# Root estimation

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?

• Is $k$ a positive integer? What kind of estimates are you looking for? – Aryabhata Apr 7 '11 at 16:37
• 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). – ogn Apr 7 '11 at 16:44
• is it possible to verify this equation on a contraction property? – Ilya Apr 7 '11 at 16:53

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

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

• 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. – ogn Apr 8 '11 at 8:25
• 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. – ogn Apr 8 '11 at 11:59
• @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. $\theta$). – Aryabhata Apr 8 '11 at 14:15
• @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). – Aryabhata Mar 6 '17 at 19:29
• @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. – Aryabhata Mar 16 '17 at 22:57