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Let $n\in\mathbb{N}$ and $u_1,u_2,\ldots ,u_n,t_1,t_2\geq 0$ be constants. I'm interested in finding the numerical solution in relation to $\alpha$ and $\beta$ to the following system of equations $$\begin{cases} \sum_{i=1}^n \left( \frac{u_i}{\beta}\right)^\alpha=t_1\\ \sum_{i=1}^n \ln\left[ \left( \frac{u_i}{\beta}\right)^\alpha\right]=t_2 \end{cases}.$$ My current solution is to extract $\beta$ from the second equation, insert it into the first and find the solution $\alpha$ with the halving algorithm. We get $$\begin{cases} \sum_{i=1}^n \left( u_i/\exp\left( \frac1n \sum_{i=1}^n\ln u_i-\frac{t_2}{n\alpha}\right)\right)^\alpha=t_1\\ \beta = \exp\left( \frac1n \sum_{i=1}^n\ln u_i-\frac{t_2}{n\alpha}\right)\end{cases}.$$ For values $$u_1=1.20167063$$ $$u_2=2.30434494$$ $$u_3=1.20587080$$ $$u_4=0.59277441$$ $$u_5=0.06592318$$ $$t_1=12.5$$ $$t_2=37.5$$ The solution is $\alpha\approx 10^{-4}$ and $\beta \approx \exp(10^4)$. Any ideas on how I can avoid $\beta$ blowing up?

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  • $\begingroup$ What is wrong with "$\beta$ blowing up" ? $\endgroup$ – Yves Daoust Jan 14 at 14:20
  • $\begingroup$ The obvious reason is that using numbers of that magnitude is not practical in a programming language like R. $\endgroup$ – Scippy Jan 14 at 15:18
  • $\begingroup$ I can't make sense of what you say. You provide an equation and don't want to accept its solution value ?! By the way, $10^4$ is not a large number. $\endgroup$ – Yves Daoust Jan 14 at 16:38
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After reduction the system is equivalent to

$$ \frac{t_2}{\alpha}+5\ln \beta =\sum_{k=1}^5\ln u_k\\ \alpha\ln \beta -\ln\left(\sum_{k=1}^5 u_k^{\alpha}\right)=-\ln t_1 $$

and after the ellimination of $\ln \beta$

$$ \frac {\alpha}{5}\sum_{k=1}^5\ln u_k-t_2 = \ln\left(\sum_{k=1}^5 u_k^{\alpha}\right)-\ln t_1 $$

calling

$$ f(\alpha) = \frac {\alpha}{5}\sum_{k=1}^5\ln u_k-t_2 -\left( \ln\left(\sum_{k=1}^5 u_k^{\alpha}\right)-\ln t_1\right) $$

at a solution $f(\alpha)$ must cross the horizontal axis. Calculating the stationary points to $f(\alpha)$ or $f'(\alpha) = 0$ we can easily verify that $\alpha = 0$ is the solution but $f(0) = -36.5837$ hence the former system doesn't have real solution.

Attached the plot for $f(\alpha)$

enter image description here

Note the two asymptotes given by

$$ a_1(\alpha) = \frac {\alpha}{5}\sum_{k=1}^5\ln u_k-t_2 - \alpha\ln\left(\min (u_k)\right)+\ln t_1\\ a_2(\alpha) = \frac {\alpha}{5}\sum_{k=1}^5\ln u_k-t_2 - \alpha\ln\left(\max (u_k)\right)+\ln t_1 $$

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