# Numerical solution to a system of equations

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?

• What is wrong with "$\beta$ blowing up" ? – Yves Daoust Jan 14 at 14:20
• The obvious reason is that using numbers of that magnitude is not practical in a programming language like R. – Scippy Jan 14 at 15:18
• 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. – Yves Daoust Jan 14 at 16:38

## 1 Answer

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 $$f(\alpha)$$

Note the two asymptotes given by

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