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I am trying to find the value(s), if any, of $ x $ in the following equation:

$$ \theta \frac{\mu^\beta}{\delta(1-\theta)} x^\beta = x + \left(\frac{1}{A}\right)^{1 / \alpha}\left(\frac{\mu x^\beta}{\delta(1-\theta)}\right)^{1 / \alpha} $$

where $ \beta > 1 $ and $ 0 < \alpha < 1 $ and $\theta, \mu, \delta \in (0,1)$.

I have attempted to solve this algebraically but have not been successful. I also tried to utilize convexity and concavity properties to simplify the problem, but it did not work out.

Any help or suggestions on how to approach this problem would be greatly appreciated.

bruno
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  • Can $\ \alpha\ $ be negative? – lonza leggiera Jun 18 '24 at 12:42
  • @lonzaleggiera no, sorry forgot to mention it: $\alpha \in (0,1).$ – bruno Jun 18 '24 at 12:45
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    Why give us all those complicated terms? This is just $ax^\beta=x+bx^{\beta/\alpha}$. And you will need to solve this numerically, I'm afraid. – TonyK Jun 18 '24 at 12:57
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    More simply, ignoring the trivial solution $x=0$, you have $$ P x^{p-1} = 1 + Qx^{pq-1} $$ where $p>1$ and $q>1$. (My $p$ and $q$ are your $\beta$ and $q=1/\alpha$.) For arbitrary $p$ and $q$, the only workable strategy is to use numerical methods. Even if $p$ and $q$ are integers, once $pq$ exceeds $5$, the problem typically won't be algebraically solvable (ie, using radicals). – Blue Jun 18 '24 at 13:07
  • @TonyK all of the parameters are between 0 and 1, i thought this may affect the solution. – bruno Jun 18 '24 at 13:42
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    There are analytic solutions using special functions as discussed here – Тyma Gaidash Jun 18 '24 at 14:07
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    @bruno: that is irrelevant $-$ the coefficients themselves ($\theta \frac{\mu^\beta}{\delta(1-\theta)}$ etc) can take any positive values. – TonyK Jun 18 '24 at 14:27
  • @ТymaGaidash Thank you so much for the link, it is invaluable. I think the last method is suitable for my problem, can you give some clues as to how to use it in this case, because i cannot see. Also, my equation does not need to be a polynomial, what can i do in that case? – bruno Jun 18 '24 at 15:06
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    @bruno The method works for $x^v+ax+b=0;a,b,v\in\Bbb C$ – Тyma Gaidash Jun 18 '24 at 16:13
  • @ТymaGaidash so does not hold in this case? – bruno Jun 18 '24 at 16:16
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    @bruno Sorry, you can substitute $x^\beta=y$ or $x^\frac\beta\alpha=y$ and get it into $y^r+ay+b=0$ form. – Тyma Gaidash Jun 18 '24 at 16:24
  • @ТymaGaidash I think I am missing something but I cannot get how to get rid of the term $x$ when I do the substitution. – bruno Jun 18 '24 at 18:15
  • @bruno You can factor $x$ out. – Тyma Gaidash Jun 18 '24 at 19:01
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    If your intent is to calculate actual real number values for $x$, given values for the parameters, then Newton's Method will do so faster than using special functions (where one must also resort to numerical methods to calculate). Analysis may be helped by having functions with known properties to work with, but in general, they are not that useful for calculation. And if you want to perform analysis on the function, then "algebra-precalculus" was not the right tag. – Paul Sinclair Jun 19 '24 at 18:23
  • @PaulSinclair thank you for your words first of all. Which tag should I use? – bruno Jun 19 '24 at 23:26
  • "Analysis" is a post-calculus subject, so not "algebra-precalculus" if that is your purpose, but I cannot say more as I do not know what your actual intent is. This is an example of an "XY question": one where the asker has some purpose "Y" that they want to accomplish, and thinks that "X" is the required next step in that direction, so they ask about "X". But occasionally the asker is wrong about X being useful for Y, so a lot of effort is wasted on explaining and accomplishing X, whereas if they had asked about Y directly, they would have gotten a much better answer. – Paul Sinclair Jun 20 '24 at 11:33
  • My comment was to point out that if calculating the value of $x$ is your Y, and finding a functional solution for the equation is the X that you think it is needed, then you are mistaken, and numerical methods such as Newton's will do the calculation more easily. This was just a guess only my part. I do not know your ultimate intent. I just wanted to point it out in case. Without knowing for sure what your actual intent is, I cannot suggest what other tags would be appropriate. – Paul Sinclair Jun 20 '24 at 11:38
  • Dear, you can find a closed and analytical expression to $x$. The solution is quite simple and it does not involver power series, convergence, hypergeometric function. If you are interested, please provides us with some numbers to all unknowns and I can write the closed expression and all results, e.g., real and complexes. – ZKZ Aug 09 '24 at 20:24
  • @ZKZ Okay then let us assume $\alpha=1/2$ and $\beta = 3/2$, i need the other parameters to remain unknown to be able to do comperative analysis on the solution. – bruno Aug 13 '24 at 12:11
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    You can simplify the equation a lot. Divide by $x$ and set $t:=\text{LHS}$. This reduces it to $t=1+ct^r$ where $r:=\frac{\beta-\alpha}{\alpha(\beta-1)}>0$. –  Aug 13 '24 at 16:40

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This problem can be written using a new special function I'be working on, the so-called Lambert-Tsallis function. For a good description I suggest the reading of Solutions 1 and 2 because there you can find its definition as well as properties used here.

After this brief presentation, I have considered the following equation: $$ax^{\beta} = x +bx^{\beta/\alpha}$$ $$\therefore ax^{\beta-1} -bx^{\beta/\alpha-1}= 1 $$ $$\therefore -bx^{\beta/\alpha-1} \bigg( 1 - \frac{a}{b}x^{\frac{\alpha\beta-\beta}{\alpha}}\bigg)= 1 $$ $$\therefore x^{\beta/\alpha-1} e_1\bigg( -\frac{a}{b}x^{\frac{\alpha\beta-\beta}{\alpha}}\bigg)= -\frac{1}{b} $$ raising to $r=\frac{\alpha\beta-\beta}{\beta -\alpha},$ one obtains $$\therefore x^{\big(\beta/\alpha-1\big)r} e_r\bigg( -r\frac{a}{b}x^{\frac{\alpha\beta-\beta}{\alpha}}\bigg)= \bigg(\frac{-1}{b}\bigg)^r $$ $$\therefore x^{\big(\beta/\alpha-1\big)r} = \frac{W_r{\bigg[-r\frac{a}{b}\bigg(\frac{-1}{b}\bigg)^r\bigg]}}{-r\frac{a}{b}} $$ $$\implies x^{\big(\beta/\alpha-1\big)} = \frac{-1/b} {1 + \frac{W_r{\bigg[ -r\frac{a}{b}\big(\frac{-1}{b}\big)^r \bigg]}} {r} } $$

To get the final expression to $x$ we will take $\alpha=1/2, \beta=3/2$ and two arbitrary values of $a=4$ and $b=1$. Substituting the values one obtains $r=-3/4$ and

$$\implies x =\pm \sqrt{ \frac{-1} {1 - \frac{4}{3} W_r{\bigg[ \frac{3a}{4b^{1/4}}\big(-1\big)^r \bigg]}} }$$

To these constants one has $W_r(z)={ 189.7320938682204 - 0.0000000000001i \\ 0.9008012302732 + 0.0000000000000i\\ 0.6835524507532 - 0.0817752453155i\\ 0.6835524507532 + 0.0817752453155i\\ }$

resulting on $x = { 0.0630 + 0.0000i \\ -0.0630 + 0.0000i\\ 2.2301 + 0.0000i\\ -2.2301 + 0.0000i\\ 1.1466 + 2.4090i\\ -1.1466 - 2.4090i\\ 1.1466 - 2.4090i\\ -1.1466 + 2.4090i\\ }$

The original problem is a "polynomial" of fractional degree. Due to this particularity, some fake solutions are introduced. The above list to $x$ includes all of them but only $x={0.0630 ; 2.2301}$ are the true solutions.

ZKZ
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