Invert an odd-symmetry function of one variable (with a parameter) using two terms and Lagrange Inversion Theorem.

Question

First of all, I don't have much rep to spend over here, but I gotta lot more at the DSP SE. So if you wanna earn some rep at the DSP SE, please go to my question over there because tomorrow I am gonna hang the maximum of 500 point bounty onto it. I can't get away with that here although if someone votes me up on this question (or something else I have done here), I can hang a bounty of 350 onto this question in a couple days.

So I have a function of the form:

$$ f(x) = \ln\left(\arctan\left(\alpha \, e^{x} \right) \right) - \ln\left(\arctan\left( \alpha \, e^{-x}\right) \right) $$

What I want to do is invert $f(x)$ (but I know I can't do it exactly with a nice closed form). I have already done a first-order approximation and i want to bump it up to a third-order approximation. And this has become a sorta copulating female canine, even though it should be straight forward.

Now this has something to do with the Lagrange Inversion Theorem which I asked about before and I only want to take it to one more term than I have.

We know from above that $f(x)$ is an odd-symmetry function:

$$ f(-x) = -f(x) $$

This means $f(0)=0$ and all even-order terms of the Maclaurin series will be zero:

$$ y = f(x) = a_1 x + a_3 x^3 + ... $$

The inverse function is also odd symmetry, goes through zero, and can be expressed as a Maclaurin series

$$ x = g(y) = b_1 y + b_3 y^3 + ... $$

and if we know what $a_1$ and $a_3$ are of $f(x)$, then we have a good idea what $b_1$ and $b_3$ must be:

$$ b_1=\frac{1}{a_1} \quad \quad \quad \quad b_3 = -\frac{a_3}{a_1^4} $$

Now, I am able to calculate the derivative of $f(x)$ and evaluate it at zero and I get $a_1$ and $b_1$ as a nice function of $\alpha$. But I am having a bitch of a time getting $a_3$ and therefore $b_3$. Can someone do this? I'd even settle for a solid expression for the third derivative of $f(x)$ evaluated at $x=0$.

Feel free to go over to the DSP site and earn some rep over there.

Answer 1

Here is an answer based upon a direct expansion of the odd function $f(x)$ \begin{align*} f(x)&=\ln\left(\arctan\left(\alpha e^x\right)\right)-\ln\left(\arctan\left(\alpha e^{-x}\right)\right)\\ &=f_1x+f_3x^3+O\left(x^5\right)\tag{1} \end{align*} into a series up to the third order. We will see this approach is albeit somewhat cumbersome, quite manageable.

Note: According to the series expansions \begin{align*} e^x&=\sum_{n=0}^\infty \frac{x^n}{n!}& &x\in\mathbb{C}\\ \arctan(x) &=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} & &|x|<1\\ \ln(x)&=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(x-1)^n & &|x-1|<1 \end{align*} the MacLaurin expansion (1) of $f(x)$ is absolut convergent for $x\in\mathbb{C}$ with \begin{align*} |\arctan(\alpha e^x)-1|&<1\\ |\arctan(\alpha e^{-x})-1|&<1\\ \alpha

At first we focus at the left-hand term $$\ln\left(\arctan\left(\alpha e^x\right)\right)$$ of $f(x)$ and start with

$$ $$

Series expansion of $\arctan$:

We obtain \begin{align*} \arctan(\alpha e^x)&=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}\alpha^{2n+1} e^{(2n+1)x}\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}\alpha^{2n+1}\sum_{j=0}^\infty \frac{1}{j!}(2n+1)^jx^j\\ &=\sum_{j=0}^\infty\frac{1}{j!}\sum_{n=0}^\infty(-1)^n(2n+1)^{j-1}\alpha^{2n+1}x^j\tag{2} \end{align*}

We now derive from (2) the coefficients up to $x^3$. Using the coefficient operator $[x^k]$ to denote the coefficient of $x^k$ in a series we obtain \begin{align*} [x^0]\arctan\left(\alpha e^x\right)&=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}\alpha^{2n+1}=\arctan \alpha\\ [x^1]\arctan\left(\alpha e^x\right)&=\sum_{n=0}^\infty (-1)^n\alpha^{2n+1}=\frac{\alpha}{1+\alpha ^2}\\ [x^2]\arctan\left(\alpha e^x\right)&=\frac{1}{2}\sum_{n=0}^\infty (-1)^n(2n+1)\alpha^{2n+1}\\ &=\frac{\alpha}{2}\cdot\frac{d}{d\alpha}\left(\sum_{n=0}^\infty (-1)^n\alpha^{2n+1}\right)\\ &=\frac{\alpha}{2}\cdot\frac{d}{d\alpha}\left(\frac{\alpha}{1+\alpha^2}\right)\\ &=\frac{\alpha(1-\alpha^2)}{2\left(1+\alpha^2\right)^2}\\ [x^3]\arctan\left(\alpha e^x\right)&=\frac{1}{6}\sum_{n=0}^\infty (-1)^n(2n+1)^2\alpha^{2n+1}\\ &=\frac{\alpha^2}{6}\sum_{n=0}^\infty (-1)^n(2n+1)(2n)\alpha^{2n-1} +\frac{\alpha}{6}\sum_{n=0}^\infty (-1)^n(2n+1)\alpha^{2n}\\ &=\frac{\alpha^2}{6}\cdot\frac{d^2}{d\alpha^2}\left(\sum_{n=0}^\infty(-1)^n\alpha^{2n+1}\right) +\frac{\alpha}{6}\cdot\frac{d}{d\alpha}\left(\sum_{n=0}^\infty(-1)^n\alpha^{2n+1}\right)\\ &=\left(\frac{\alpha^2}{6}\cdot\frac{d^2}{d\alpha^2}+\frac{\alpha}{6}\cdot\frac{d}{d\alpha}\right)\left(\frac{\alpha}{1+\alpha^2}\right)\\ &=\frac{\alpha^2}{6}\cdot\frac{2\alpha\left(\alpha^2-3\right)}{(1+\alpha^2)^3} +\frac{\alpha^2}{6}\cdot\frac{1-\alpha^2)}{(1+\alpha^2)^2}\\ &=\frac{\alpha^5-6\alpha^3+\alpha}{6(1+\alpha^2)^3} \end{align*}

We conclude \begin{align*} \arctan\left(\alpha e^x\right)&=\arctan(\alpha)+\frac{\alpha}{1+\alpha^2}x+\frac{\alpha(1-\alpha^2)}{2\left(1+\alpha^2\right)^2}x^2\\ &\qquad+\frac{\alpha^5-6\alpha^3+\alpha}{6(1+\alpha^2)^3}x^3+O(x^4)\tag{3} \end{align*}

$$ $$

Powers in logarithmic series:

In order to derive the coefficients of the logarithmic series \begin{align*} \ln\left(\arctan(\alpha e^x)\right)&=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}(\arctan(\alpha e^x)-1)^n \end{align*} we write the expression (3) as \begin{align*} \arctan\left(\alpha e^x\right)&=a_0+a_1x+a_2x^2+a_3x^3+O(x^4) \end{align*} and we consider \begin{align*} \ln\left(\arctan(\alpha e^x)\right)&=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}((a_0-1)+a_1x+a_2x^2+a_3x^3)^n+O(x^4)\tag{4} \end{align*}

We now set $A(x)=(a_0-1)+a_1x+a_2x^2+a_3x^3$ and extract the coeffcients of $x^0$ to $x^3$ from \begin{align*} \left(A(x)\right)^n&=((a_0-1)+a_1x+a_2x^2+a_3x^3)^n\\ &=\sum_{j=0}^n\binom{n}{j}(a_0-1)^j\left(a_1x+a_2x^2+a_3x^3\right)^{n-j}\\ &=\sum_{j=0}^n\binom{n}{j}(a_0-1)^j\sum_{k=0}^{n-j}\binom{n-j}{k}a_1^kx^k\left(a_2x^2+a_3x^3\right)^{n-j-k}\tag{5}\\ \end{align*}

We obtain from (5) \begin{align*} [x^0]\left(A(x)\right)^n&=\binom{n}{0}(a_0-1)^n=(a_0-1)^n\\ [x^1]\left(A(x)\right)^n&=\binom{n}{1}(a_0-1)^{n-1}\binom{1}{0}a_1=a_1n(a_0-1)^{n-1}\\ [x^2]\left(A(x)\right)^n&=\binom{n}{1}(a_0-1)^{n-1}\binom{1}{1}a_2+\binom{n}{2}(a_0-1)^{n-2}a_1^2\\ &=a_2n(a_0-1)^{n-1}+\frac{1}{2}n(n-1)a_1^2(a_0-1)^{n-2}\\ [x^3]\left(A(x)\right)^n&=\binom{n}{1}(a_0-1)^{n-1}\binom{1}{1}a_3+\binom{n}{2}(a_0-1)^{n-2}\binom{2}{1}a_1a_2\\ &\qquad+\binom{n}{3}(a_0-1)^{n-3}\binom{3}{0}a_1^3\\ &=na_3(a_0-1)^{n-1}+a_1a_2n(n-1)(a_0-1)^{n-2}\\ &\qquad+\frac{1}{6}n(n-1)(n-2)a_1^3(a_0-1)^{n-3}\tag{6} \end{align*}

Series expansion of logarithm:

We calculate using (6) the coefficients of $\ln(\arctan(\alpha e^x))$ in terms of $a_j, 0\leq j\leq 3$

\begin{align*} [x^0]&\ln(\arctan(\alpha e^x))=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}[x^0]A(x)\\ &=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}[x^0](a_0-1)^n\\ &=\ln(a_0-1)\\ [x^1]&\ln(\arctan(\alpha e^x))=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}[x^1]A(x)\\ &=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}[x^0]a_1n(a_0-1)^{n-1}\\ &=a_1\sum_{n=0}^\infty(-1)^n(a_0-1)^n\\ &=\frac{a_1}{a_0}\\ [x^2]&\ln(\arctan(\alpha e^x))=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}[x^2]A(x)\\ &=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\left(a_2n(a_0-1)^{n-1}+\frac{1}{2}n(n-1)a_1^2(a_0-1)^{n-2}\right)\\ &=\sum_{n=0}^\infty (-1)^{n}\left(a_2(a_0-1)^{n}+\frac{1}{2}(n-1)a_1^2(a_0-1)^{n-1}\right)\\ &=\left(a_2+\frac{a_1^2}{2}\frac{d}{da_0}\right)\left(\sum_{n=0}^\infty(-1)^n(a_0-1)^n\right)\\ &=\left(a_2+\frac{a_1^2}{2}\frac{d}{da_0}\right)\left(\frac{1}{a_0}\right)\\ &=\frac{a_2}{a_0}-\frac{a_1^2}{2a_0^2}\\ [x^3]&\ln(\arctan(\alpha e^x))=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}[x^3]A(x)\\ &=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\left(na_3(a_0-1)^{n-1}+a_1a_2n(n-1)(a_0-1)^{n-2}\right.\\ &\qquad\qquad\qquad\qquad\left.+\frac{1}{6}n(n-1)(n-2)a_1^3(a_0-1)^{n-3}\right)\\ &=\sum_{n=0}^\infty (-1)^{n}\left(a_3(a_0-1)^{n}+a_1a_2n(a_0-1)^{n-1}+\frac{1}{6}n(n-1)a_1^3(a_0-1)^{n-2}\right)\\ &=\left(a_3+a_1a_2\frac{d}{da_0}+\frac{a_1^3}{6}\frac{d^2}{da_0^2}\right)\left(\sum_{n=0}^\infty(-1)^n(a_0-1)^n\right)\\ &=\left(a_3+a_1a_2\frac{d}{da_0}+\frac{a_1^3}{6}\frac{d^2}{da_0^2}\right)\left(\frac{1}{a_0}\right)\\ &=\frac{a_3}{a_0}-\frac{a_1a_2}{a_0^2}+\frac{a_1^3}{3a_0^3}\tag{7} \end{align*}

$$ $$

Series expansion of $f(x)$:

Now it's time to harvest. We finally obtain with (3) and (7) respecting that $f(x)$ is odd

\begin{align*} \color{blue}{f(x)}&\color{blue}{=\ln\left(\arctan\left(\alpha e^x\right)\right)-\ln\left(\arctan\left(\alpha e^{-x}\right)\right)}\\ &=\ln(a_0-1)+\frac{a_1}{a_0}x+\left(\frac{a_2}{a_0}-\frac{a_1^2}{2a_0^2}\right)x^2 +\left(\frac{a_3}{a_0}-\frac{a_1a_2}{a_0^2}+\frac{a_1^3}{3a_0^3}\right)x^3\\ &\qquad-\ln(a_0-1)+\frac{a_1}{a_0}x-\left(\frac{a_2}{a_0}-\frac{a_1^2}{2a_0^2}\right)x^2\\ &\qquad\qquad +\left(\frac{a_3}{a_0}-\frac{a_1a_2}{a_0^2}+\frac{a_1^3}{3a_0^3}\right)x^3+O(x^5)\\ &=\frac{2a_1}{a_0}x+2\left(\frac{a_3}{a_0}-\frac{a_1a_2}{a_0^2}+\frac{a_1^3}{3a_0^3}\right)x^3+O(x^5)\\ &\color{blue}{=\frac{2\alpha}{(1+\alpha^2)\arctan(\alpha)}x +\frac{\alpha}{3(1+\alpha^2)^3\arctan(\alpha)}\Bigg(\alpha^4-6\alpha^2+1}\\ &\qquad\color{blue}{\left.-\frac{3\alpha(1-\alpha^2)}{\arctan(\alpha)}+\frac{2\alpha^2}{\left(\arctan(\alpha)\right)^2}\right)x^3+O(x^5)} \end{align*}