If f(x)=(e^x+e^-x)/2, what is the inverse function?

AI Thread Summary
The discussion centers on finding the inverse of the function f(x) = (e^x + e^-x)/2, which is identified as the hyperbolic cosine function, cosh. Participants clarify that the inverse function is arccosh(x), emphasizing that cosh is not injective over the entire real line and must be restricted to either x<0 or x>0 for a proper inverse. The quadratic equation derived from the function is discussed, with participants noting the importance of selecting the positive root to maintain the function's invertibility. There is also a debate over the notation used for the inverse, with some preferring "arcosh" over "arccosh," though both terms are recognized in different contexts. The conversation highlights the mathematical intricacies involved in defining and working with inverse functions in this context.
Darkmisc
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Homework Statement
If f(x)=(e^x+e^-x)/2, what is the inverse function?
Relevant Equations
f(x)=(e^x+e^-x)/2
Hi everyone

This is the solution for the problem.

image_2022-06-18_142157071.png


I don't understand how they got from
image_2022-06-18_142300299.png

To
1655526208447.png


This was my attempt at a solution

WIN_20220618_14_09_15_Scan.jpg


I can't seem to get rid of one of the y terms and am left with one on each side.

Could someone explain the solution to me please?

Thanks
 
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##e^x + e^{-x}## is not injective in ##\mathbb R## (even function). You should assume either ##x<0## or ##x>0##.
 
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note: x = (1/2)(e^y + e^-y) = cos(iy). so the inverse is -i.arccos(x).
 
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mathwonk said:
note: x = (1/2)(e^y + e^-y) = cos(iy). so the inverse is -i.arccos(x).

We call that function "cosh", so the inverse is \operatorname{arccosh}(x).
 
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pasmith said:
We call that function "cosh", so the inverse is \operatorname{arccosh}(x).
Isn't it just a local inverse, as @nuuskur pointed out?
 
er sorry $$\operatorname{arccosh}(x)=-i\arccos(x)$$?

Btw there is a silly typo that was spotted successfully by @FactChecker the quadratic with respect to ##e^y## is ##(e^y)^2-2xe^y+1=0## for which you can set ##z=e^y## solve it for z(x) as a normal quadratic and then take $$e^y=z(x)\Rightarrow y=\ln z(x)$$.
 
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Delta2 said:
er sorry $$\operatorname{arccosh}(x)=-i\arccos(x)$$?

Btw there is a silly typo that was spotted successfully by @FactChecker the quadratic with respect to ##e^y## is ##(e^y)^2-2xe^y+1=0## for which you can set ##z=e^y## solve it for z(x) as a normal quadratic and then take $$e^y=z(x)\Rightarrow y=\ln z(x)$$.
With the caveats that ##z(x) >0##, and, for well definedness, its an injection, I guess.
 
Delta2 said:
$$\operatorname{arccosh}(x)=-i\arccos(x)$$?
You're probably confused because you're thinking of cosine defined for reals. If you consider the cosine function defined for complex numbers, it makes sense.

WWGD said:
With the caveats that ##z(x) >0##, and, for well definedness, its an injection, I guess.
If ##y## is real, then ##e^y>0##.

Note that the quadratic has two solutions, ##y_+## and ##y_-##. To make the inverse a function, we arbitrarily choose the positive root and map ##x## to ##y_+##.

It's not obvious, but you can show that ##y_- = -y_+##.
 
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  • #10
Am I the only one reacting on the use of ”arccosh” instead of the standard ”arcosh”?
 
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vela said:
It's not obvious, but you can show that ##y_- = -y_+##.
Is that true? I don't see it.
 
  • #12
Hint: ##(x + \sqrt{x^2-1})(x-\sqrt{x^2-1})=1##.
 
  • #13
Orodruin said:
Am I the only one reacting on the use of ”arccosh” instead of the standard ”arcosh”?
I think the "arc" notation is pretty common. Mathematica, for example, calls the function ArcCosh.

In fact, until you pointed it out, I didn't know "arcosh" was the standard way to write it. :smile:
 
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  • #14
vela said:
Mathematica, for example, calls the function ArcCosh.
Mathematica is then not up to ISO 80000-2 standard. 🤔

But yes, it is a fairly common mistake but arccosh would be a misnomer as the function does not relate to arc length but to area (unless measuring in Minkowski geometry….)
 
  • #15
vela said:
You're probably confused because you're thinking of cosine defined for reals. If you consider the cosine function defined for complex numbers, it makes sense.If ##y## is real, then ##e^y>0##.

Note that the quadratic has two solutions, ##y_+## and ##y_-##. To make the inverse a function, we arbitrarily choose the positive root and map ##x## to ##y_+##.

It's not obvious, but you can show that ##y_- = -y_+##.
Ah, I missed that result before log(z(x)).
 
  • #16
vela said:
Hint: ##(x + \sqrt{x^2-1})(x-\sqrt{x^2-1})=1##.
Ehm that is correct but all you can actually prove with this is that ##y_{+}=\frac{1}{y_{-}}##.

Ah ok you mean the solutions to ##e^y=z(x)## or ##e^y=\frac{1}{z(x)}##
 
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  • #17
For invertibility of ##\cosh x## (in ##\mathbb R##), we usually mean
<br /> \cosh : [0,\infty) \to [1,\infty),\quad \cosh(x) := \frac{e^x+e^{-x}}{2}.<br />
Putting
<br /> g : [1,\infty ) \to [0,\infty ),\quad g(x) := \ln (x+\sqrt{x^2-1})<br />
one can verify that ##g\circ \cosh## and ##\cosh \circ g## are the identities. From ##y_+y_- = 1## it follows that ## \ln y_- < 0##, so it couldn't be the correct choice in this case.

But it's completely fine to define ##\cosh## on the negative side. In that case, for the inverse, one also opts for ##\ln y_-##.
 
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