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Let $n\ge 1$ be a positive integer. The question is to solve the following transcendental equation: \begin{equation} \left(1+q\right)^{2 n} = \frac{\sqrt{\pi}}{2} \frac{1-q}{\sqrt{q}} \sqrt{n} \end{equation}

where $q\in (0,1)$.It is easily seen that for every $n$ the solution is unique. Indeed, the left-hand-side is a monotonically increasing curve that starts at unity at $q=0$ and reaches $2^{2 n}$ at $q=1$. On the other hand the right-hand-side is a monotonically decreasing curve that starts at infinity at $q=0$ and reaches zero at $q=1$. The two curves must have an intersection somewhere inside $(0,1)$. The equation is transcendental so it probably cannot be solved in closed form. The question is whether a closed form solution can be found in the limit of large values of $n$? Below I plot the numerical solution as a function of $n$ in a semi-logarithmic scale.As we can see for big values of $n$ the solution seems to behave as : \begin{equation} q_0 \simeq \frac{A}{n^\theta} \end{equation} Here $A$ and $\theta$ are the intercept and the slope of the straight line in the right part of the Figure. The question is can we find those values $A$ and $\theta$? enter image description here

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  • $\begingroup$ I'd call such a "limit solution" an "asymptotic solution". $\endgroup$
    – rajb245
    Aug 1 '14 at 18:53
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Note that as $n\rightarrow \infty$ the value of $q$ tends to zero. for that reason we can neglect the term $-q$ in the numerator in the right hand side. If we take a careful look at the plot above we will see that the slope of the line is not constant but it slightly increases with $n$. Therefore we alter our ansatz slightly and write:

\begin{equation} q = A \frac{\log(n)}{n} \end{equation} We multiply our original equation by $\sqrt{q}$ and insert the ansatz into it. We have: \begin{eqnarray} \sqrt{q} \left(1+q\right)^{2 n} - \frac{\sqrt{\pi}}{2} \sqrt{n} &=& 0 \\ \sqrt{A} \sqrt{\frac{\log(n)}{n}} \left(1 + A \frac{\log(n)}{n}\right)^{2 n} - \frac{\sqrt{\pi}}{2} \sqrt{n} &=& 0 \end{eqnarray} Now, by expressing the power term in brackets through an exponential from a logarithm and expanding the later in a Taylor series we readily see that the power term in brackets behaves like $n^{2 A}$. Thus we have \begin{eqnarray} \sqrt{A} \sqrt{\log(n)} n^{2 A-1/2} - \frac{\sqrt{\pi}}{2} \sqrt{n} &=& 0 \\ \end{eqnarray} For the above equailty to hold at least approximately $A$ has to be equal to $1/2$. Then we have: \begin{equation} \left(\frac{1}{\sqrt{2}} \sqrt{\log(n)} - \frac{\sqrt{\pi}}{2}\right) \sqrt{n} = 0 \end{equation} We easily check that the left hand side is zero for $n=4.8105$. It is equal to one for $n=1231.15$ and it is equal to two for $n=1 .72033 \cdot 10^7$. All this suggests that the ansatz is a good approximation to the solution at least for intermediate values of $n$ . Below I plotted the true solution to our equation (Purple) along with our ansatz (Blue) as a function of $n$.As you can see the ansatz indeed provides a very good approximation over a large range of $n$ values. enter image description here

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