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Proof of Hyperbolic Functions

by Calculuser
Tags: functions, hyperbolic, proof
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Calculuser
#1
Feb28-13, 09:59 AM
P: 26
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. [itex]sinh=\frac{e^{x}-e^{-x}}{2},cosh=\frac{e^{x}+e^{-x}}{2}[/itex] How do I get these formulas mathematically??
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mfb
#2
Feb28-13, 11:51 AM
Mentor
P: 12,081
This is a common definition of them, and you cannot prove definitions.
If you want to use another definition, there is some way to relate them, but that depends on your favorite definition.
eyesontheball1
#3
Feb28-13, 04:23 PM
P: 16
cos(ix)=cosh(x), sin(ix)=isinh(x); note that even when x is complex-valued (if you've any familiarity with complex analysis), both of these identities still hold. this illustrates the basic relation between circular functions (trig functions) and the hyperbolic functions cosh, sinh, tanh, etc.

Moreover, if y=ix, cos^2(y)+sin^2(y) = 1 => cos^2(ix)+sin^2(ix) = 1 => cosh^2(x) + (i^2)*sinh^2(x) = 1 => cosh^2(x) + (-1)*sinh^2(x) = 1 => cosh^2(x) - sinh^2(x) = 1, as expected.

rbj
#4
Mar1-13, 11:06 PM
P: 2,251
Proof of Hyperbolic Functions

Quote Quote by mfb View Post
This is a common definition of them, and you cannot prove definitions.
If you want to use another definition, there is some way to relate them, but that depends on your favorite definition.
there is a reason these became known as hyperbolic trig functions in contrast to the circular trigonometric functions.

take a look at that figure at http://en.wikipedia.org/wiki/Hyperbolic_function .

if it's the circular trig functions, you sweep a sector with area [itex]\frac{a}{2}[/itex] under the circular curve of

[tex] x^2 + y^2 = 1 [/tex]

after sweeping out a pie-shaped sector of that area with that circular curve, the point on the curve has coordinates of [itex]x=\cos(a)[/itex] and [itex]y=\sin(a)[/itex] .

with hyperbolic trig functions, you sweep a "sector" with the same area [itex]\frac{a}{2}[/itex] over the hyperbola curve of

[tex] x^2 - y^2 = 1 [/tex]

after sweeping out that sector of that area with that hyperbolic curve, the point on the curve has coordinates of [itex]x=\cosh(a)[/itex] and [itex]y=\sinh(a)[/itex] .

see the similarity?
mfb
#5
Mar2-13, 12:58 PM
Mentor
P: 12,081
Sure, but where is the point? This was a historic definition of the hyperbolic definitions. It is possible to show that they are identical with modern, algebraic definitions of those functions.
rbj
#6
Mar2-13, 01:57 PM
P: 2,251
well, try doing a couple of integrals.

this might be the reverse of what is defined and what is the consequence, but suppose you pick any [itex]a>0[/itex] and define

[tex] \cosh(a) = \frac{e^{a} + e^{-a}}{2} [/tex]

[tex] \sinh(a) = \frac{e^{a} - e^{-a}}{2} [/tex]

then find the area under this curve:

[tex] \int_{0}^{\cosh(a)} \frac{\sinh(a)}{\cosh(a)} \ x \ dx [/tex]

(that one is easy) and then subtract the area under this curve:

[tex] \int_{1}^{\cosh(a)} \sqrt{x^2 - 1} \ dx [/tex]

and do some considerable simplification, what do you think the net area would turn out to be?
rbj
#7
Mar2-13, 10:45 PM
P: 2,251
nobody took me up on my question.

the answer is [itex]\frac{a}{2}[/itex].


[tex] \int_{0}^{\cosh(a)} \frac{\sinh(a)}{\cosh(a)} \ x \ dx \ - \ \int_{1}^{\cosh(a)} \sqrt{x^2 - 1} \ dx \ = \ \frac{a}{2} [/tex]

when [itex]a > 0[/itex] and

[tex] \cosh(a) \triangleq \frac{e^{a} + e^{-a}}{2} [/tex]

[tex] \sinh(a) \triangleq \frac{e^{a} - e^{-a}}{2} [/tex]


i hope that satisfies the OP's request for "proof".
morrobay
#8
Mar3-13, 08:28 PM
P: 382
You can prove them by working out the areas like I did here:
http://www.physicsforums.com/showthread.php?t=336897
rbj
#9
Mar4-13, 11:24 AM
P: 2,251
it states a formula for the tough integral (which can be checked by differentiation) and that is good, but it still doesn't show that the net area (the difference of two integrals) is [itex]\frac{a}{2}[/itex].

someone should do that here using [itex]\LaTeX[/itex].
morrobay
#10
Mar5-13, 10:00 PM
P: 382
Quote Quote by rbj View Post
nobody took me up on my question.

the answer is [itex]\frac{a}{2}[/itex].


[tex] \int_{0}^{\cosh(a)} \frac{\sinh(a)}{\cosh(a)} \ x \ dx \ - \ \int_{1}^{\cosh(a)} \sqrt{x^2 - 1} \ dx \ = \ \frac{a}{2} [/tex]

when [itex]a > 0[/itex] and

[tex] \cosh(a) \triangleq \frac{e^{a} + e^{-a}}{2} [/tex]

[tex] \sinh(a) \triangleq \frac{e^{a} - e^{-a}}{2} [/tex]


i hope that satisfies the OP's request for "proof".
The question on the above is : How and why does ' e ' get into the definition.
Besides the fact that it is a parameter.
rbj
#11
Mar6-13, 01:39 PM
P: 2,251
Quote Quote by morrobay View Post
The question on the above is : How and why does ' [itex]e[/itex] ' get into the definition.
Besides the fact that it is a parameter.
i dunno where the original insight into this relationship comes from. i learned out of a textbook like anyone else.

it's sorta like this: a perfectly legitimate technique to solve differential equations or to integrate functions (which is really just solving a differential equation) is to guess at the answer, calculate the derivatives plug it back into the diff. eq. and see if equality results. if equality results, you have found a solution (or anti-derivative) and, depending on the order of the diff. eq., you might be done.

so here some author in some textbook is saying:

If you define the hyperbolic cosine and sine as thus:

[tex] \cosh(x) \triangleq \frac{e^{x} + e^{-x}}{2} [/tex]

[tex] \sinh(x) \triangleq \frac{e^{x} - e^{-x}}{2} [/tex]

with [itex]e[/itex] being the base to the natural logarithm, then the following are true:

[tex] \left( \cosh(a) \right)^2 - \left( \sinh(a) \right)^2 = 1[/tex]

and

[tex] \int_{0}^{\cosh(a)} \frac{\sinh(a)}{\cosh(a)} \ x \ dx \ - \ \int_{1}^{\cosh(a)} \sqrt{x^2 - 1} \ dx \ = \ \frac{a}{2} [/tex]

the latter, at least for [itex]a > 0[/itex] . we can work out the integrals and see that this is true.

use that fact to help you interpret the drawing at http://en.wikipedia.org/wiki/File:Hy...unctions-2.svg .

now, if we changed the base of the [itex]\cosh[/itex] and [itex]\sinh[/itex] definitions to something else, the first fact would continue to be true (but for another [itex]a[/itex]) since all it does is scale the x-axis. but the second fact (with the integrals) would no longer be true. it would be off by a scaling factor.

but somehow, someone had the insight to see this and guess at the relationship, and then it's just a matter of checking it to see that the guess is correct.

afterthought: other facts that are true with that definition is:

[tex] \frac{d}{dx}\cosh(x) = \sinh(x) [/tex]

and

[tex] \frac{d}{dx}\sinh(x) = \cosh(x) [/tex]

and this would not be true with just any definition.
Goa'uld
#13
Aug25-13, 11:16 AM
P: 25
Here is also something interesting to consider. It can be shown that [tex]\text{(1) }e^{ix}=\cos{x}+i\cdot\sin{x}[/tex] Also, [tex]\text{(2) }e^{-ix}=\cos{x}-i\cdot\sin{x}[/tex] Adding (1) and (2) together and dividing by 2 you get [tex]\cos{x}=\frac{e^{ix}+e^{-ix}}{2}[/tex] Subtracting (2) from (1) and dividing by 2 you get [tex]i\cdot\sin{x}=\frac{e^{ix}-e^{-ix}}{2}[/tex] Now, replacing x with i*x you will get [tex]\text{(3) }\cos{(ix)}=\frac{e^{x}+e^{-x}}{2}=\cosh{x}[/tex] and [tex]\text{(4) }i\cdot\sin{(ix)}=\frac{e^{-x}-e^{x}}{2}=-\sinh{x}[/tex]

If you take the base case of the hyperbola in Cartesian coordinates [itex]x^2-y^2=1[/itex] and substitute [itex]x=\cosh{\theta}=\cos(i\theta)[/itex] and [itex]y=\sinh{\theta}=-i\cdot\sin(i\theta)[/itex] you end up with the base case of the circle, [itex]\cos^{2}(i\theta)+\sin^{2}(i\theta)=1[/itex], which is always true. It also works for the exponential representation of the hyperbolic trig functions. Also, If you take the derivative of (3) with respect to x you will get (4), and if you take the derivative of (4) with respect to x you will get (3) further proving their validity. This post, along with what is shown previously in this thread show that the hyperbolic trig functions exist, are valid, and have a connection to the normal trig functions.


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