1.7 Hyperbolic Functions

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\ffrac }[2]{\frac {\text {\footnotesize $#1$}}{\text {\footnotesize $#2$}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\C }[0]{\mathbb C} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\dis }[0]{\displaystyle } \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\ppx }[0]{\frac {\partial }{\partial x}} \newcommand {\ppy }[0]{\frac {\partial }{\partial y}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{ #1 \bigg |} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{proj} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\cis }{cis} \DeclareMathOperator {\Arg }{Arg} \DeclareMathOperator {\Rep }{Re} \DeclareMathOperator {\Imp }{Im} \DeclareMathOperator {\sech }{sech} \DeclareMathOperator {\csch }{csch} \DeclareMathOperator {\Log }{Log} \newcommand {\tightoverset }[2]{\mathop {#2}\limits ^{\vbox to -.5ex{\kern -0.75ex\hbox {$#1$}\vss }}} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\utan }[0]{\vec {\hat {t}}} \newcommand {\unormal }[0]{\vec {\hat {n}}} \newcommand {\ubinormal }[0]{\vec {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\vector }[1]{\left \langle #1\right \rangle }$

We learn the basic properties of the hyperbolic functions.

The hyperbolc sine and hyperbolic cosine functions are defined as $\sinh x = \frac {e^x - e^{-x}}{2} \;\; \mbox {and} \;\; \cosh x = \frac {e^x + e^{-x}}{2}$

example 1 Solve the equation $\cosh x = 6$ for $x$ .
We use the definition of the hyperbolic cosine: $\cosh x = \frac {e^x + e^{-x}}{2} = 6$ Ergo $e^x +e^{-x} = 12$ Multiplying by $e^x$ , we get $e^{2x} - 12e^x +1 = 0 \;\; \mbox {or} \;\; u^2 - 12u + 1 = 0$ where $u = e^x$ . Now the quadratic formula gives $u = \frac {12 \pm \sqrt {144 - 4}}{2} = 6 \pm \sqrt {35}$ Finally, we get the two solutions by taking the natural logarithm of $u$ : $x = \ln \left (6 \pm \sqrt {35}\right )$ These answers are negatives of one another (verify).

[problem 1a] Solve the following equation $\cosh x = 3$
The answers are (in increasing order):
$x = \answer {\ln (3-2\sqrt 2)}$ and $x = \answer {\ln (3+2\sqrt 2)}$

[problem 1b] Solve the following equation $\sinh x = 3$
The answer is:
$x = \answer {\ln (3+\sqrt {10})}$

(problem 1c) Find inverses for the hyperbolic sine and hyperbolic cosine functions.

Here is a video solution for problem 1a:

The hyperbolic sine function is an odd function: $\sinh (-x) = -\sinh (x)$ and the hyperbolic cosine function is even: $\cosh (-x) = -\cosh (x)$

The hyperbolic sine and cosine satisfy the fundamental identity $\cosh ^2 x - \sinh ^2 x =1$ which means that the point $(\cosh t, \sinh t)$ lies on the (right branch of) the hyperbola $x^2 -y^2 = 1$ This is why the functions are referred to as the hyperbolic functions.

The other four hyperbolic functions can be created from the hyperbolic sine and hyperbolic cosine functions: $\tanh x = \frac {\sinh x}{\cosh x},\;\; \coth x = \frac {\cosh x}{\sinh x},$ $\sech x = \frac {1}{\cosh x}, \;\; \mbox {and}\;\; \csch x = \frac {1}{\sinh x}$

example 2 Find the derivative of the hyperbolic sine function.
The derivative of the hyperbolic sine function is given by $\frac {d}{dx} \sinh (x) = \frac {d}{dx} \left ( \frac {e^x -e^{-x}}{2} \right ) = \frac {e^x + e^{-x}}{2} = \cosh x$

(problem 2) Find the derivatives of the other 5 hyperbolic functions.
$\frac {d}{dx} \cosh x = \answer {\sinh x}$
$\frac {d}{dx} \tanh x = \answer {\sech ^2 x}$
$\frac {d}{dx} \sech x = \answer {-\sech x \tanh x}$
$\frac {d}{dx} \coth x = \answer {-\csch ^2 x}$
$\frac {d}{dx} \csch x = \answer {-\csch x \coth x}$

example 3 Find the Maclaurin series representation for the hyperbolic cosine.
Recall that the Maclaurin series for $e^x$ is given by $e^x = \sum _{n=0}^\infty \frac {x^n}{n!}$ Thus $\cosh x = \frac {e^x + e^{-x}}{2} = \frac 12 \sum _{n=0}^\infty \frac {x^n}{n!} + \frac 12 \sum _{n=0}^\infty \frac {(-x)^n}{n!}$ $= \sum _{\substack {n=0\\ n\; even}}^\infty \frac {x^n}{n!} = \sum _{n=0}^\infty \frac {x^{2n}}{(2n)!}$

(problem 3) Find the Maclaurin series for the hyperbolic sine function.

$\sinh x=\displaystyle {\sum _{n=0}^\infty (-1)^n \frac {x^{2n+1}}{(2n+1)!}}$ $\sinh x=\displaystyle {\sum _{n=0}^\infty \frac {x^{2n+1}}{ n!}}$ $\sinh x=\displaystyle {\sum _{n=0}^\infty \frac {x^{2n+1}}{(2n+1)!}}$

Here is a video solution for problem 3:

2024-09-27 14:06:37