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How to compute $x$ from

$$q x^p = 1 - x$$

where $x$ and $q$ are positive, while $p$ is a real number?

When $p > 0$: it's two monotonic functions, one increasing and one decreasing, and having opposite relations on the ends - single solution. When $p \le 0$: second derivative is positive - up to two solutions.

And, since this equation is not even algebraic, I'm almost sure that $x$ may only be found, at best, as a power series, but not expressed finitely through elementary functions. If so, then what is the series? (Sure binary search may be employed, but a more elegant method is preferable.)

Alternatively, how to compute $x$ and $y$ from

$$a \log|x| + b \log|y| = c$$

$$x + y = 1$$

where $\log x, \log y$, $a, b$, and $c$ are real?

Sure I could expand the logarithms into the series and perform something weird with the coefficients, but isn't there an easier and cheaper (and well-known) way? Since the problem may be so easily formulated, someone probably already solved it, but I couldn't find anything related.

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  • $\begingroup$ If $p>0$, the fixed point iteration $x=1-q x^p, x_0 = 0.5$ seems to converge fast to the answer. If $p<0$, you could clarify if/when $x^p, x<0$ is real number? $\endgroup$ Commented Feb 26, 2016 at 2:50
  • $\begingroup$ It seems to be that. Thanks! I'm only interested in positive $x$ right now. $\endgroup$ Commented Feb 26, 2016 at 3:33
  • $\begingroup$ "And cheaper" Making a cool program to solve it? $\endgroup$ Commented Mar 23, 2016 at 0:05

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Since both sides of the equation have a simple derivative, The Newton-Raphson method would give a "simple" recursive law for x approaching the solution. I think that is the most one can look for.

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  • $\begingroup$ Did it. It can also be applied to the more general case $y:=q_1 x^{p_1}+q_2 x^{p_2}+\cdots+x-1=0$. And fixed-point iteration too. But there may be no or two, or maybe even more solutions for $x$. $\endgroup$ Commented Mar 23, 2016 at 9:28
  • $\begingroup$ @AndrewMiloradovsky From looking at it through the fundamental theorem of algebra, there should be a $p$ amount of solutions, if $p\in\mathbb{N}$, or an infinite amount, I think. $\endgroup$ Commented Mar 23, 2016 at 11:40
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Simply put, there is no 'general closed-form solution' to a polynomial of degree $5$ or greater.

Since $p>5$ is more than possible, or even where $p$ is not a whole number, the general solution to this problem cannot be solved for.

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  • $\begingroup$ Yes. But is there a qualitative theory for this type of equations? E.g., what is the form of the solution space for $q_1 x^{p_1}+q_2 x^{p_2}+\cdots=1-x$, where $p_1,p_2,\dots$ are not necessarily integers? To apply the numerical methods more reliably. (It's part of an optimization problem.) $\endgroup$ Commented Mar 23, 2016 at 9:36
  • $\begingroup$ The original problem is to find maximum of $\sum_i A_i x_i^{\alpha_i}$ under $x\ge 0$ and $\sum_i x_i \le X$. $\endgroup$ Commented Mar 23, 2016 at 9:44
  • $\begingroup$ @AndrewMiloradovsky Oh, my bad. $\endgroup$ Commented Mar 23, 2016 at 11:39
  • $\begingroup$ If we include complex solutions, then it's something mind-blowing... even in $x^p=1$ where $p\in\mathbb{R}$... But when $x\in\mathbb{R}$, even if $p$ is not rational, as in the question, there's from zero to two solutions (or at least it looks so, for $p < 0$). So I think there must be a theory for this kind of thing. $\endgroup$ Commented Mar 23, 2016 at 12:39
  • $\begingroup$ @AndrewMiloradovsky In the scenario $x^p=1$, where $p\in\mathbb{R}$, the solution may be found be rewriting $$x^p=e^{\pm2\pi in}$$ where $n=0,1,2,3,\dots$ I can say that much. For it to be real, $\sin(2\pi\frac np)=0\implies\frac np\in\mathbb{Z}$, if I did that right. Not so easily done if you put another $x$ on the right side. $\endgroup$ Commented Mar 23, 2016 at 22:45

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