Asymptotic expansion of the inverse of $x\mapsto x+x^{\small\sqrt2}+x^2$ near zero

Question

^{This is a follow-up to my previous question "Asymptotic expansion of the inverse of $x\mapsto x+x^\phi$ near zero".}

Consider a continuous real-valued monotone increasing function $f:\mathbb R^+\to\mathbb R^+$ satisfying $f\big(x+x^{\small\sqrt2}+x^2\big)=x.$

I am interested in an asymptotic expansion of $f(z)$ for $z\to0^+$ in terms of powers of $z$. I was able to find a few initial terms by manually balancing coefficients: $$f(z)=z-z^{\small\sqrt2}+\sqrt2\;z^{\small\unicode{x202f}2\unicode{x202f}\sqrt2-1}-z^2+\mathcal O\big(z^{\small\unicode{x202f}3\unicode{x202f}\sqrt2-2}\big), \quad z\to0^+.\tag{$\diamond$}$$ Computing next terms in an ad hoc fashion quickly becomes tedious, so I am looking for a more systematic approach that would allow to obtain a general formula for the terms of this series. I expect it to be a mix of integer powers of $z$ and irrational powers involving $\sqrt2$. Also, I would like to know the radius of convergence of that series.

More generally, I am looking for a uniform approach for inverting generalized polynomials of a single variable that may contain both rational and irrational powers.

Answer 1

Too long for a comment.

Look at $$f(x)=x(1+x^{\sqrt2-1}+x)\in S=\Bbb{R}[[x,x^{\sqrt2-1}]]$$ The exponents are well-ordered.

Let $$\phi : x\to f(x), \ x^{\sqrt2-1} \to x^{\sqrt2-1} \sum_{k\ge 0} {\sqrt2-1 \choose k} (x^{\sqrt2-1}+x)^k$$ Then $\phi$ is an injective endomorphism of $S$, since $\phi(x^a)= x^a(1+O(x^{\sqrt2-1})+O(x))$ then $\phi$ is surjective. Whence $\phi(f^{-1}(x))=x$ for some $f^{-1}(x)\in S$.

It remains to find the coefficients one by one as you did. I don't see much reason to believe that $f^{-1}(x)$ converges for some $x>0$.

Answer 2

A naive approach:

We have to solve ( with $y=f$)

$$y+ y^{\sqrt{2}} + y^2 = x$$

Let's take a simpler equation $y - y^2 = x$, equivalent to $y = x + F(y)$, so should have $$y = x + F(x+F(y)) = x+F(x+F(x+F(y)))= x+F(x+F(x+\ldots )))$$ (note that $F$ is not a linear operator, if it were, we would have the Neumann series for $(I-F)^{-1}$), still we have a fixed point problem that is solved with iterations. Using this iteration approach we get the Taylor series for the solution $y = \frac{1 - \sqrt{1 - 4 x}}{2}$.

We might use this approach for $$y = x - y^{\sqrt{2}}- y^2 = x - ( x- y^{\sqrt{2}}- y^2)^{\sqrt{2}} -( x- y^{\sqrt{2}}- y^2)^2= \cdots $$

$\bf{Added:}$

The equation $y = T(y)= x - y^{\sqrt{2}} - y^2$ can be solve by successive approximations $y_{n+1} = T(y_n)$. For instance, we can consider $y_0=x$, $y_1 = x - x^{\sqrt{2}} - x^2$, $y_2 = x - (x - x^{\sqrt{2}} - x^2)^{\sqrt{2}} - (x - x^{\sqrt{2}} - x^2)^2$, etc. One should figure out (easy) how to expand
$$( x+ \sum_{\alpha>1} a_{\alpha}x^{\alpha} )^{\beta} = x^{\beta} ( 1 + \sum_{\gamma >0} c_{\gamma} x^{\gamma})^{\beta}$$ We use the expansion $$(1+ \sum_{\gamma} t_{\gamma})^{\beta} = \sum_{\beta_1 < \ldots \beta_r}\sum_l \frac{\beta(\beta-1)\cdots (\beta- |l|+1)}{l!} t_{\gamma_1}^{l_1}\cdots t_{\gamma_r}^{l_r} $$ We are dealing with formal series with exponents positive and $\to \infty$.

To be continued...

Answer 3

Some thoughts

Conjecture: When $y>0$ is sufficiently small, the unique positive real solution of $x + x^{\sqrt{2}} + x^2 = y$ is given by \begin{align} x &= y + \sum_{k=1}^\infty \frac{(-1)^k}{k!} \frac{\mathrm{d}^{k-1}}{\mathrm{d} y^{k-1}}\Big[(y^{\sqrt{2}} + y^2)^k\Big]\\ &= y - (y^{\sqrt{2}} + y^2) + (\sqrt{2}y^{2\sqrt{2}-1} + (2+\sqrt{2})y^{\sqrt{2}+1} + 2y^3) - \cdots \tag{1} \end{align}

Motivation: The equation is written as $x = y + (-1)(x^{\sqrt{2}} + x^2)$. By the Lagrange reversion theorem (https://en.wikipedia.org/wiki/Lagrange_reversion_theorem), we guess (1).

Numerical experiment using Maple: When $y = \frac{1}{100}$, the solution of $x + x^{\sqrt{2}} + x^2 = y$ is $x_0 \approx 0.008704284007$, while $y + \sum_{k=1}^{20} \frac{(-1)^k}{k!} \frac{\mathrm{d}^{k-1}}{\mathrm{d} y^{k-1}}\Big[(y^{\sqrt{2}} + y^2)^k\Big] \approx 0.008704284006$.

However, when $y$ is larger, it may fail. For example, $y = 1/2$, $y + \sum_{k=1}^{5} \frac{(-1)^k}{k!} \frac{\mathrm{d}^{k-1}}{\mathrm{d} y^{k-1}}\Big[(y^{\sqrt{2}} + y^2)^k\Big]\approx -19.46221804$.

So I guess we need sufficiently small $y$.

Further work: 1) What is the ROC of the series (1)? 2) Prove (or disprove) if it is the solution.