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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 $k$ a positive integer? What kind of estimates are you looking for?2011-04-07
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    Yes, $k$ is positive integer. If $a = 1$ then an estimation on $x$ is $\Theta(k/\log k)$. I am looking for a similar kind of estimation where $a$ is fixed (an estimation as a function of $k$ (of course that depends also on $a$) where $k$ is large).2011-04-07
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    is it possible to verify this equation on a contraction property?2011-04-07

1 Answers 1

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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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    Nice! But I am wondering what happens if $a$ is a function of $n$, say $a = 2^{-n}$. Now $\log(na)$ is negative, but the equation has positive root.2011-04-08
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    Put $y = ax$, we have $(ax)^n = (ax + a)^{n-1} \leq (ax + 1)^{n-1}$. Thus, $x \leq c \cdot \frac{n}{a \log n}$ for some constant $c$. % However, I think the bound could be strengthened.2011-04-08
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    @ogn: Even if $a$ is $2^{-n}$, $W((n-1)a)$ will be positive, you will just have to pick the asymptotics for $W(x)$ as $x \to 0$ (here we picked the one where $x \to \infty$). So I don't see the problem. For constant $a$, the bound I gave is tight (i.e. $\theta$).2011-04-08
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    Of course, I agree we can probably give more accurate bounds, like giving the lower order terms and the constants involved, but that wasn't clear from your comment.2011-04-08
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    @Moron: I am wondering when $a$ is small, say $a = 2^{-n}$, whether there is a good bound for $W(na)$.2011-04-09
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    @ogn: You can use the Taylor series for $W$.2011-04-09
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    Dont know if I should be impressed or not. Having trouble following the logic, personally. Other people are giving congrats though. But after actually running a series of example problems with Wolfram, I find the relationship fails time and time again. All that work and all you had to do was try a case to see that its total bunk.2017-03-06
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    @CogitoErgoCogitoSum. It is an asymptotic estimate. That itself should tell you that plugging in values is just wasting your time (at least for small n).2017-03-06
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    I know. Didnt say it was an equality. I said "the relationship fails". You have just proven that you did not even bother to test my claim before criticizing it. Which is *precisely* what my original criticism was. You went on and repeated the same ineptitude. Its perfectly plausible that my testing was faulty and that Im mistaken, but if youre not even going to check the work then I dont hold you in high enough regard to debate it with you any further.2017-03-16
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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