Chapter 3.2, Problem 61E

Solution

a.

To prove: The given identity.

  $\tanh \left(x\right)=\frac{\sinh \left(x\right)}{\cosh \left(x\right)}$ .

Given:

The identity is $\tanh \left(x\right)=\frac{\sinh \left(x\right)}{\cosh \left(x\right)}$ .

Formula used:

  $\sinh \left(x\right)=\frac{e^x-e^{-x}}{2},\cosh \left(x\right)=\frac{e^x+e^{-x}}{2},\tanh \left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$ .

Calculation:

Consider,

  $\begin{array}{c}\frac{\sinh \left(x\right)}{\cosh \left(x\right)}=\frac{\frac{e^x-e^{-x}}{2}}{\frac{e^x+e^{-x}}{2}}\\ =\frac{e^x-e^{-x}}{e^x+e^{-x}}\\ =\tanh \left(x\right)\end{array}$

Hence, $\tanh \left(x\right)=\frac{\sinh \left(x\right)}{\cosh \left(x\right)}$ .

b.

To prove: That the hyperbolic tangent function is an odd function.

The hyperbolic tangent function is an odd function.

Given:

  $\tanh \left(x\right)=\frac{\sinh \left(x\right)}{\cosh \left(x\right)}$ .

Concept used:

A function $f\left(x\right)$ is said to be an odd function if $f\left(-x\right)=-f\left(x\right)$ .

Calculation:

Consider,

  $\begin{array}{c}\tanh \left(-x\right)=\frac{\sinh \left(-x\right)}{\cosh \left(-x\right)}\\ =\frac{-\sinh \left(x\right)}{\cosh \left(x\right)}\\ =-\tanh \left(x\right)\end{array}$

Hence, the hyperbolic tangent function is an odd function.

c.

To prove: That the given function is a logistic function.

  $f\left(x\right)=1+\tanh \left(x\right)$ is a logistic function.

Given:

  $f\left(x\right)=1+\tanh \left(x\right)$ .

Formula used:

  $\tanh \left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$ .

Calculation:

Consider,

  $\begin{array}{c}f\left(x\right)=1+\tanh \left(x\right)\\ =1+\frac{e^x-e^{-x}}{e^x+e^{-x}}\\ =\frac{\left(e^x+e^{-x}\right)+\left(e^x-e^{-x}\right)}{e^x+e^{-x}}\\ =\frac{2e^x}{e^x+e^{-x}}\\ =\frac{2e^x}{e^x+e^{-x}}\cdot \frac{e^{-x}}{e^{-x}}\\ =\frac{2}{1+e^{-2x}}\end{array}$

Since it is in the form of $f\left(x\right)=\frac{a}{1+e^{-kx}}$ , therefore it is a logistic function.

Hence, $f\left(x\right)=1+\tanh \left(x\right)$ is a logistic function.