MHB How Is the Inverse Hyperbolic Tangent Derived from Its Definition?

[Guest2]

Show from the definition of arctanh as the inverse function of tanh that, for $x \in (-1, 1)$

$$\tanh^{-1}{x} = \frac{1}{2}\log\left(\frac{1+x}{1-x}\right)$$

The definition of hyperbolic tangent is $\displaystyle \tanh{h} = \frac{e^x-e^{-x}}{e^{x}+e^{-x}}$

Let $\displaystyle y = \frac{e^x-e^{-x}}{e^x+e^{-x}} =\frac{e^x+e^{-x}-2e^{-x}}{e^x+e^{-x}}= 1-\frac{2e^{-x}}{e^x+e^{-x}} $

So $\displaystyle 1-y = \frac{2e^{-x}}{e^x+e^{-x}}$ so $\frac{1-y}{1+y} = 2e^{-2x}$ therefore $\displaystyle \log(\frac{1-y}{1+y}) = \log(e^{-2x}) = -2x$, so $x = -\log\left(\frac{1-y}{1+y}\right) = \frac{1}{2}\log\left(\frac{1+y}{1-y}\right)$

Therefore $\tanh^{-1}{x} = \frac{1}{2}\log\left(\frac{1+x}{1-x}\right)$. Is this correct? How do I find the expansion of $\tanh^{-1}{x}$ upto and including the term containing $x^5$ from this?

 

[Guest2]

I get it I think

$\displaystyle \frac{1}{1-x} = \sum_{k=0}^{\infty} x^k \implies \log(1-t) = -\sum_{k =0}^{\infty} \frac{x^{k+1}}{k+1}$

Similarly, $\displaystyle \frac{1}{1+x} = \sum_{k=0}^{\infty} (-1)^kx^k \implies \log(1+t) = \sum_{k =0}^{\infty} \frac{(-1)^kx^{k+1}}{k+1}$

Therefore $\displaystyle \frac{1}{2}\log\left(\frac{1+x}{1-x}\right) = \frac{1}{2}\sum_{k =0}^{\infty} \frac{(-1)^kx^{k+1}}{k+1}+\frac{1}{2}\sum_{k =0}^{\infty} \frac{x^{k+1}}{k+1}$

So the answer is $\displaystyle x+\frac{x^3}{3}+\frac{x^5}{5}+\mathcal{O}(x^6)$ (Happy)

 

[Prove It]

Show from the definition of arctanh as the inverse function of tanh that, for $x \in (-1, 1)$

$$\tanh^{-1}{x} = \frac{1}{2}\log\left(\frac{1+x}{1-x}\right)$$

The definition of hyperbolic tangent is $\displaystyle \tanh{h} = \frac{e^x-e^{-x}}{e^{x}+e^{-x}}$

Let $\displaystyle y = \frac{e^x-e^{-x}}{e^x+e^{-x}} =\frac{e^x+e^{-x}-2e^{-x}}{e^x+e^{-x}}= 1-\frac{2e^{-x}}{e^x+e^{-x}} $

So $\displaystyle 1-y = \frac{2e^{-x}}{e^x+e^{-x}}$ so $\frac{1-y}{1+y} = 2e^{-2x}$ therefore $\displaystyle \log(\frac{1-y}{1+y}) = \log(e^{-2x}) = -2x$, so $x = -\log\left(\frac{1-y}{1+y}\right) = \frac{1}{2}\log\left(\frac{1+y}{1-y}\right)$

Therefore $\tanh^{-1}{x} = \frac{1}{2}\log\left(\frac{1+x}{1-x}\right)$. Is this correct? How do I find the expansion of $\tanh^{-1}{x}$ upto and including the term containing $x^5$ from this?

I suspect that what you have written is correct, but it is easier to resolve this into a quadratic equation...

$\displaystyle \begin{align*} x &= \textrm{artan}\,\left( y \right) \\ y &= \tanh{(x)} \\ y &= \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}} \\ \left( \mathrm{e}^x + \mathrm{e}^{-x} \right) \, y &= \mathrm{e}^x - \mathrm{e}^{-x} \\ \mathrm{e}^x\,y + \mathrm{e}^{-x}\,y &= \mathrm{e}^x - \mathrm{e}^{-x} \\ \mathrm{e}^x \, \left( \mathrm{e}^x\,y + \mathrm{e}^{-x}\,y \right) &= \mathrm{e}^x \,\left( \mathrm{e}^x - \mathrm{e}^{-x} \right) \\ \left( \mathrm{e}^{x} \right) ^2 \,y + y &= \left( \mathrm{e}^{x} \right) ^2 - 1 \\ 1 + y &= \left( \mathrm{e}^x \right) ^2 - \left( \mathrm{e}^x \right) ^2 \, y \\ 1 + y &= \left( \mathrm{e}^x \right) ^2 \left( 1 - y \right) \\ \left( \mathrm{e}^x \right) ^2 &= \frac{1 + y }{1 - y} \\ \mathrm{e}^x &= \sqrt{ \frac{1 + y}{1 - y} } \\ x &= \ln{ \left( \sqrt{ \frac{1 + y}{1 - y} } \right) } \\ x &= \ln{ \left[ \left( \frac{1 + y}{1 - y} \right) ^{\frac{1}{2}} \right] } \\ x &= \frac{1}{2}\ln{ \left( \frac{1 + y}{1 - y} \right) } \end{align*}$

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I get it I think

$\displaystyle \frac{1}{1-x} = \sum_{k=0}^{\infty} x^k \implies \log(1-t) = -\sum_{k =0}^{\infty} \frac{x^{k+1}}{k+1}$

Similarly, $\displaystyle \frac{1}{1+x} = \sum_{k=0}^{\infty} (-1)^kx^k \implies \log(1+t) = \sum_{k =0}^{\infty} \frac{(-1)^kx^{k+1}}{k+1}$

Therefore $\displaystyle \frac{1}{2}\log\left(\frac{1+x}{1-x}\right) = \frac{1}{2}\sum_{k =0}^{\infty} \frac{(-1)^kx^{k+1}}{k+1}+\frac{1}{2}\sum_{k =0}^{\infty} \frac{x^{k+1}}{k+1}$

So the answer is $\displaystyle x+\frac{x^3}{3}+\frac{x^5}{5}+\mathcal{O}(x^6)$ (Happy)

That is correct, nice job. You should also say what the region of convergence is though.

 

[Guest2]

I suspect that what you have written is correct, but it is easier to resolve this into a quadratic equation...

$\displaystyle \begin{align*} x &= \textrm{artan}\,\left( y \right) \\ y &= \tanh{(x)} \\ y &= \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}} \\ \left( \mathrm{e}^x + \mathrm{e}^{-x} \right) \, y &= \mathrm{e}^x - \mathrm{e}^{-x} \\ \mathrm{e}^x\,y + \mathrm{e}^{-x}\,y &= \mathrm{e}^x - \mathrm{e}^{-x} \\ \mathrm{e}^x \, \left( \mathrm{e}^x\,y + \mathrm{e}^{-x}\,y \right) &= \mathrm{e}^x \,\left( \mathrm{e}^x - \mathrm{e}^{-x} \right) \\ \left( \mathrm{e}^{x} \right) ^2 \,y + y &= \left( \mathrm{e}^{x} \right) ^2 - 1 \\ 1 + y &= \left( \mathrm{e}^x \right) ^2 - \left( \mathrm{e}^x \right) ^2 \, y \\ 1 + y &= \left( \mathrm{e}^x \right) ^2 \left( 1 - y \right) \\ \left( \mathrm{e}^x \right) ^2 &= \frac{1 + y }{1 - y} \\ \mathrm{e}^x &= \sqrt{ \frac{1 + y}{1 - y} } \\ x &= \ln{ \left( \sqrt{ \frac{1 + y}{1 - y} } \right) } \\ x &= \ln{ \left[ \left( \frac{1 + y}{1 - y} \right) ^{\frac{1}{2}} \right] } \\ x &= \frac{1}{2}\ln{ \left( \frac{1 + y}{1 - y} \right) } \end{align*}$

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That is correct, nice job. You should also say what the region of convergence is though.

Thank you. In speaking of the range of convergence, how do I find the radius converges when I've the sum of two series as in above?

 

[Prove It]

Thank you. In speaking of the range of convergence, how do I find the radius converges when I've the sum of two series as in above?

Any series which is generated from combining two or more other series only converges where the original series all converged.