What is Hyperbolic functions: Definition and 81 Discussions
In mathematics, hyperbolic functions are analogues of the ordinary trigonometric functions, but defined using the hyperbola rather than the circle. Just as the points (cos t, sin t) form a circle with a unit radius, the points (cosh t, sinh t) form the right half of the unit hyperbola. Also, just as the derivatives of sin(t) and cos(t) are cos(t) and –sin(t), the derivatives of sinh(t) and cosh(t) are cosh(t) and +sinh(t).
Hyperbolic functions occur in the calculations of angles and distances in hyperbolic geometry. They also occur in the solutions of many linear differential equations (such as the equation defining a catenary), cubic equations, and Laplace's equation in Cartesian coordinates. Laplace's equations are important in many areas of physics, including electromagnetic theory, heat transfer, fluid dynamics, and special relativity.
The basic hyperbolic functions are:
hyperbolic sine "sinh" (),
hyperbolic cosine "cosh" (),from which are derived:
hyperbolic tangent "tanh" (),
hyperbolic cosecant "csch" or "cosech" ()
hyperbolic secant "sech" (),
hyperbolic cotangent "coth" (),corresponding to the derived trigonometric functions.
The inverse hyperbolic functions are:
area hyperbolic sine "arsinh" (also denoted "sinh−1", "asinh" or sometimes "arcsinh")
area hyperbolic cosine "arcosh" (also denoted "cosh−1", "acosh" or sometimes "arccosh")
and so on.
The hyperbolic functions take a real argument called a hyperbolic angle. The size of a hyperbolic angle is twice the area of its hyperbolic sector. The hyperbolic functions may be defined in terms of the legs of a right triangle covering this sector.
In complex analysis, the hyperbolic functions arise as the imaginary parts of sine and cosine. The hyperbolic sine and the hyperbolic cosine are entire functions. As a result, the other hyperbolic functions are meromorphic in the whole complex plane.
By Lindemann–Weierstrass theorem, the hyperbolic functions have a transcendental value for every non-zero algebraic value of the argument.Hyperbolic functions were introduced in the 1760s independently by Vincenzo Riccati and Johann Heinrich Lambert. Riccati used Sc. and Cc. (sinus/cosinus circulare) to refer to circular functions and Sh. and Ch. (sinus/cosinus hyperbolico) to refer to hyperbolic functions. Lambert adopted the names, but altered the abbreviations to those used today. The abbreviations sh, ch, th, cth are also currently used, depending on personal preference.
I was able to solve with a rather longer way; there could be a more straightforward approach;
My steps are along these lines;
##\sinh^{-1} x = 2 \ln (2+ \sqrt{3})##
##\sinh^{-1} x = \ln (7+ 4\sqrt{3})##
##x = \sinh[ \ln (7+ 4\sqrt{3})]##
##x = \dfrac {e^{\ln (7+ 4 \sqrt{3})} - e^{-[\ln 7+ 4...
My question is on the highlighted part (circled in red);
Why is it wrong to pre-multiply each term by ##e^x##? to realize ,
##5e^{2x} -2-9e^x=0## as opposed to factorising by ##e^{-x} ## ?
The other steps to required solution ##x=\ln 2## is quite clear and straightforward to me.
I found one ages ago about the hyperbolic functions, but it hadn’t been translated to English from German yet. Anyone know of a good book on hyperbolic functions and other transcendental functions besides the circular functions (trigonometric)?
My take;
##2\cosh x = e^x +e^{-x}##
I noted that i could multiply both sides by ##e^x## i.e
##e^x⋅2\cosh x = e^x(e^x +e^{-x})##
##e^x⋅2\cosh x = e^{2x}+1##
thus,
##\dfrac{e^x}{1+e^{2x}}=\dfrac{\cosh x + \sinh x}{e^x⋅2\cosh x}##
##= \dfrac{\cosh x +...
This is a textbook question and i have no solution. My attempt:
We know that ##\cosh x = \dfrac{e^x + e^{-x}}{2}##
and ##\cosh u = \dfrac{{x^2 + 1}}{2x}## it therefore follows that;
##e^{2u} = x^2##
##⇒u = \dfrac {2\ln...
I came across this question; i noted that the hyperbolic trigonometry properties are somewhat similar to what i may call cartesian trigonometry properties...
My approach on this;
##\tanh x = \sinh y##
...just follows from
##y=\sin^{-1}(\tan x)##
##\tan x = \sin y##
Therefore...
While deriving Lorentz transformation equations, my professor assumes the following:
As ##\beta \rightarrow 1,##
$$-c^2t^2 + x^2 = k$$
approaches 0. That is, ##-c^2t^2 + x^2 = 0.## But the equation of the hyperbola is preserved in all inertial frames of reference. Why would ##-c^2t^2 + x^2##...
Hello everyone first time here. don't know if it's the correct group... Am having some issues wiz my maths homework that going to count as a final assessment. Really Really need help.
The function (f), with a period of 2π is : f(x) = cosh(x-2π) if x [π;3π]..
I had to do a graph as the first...
The hyperbolic function ##\cosh t \text{ and } \sinh t## respectively represent the x and y coordinate of the parametric equation of the parabola ##x^2-y^2=1##. The exponential expressions of these hyperbolic functions are
$$
\begin{cases}
\sinh x = {e^x-e^{-x}} /2; \: x \in \mathbb R, \: f(x)...
Here is a tough integral that I'm not quite sure how to do. It's the Gaussian average:
$$
I = \int_{-\infty}^{\infty}dx\, \frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}\sqrt{1+a^2 \sinh^2(b x)}
$$
for ##0 < a < 1## and ##b > 0##. Obviously the integral can be done for ##a = 0## (or ##b=0##) and for...
I am trying to understand why maxwell equations are correct in any reference frames? While i started to understand of his laws of physics a bit i could not imagine why he uses hyperbolic functions such as coshw instead of spherical ones in position and time relation between moving frames...
Homework Statement
For the expression
$$r = \frac{i\kappa\sinh(\alpha L)}{\alpha\cosh(\alpha L)-i\delta\sinh(\alpha L)} \tag{1}$$
Where ##\alpha=\sqrt{\kappa^{2}-\delta^{2}}##, I want to show that:
$$\left|r\right|^{2} = \left|\frac{i\kappa\sinh(\alpha L)}{\alpha\cosh(\alpha...
Mod note: Because his caps-lock key is stuck, it's OK for this post to be in all caps.
FIRSTLY, MY LAPTOP'S CAPS LOCK IS BEHAVING REALLY WEIRD AND I HAVE NO CONTROL ON IT WHATSOEVER. SO SORRY FOR POSTING IN ALL CAPS/ALL SMALL LETTERS
I HAVE RECENTLY LEARNED HYPERBOLIC FUNCTIONS. HOWEVER, I AM...
Homework Statement
Can this function be integrated analytically?
##f=\exp \left(-\frac{e^{-2 \theta } \left(a \left(b^2 \left(e^{2 \theta }-1\right)^2 L^2+16\right)-32
\sqrt{a} e^{\theta }+16 e^{2 \theta }\right)}{b L^4}\right),##
where ##a##, ##b## and ##L## are some real positive...
Homework Statement
Show that the tangent to ##x^2-y^2=1## at points ##x_1=\cosh (u)## and ##y_1=\sinh(u)## cuts the x-axis at ##{\rm sech(u)}## and the y-axis at ##{\rm -csch(u)}##.
Homework Equations
Hyperbolic sine: ##\sinh (u)=\frac{1}{2}(e^u-e^{-u})##
Hyperbolic...
I noticed the graphs of ##y=\cos^{-1}x## and ##y=\cosh^{-1}x## are similar in the sense that the real part of one is the imaginary part of the other. This is true except when ##x<-1## where the imaginary part of ##y=\cos^{-1}x## is negative but the real part of ##y=\cosh^{-1}x## is positive.
I...
Homework Statement
Split the function f(x) = ex + πe−x into odd and even parts, and express your result in terms of cosh x and sinh x.
Homework Equations
f(x) = 0.5[f(x) + f(-x)] +0.5[f(x) - f(-x)]
The Attempt at a Solution
So i know that:
ex = 0.5[ex - e-x] + 0.5[ex + e-x] = sinh(x) +...
Good afternoon,
i was just wondering if this equation is possibly solvable where I(z) is a function of z. The equation is:
I(z)=cosh(1/2 ∫I(z)dz)
I know it looks stupid but is it possible? How would you approach this problem?
Thank you.
What is the general solution of the following hyperbolic partial differential equation:
The head (h) at a specified distance (x) is a sort of a damping function in the form:
Where, a, b, c and d are constants. And the derivatives are with respect to t (time) and x (distance).
Thanks in advance.
To express the ##\cosh^{-1}## function as a logarithm, we start by defining the variables ##x## and ##y## as follows:
$$y = \cosh^{-1}{x}$$
$$x = \cosh{y}$$
Where ##y ∈ [0, \infty)## and ##x ∈ [1, \infty)##.
Using the definition of the hyperbolic cosine function, rearranging, and multiplying...
Homework Statement
Attached is the problem
Homework Equations
My question is am i going about it the right way for question C). I have done A and B and am sure they are correct.
The Attempt at a Solution
Attached
Homework Statement
I thought it would be better to attach it.
Homework Equations
The Attempt at a Solution
So for the first part I've found that A^2=the Identity matrix, but from there I don't have much of an idea on how to substitute that into the equation for M and end up with...
show that the locus of the point \left(\dfrac{a(cosh\theta + 1)}{2cosh\theta},\dfrac{b(cosh\theta - 1)}{2sinh\theta}\right)
has equation x(4y^2 + b^2) = ab^2
working: http://gyazo.com/4c96af128d0293bce7f18029c2f54b0d
where have I gone wrong :(
[SOLVED] Fourier Series Involving Hyperbolic Functions
Hello everyone!
Sorry if this isn't the appropriate board, but I couldn't think of which board would be more appropriate. I was running through some problems I have to do as practice for a test and I got stuck on one I'm 99% sure they'll...
I've searched and thought on it for a long time but I couldn't find any mathematical proof or something else about the formula of hyperbolic functions. sinh=\frac{e^{x}-e^{-x}}{2},cosh=\frac{e^{x}+e^{-x}}{2} How do I get these formulas mathematically??
Hi,
I am trying to integrate (tanh(x)+coth(x))/((cosh(x))^2)
I am substituting u=tanh(x), du=dx/((cosh(x))^2)
and end up with 1/2(tanh(x))^2 + ln |tanh(x)| + C
which is incorrect. What am I doing wrong??
Homework Statement
Compute the following:
\int \frac{cosh(x)}{cosh^2(x) - 1}\,dx
Homework Equations
\int cosh(x)\,dx = sinh(x) + C
The Attempt at a Solution
I had no clue where to start, so I went to WolfRamAlpha, and it used substitution but it made u = tanh(\frac{x}{2})...
Homework Statement
Prove sin(x-iy) = sin(x) cosh(y) - i cos(x) sinh(y)
Homework Equations
The Attempt at a Solution
I tried to prove it by developing sinh into it's exponential form, but I get stuck.
sinh(x-iy) = [ ei(x-iy) - e-i(x-iy) ] /2i
= [ eixey - e-ix e-y ] /2i...
Homework Statement
I am a little confused on the steps to take to solve these kinds of functions.
Solve:
cosh z = 2i
The Attempt at a Solution
We were given identities for sinh z = 0 and cosh z = 0 and also other identities like
cosh(z) = cos (iz)
So cos (iz) = 2i
cos...
Homework Statement
Find \int \frac{x}{\sqrt{2x^2-2x+1}}\,dx
The attempt at a solution
First, i complete the square for the quadratic expression:
2x^2-2x+1=2((x-\frac{1}{2})^2+\frac{1}{4})
\int \frac{x}{\sqrt{2x^2-2x+1}}\,dx=\int \frac{x}{\sqrt 2 \sqrt{(x-\frac{1}{2})^2+\frac{1}{4}}}\,dx...
Homework Statement
Express the function cosh(6x) in terms of powers of cosh(x)
Homework Equations
The Attempt at a Solution
Okay the problem booklet also asks me to do the opposite. Express cosh(x)^6 as mutiples of cosh(x). I can do that fine, I just simply write it out as [1/2(e^x + e^-x)]^6...
Hi,
My brain is not working today. So can someone please tell me what I am doing wrong.
(^2 = squared)
coshy^2 - sinhy^2 = 1, how do I rearrange this for coshy^2
I keep getting: coshy^2 = 1 + Sinhy^2
The book that I'm looking at has it this way: coshy^2 = Sinhy^2 + 1
Thanks
Obs
Homework Statement
\int cosh(2x)sinh^{2}(2x)dx
Homework Equations
Not sure
The Attempt at a Solution
This was an example problem in the book and was curious how they got to the following answer:
\int cosh(2x)sinh^{2}(2x)dx = \frac{1}{2}\int sinh^{2}(2x)2cosh(2x) dx
=...
Homework Statement
The area bounded by y=2 coshx, the x-axis, the y-axis, and the line x=4 is revolved about the x-axis. Find the volume of the solid generated.
Homework Equations
I sliced the area along the axis of revolution. That is the strip is dx. So the equation necessary is...
Hyperbolic functions :(((
Homework Statement
Question is:
and
Homework Equations
Now from what i can recall the formula for sin(A+B) = sinAcosB+sinbcosA
so same goes for hyperbolic function i suppose?
sinh(2x+x) = sinh2xcoshx+cosh2xsinx
(2sinhxcoshx)coshx +...
1. Differentiate cosh(x) using first principles
2. cosh(x) = (e^x+e^-x)/2
From previous exercises, I know the answer will be sinh(x)= (e^x-e^-x)/2 but I cannot get to the answer.
I seem to be left with the equation: lim h ---> 0 (e^2x*e^2h +1-e^h*2e^x +e^h)/(2h*e^x*e^h)
But when...