Problem 64 - Thomas Calculus in SI Units

Question

When hyperbolic function keys are not available on a calculator, it is still possible to evaluate the inverse hyperbolic functions by expressing them as logarithms, as shown here. (We give one derivation in Section 8.4.) $$\begin{aligned}&\sinh ^{-1} x=\ln (x+\sqrt{x^{2}+1}), \quad-\infty < x<\infty\\\&\cosh ^{-1} x=\ln (x+\sqrt{x^{2}-1}), \quad x \geq 1\\\ &\tanh ^{-1} x=\frac{1}{2} \ln \frac{1+x}{1-x}, \quad|x| < 1\\\&\operatorname{sech}^{-1} x=\ln \left(\frac{1+\sqrt{1-x^{2}}}{x}\right), \quad0 < x \leq 1\\\&\operatorname{csch}^{-1} x=\ln \left(\frac{1}{x}+\frac{\sqrt{1+x^{2}}}{|x|}\right), \quad x \neq 0\\\&\operatorname{coth}^{-1} x=\frac{1}{2} \ln \frac{x+1}{x-1}, \quad |x|>1\end{aligned}$$ Use the formulas in the box here to express the numbers in terms of natural logarithms. $$\operatorname{coth}^{-1}(5 / 4)$$

Step-by-Step Solution

Answer

\( \operatorname{coth}^{-1} \left( \frac{5}{4} \right) = \frac{1}{2} \ln 9 \).

Step 1: Identify the appropriate formula

We need to express \( \operatorname{coth}^{-1}(5 / 4) \) in terms of natural logarithms. The formula provided for \( \operatorname{coth}^{-1} x \) is \( \frac{1}{2} \ln \frac{x+1}{x-1} \), which applies when \( |x| > 1 \). Confirm that \( \frac{5}{4} > 1 \), so this formula is applicable.

Step 2: Substitute the value into the formula

Substitute \( x = \frac{5}{4} \) into the formula: \( \operatorname{coth}^{-1} x = \frac{1}{2} \ln \frac{x+1}{x-1} \). This gives us: \[ \operatorname{coth}^{-1} \left( \frac{5}{4} \right) = \frac{1}{2} \ln \frac{\frac{5}{4}+1}{\frac{5}{4}-1} \]

Step 3: Simplify the expression inside the logarithm

First, calculate \( \frac{5}{4} + 1 \) and \( \frac{5}{4} - 1 \). We find: \( \frac{5}{4} + 1 = \frac{9}{4} \) and \( \frac{5}{4} - 1 = \frac{1}{4} \). Substitute these values back into the expression: \[ \frac{9/4}{1/4} = 9 \]

Step 4: Complete the expression

Now substitute the simplified fraction back into the expression for \( \operatorname{coth}^{-1} \): \[ \operatorname{coth}^{-1} \left( \frac{5}{4} \right) = \frac{1}{2} \ln 9 \]

Step 5: Final Result

Thus, \( \operatorname{coth}^{-1}(5 / 4) \) is expressed in natural logarithms as \( \frac{1}{2} \ln 9 \).

Key Concepts

Natural Logarithms

Natural logarithms, denoted as \( \ln \), are a fundamental concept in mathematics, specifically within calculus and algebra. A natural logarithm uses the base \( e \), where \( e \approx 2.71828 \) is an irrational number, similar to \( \pi \).
When we say \( \ln(x) \), it means we're finding the power to which \( e \) must be raised to get \( x \). For example, \( \ln(e) = 1 \) because \( e^1 = e \).
Natural logarithms are interconnected with exponential functions, which makes them extremely powerful and essential in various mathematical formulas. They appear frequently in logarithmic expressions of inverse hyperbolic functions and can help simplify complex numbers.
Understanding how to manipulate natural logarithms is crucial, especially for students dealing with advanced math topics, as they allow inverse hyperbolic functions to be expressed without the direct need for hyperbolic function keys on a calculator. This makes computations more accessible when tools are limited.

Coth Inverse

The inverse hyperbolic cotangent function, denoted as \( \operatorname{coth}^{-1}(x) \), plays a critical role in mathematics, especially when discussing inverse hyperbolic functions. It's defined only for values where the absolute value of \( x \), \( |x| \), is greater than 1. This domain ensures the function behaves accurately.
The inverse cotangent can be expressed using natural logarithms as \( \operatorname{coth}^{-1} x = \frac{1}{2} \ln \frac{x+1}{x-1} \). This relationship helps perform calculations without directly using hyperbolic function calculators.
For clarity, when finding the \( \operatorname{coth}^{-1} \) of a number, like \( \operatorname{coth}^{-1}(5/4) \), substitute \( x \) into the equation:

• Check if \( x > 1 \) or \( x < -1 \) to ensure it's within the domain.
• Substitute the value into the formula.
• Simplify using basic arithmetic to make it easy to calculate the logarithm.

Once simplified, like in the example \( \operatorname{coth}^{-1}(5/4) = \frac{1}{2} \ln 9 \), the outcome is a form easily computable using any basic calculator, thanks to its expression in terms of \( \ln \).

Hyperbolic Identities

Hyperbolic identities serve as the backbone for understanding hyperbolic functions and their inverses. These identities are analogous to trigonometric identities, providing relationships among hyperbolic functions that simplify problem-solving.
The hyperbolic sine function \( \sinh(x) \), hyperbolic cosine \( \cosh(x) \), and hyperbolic tangent \( \tanh(x) \), among others, form a set of connected functions similar to sine, cosine, and tangent in trigonometry.
These functions have identities such as:

• \( \cosh^2(x) - \sinh^2(x) = 1 \)
• \( 1 - \tanh^2(x) = \text{sech}^2(x) \)

Inverse hyperbolic functions extend these identities into avenues where one can express them using logarithmic expressions. This is crucial for scenarios where calculators restrict hyperbolic keys, as these identities ensure one can still calculate properties of functions that involve cosh, sinh, tanh, etc.
Understanding these relationships allows students to manipulate and convert expressions to their logarithmic forms, ultimately simplifying computations in their mathematical studies and broader applications.