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

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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. 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.

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