So, for tanh(ax),y = tanh(ax)x = tanh-1yx = (tanh-1y)/a

[Char. Limit]

Homework Statement

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.

Homework Equations

\frac{d tanh^{-1}\left(ax\right)}{dx} = \frac{a}{1-\left(ax\right)^2}

\frac{1}{1-u^2} = \frac{1}{2} \left(\frac{1}{u+1} - \frac{1}{u-1}\right)

The Attempt at a Solution

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

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

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

\int \frac{1}{1-u^2} du

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

\frac{1}{2} \left( \int \frac{du}{u+1} - \int \frac{du}{u-1} \right)

Next came the actual integration, of course...

\frac{1}{2} \left( log\left(u+1\right) - log\left(u-1\right) \right)

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

\frac{1}{2} log\left(\frac{ax+1}{ax-1}\right)

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?

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

Rearranging to isolate the logarithm...

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

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

\frac{ax+1}{ax-1} = e^{2y}

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.

a^2 x^2 - 1 = a e^{2y} x - e^{2y}

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

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

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

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

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

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

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

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

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

x= \frac{e^{2y} + e^{2y} - 2}{2a} OR x = \frac{e^{2y} - e^{2y} + 2}{2a}

And then I'll simplify that.

x = \frac{e^{2y} - 1}{a} OR x = \frac{1}{a}

Now, neither of those seem right. What did I do wrong?

 

[Bohrok]

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

\frac{ax+1}{ax-1} = e^{2y}

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.

a^2 x^2 - 1 = a e^{2y} x - e^{2y}

You multiplied the left side by ax-1 twice instead of once to get rid of just the denominator.

 

[Char. Limit]

You multiplied the left side by ax-1 twice instead of once to get rid of just the denominator.

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...

ax+1 = a e^{2y} x - e^{2y}

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...

ax - a e^{2y} x = -1 - e^{2y}

Multiply by -1...

a e^{2y} x - ax = e^{2y} + 1

Factor out the ax...

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

Divide and conquer!

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

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...

 

[Bohrok]

I think that 1/a does belong there. Apparently
\frac{e^{2y} + 1}{e^{2y} - 1} = \frac{1 + e^{2y}}{1 - e^{2y}}

and this is what I'm trying to figure out. The second "version" comes from writing 1/(1 - u²) as
\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)
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.

 

[Mark44]

I think that 1/a does belong there. Apparently
\frac{e^{2y} + 1}{e^{2y} - 1} = \frac{1 + e^{2y}}{1 - e^{2y}}

Not true. Each one is the negative of the other.

and this is what I'm trying to figure out. The second "version" comes from writing 1/(1 - u²) as
\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)
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.

See above.

 

[Bohrok]

Not true. Each one is the negative of the other.

See above.

Ok, I had overlooked the absolute values required when you integrate and get natural logs. Still using u:

\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|

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

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...

Factor e^x 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
Inverse:
x = (ay)²
√x = ay
y = (√x)/a

y = g(ax) = sin(ax)
Inverse:
x = sin(ay)
sin^-1x = ay
y = (sin^-1x)/a