Let $c\in(0,1)$ be a constant and let $k$ be a positive odd integer, and let $a(k)$ denote the value of $a$ that satisfies the equation $$(1-a)^kk\sqrt{a}=c$$. As $k\rightarrow\infty$, what can we say about the asymptotic behavior of $a(k)$? Obviously, $a(k)$ must approach zero, but at what rate?

  • $\begingroup$ If $k$ is even then there are three positive solutions $a(k)$, and if $k$ is odd there are two. Are you only interested in the smallest one? $\endgroup$ – Antonio Vargas Sep 5 '14 at 17:03
  • $\begingroup$ Good point! I hadn't realized that -- let's assume $k$ is odd and look for the largest solution, in that case. Question edited accordingly. $\endgroup$ – Whitacre Sep 5 '14 at 17:49

If we divide through by $k$ the equation becomes

$$ (1-a)^k \sqrt{a} = c/k. $$

If we suppose that $a = o(1/k)$ then $(1-a)^k \sim 1$, so we get $\sqrt{a} \sim c/k$ and thus $a \sim c^2/k^2$. This is the smaller of the two positive solutions for odd $k$.

To find the larger solution, take logs of both sides of the equation to get

$$ k \log(1-a) + \tfrac{1}{2}\log a = \log c - \log k. \tag{$*$} $$

Since $a \to 0$ we have $\log(1-a) \approx -a$, so this equation is analogous to something like

$$ -ka + \tfrac{1}{2}\log a \asymp -\log k. $$

We can find two solutions to this equation by letting one or the other of the terms on the left-hand side balance with the right-hand side; either

  • $\log a \asymp -2\log k$, which is the solution $a \sim c^2/k^2$ we found earlier, or

  • $ka \asymp \log k$, which is the larger solution.

Note that in the second case we also have a balance between the two terms on the left, $-ka \asymp \log a$, so we won't get the correct leading order term by just supposing that $ka \sim \log k$. We need to also consider the contribution from the $\log a$ term.

Let's formalize this to explicitly calculate some terms of the asymptotic for $a$.

If we suppose $a \sim b \log k/k$, where $b$ is some constant to be determined, then

$$ \log(1-a) = -a + O(a^2) = -a + o(1/k) $$


$$ \log a = -\log k + \log\log k + \log b + o(1), $$

so upon substituting these into $(*)$ we obtain

$$ -ka - \tfrac{1}{2} \log k + \tfrac{1}{2}\log\log k + \tfrac{1}{2}\log b + o(1) = -\log k + \log c $$

or, rearranging,

$$ a = \frac{\log k}{2k} + \frac{\log\log k}{2k} + \frac{\log(\sqrt{b}/c)}{k} + o\left(\frac{1}{k}\right). $$

We assumed $a \sim b\log k/k$, so from this we can see that $b = 1/2$, and thus

$$ a = \frac{\log k}{2k} + \frac{\log\log k}{2k} - \frac{\log(c\sqrt{2})}{k} + o\left(\frac{1}{k}\right). $$

Of course this is pretty non-rigorous, but it should be relatively straightforward to justify the error bounds along the way.


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