Vrftsjtryk

Suppose $a,binmathbb{N}$ and, moreoever, $1leqslant a$ and $bgeqslant a+2$.

I am considering the polynomial
$$
f_{a,b}(x):=x^{2b}-frac{x^{b-a}-1}{x-1}
$$
which has exactly one positive (simple) root $x_{a,b}$ and, moreover, $x_{a,b}>1$. In particular, $lim_{atoinfty}x_{a,b}=1$.

I am trying to analyse at which rate $x_{a,b}$ tends to $1$ as $atoinfty$. To this end, I make the ansatz
$$
x_{a,b}=1+y_{a,b}
$$
and now try to analyse at which rate $y_{a,b}to 0$ as $atoinfty$, say, $y_{a,b}=frac{1}{a}+o(1/a)$ or whatever the correct rate might be.

Do you have any idea how to get this?

My first attempt was to plug the ansatz for $x_{a,b}$ in the polynomial:
begin{align*}
&(1+y_{a,b})^{2b}-frac{(1+y_{a,b})^{b-a}-1}{(1+y_{a,b})-1}=0\
&Leftrightarrow (1+y_{a,b})^{2b+1}-(1+y_{a,b})^{2b}-(1+y_{a,b})^{b-a}+1=0
end{align*}

Using the binomial theorem, I wrote the last equation as
begin{align*}
&sum_{k=1}^{2b+1}binom{2b+1}{k}y_{a,b}^k-sum_{k=1}^{2b}binom{2b}{k}y_{a,b}^k-sum_{k=1}^{b-a}binom{b-a}{k}y_{a,b}^k\
&=y_{a,b}(1-(b-a))+sum_{k=2}^{2b+1}binom{2b+1}{k}y_{a,b}^k-sum_{k=2}^{2b}binom{2b}{k}y_{a,b}^k-sum_{k=2}^{b-a}binom{b-a}{k}y_{a,b}^k\
&=0
end{align*}

Factoring out $y_{a,b}$, what I get is
begin{equation*}
y_{a,b}cdot left(1-b+a+sum_{k=1}^{2b}binom{2b+1}{k+1}y_{a,b}^k-sum_{k=1}^{2b-1}binom{2b}{k+1}y_{a,b}^k-sum_{k=1}^{b-a-1}binom{b-a}{k+1}y_{a,b}^kright)=0.
end{equation*}

This equation is fulfilled exactly if $y_{a,b}=0$ (what seems not to be helpful for my purpose) or if
begin{equation}
sum_{k=1}^{2b}binom{2b+1}{k+1}y_{a,b}^k-sum_{k=1}^{2b-1}binom{2b}{k+1}y_{a,b}^k-sum_{k=1}^{b-a-1}binom{b-a}{k+1}y_{a,b}^k=b-a-1.
end{equation}

Maybe this last equation can help to get the desired Information about $y_{a,b}$; however, I don’t see how.

I am thankful for your ideas.

For brevity, I will denote $n = 2b$, $k = b-a$ and drop the index $a,b$ in $y := y_{a,b}$. Our equation is
$$(1+y)^n = frac{(1+y)^k - 1}{y}$$
which we can rewrite as
$$(1+y)^k = 1 + y (1+y)^n$$
and taking logarithms, we get
$$k ln(1+y) = ln(1 + y (1+y)^n)$$
Now, assuming $y (1+y)^n underset{a to infty}{longrightarrow} 0$, we get the equivalents
$$k y underset{a to infty}{sim} y (1+y)^n$$
dividing by $y$ and taking logarithms yields
$$ln(k) underset{a to infty}{sim} n ln(1+y) underset{a to infty}{sim} n y$$
and finally we get
$$y underset{a to infty}{sim} frac{ln(k)}{n} = frac{ln(b-a)}{2b}$$

EDIT : There was a mistake initially in assuming the condition $y (1+y)^n underset{a to infty}{longrightarrow} 0$ was automatic, but it's only true under the assumption that $k ln(k) = o(n)$. In this section, we prove the following three properties are equivalent :

1) $y (1+y)^n underset{a to infty}{longrightarrow} 0$

2) $y sim frac{ln(k)}{n}$

3) $k ln(k) = o(n)$

We have just proved that $1) Longrightarrow 2)$.

Let's prove $2) Longrightarrow 3)$. We assume 2), that is $y sim frac{ln(k)}{n}$. We can then compute that
begin{eqnarray*}
y (1+y)^n &=& frac{ln(k)}{n} e^{n ln(1 + frac{ln(k)}{n})} \
&=& frac{ln(k)}{n} e^{ln(k) - frac{(ln(k))^2}{2n} + oleft(frac{(ln(k))^2}{n}right)} \
&=& frac{ln(k)}{n} k , e^{- frac{(ln(k))^2}{2n} + oleft(frac{(ln(k))^2}{n}right)} \
&sim& frac{k ln(k)}{n}
end{eqnarray*}
Which implies that $frac{k ln(k)}{n} to 0$, which is property 3).

And finally, we prove that $3) Longrightarrow 1)$. We assume $k ln(k) = o(n)$. Let $epsilon > 0$ be a fixed parameter. We prove that for $a$ sufficiently large
$$frac{ln(k) - epsilon}{n} < y < frac{ln(k) + epsilon}{n}$$
To do that, we compute the asymptotic expansion of $f_{a,b}left(1+ frac{ln(k) pm epsilon}{n}right)$ and look at its sign. Indeed we have
begin{eqnarray*}
left(1 + frac{ln(k) pm epsilon}{n}right)^n &=& e^{n lnleft(1+ frac{ln(k) pm epsilon}{n}right)}\
&=& e^{(ln(k) pm epsilon) - frac{(ln(k) pm epsilon)^2}{2n} + oleft(frac{(ln k pm epsilon)^2}{2n}right)}\
&=& k , e^{pmepsilon} e^{- frac{(ln(k) pm epsilon)^2}{2n} + oleft(frac{(ln k pm epsilon)^2}{2n}right)}\
&=& k , e^{pmepsilon} + o(k)
end{eqnarray*}
And
begin{eqnarray*}
frac{left(1 + frac{ln(k) pm epsilon}{n}right)^k-1}{frac{ln(k) pm epsilon}{n}} &=& frac{n}{ln(k) pm epsilon} left(e^{k lnleft(1+ frac{ln(k) pm epsilon}{n}right)}-1right) \
&=& frac{n}{ln(k) pm epsilon} left(e^{frac{k(ln(k) pm epsilon)}{n} + oleft(frac{k(ln(k) pm epsilon)}{n}right)} - 1 right)\
&=& k + o(k)
end{eqnarray*}
So we finally get
$$f_{a,b}left(1+ frac{ln(k) pm epsilon}{n}right) sim (e^{pmepsilon}-1) , k $$
This means that for $a$ sufficiently large, we have $f_{a,b}left(1+ frac{ln(k) - epsilon}{n}right) < 0$ and $f_{a,b}left(1+ frac{ln(k) + epsilon}{n}right) > 0$ and so there is a root of $f_{a,b}$ between those two boundaries. But because there is only one positive root, it must be $y$, and we deduce
$$frac{ln(k) - epsilon}{n} < y < frac{ln(k) + epsilon}{n}$$

And now we can asymptotically bound the quantity $y (1+y)^n$ by

$$ underbrace{frac{ln(k) - epsilon}{n} left(1 + frac{ln(k) - epsilon}{n}right)^n}_{sim frac{k (ln(k)-epsilon) e^{-epsilon}}{n}} < y (1+y)^n < underbrace{frac{ln(k) + epsilon}{n} left(1 + frac{ln(k) + epsilon}{n}right)^n }_{sim frac{k (ln(k)+epsilon) e^{epsilon}}{n}}$$

where both bounds converge to $0$ from the assumption that $k ln(k) = o(n)$. Which proves 1).