Finding the Inverse Function of tanh(x) in the Interval (-1,1)

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SUMMARY

The inverse function of the hyperbolic tangent, defined as ##f:= \tanh(x) = \frac{e^x-e^{-x}}{e^x+e^{-x}}##, is proven to be ##f^{-1}(x) = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)## for all x in the interval (-1, 1). Additionally, the series representation of the inverse function is given by ##f^{-1}(x) = \sum\limits_{k=0}^{\infty} \frac{x^{2k+1}}{2k+1}##. The discussion highlights the relevance of the Taylor expansion for logarithmic functions, specifically ##\ln(1+x)## and ##\ln(1-x)##, in deriving these results.

PREREQUISITES
  • Understanding of hyperbolic functions, specifically the hyperbolic tangent function.
  • Familiarity with logarithmic functions and their properties.
  • Knowledge of Taylor series expansions and their applications.
  • Basic calculus concepts, including limits and convergence of series.
NEXT STEPS
  • Study the Taylor series expansion of ##\ln(1+x)## and ##\ln(1-x)##.
  • Explore the properties and applications of hyperbolic functions in calculus.
  • Investigate the convergence criteria for power series within specific intervals.
  • Learn about the relationship between inverse functions and their derivatives.
USEFUL FOR

Students studying calculus, mathematicians interested in hyperbolic functions, and anyone seeking to understand the properties of inverse functions within the context of real analysis.

annaphys
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Homework Statement



##f:= tanh = \frac{e^x-e^{-x}}{e^x+e^{-x}}##

Prove that
##f^{-1}(x)= \sum\limits_{k=0}^{\infty} \frac{x^{2k+1}}{2k+1}## for all x in (-1,1)

The Attempt at a Solution


I also found the inverse function to be:

##f^{-1}(x)= \frac{1}{2}ln(\frac{1+x}{1-x})##

I tried working with the taylor polynomial but unfortunately nothing came out of it. Could someone point me in the right direction?
 
Last edited:
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annaphys said:

Homework Statement



##f:= tanh = \frac{e^x-e^{-x}}{e^x+e^{-x}}##

Prove that
##f^{-1}(x)= \sum\limits_{k=0}^{\infty} \frac{x^{2k+1}}{2k+1}##

The Attempt at a Solution


I also found the inverse function to be:

##f^{-1}(x)= \frac{1}{2}ln(\frac{1+x}{1-x})##

I tried working with the taylor polynomial but unfortunately nothing came out of it. Could someone point me in the right direction?

Do you know the Taylor expansion of ##ln(1+x)##? What about ##ln(1-x)##? Then you can use a fact about ##ln(a/b)##.
 
Hey thanks for the response! Does that also hold for all x's in (-1,1)? I know that holds for x<<1.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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