Using an asymptotic approximation to obtain a closed form solution

I needed to find an extremum of a function of the form

$f(\bar{T}_1) = \big[\mathcal{N}_1!\big]^{\gamma - 1}(\bar{T}_1)^{\mathcal{N}_1} + \big[(\mathcal{N} - \mathcal{N}_1)!\big]^{- 1}\bigg(\frac{\mathcal{N}!}{\mathcal{N}_1!}\bigg)^{\gamma}(\bar{T} - \bar{T}_1)^{\mathcal{N}-\mathcal{N}_1}$

by choice of $\bar{T}_1$ , ideally obtaining a closed form solution for the critical value $\bar{T}_1^{\ast}$ . The precise context is not relevant here – I just want to record an asymptotic approximation trick I was able to use to solve this problem. Differentiating and setting equal to zero we obtain the first-order condition

$\mathcal{N}_1 \big[\mathcal{N}_1!\big]^{\gamma - 1}(\bar{T}_1^{\ast})^{\mathcal{N}_1-1}$

$- (\mathcal{N} - \mathcal{N}_1)\big[(\mathcal{N} - \mathcal{N}_1)!\big]^{- 1}\bigg(\frac{\mathcal{N}!}{\mathcal{N}_1!}\bigg)^{\gamma}(\bar{T} - \bar{T}_1^{\ast})^{\mathcal{N}-\mathcal{N}_1 - 1} = 0$

which implies

$\frac{(\bar{T}_1^{\ast})^{\mathcal{N}_1-1}}{(\bar{T} - \bar{T}_1^{\ast})^{\mathcal{N}-\mathcal{N}_1 - 1}} = \frac{(\mathcal{N} - \mathcal{N}_1)\big[(\mathcal{N} - \mathcal{N}_1)!\big]^{- 1}\bigg(\frac{\mathcal{N}!}{\mathcal{N}_1!}\bigg)^{\gamma}}{\mathcal{N}_1 \big[\mathcal{N}_1!\big]^{\gamma - 1}} = \frac{\big[\mathcal{N}!\big]^{\gamma}(\mathcal{N}_1-1)!}{\big[\mathcal{N}_1!\big]^{2\gamma}(\mathcal{N}-\mathcal{N}_1-1)!} \qquad \qquad \qquad (1)$

This does not immediately yield a closed form solution for $\bar{T}_1^{\ast}$ , but I realised that in the present context, with $\mathcal{N} \gg \mathcal{N}_1 \gg 1$ , and $\bar{T} \gg \bar{T}_1^{\ast}$ , we can in fact satisfactorily approximate the left-hand side as

$\frac{(\bar{T}_1^{\ast})^{\mathcal{N}_1-1}}{(\bar{T} - \bar{T}_1^{\ast})^{\mathcal{N}-\mathcal{N}_1 - 1}} = \bigg(\frac{(\bar{T}_1^{\ast})^{\mathcal{N}_1-1}}{\bar{T}^{\mathcal{N}-\mathcal{N}_1 - 1}}\bigg) \cdot \bigg \{ \bigg(1 - \frac{\bar{T}_1^{\ast}}{\bar{T}}\bigg)^{-(\mathcal{N}-\mathcal{N}_1 - 1)}\bigg \}$

$\approx \bigg(\frac{(\bar{T}_1^{\ast})^{\mathcal{N}_1-1}}{\bar{T}^{\mathcal{N}-\mathcal{N}_1 - 1}}\bigg) \cdot \bigg(1 + (\mathcal{N}-\mathcal{N}_1 - 1)\frac{\bar{T}_1^{\ast}}{\bar{T}}\bigg)$

$\approx \frac{(\bar{T}_1^{\ast})^{\mathcal{N}_1-1}}{\bar{T}^{\mathcal{N}-\mathcal{N}_1 - 1}} + (\mathcal{N}-\mathcal{N}_1 - 1) \cdot \frac{(\bar{T}_1^{\ast})^{\mathcal{N}_1}}{\bar{T}^{\mathcal{N}-\mathcal{N}_1}}$

$\approx (\mathcal{N}-\mathcal{N}_1) \cdot \frac{(\bar{T}_1^{\ast})^{\mathcal{N}_1}}{\bar{T}^{\mathcal{N}-\mathcal{N}_1}}$

Then we can write (1) using the gamma function $n! = \Gamma(n+1)$ as

$(\mathcal{N}-\mathcal{N}_1) \cdot \frac{(\bar{T}_1^{\ast})^{\mathcal{N}_1}}{\bar{T}^{\mathcal{N}-\mathcal{N}_1}} = \frac{[\Gamma(\mathcal{N}+1)]^{\gamma}\cdot [\Gamma(\mathcal{N}_1)]}{[\Gamma(\mathcal{N}_1+1)]^{2\gamma} \cdot [\Gamma(\mathcal{N}-\mathcal{N}_1)]}$

and thus

$T_1^{\ast} = \frac{ [\Gamma(\mathcal{N}+1)]^{\frac{\gamma}{\mathcal{N}_1}} \cdot [\Gamma(\mathcal{N}_1)]^{\frac{1}{\mathcal{N}_1}} \cdot \bar{T}^{\ \frac{(\mathcal{N}-\mathcal{N}_1)}{\mathcal{N}_1}}}{(\mathcal{N} - \mathcal{N}_1)^{\frac{1}{\mathcal{N}_1}}\cdot [\Gamma(\mathcal{N}_1+1)]^{\frac{2\gamma}{\mathcal{N}_1}}\cdot [\Gamma(\mathcal{N}-\mathcal{N}_1)]^{\frac{1}{\mathcal{N}_1}}}$

which is the kind of closed form solution I wanted.

Using an asymptotic approximation to obtain a closed form solution

Published by Dr Christian P. H. Salas

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