Finding roots of certain integer exponent polynomial

Question

I am looking to find the roots of $f(x) = x^{2a+1}+x^{a+1}+x-x^a$ where $a$ is a rational with $0<a<1$. With $a=p/q, p<q$ and lettings $z^q=x$ this can be transformed into the polynomial

$$p(z) = z^{2p+q}+z^{p+q}+z^q-z^p$$ As I am looking for roots, this could also be expressed as $$0 = z^{2p}+(1-\frac{1}{z^q})z^p+1, z\not= 0$$ which looks similar to a quadratic in $z^p$. Might be extraneous, but $q=100$ and $90<p<99$. Is there any guidance into how to solve this? I have attempted expressing as exponential polynomials and Taylor series expansion but have become stuck.

Answer 1

It seems that, beside the trivial $x=0$, there only one zero for the function $$f(x) = x^{2a+1}+x^{a+1}+x-x^a$$ and the solution is a nice function of $a$ (have a look here). If the link is broken, type in Wolfram Alpha

ContourPlot[x^(2a+1)+x^(a+1)+x-x^a == 0,{a,0,1},{x,0,0.35}]

A model of the type $$\color{blue}{x=\frac 13 +\sum_{n=1}^{p-1}\alpha_n\,a^{\beta n}-\left(\frac 13+\sum_{n=1}^{p-1}\alpha_n \right)\,a^{\beta p}}$$ can do a quite decent job for an approximation. For sure, the parameters $\alpha_n$ and $\beta$ have to be adjusted in order to match the data.

To show the impact of $p$ on the sum of squares ($99$ data points generated from $a=0.01$ up to $a=0.99$ by steps of $0.01$)

$$\left( \begin{array}{cc} \text{ p} & \text{SSQ(p)} \\ 2 & 1.63987\times 10^{-2} \\ 3 & 1.09277\times 10^{-3} \\ 4 & 7.57307\times 10^{-4} \\ 5 & 1.27689\times 10^{-4} \\ 6 & 9.27178\times 10^{-5} \\ \end{array} \right)$$

Trying to avoid too many terms, with $p=4$, $\beta \sim 2.578$. So, selecting $\beta=\frac 52$ and making the coefficients rational, the estimate is

$$x_0=\frac 13-\frac{23}{117}a^{\frac 5 2}-\frac{19}{104}a^5+\frac{82}{145}a^{\frac{15} 2}-\frac{40}{77}a^{10}$$ This corresponds to $R^2=0.99990$.

Making one or two iterations of Newton method will probably be sufficient (for sure, better would be Halley or Householder mathods).

Some numbers $$\left( \begin{array}{cccc} a & x_0 & x_1 & \text{solution} \\ 0.05 & 0.333223 & 0.332998 & 0.332998\\ 0.10 & 0.332710 & 0.331987 & 0.331986\\ 0.15 & 0.331607 & 0.330287 & 0.330286\\ 0.20 & 0.329762 & 0.327877 & 0.327875\\ 0.25 & 0.327029 & 0.324722 & 0.324718\\ 0.30 & 0.323264 & 0.320777 & 0.320771\\ 0.35 & 0.318328 & 0.315981 & 0.315975\\ 0.40 & 0.312101 & 0.310255 & 0.310251\\ 0.45 & 0.304499 & 0.303503 & 0.303501\\ 0.50 & 0.295490 & 0.295598 & 0.295598\\ 0.55 & 0.285107 & 0.286376 & 0.286373\\ 0.60 & 0.273431 & 0.275619 & 0.275609\\ 0.65 & 0.260532 & 0.263023 & 0.263008\\ 0.70 & 0.246329 & 0.248168 & 0.248159\\ 0.75 & 0.230337 & 0.230468 & 0.230468\\ 0.80 & 0.211237 & 0.209050 & 0.209032\\ 0.85 & 0.186197 & 0.182425 & 0.182362\\ 0.90 & 0.149869 & 0.147709 & 0.147682\\ 0.95 & 0.092937 & 0.098827 & 0.098517\\ \end{array} \right)$$

Answer 2

I prefer to make a separate answer for this other approach.

In fact, we are looking for the inverse of $$\color{red}{a=\frac{1}{\log (x)}\,\log \left(\frac{(1-x)-\sqrt{(1-3 x) (1+x)}}{2 x}\right)}$$ If $a$ is "small", we can expand the rhs as a series around $x=\frac 13$ and then use power series reversion and find $$x=\frac 13+\sum_{n=1}^\infty \alpha_n\,a^{2n}$$ all coefficient being explicit and, apparently, all negative.

Then, similar to the previous approach, write

$$\color{blue}{x=\frac 13 +\sum_{n=1}^{p-1}\alpha_n\,a^{2 n}-\left(\frac 13+\sum_{n=1}^{p-1}\alpha_n \right)\,a^{2 p}}$$

Since the exponent has not been "optimized", we need more terms for the same quality of approximation; this does not make problem since it is just a linear regression.

For example $$x_0=\frac{1}{3}-\frac{4}{425}a^2-\frac{1905 }{1039}a^4+\frac{1590 }{187}a^6-\frac{5207 }{285}a^8+\frac{14423 }{805}a^{10}-\frac{1640 }{247}a^{12}$$

Edit

Another thing which could be done is to let $x=\frac{1-t^2}{3}$ and face the problem of the inverse of $$\color{red}{a=\frac{1}{\log \left(\frac{1-t^2}{3}\right)}\log \left(\frac{(2+t^2)-t \sqrt{3(4-t^2)} }{2 \left(1-t^2\right)}\right)}$$ which gives a much better conditioned function.

Expanding the result as a series around $t=0$ gives a good approximation of it and using power series reversion gives $$t=\sum_{n=0}^\infty \alpha_n \, a^{2n+1}$$ On this basis, a curve fit gives $$t_0=\frac{79 }{121}a-\frac{13 }{72}a^3+\frac{109 }{73}a^5-\frac{195 }{83}a^7+\frac{93}{68}a^9$$

What could be interesting is that expanding around $t=1$, we have $$a=1+\frac{2 (1-t)}{3 \log \left(\frac{2 (1-t)}{3}\right)}+O\left((t-1)^2\right)$$ which gives, as an approximation for $a$ close to $1$, we have as an estimate $$\color{blue}{t_0=1-\frac{3}{2} (1-a)\,\, W\left(\frac{1}{1-a}\right)}$$ where $W(.)$ is Lambert function.

So, if we consider that we look for the zero of function $$f(t)=\frac{1}{\log \left(\frac{1-t^2}{3}\right)}\log \left(\frac{(2+t^2)-t \sqrt{3(4-t^2)} }{2 \left(1-t^2\right)}\right)-a$$ making one single iteration of Halley's method $$t_1= t_0 - \frac {2 f(t_0)\, f'(t_0)} {2 {[f'(t_0)]}^2 - f(t_0)\, f''(t_0)}$$ should provide almost the solution.

$$\left( \begin{array}{cccc} a & t_0 & t_1 & \text{solution} \\ 0.90 & 0.73817080 & 0.74629177 & 0.74629272 \\ 0.91 & 0.75524825 & 0.76286375 & 0.76286461 \\ 0.92 & 0.77326656 & 0.78028635 & 0.78028709 \\ 0.93 & 0.79236762 & 0.79870056 & 0.79870117 \\ 0.94 & 0.81273661 & 0.81829202 & 0.81829250 \\ 0.95 & 0.83462475 & 0.83931465 & 0.83931500 \\ 0.96 & 0.85839097 & 0.86213428 & 0.86213450 \\ 0.97 & 0.88458797 & 0.88731845 & 0.88731857 \\ 0.98 & 0.91417329 & 0.91585665 & 0.91585670 \\ 0.99 & 0.94921555 & 0.94989386 & 0.94989212 \\ \end{array} \right)$$