}{2} = \frac{x^2 - 1}{2x} \).
03

Rewrite \( \tanh(2 \ln x) \) using exponential form

The hyperbolic tangent function \( \tanh(y) \) is \( \frac{\sinh(y)}{\cosh(y)} \). Using \( \tanh(2 \ln x) \) equals \( \frac{\sinh(2 \ln x)}{\cosh(2 \ln x)} \), and applying double angle identities: \( \sinh(2y) = 2\sinh(y)\cosh(y) \) and \( \cosh(2y) = \cosh^2(y) + \sinh^2(y) \). Substitute \( y = \ln x \): \( \tanh(2 \ln x) = \frac{2\cdot \frac{x^2 - 1}{2x} \cdot \frac{x^2 + 1}{2x}}{\left(\frac{x^2 + 1}{2x}\right)^2 + \left(\frac{x^2 - 1}{2x}\right)^2} \). Simplifying, we obtain \( \tanh(2 \ln x) = \frac{4x^2(x^4 - 1)}{2(x^4 + 6x^2 + 1)} \) which further reduces to \( \frac{x^4 - 1}{x^4 + 6x^2 + 1} \).
04

Rewrite \( \cosh(-\ln x) \) using exponential form

The function \( \cosh(-y) \) is equal to \( \cosh(y) \). Therefore, \( \cosh(-\ln x) = \cosh(\ln x) \). From Step 1, this is \( \frac{x^2 + 1}{2x} \).

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Key Concepts

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

Exponential Form
Hyperbolic functions, such as the hyperbolic cosine, sine, and tangent, are closely related to exponential functions. Their definitions rely on the fundamental properties of exponentials. Understanding this "exponential form" is key to transforming and simplifying expressions in calculus, algebra, and other mathematical settings.

For example, the hyperbolic cosine function, written as \( \cosh(y) \), has an exponential form: \[ \cosh(y) = \frac{e^y + e^{-y}}{2} \].

Similarly, the hyperbolic sine \( \sinh(y) \) can be expressed as:\[ \sinh(y) = \frac{e^y - e^{-y}}{2} \],
and the hyperbolic tangent \( \tanh(y) \) is the ratio of these two functions:\[ \tanh(y) = \frac{\sinh(y)}{\cosh(y)} \].

These expressions use the natural exponent \( e \), where \( e^y \) and \( e^{-y} \) are exponential functions. Usually, for practical purposes, substituting specific values or expressions, like \( y = \ln x \), allows us to rewrite hyperbolic functions in simpler or polynomial forms.
Polynomial Ratio
When hyperbolic functions are expressed in exponential forms, it often becomes possible to rewrite them as ratios of polynomials. This transformation simplifies many mathematical problems, especially those involving calculus.

Consider the transformation of \( \cosh(\ln x) \):
  • Start with \( \cosh(y) = \frac{e^y + e^{-y}}{2} \).
  • Substitute \( y = \ln x \) to get \( \cosh(\ln x) = \frac{x + \frac{1}{x}}{2} \).
  • This simplifies to \( \frac{x^2 + 1}{2x} \), which is a polynomial ratio.
Similarly, for \( \sinh(\ln x) \):
  • Using \( \sinh(y) = \frac{e^y - e^{-y}}{2} \), substitute to get \( \frac{x - \frac{1}{x}}{2} \).
  • This simplifies further to \( \frac{x^2 - 1}{2x} \).
With these transformations, you can handle otherwise complex expressions with much simpler polynomial ratios which are easier to manipulate in algebraic calculations.
Hyperbolic Cosine
The hyperbolic cosine function, denoted as \( \cosh(y) \), is akin to its trigonometric counterpart but derived from exponential functions.

In exponential form, it is defined by: \[ \cosh(y) = \frac{e^y + e^{-y}}{2} \],where \( e \) is the base of the natural logarithm. This symmetric form allows us to easily derive properties of the function.

For solving problems, it's often useful to simplify into a ratio of polynomials. For instance, \( \cosh(\ln x) \) simplifies to a function of \( x \), given by:
  • \( \frac{x^2 + 1}{2x} \).
One key property is that \( \cosh(-y) = \cosh(y) \), which simplifies calculations when negative arguments are involved, as seen in \( \cosh(-\ln x) = \cosh(\ln x) \).

Hyperbolic cosine is crucial in many mathematical areas, including hyperbolic geometry and complex analysis, giving it broad applicability beyond just algebraic expressions.
Hyperbolic Sine
The hyperbolic sine, represented as \( \sinh(y) \), is another fundamental hyperbolic function that plays a significant role in both pure and applied mathematics.

Its exponential form: \[ \sinh(y) = \frac{e^y - e^{-y}}{2} \]uniquely defines its behaviors and properties. This function represents the odd component in hyperbolic equations, given that \( \sinh(-y) = -\sinh(y) \).

When applying transformations like \( y = \ln x \), these functions convert beautifully into polynomials, such as:
  • \( \sinh(\ln x) = \frac{x^2 - 1}{2x} \).
Understanding these transformations can help solve differential equations and model real-world problems like heat transfer or relativistic physics, where \( \sinh \) naturally arises.
Hyperbolic Tangent
Hyperbolic Tangent, symbolized by \( \tanh(y) \), is among the most applied hyperbolic functions, particularly due to its role in sigmoid and logistic functions in statistics and neural networks.

Defined as the ratio of \( \sinh(y) \) to \( \cosh(y) \), it is given by:\[ \tanh(y) = \frac{\sinh(y)}{\cosh(y)} = \frac{e^y - e^{-y}}{e^y + e^{-y}} \].

In practical transformations, like converting \( \tanh(2 \ln x) \), it becomes crucial to use trigonometric identities:
  • Using double angle formulas, \( \tanh(2 \ln x) = \frac{x^4 - 1}{x^4 + 6x^2 + 1} \).
This polynomial form helps analyze hyperbolic identities or integrate \( \tanh \) in calculus problems better.

Its bounded range between -1 and 1 makes it suitable for normalization processes, enhancing its application scope beyond theoretical mathematics into practical computations in engineering and data science.

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Most popular questions from this chapter

Writing Faced with the problem of computing the volume of a solid of revolution, how would you go about deciding whether to use the method of disks/washers or the method of cylindrical shells?

Use cylindrical shells to find the volume of the torus obtained by revolving the circle \(x^{2}+y^{2}=a^{2}\) about the line \(x=b,\) where \(b>a>0 .\) [Hint: It may help in the integration to think of an integral as an area.]

Suppose that a hollow tube rotates with a constant angular velocity of \(\omega\) rad/s about a horizontal axis at one end of the tube, as shown in the accompanying figure. Assume that an object is free to slide without friction in the tube while the tube is rotating. Let \(r\) be the distance from the object to the pivot point at time \(t \geq 0,\) and assume that the object is at rest and \(r=0\) when \(t=0\). It can be shown that if the tube is horizontal at time \(t=0\) and rotating as shown in the figure, then $$ r=\frac{g}{2 \omega^{2}}[\sinh (\omega t)-\sin (\omega t)] $$ during the period that the object is in the tube. Assume that \(t\) is in seconds and \(r\) is in meters, and use \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) and \(\omega=2 \mathrm{rad} / \mathrm{s}\) $$ \begin{array}{l}{\text { (a) Graph } r \text { versus } t \text { for } 0 \leq t \leq 1} \\ {\text { (b) Assuming that the tube has a length of } 1 \mathrm{m} \text { , approxi- }} \\ {\text { mately how long does it take for the object to reach the }} \\ {\text { end of the tube? }} \\ {\text { (c) Use the result of part (b) to approximate } d r / d t \text { at the }} \\\ {\text { instant that the object reaches the end of the tube. }}\end{array} $$

Find \(d y / d x\) $$ y=e^{x} \operatorname{sech}^{-1} \sqrt{x} $$

In each part, a value for one of the hyperbolic functions is given at an unspecified positive number \(x_{0} .\) Use appropriate identities to find the exact values of the remaining five hyperbolic functions at \(x_{0}\). $$ \text { (a) } \sinh x_{0}=2 \quad \text { (b) } \cosh x_{0}=\frac{5}{4} \quad \text { (c) } \tanh x_{0}=\frac{4}{5} $$

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