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What is the estimation for the positive root of the following equation $ ax^k = (x+1)^{k-1} $ where $a > 0$ (specifically $0 < a \leq 1$).

Could you point out some reference related to the question?

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    is it possible to verify this equation on$a$contraction property?2011-04-07

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Put $\displaystyle z = 1 + \frac{1}{x}$ and we get the equation

$z^{n-1}(z-1) = a$

(I prefer to use $\displaystyle n$ instead of $\displaystyle k$)

We can easily see that $\displaystyle z \in (1,2)$

Assume that $\displaystyle z = 1 + \frac{g(n)}{n-1}$

Thus we have that

$\left(1 + \frac{g(n)}{n-1}\right)^{n-1} g(n) = (n-1)a$

Now we have that $\displaystyle e^{x/2} \lt 1 + x \lt e^x$ for $\displaystyle x \in (0,1)$

And so we get $e^{g(n)} g(n) \gt (n-1)a \gt e^{g(n)/2} g(n)$

Now since $\displaystyle xe^x$ is increasing, and the root of $\displaystyle xe^x = y$ is given by the LambertW function: $\displaystyle W(y)$.

Thus we get that

$\displaystyle g(n) \gt W((n-1)a)$

and

$\displaystyle g(n) \lt 2 W\left(\frac{(n-1)a}{2}\right)$

It is well known that $\displaystyle W(x) = \theta(\log x)$ (as $\displaystyle x \to \infty$) and so we get that

$\displaystyle g(n) = \theta (\log (n-1)a) = \theta(\log na)$

Thus the root $\displaystyle r(n)$ is $\theta\left(\frac{n}{\log na}\right)$

In fact, I would go so far as to guess that

$ \lim_{n \to \infty} \dfrac{r(n)W((n-1)a)}{n-1} = 1$

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    @CogitoErgoCogitoSum: You haven't shown any work at all and don't seem to have understood what I was saying (in the answer and comment). I suspect you are doing this deliberately and it is going to be hard for me to be polite if I continue this discussion. So bye.2017-03-16