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I have the below function:

$$\pi(x) = \frac{s_0\cdot \left(1-\left(\frac{s_1}{s_1+x \cdot \lambda}\right)^{k}\right) \cdot r_1}{s_0\cdot \left(1-\left(\frac{s_1}{s_1+x \cdot \lambda}\right)^{k}\right) \cdot \gamma+ r_0}- x$$ with:

  • $x, s_0, s_1, r_0, r_1 \in R^+$
  • $\lambda, \gamma \in [0,1]$
  • $k \in \{0.25, 4\}$

I am trying to find the $x$ that maximises $\pi(x)$ for both k values (by graphing the function on Desmos, one can see there exists a maximum).

We solve $\frac{d \pi(x)}{dx} = 0$ i.e. $$ \frac{\left(s_{0}\cdot r_{1}\cdot r_{0}\cdot\lambda\cdot k\cdot\left(\frac{s_{1}}{s_{1}+\lambda\cdot x}\right)^{1+k}\right)}{s_{1}\cdot\left(s_{0}\cdot\gamma\cdot\left(\left(\frac{s_{1}}{s_{1}+\lambda\cdot x}\right)^{k}-1\right)-r_{0}\right)^{2}}-1=0$$

$$\left(s_{0}\cdot r_{1}\cdot r_{0}\cdot\lambda\cdot k\cdot\left(\frac{s_{1}}{s_{1}+\lambda\cdot x}\right)^{1+k}\right) = s_{1}\cdot\left(s_{0}\cdot\gamma\cdot\left(\left(\frac{s_{1}}{s_{1}+\lambda\cdot x}\right)^{k}-1\right)-r_{0}\right)^{2}$$

We introduce:

  • $a=s_{0}\cdot r_{1}\cdot r_{0}\cdot\lambda \cdot k$
  • $y=s_1+\lambda \cdot x$

and rewrite $$ \pm\sqrt{a} \cdot s_1^{\frac{k}{2}} \cdot y^{-\frac{k+1}{2}} = s_0 \cdot \gamma \cdot \left(\left(\frac{s_{1}}{y}\right)^{k}-1\right)-r_{0} $$ $$ \pm \sqrt{a} \cdot s_1^{\frac{k}{2}} \cdot y^{-\frac{k+1}{2}} - s_0 \cdot \gamma \cdot s_{1}^k \cdot y^{-k} + s_0 \cdot \gamma +r_{0} =0 $$

and going one step further:

$$A \cdot y^{-\frac{k+1}{2}} + B \cdot y^{-k} + C= 0$$ with:

  • $A=\pm \sqrt{s_{0} \cdot r_{1} \cdot r_{0} \cdot \lambda \cdot k \cdot s_1^k}$
  • $B = - s_0 \cdot \gamma \cdot s_{1}^k$
  • $C=s_0 \cdot \gamma +r_{0}$

How can we go from here? One solution to make things easier could be to rewrite the equation with $t=\frac{1}{\sqrt{y}}$ but not sure that leads anywhere.

NB: this is a repost from maths.stackexchange as it seems to be more appropriate for mathoverflow as more research oriented

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  • $\begingroup$ mathoverflow.net/questions/418392/… $\endgroup$ Commented Mar 25, 2024 at 9:38
  • $\begingroup$ @CaveJohnson Does that work for a pseudo trinomial in the sense that here the exponents being negative, this is not a real trinomial? $\endgroup$ Commented Mar 25, 2024 at 9:43
  • $\begingroup$ Hi Arthur! I think you wrote to me on April 15 asking for this problem, and I sent you a more or less complete solution in terms of hypergeometric series, and by Newton iteration for comparison. Was it ok? $\endgroup$ Commented Aug 23, 2024 at 13:30

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After spending some time on it, below is my proposed solution.

We can undergo another change of variables to make it a trinomial: $t=\frac{1}{\sqrt{y}}$ and we obtain: \begin{align} A \cdot t^{k+1} + B \cdot t^{2k} +C =0 \ \ (1) \end{align}

Case of k=4

For $k=4$, the exponents $k+1=5$ and $2k=8$ are integers, making (1) a proper trinomial: \begin{align} A \cdot t^{5} + B \cdot t^{8} +C =0 \ \ (2) \end{align}

There is probably a Lagrange inversion theorem series solution to it, we try to formalize it: The problem has the form $z=f(w)$ with:

  • $f(w)=A \cdot w^5+B \cdot w^8$
  • $f'(w) = 5A\cdot w^4+8B\cdot w^7$
  • $z=-C$

$f$ is analytical in $w_0=1$ and $f'(w_0)=5A+8B \neq 0$.

We can now express the form $w=h(z)$ with: $$h(z) = w_0 + \sum_{n=1}^\infty h_n \cdot \frac{\left(z-f(w_0)\right)^n}{n!}$$

where:

$$h_n=\lim_{w \rightarrow w_0} \frac{d^{n-1}}{dw^{n-1}}\left[\left(\frac{w-w_0}{f(w) - f(w_0)}\right)^n\right]$$

Case of k=1/4

For $k=0.25$, the exponents are $k+1=1.25$ and $2k=0.5$ which are not integers, so we could make another change of variable so that solving (1) is equivalent to solving: \begin{align} A \cdot t^{5} + B \cdot t^{2} +C =0 \ \ (3) \end{align}

And we can follow a similar reasoning as in the case of $k=4$ to find a Lagrange inversion theorem series solution

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