1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Inverse Hyperbolic Tangent

  1. Jul 7, 2010 #1

    Char. Limit

    User Avatar
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook