Problem 14 - Essential Calculus Early Transcendentals

Question

Prove the identity. $$ \frac{1+\tanh x}{1-\tanh x}=e^{2 x} $$

Step-by-Step Solution

Answer

The identity is proven: \( \frac{1 + \tanh x}{1 - \tanh x} = e^{2x} \).

Step 1: Recall the Definition of Hyperbolic Tangent

The hyperbolic tangent function is defined as \( \tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}} \). Use this definition to express \( 1 + \tanh x \) and \( 1 - \tanh x \).

Step 2: Simplify the Numerator and Denominator

Calculate \( 1 + \tanh x \) and \( 1 - \tanh x \):\[1 + \tanh x = 1 + \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{(e^x + e^{-x}) + (e^x - e^{-x})}{e^x + e^{-x}} = \frac{2e^x}{e^x + e^{-x}}\]\[1 - \tanh x = 1 - \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{(e^x + e^{-x}) - (e^x - e^{-x})}{e^x + e^{-x}} = \frac{2e^{-x}}{e^x + e^{-x}}\]

Step 3: Form the Fraction

Substitute the results from Step 2 into the expression \( \frac{1 + \tanh x}{1 - \tanh x} \):\[\frac{1 + \tanh x}{1 - \tanh x} = \frac{\frac{2e^x}{e^x + e^{-x}}}{\frac{2e^{-x}}{e^x + e^{-x}}}\]

Step 4: Simplify the Expression

Simplify the fraction by cancelling \( e^x + e^{-x} \) in the numerator and denominator:\[= \frac{2e^x}{2e^{-x}} = \frac{e^x}{e^{-x}} = e^{x - (-x)} = e^{2x}\]

Step 5: Final Result Confirmation

The left-hand side simplifies to \( e^{2x} \), which matches the right-hand side of the equation. Thus, the identity \( \frac{1+\tanh x}{1-\tanh x}=e^{2 x} \) is proven.

Key Concepts

Hyperbolic Tangent

The hyperbolic tangent, denoted as \( \tanh x \), is a fundamental function in hyperbolic trigonometry. It bears resemblance to the traditional tangent function in circular trigonometry but is based on hyperbolas rather than circles. The definition of \( \tanh x \) is given by the ratio:

• \( \tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}} \)

This formula uses the exponential function to describe a hyperbolic feature. The hyperbolic tangent relates to many areas in mathematics and physics, especially where modeling of real-world phenomena involving hyperbolas is needed. For our problem, it's important to understand how \( \tanh x \) can be expressed in terms of exponential functions for further simplification.

Exponential Function

The exponential function, \( e^x \), is a powerful mathematical construct with unique properties. It forms the basis for defining the hyperbolic functions, including \( \tanh x \). Essentially, \( e^x \) describes continuous growth or decay and is defined as:

• Continuous and can be raised to any real number \( x \).
• Its derivative and integral are both itself, which is a unique property.

In solving identities involving hyperbolic functions, the manipulation of exponential terms like \( e^x \) and \( e^{-x} \) plays a pivotal role. Expressions like \( \frac{1+\tanh x}{1-\tanh x} \) can be simplified by converting \( \tanh x \) into its exponential form, ultimately leading to simplified equations.

Algebraic Simplification

Algebraic simplification involves rearranging and reducing expressions into their simplest form. In the problem at hand, we start by handling complex fractions containing the hyperbolic tangent. The steps are:

• Combine like terms to simplify top and bottom parts separately.
• Use common denominators to achieve a single fraction representation.
• Cancel common terms to reduce complexity.

In our step-by-step proof, we condense the expression \( 1 + \tanh x \) and \( 1 - \tanh x \) into simpler forms by utilizing shared terms and cancelling them out. This results in a straightforward expression \( \frac{e^{2x}}{1} = e^{2x} \), completing the simplification process.

Identity Proof

In mathematics, an identity is an equation that is true for all permissible values of its variable. Proving an identity, like \( \frac{1+\tanh x}{1-\tanh x} = e^{2x} \), requires showing that two expressions are equivalent. This involves:

• Understanding the functions involved: here hyperbolic and exponential.
• Using algebraic simplification to transition from a complex form to a known result.
• Ensuring each step logically follows to maintain equivalence.

By substituting definitions and simplifying expressions carefully, the identity proof shows that both sides of the equation behave identically for any value of \( x \). This demonstrates the power of mathematical manipulation in exploring equivalences.