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Peace be upon you,

I want to solve a system of two equations in which the existence of $ln\left(\frac{\alpha}{\alpha+\beta}\right)$ function makes some limitations in iterations of the Newton-Raphson method. ($c$ letters represent some constants) \begin{align*} \begin{cases} ln\left(\frac{\alpha}{\alpha+\beta}\right)-c_{s1}+\sum_{i\in A}{c_i\left( \alpha^{-i}-(\alpha+\beta)^{-i}\right)}\\ ln\left(\frac{\beta}{\alpha+\beta}\right)-c_{s1}+\sum_{i\in A}{c_i\left( \beta^{-i}-(\alpha+\beta)^{-i}\right)} \end{cases} \end{align*} I tried to replace the unknown variables ($\alpha$ and $\beta$) by $exp(\alpha)$ and $exp(\beta)$; then, after finding the answer replace the answer ($\alpha_{final}$ and $\beta_{final}$) by $exp(\alpha_{final})$ and $exp(\beta_{final})$. but, the treatment of the system became bad and it diverged.

Does anyone have any idea to pass up the negative values or to use another appropriate method for solving the 2-equation system?

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2 Answers 2

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Put:

$$X = \log\left(\frac{\alpha}{\alpha + \beta}\right) = \log\left(\frac{1}{1 + \frac{\beta}{\alpha}}\right)$$

Then working in terms of X will ensure that the logarithm is always well defined. You then need to define another independent variable Y such that a linarization of Y in terms of small changes in $\alpha$ and $\beta$ does not become almost linearly dependent on the way the change in X depends on small changes in $\alpha$ and $\beta$. Since X depends on the ratio of $\alpha$ and $\beta$, you can choose Y to be a function of the product of $\alpha$ and $\beta$, so:

$$Y = \alpha\beta$$

might work well, at least you'll have elminated two potential problems with Newton-Raphson.

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We can simply substitute $\alpha$ by $a^2$ and $\beta$ by $b^2$.

I did it before but ignored (or neglected) posting the answer up to now.

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