Natural Log Composed with Hyperbolic Tangent & this Ratio

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SUMMARY

The discussion clarifies the mathematical identity involving the hyperbolic tangent function and natural logarithm. The correct expression is \(\frac{x-1}{x+1} = \tanh\left(\frac{\ln x}{2}\right)\), not \(\frac{x-1}{x+1} = \tanh\left(\ln\left(\frac{x}{2}\right)\right)\). This correction stems from the identity \(\text{artanh}(x) = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)\), which can be derived by substituting \(y = \frac{1+x}{1-x}\) and solving for \(w\) in the equation \(z = \tanh(w)\).

PREREQUISITES
  • Understanding of hyperbolic functions, specifically \(\tanh\) and \(\text{artanh}\)
  • Familiarity with natural logarithms and their properties
  • Basic algebraic manipulation skills
  • Knowledge of mathematical identities and their derivations
NEXT STEPS
  • Study the derivation of the identity \(\text{artanh}(x) = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)\)
  • Explore applications of hyperbolic functions in calculus and physics
  • Learn about the properties and applications of logarithmic functions
  • Investigate further mathematical identities involving hyperbolic and logarithmic functions
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Mathematicians, students studying calculus or advanced algebra, and anyone interested in the properties of hyperbolic functions and logarithms.

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Hello,

Consider x \in (0,1), that is x between 0 and 1. Can someone explain why the following is true:
\frac{x-1}{x+1} = \tanh \left( \ln \left( \frac{x}{2} \right) \right)
 
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It's not true. That equality doesn't hold. The correct expression is

\frac{x-1}{x+1} = \mbox{tanh}\left(\frac{\ln x}{2}\right)

This follows from the identity

\mbox{artanh}(x) = \frac{1}{2} \ln \left( \frac{1+x}{1-x}\right)

You can get from this to the other form by making the replacement y = (1+x)/(1-x). To derive this identity, solve the following for w:

z = \mbox{tanh}(w) = \frac{e^w-e^{-w}}{e^w+e^{-w}}
 


Ah thank you!

Sorry about that. I made a typo.
 

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