Question

Prove the expression for \(\sinh ^{-1}(x)\). Multiply \(x=\sinh (y)=(1 / 2)\left(e^{y}-e^{-y}\right)\) by \(2 e^{y}\) and solve for \(y\). Does your expression match the textbook?

Short answer

The expression matches: \(\sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1})\).

Step-by-step solution

Define Hyperbolic Sine Function

The hyperbolic sine function is defined as \(\sinh(y) = \frac{1}{2}(e^y - e^{-y})\). We know, \(x = \sinh(y)\).

Multiply by Exponential Term

Multiply both sides of the equation \(x = \frac{1}{2}(e^y - e^{-y})\) by \(2e^y\) to eliminate the fraction:\[2e^y x = e^{2y} - 1\].

Rearrange Equation to Form a Quadratic

Rearrange the equation \(2e^y x = e^{2y} - 1\) as a quadratic in terms of \(e^y\):\[e^{2y} - 2xe^y - 1 = 0\].

Solve the Quadratic Equation

Solve the quadratic \(e^{2y} - 2xe^y - 1 = 0\) using the quadratic formula:\\[e^y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].Here, \(a = 1\), \(b = -2x\), and \(c = -1\).Plug in the values:\[e^y = \frac{-(-2x) \pm \sqrt{(-2x)^2 - 4(1)(-1)}}{2(1)}\].Simplifying gives:\[e^y = \frac{2x \pm \sqrt{4x^2 + 4}}{2}\] which simplifies further to:\[e^y = x \pm \sqrt{x^2 + 1}\].

Choose the Appropriate Solution

Since \(e^y\) is positive, choose the positive solution:\[e^y = x + \sqrt{x^2 + 1}\].

Solve for \(y\)

Take the natural logarithm on both sides to solve for \(y\):\[y = \ln(x + \sqrt{x^2 + 1})\].

Verify Expression Match with Textbook

The expression for \(y\) is \(\sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1})\). This matches the textbook formula for the inverse hyperbolic sine function.

Key concepts

Hyperbolic Sine

The hyperbolic sine function, denoted as \( \sinh(y) \), is akin to the well-known sine function but is used in hyperbolic geometry. It is defined by the expression \( \sinh(y) = \frac{1}{2}(e^y - e^{-y}) \). This function takes into account exponential growth and decay, making it quite useful in various mathematical contexts beyond simple trigonometry.

Hyperbolic sine relates to the exponential function which involves the constant \( e \), necessary in calculus due to its naturally occurring properties. The constants in this function, namely the exponents, drive the behavior of hyperbolic functions. It's important to note that the inverse of \( \sinh(y) \), which you might encounter as \( \sinh^{-1}(x) \), provides a way to "undo" the hyperbolic effect, much like the arcsine for sine functions.

Quadratic Equation

A quadratic equation is fundamental in algebra and has the general form \( ax^2 + bx + c = 0 \). In the context of our exercise, this equation arises naturally when manipulating expressions in hyperbolic functions. Here, the transformation \( 2e^y x = e^{2y} - 1 \) is rearranged into the quadratic equation \( e^{2y} - 2xe^y - 1 = 0 \).

Dealing with quadratics involves the well-known quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which provides solutions for \( x \) when \( a \), \( b \), and \( c \) are constants. For our problem, the values are \( a = 1 \), \( b = -2x \), and \( c = -1 \). Solving these quadratics is crucial in many fields, including engineering and physics, due to their predictability and precision.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is the inverse of the exponential function \( e^x \). Basically, it answers the question: "To what power must \( e \), the base of natural logarithms, be raised, to get \( x \)?" In the solution provided, taking the natural logarithm of both sides helps to solve for \( y \) after finding the expression \( e^y \).

When the equation \( e^y = x + \sqrt{x^2 + 1} \) is confronted with \( \ln \), it simplifies finding \( y \). The expression \( y = \ln(x + \sqrt{x^2 + 1}) \) ultimately tells us that the inverse hyperbolic sine, \( \sinh^{-1}(x) \), is expressed using a natural logarithm for clarity and simplicity in mathematical writing. This step is crucial because logarithms are essential in solving equations involving exponential growth, decay, and time-related applications.

Proof Techniques

Proof techniques are methods used to demonstrate the truth of a given statement in mathematics. In this exercise, several proof techniques are used in combination. The first is algebraic manipulation which transforms the given expression into a quadratic equation.

After forming the quadratic, solving it through substitution and application of the quadratic formula is a systematic way to find solutions. The choice of solution, dictated by the requirement that \( e^y \) must be positive, employs a logical proof reasoning called deduction. By ensuring the solution aligns with known properties of exponentials and logarithms, the final expression is verified against the textbook, demonstrating the concept of verification in proof. Through these methods, accuracy and understanding of the proof are achieved, showcasing why such techniques are a pillar in mathematical problem-solving.