Question

Verify the formulas. $${\tanh }^{-1}z=\frac{1}{2}\ln \frac{1+z}{1-z}$$

Short answer

${\tanh }^{-1}z=\frac{1}{2}\ln \frac{1+z}{1-z}$

Step-by-step solution

Recall the definition of hyperbolic tangent

The hyperbolic tangent function is defined as \ $\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}$. To find the inverse, we want to express $x$ in terms of $z$.

Set $z=\tanh x$

Given $z=\tanh x$, substitute $z$ in the hyperbolic tangent definition: $z=\frac{e^x-e^{-x}}{e^x+e^{-x}}$.

Solve for $e^x$

Multiply both sides by $e^x+e^{-x}$: \ $z(e^x+e^{-x})=e^x-e^{-x}$. Next, collect all terms involving $e^x$ on one side: \ $z{e}^x+z{e}^{-x}=e^x-e^{-x}$ \ Then rearrange to get: \ $e^x(z-1)=-e^{-x}(z+1)$.

Simplify the equation

Multiply both sides by $e^x$: \ $e^{2x}(z-1)=-(z+1)$. Then solve for $e^{2x}$: \ $e^{2x}=\frac{-(z+1)}{z-1}$.

Take the natural logarithm

Taking the natural logarithm of both sides, we get: \ $2x=\ln \left(\frac{-(z+1)}{z-1}\right)$. To remove the negative sign, \ $2x=\ln \left(\frac{1+z}{1-z}\right)$, since $z$ is assumed real and $zeq\pm 1$.

Solve for $x$

Finally, solve for $x$ by dividing both sides by 2: \ $x=\frac{1}{2}\ln \left(\frac{1+z}{1-z}\right)$.

Key concepts

hyperbolic tangent

The hyperbolic tangent function, often denoted as $\tanh$, is similar to the tangent function but for hyperbolic angles. Just like sine and cosine have their hyperbolic counterparts $\sinh$ and $\cosh$, tangent has $\tanh$. The hyperbolic tangent is defined as follows:
$\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}$.

Where:

• $e$: The base of the natural logarithm, approximately equal to 2.718.
• $x$: The variable or angle in hyperbolic terms.

Using this definition, you can see how it behaves similarly to the regular tangent function but in an exponential context. The range of $\tanh x$ is between -1 and 1, which makes it useful for various applications in mathematics and physics.

natural logarithm

The natural logarithm, denoted as $\ln$, is a fundamental concept in mathematics, especially in calculus and algebra. It is the inverse operation of the exponential function involving the constant $e$. Mathematically, it is represented as:
$\ln x$ where $e^{\ln x}=x$

Important properties of the natural logarithm include:

• $\ln 1=0$: Because $e^0=1$.
• $\ln (xy)=\ln x+\ln y$: Logarithms convert multiplication into addition.
• $\ln \left(\frac{x}{y}\right)=\ln x-\ln y$: Logarithms convert division into subtraction.
• $\ln (x^n)=n\ln x$: Logarithms convert exponentiation into multiplication.

Natural logarithms are particularly useful in solving equations involving exponential functions, which is exactly what we do when finding the inverse of the hyperbolic tangent function.

inverse functions

An inverse function reverses the operation done by a specific function. If you have a function $f$ that maps an input $x$ to an output $y$, then the inverse function, denoted as $f^{-1}$, maps $y$ back to $x$. For the hyperbolic tangent function $\tanh$, its inverse is denoted as ${\tanh }^{-1}$ or arctanh.

To find the inverse of $\tanh x$, we start with: $z=\tanh x$
which can be rewritten using the definition:
$z=\frac{e^x-e^{-x}}{e^x+e^{-x}}$
We then solve this equation for $x$ in terms of $z$. The steps involve algebraic manipulation (like multiplying both sides by $e^x+e^{-x}$, and using logarithms) to isolate $x$. Finally, we get:
$x={\tanh }^{-1}z=\frac{1}{2}\ln \left(\frac{1+z}{1-z}\right)$.

Inverse functions are crucial in many mathematical operations, especially when you need to 'undo' a function.

algebraic manipulation

Algebraic manipulation involves rearranging and simplifying equations using fundamental algebraic operations like addition, subtraction, multiplication, and division. In the context of finding the inverse of the hyperbolic tangent function, several steps leverage these techniques. Let's break down a few key manipulations:

• Multiplying both sides by common terms to eliminate denominators.
• Combining like terms and isolating variables.
• Utilizing exponential rules, such as $e^x\cdot e^{-x}=1$.
• Applying properties of logarithms to simplify expressions.

For example, in the step to solve for $e^x$:
$z=\frac{e^x-e^{-x}}{e^x+e^{-x}}$
we multiply both sides by $e^x+e^{-x}$ to clear the denominator, and finally use logarithms to express the solution in a more simplified form.

Mastering algebraic manipulation is essential for solving complex equations and understanding higher-level mathematics.