Understanding the Math Behind coth(y) = x

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SUMMARY

The discussion centers on the mathematical relationship defined by the equation \( coth(y) = x \), leading to the derivation of \( y = \frac{1}{2} \ln \frac{x+1}{x-1} \). Participants explore the transformation of the equation \( x = \frac{e^{2y}+1}{e^{2y}-1} \) and its implications for understanding inverse hyperbolic functions. The conversation highlights the need for a deeper intuition regarding hyperbolic functions, particularly their non-periodic nature compared to trigonometric functions. The discussion emphasizes the importance of visualizing these functions to grasp their behavior and applications.

PREREQUISITES
  • Understanding of hyperbolic functions, specifically \( coth \) and \( sinh \)
  • Familiarity with logarithmic identities and properties
  • Basic knowledge of inverse functions and their graphical representations
  • Ability to manipulate algebraic expressions involving exponentials
NEXT STEPS
  • Study the properties and applications of hyperbolic functions in calculus
  • Learn about the graphical representations of inverse hyperbolic functions
  • Explore the relationship between trigonometric and hyperbolic functions
  • Investigate the practical applications of hyperbolic functions in physics and engineering
USEFUL FOR

Mathematicians, students studying calculus or advanced algebra, and anyone interested in deepening their understanding of hyperbolic and inverse hyperbolic functions.

ognik
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This is one of those 'huh?' moments where I can follow what is said, but don't understand it at all.
From $ coth(y) = x = \frac{{e}^{2y+1}}{{e}^{2y-1}}, $ we extract: $
y=\frac{1}{2} ln \frac{x+1}{x-1} $.
I've looked at graphs and definitions online, I follow what is done (kind of swapping x and y) - but would like to understand the details instead of just parroting it.
So I tried: $\ln\left({x}\right) = \ln\left({{e}^{2y}+1}\right) - \ln\left({{e}^{2y}-1}\right) $ - and am stuck here...
 
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I would begin with:

$$x=\frac{e^{2y}+1}{e^{2y}-1}$$

and solve for $e^{2y}$ first:

$$x\left(e^{2y}-1\right)=e^{2y}+1$$

$$e^{2y}(x-1)=x+1$$

$$e^{2y}=\frac{x+1}{x-1}$$

Can you proceed?
 
Yes, more than enough, thanks muchly. Just frustrating that I sometimes can't see the bleedin' obvious ...:-(

Do you (or anyone) perhaps have a good way for me to get a lasting 'intuition' about what inverse hyperbolics are? I look at, for example, the well known sin x; it is periodic.

Then, it seems, sinh x is a reflection of sin x about the line y=x. It ends up not very dissimilar from sin x (I looked at 7. The Inverse Trigonometric Functions) , but with a limited range - it is not periodic?

Then arcsin x is again a reflection of sinh x about y=x. It looks closer to what sinx was (Inverse Hyperbolic Functions) , also not periodic?

But what do hyperbolic and inverse hyperbolic functions do - apart from causing me to see double after a while ...Sin is a wave, I can look at ripples in a pond etc. The others?
 

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