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Inverse Hyperbolic Tangent

  1. Jul 7, 2010 #1

    Char. Limit

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    Gold Member

    1. The problem statement, all variables and given/known data
    Now, I decided for no real reason to derive a formula for the hyperbolic tangent using only what I know about the derivative of the inverse hyperbolic tangent. However, what I have looks wrong, and I'd like to check it here.

    2. Relevant equations
    [tex]\frac{d tanh^{-1}\left(ax\right)}{dx} = \frac{a}{1-\left(ax\right)^2}[/tex]

    [tex]\frac{1}{1-u^2} = \frac{1}{2} \left(\frac{1}{u+1} - \frac{1}{u-1}\right)[/tex]

    3. The attempt at a solution

    So, I started with the derivative, of course...

    [tex]\int \frac{a}{1-\left(ax\right)^2} dx[/tex]

    I did a quick u-substitution, with u = a x and du = a dx:

    [tex]\int \frac{1}{1-u^2} du[/tex]

    I used the second equation in part 2, which I derived seperately:

    [tex]\frac{1}{2} \left( \int \frac{du}{u+1} - \int \frac{du}{u-1} \right)[/tex]

    Next came the actual integration, of course...

    [tex]\frac{1}{2} \left( log\left(u+1\right) - log\left(u-1\right) \right)[/tex]

    I rearranged with the logarithmic rules and resubsituted ax to get...

    [tex]\frac{1}{2} log\left(\frac{ax+1}{ax-1}\right)[/tex]

    Now, this should be the formula for the inverse hyperbolic tangent. So, if I set this equal to y, and then solve for x, I should have the formula for the hyperbolic tangent, right?

    [tex]\frac{1}{2} log\left(\frac{ax+1}{ax-1}\right) = y[/tex]

    Rearranging to isolate the logarithm...

    [tex]log\left(\frac{ax+1}{ax-1}\right) = 2y[/tex]

    And now I raise e to both sides to remove the logarithm.

    [tex]\frac{ax+1}{ax-1} = e^{2y}[/tex]

    Now, I'll multiply both sides by ax-1 (why not, what else can I do?), and see what happens. Note that from here I'm pretty much doing this as I write.

    [tex]a^2 x^2 - 1 = a e^{2y} x - e^{2y}[/tex]

    I'll rearrange to set the right side equal to zero...

    [tex]a^2 x^2 + \left(-a e^{2y}\right) x + \left(e^{2y} - 1\right) = 0[/tex]

    Now I tried to use the quadratic formula to solve this, which is probably where I went wrong...

    [tex] x = \frac{a e^{2y} \pm \sqrt{a^2 e^{4y} - 4 a^2 e^{2y} + 4 a^2}}{2 a^2}[/tex]

    Simplifying the radical slightly by pulling out the a, I get...

    [tex]x = \frac{a\left( e^{2y} \pm \sqrt{e^{4y} - 4 e^{2y} + 4}\right)}{2 a^2}[/tex]

    So, the a's will cancel, and then I recognize the radical as a perfect square...

    [tex]x = \frac{e^{2y} \pm \sqrt{\left(e^{2y} - 2\right)^2}}{2a}[/tex]

    And now I'll cancel the square and square root and split this into two equations...

    [tex]x= \frac{e^{2y} + e^{2y} - 2}{2a} OR x = \frac{e^{2y} - e^{2y} + 2}{2a}[/tex]

    And then I'll simplify that.

    [tex]x = \frac{e^{2y} - 1}{a} OR x = \frac{1}{a}[/tex]

    Now, neither of those seem right. What did I do wrong?
  2. jcsd
  3. Jul 7, 2010 #2
    You multiplied the left side by ax-1 twice instead of once to get rid of just the denominator.
  4. Jul 7, 2010 #3

    Char. Limit

    User Avatar
    Gold Member

    Oh, wow, I feel like an idiot now.

    So, if we do it the right way instead (why not, right?), this is what I come up with...

    [tex]ax+1 = a e^{2y} x - e^{2y}[/tex]

    That's much easier to solve than the complicated quadratic I was working on earlier... Let's put all the x terms to one side and all of the constants to the other...

    [tex]ax - a e^{2y} x = -1 - e^{2y}[/tex]

    Multiply by -1...

    [tex]a e^{2y} x - ax = e^{2y} + 1[/tex]

    Factor out the ax...

    [tex]ax\left(e^{2y} - 1\right) = e^{2y} + 1[/tex]

    Divide and conquer!

    [tex]x = \frac{1}{a} \left( \frac{e^{2y} + 1}{e^{2y} - 1} \right)[/tex]

    That still doesn't seem right though... is that 1/a supposed to be there? I don't think that's supposed to be there...
  5. Jul 7, 2010 #4
    I think that 1/a does belong there. Apparently
    [tex]\frac{e^{2y} + 1}{e^{2y} - 1} = \frac{1 + e^{2y}}{1 - e^{2y}}[/tex]

    and this is what I'm trying to figure out. The second "version" comes from writing 1/(1 - u2) as
    [tex]\frac{1}{2}\left(\frac{1}{1 + u} + \frac{1}{1 - u}\right) \text{instead of } \frac{1}{2}\left(\frac{1}{u + 1} - \frac{1}{u - 1}\right)[/tex]
    The second one is valid, but using the first one easily works out to the formula for tanh(x) with the exponentials. I don't know why, but it's tricky...

    [Edit] I hope those first two fractions are equal, I could have been too tired last night and missed something. I'll take a look at it again later.
    Last edited: Jul 7, 2010
  6. Jul 7, 2010 #5


    Staff: Mentor

    Not true. Each one is the negative of the other.
    See above.
  7. Jul 7, 2010 #6
    Ok, I had overlooked the absolute values required when you integrate and get natural logs. Still using u:

    [tex]\ln\left|\frac{u + 1}{u - 1}\right| = \ln\left|\frac{u + 1}{(-1)(1 - u)}\right| = \ln\left|\frac{1 + u}{1 - u}\right|[/tex]

    Factor ex from the numerator and denominator, cancel them, then change the variables to get the inverse.

    As far as 1/a, consider some functions like f(x) = x2 and g(x) = sin(x) and their inverses (with restricted domain) with ax.
    y = f(ax) = (ax)2
    x = (ay)2
    √x = ay
    y = (√x)/a

    y = g(ax) = sin(ax)
    x = sin(ay)
    sin-1x = ay
    y = (sin-1x)/a
    Last edited: Jul 7, 2010
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