# Math Genius: Lower bound on the location of the root in \$(0,1)\$ of the trinomial \$z^m + z^l -1\$ involving the degrees

I will begin with my original problem: Let $$r : [0,1) to (0,infty )$$ a continuous, increasing function such that $$lim_{hto 1} r(h) = infty$$. In my case $$r$$ is explicitly given. For fixed $$xin (0,1)$$ the set of $$h$$ fulfilling
$$x^{r(h) – r(0)} + x^{8r(h)} – 1 geq 0$$
is of the form $$[0, h_0 (x)]$$. One can see that $$lim_{xto 0} h_0 (x) = 0$$. Actually, I want to say something about the speed of this convergence. For this purpose I tried to consider the problem the other way round.

This is, for fixed $$hin [0,1)$$ one can try to bound the unique root $$x_0 (h) := x_0 (a,b) in (0,1)$$ of

$$p_{a,b} (x) := x^b + x^a -1$$

in terms of $$b := r(h) – r(0)$$, $$a := 8r(h)$$. Indeed, assume $$B(r(h))$$ is a lower bound. That is $$B(r(h)) leq x_0 (h)$$. Then $$B( r(h_0 (x)) ) leq x_0 (h_0 (x)) = x$$. If $$B$$ is nice enough and increasing, we will get information about the speed of convergence.

My question is how to bound $$x_0 (a,b)$$ from below.

I tried the following:
For $$a leq a’ , b leq b’$$ we have
$$p_{a,b} geq p_{a’,b’}$$
which implies that $$x_0 (a,b) leq x_0 (a’,b’)$$. Thus I had the idea to define $$lfloor s rfloor_n := sup{ frac k n leq s : kin Bbb{N}_0 }$$. Then $$x_0 (lfloor arfloor_n , lfloor brfloor_n) leq x_0 (a,b)$$. The advantage is that
$$p_{lfloor arfloor_n , lfloor brfloor_n} (z^n) = z^{n cdot lfloor brfloor_n } + z^{n cdot lfloor arfloor_n } -1$$
is an ordinary polynomial in $$z$$. In particular it is an trinomial with coefficients $$1,1,-1$$.
The naive way to bound the roots of $$z^m + z^l -1$$ from below is to search for a upperbound of the roots of $$-z^{max (m,l)} + z^{vert m -lvert} + 1$$ (cf. https://en.wikipedia.org/wiki/Geometrical_properties_of_polynomial_roots#Bounds_on_all_roots) and take the inverse.
By Chauchy’s bound and Lagrange’s bound this would result in a lower bound $$x_0 (lfloor arfloor_n , lfloor brfloor_n) geq (frac 1 2 ) ^n to 0$$, which does not help much. This is, because the lower bound does not involve the degrees. One also sees that this is a very bad lower bound because the root of $$z^m + z^l -1$$ in $$(0,1)$$ goes to $$1$$ if $$m,l$$ are growing large.

Is there a possibility to bound the root of $$z^m + z^l -1$$ in $$(0,1)$$ from below in terms of the degrees?
Any reference on bounding positive roots is also appreciated.

Edit: For $$h$$ being small $$b$$ will be small eventually. Martin R’s comment suggested the bound $$x_0(a,b) geq 2^{-frac 1 b}$$. This can be achieved by the following: For $$a-b geq 0$$ root $$x$$ will fulfil $$x^b (1 + x^{a-b}) = 1$$. Since $$x^{a-b} in (0,1)$$ we have $$2x^b geq x^b (1 + x^{a-b}) = 1$$, thus $$x^b geq 2^{-frac 1 b}$$.

For the idea formulated above this leads to $$r(h_0 (x) ) – r(0) leq -frac{log (2)}{log (x)}$$.

I do not know if this will give the good bounds you look for.

Admitting that $$(m,l)$$ are large, the solutions are close to $$1$$. So, let $$z=1-x$$ and expand as series around $$x=0$$. This would give
$$(1-x)^m+(1-x)^l-1=1- (l+m)x+frac{1}{2} ((l-1) l+(m-1) m)x^2+-frac{1}{6} left( (l-2) (l-1) l- (m-2) (m-1) mright)x^3+Oleft(x^4right)$$ Now, using series reversion, this would give, as an approximation,
$$x=frac{1}{l+m}+frac{(l-1) l+(m-1) m}{2 (l+m)^3}+Oleft(frac 1{(l+m)^5} right)$$
Trying for $$m=12$$ and $$l=34$$, this would give $$x=frac{2743}{97336}$$ that is to say $$z=frac{94593}{97336}approx 0.9718$$ while the “exact” solution would be $$0.9676$$.

For the reversed series, I did not write the next term which is quite messy but it seems to me that, if $$l > m$$ it is positive. So, the $$x$$ will be a lower bound and the corresponding $$z$$ an upper bound.

Using the next (not written term), for the worked example, we would have $$x=frac{3152531}{102981488}$$ that is to say $$z=frac{99828957}{102981488}approx 0.9694$$ which is much better.