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Finding Inverse Hyperbolic secant in terms of logarithms ?

  1. Aug 29, 2011 #1
    The Problem is when I Compute the Inverse I have to solutions

    [itex]sech^{-1}(x) = ln(\frac{1\pm \sqrt{1-x^{2}}}{x}) : 0<x\leq 1 [/itex]

    And this not function which of them I will choose

    Another Question is how can I prove without the graph that csch (x) is one - to -one

    thanks
     
    Last edited: Aug 29, 2011
  2. jcsd
  3. Aug 29, 2011 #2
    For the second question I proved it depending on the definition of one to one function and csch(x) ,where since sinh(x) is one to one ,the whenever sinh(c)=sinh(d) , c=d
    Then it follows that whenever 1/sinh(c) = 1/sinh(d) , c=d
    And thus the result follows and that's what I did .
     
  4. Aug 29, 2011 #3
    For the first question , I tried to check my solution I tried to check the solution directly from the definition of inverse function,

    Let [itex] h(x) = ln(\frac{1-\sqrt{1-x^2}}{x}) : 0<x\leq 1 [/itex]

    [itex] g(x)=ln (\frac{1+ \sqrt{1-x^2}}{x}) : 0<x\leq 1 [/itex]

    [itex]f(x) = sech(x) : x\geq 0 [/itex]

    then we have after some Algebra ,
    [itex](f o h)(x) = x : 0<x\leq 1[/itex]
    Whereas [itex](h o f)(x) = -x : x\leq0[/itex]
    And Thus h is not the inverse of f , Checking g we will see that it do its job it is the inverse where

    [itex](f o g)(x) = x : 0<x\leq 1[/itex]
    Whereas [itex](g o f)(x) = x : x\leq0[/itex]
     
  5. Aug 30, 2011 #4

    HallsofIvy

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    Exactly what does the question say? The reason you are getting "[itex]\pm[/itex]" is because sech is NOT a one-to-one function and so does NOT HAVE an inverse over all real numbers. You can, of course, restrict the domain so that it does have an inverse. How the domain is restricted will determine the sign.
     
  6. Aug 30, 2011 #5
    I have restricted the domain such that x is greater than or equal zero and I also had Both signs So I Checked the solutions from definition of Inverse function And Choosed the positive sign As I have written above.
     
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