In an optimiation problem I came across the following daunting equation: $$ \log \left(\frac{1-t_2}{1-y}\right) \log \left(\frac{(1-x) (t_1-t_2)}{(1-t_1) (x-y)}\right) \log \left(\frac{t_2 x}{t_1 y}\right)\\=\log \left(\frac{t_2}{y}\right) \log \left(\frac{x (t_1-t_2)}{t_1 (x-y)}\right) \log \left(\frac{(1-t_2) (1-x)}{(1-t_1) (1-y)}\right) $$ in which $x,y\in[0,1]$ are given constants and $t_1,t_2\in[0,1]$ are the variables.

I ended up plotting it numerically and noting that at the solutions (over $t_1,t_2$) the logarithmic factors would be pairwise equal, which in particular implied the hyperbola $$t_1 (1-x) y=t_1 t_2 (y-x)+t_2 x (1-y)$$ along with a few trivial solutions ($t_2=y$, $t_1=x$, $x t_2 = t_1 y$ and $t_2=t_1$). Checking that these are indeed solutions is easy, but coming up with them was not.

I wonder if there was a more structural approach I could have taken to come up with this result?

In fact I have some even more formidable equations which I now hope have simple polynomial solutions, but those I haven't been lucky enough to guess.

Thomas Ahle
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  • @Saad Thank you, I have tried to clarify now. By pairwise equality, I mean that each factor on the left-hand side equals a factor on the right-hand side. It seems odd to me why this is the case. – Thomas Ahle Mar 14 '19 at 11:34

1 Answers1



$\color{brown}{\textbf{Standing of the task.}}$

Denote $$u=t_2,\quad v=t_1,$$ $$p = \log\dfrac{u}{x},\quad q = \log\dfrac{v}{y},\quad r = \log\dfrac{1-u}{1-x},\quad s = \log\dfrac{1-v}{1-y},\quad w = \log\dfrac{v-u}{y-x},\quad $$ $$f(u,v) = s(w-r)(q-p) - q(w-p)(s-r) = sr(p-q) - swp + pq(s-r) +qwr,$$ $$f(u,v) = \begin{vmatrix}w-p-s & w-q-r \\ ps & rq\end{vmatrix},$$ then \begin{align} &f_{mn}(u,v) = \dfrac{\partial^{m+n}f_{mn}(u,v)}{\partial u^m\partial v^n}\\[4pt] &= \sum\limits_{i=0}^m\sum\limits_{j=0}^n \binom mi \binom nj \begin{vmatrix} w_{ij}-\delta_{0j}\,p_i-\delta_{i0}\,s_{j} & w_{ij}-\delta_{0j}\,q_i-\delta_{i0}\,r_j \\ p_{m-i}\,s_{n-j} & r_{m-i}\,q_{n-j} \end{vmatrix},\quad\text{where}\\ &p_i = p^{(i)}(u) = \delta_{i0}\log\dfrac{u}{x} + (\delta_{i0}-1)^{i-1}\,\dfrac{(i-1)!}{u^i},\\ &q_j = q^{(j)}(v) = \delta_{0j}\log\dfrac{v}{y} + (\delta_{0j}-1)^{j-1}\,\dfrac{(j-1)!}{v^j},\\ &r_i= r^{(i)}(u) = \delta_{i0}\log\dfrac{1-u}{1-x} + (\delta_{i0}-1)\,\dfrac{(i-1)!}{(1-u)^i},\\ &s_{ij} = s^{(j)}(v) =\delta_{0j}\log\dfrac{1-v}{1-y} + (\delta_{0j}-1)\,\dfrac{(j-1)!}{(1-v)^j},\\ &w_{ij}=\dfrac{\partial^{i+j}w}{\partial u^i\partial v^j} = \delta_{i0}\delta_{0j}\log\dfrac{v-u}{y-x} + (\delta_{i0}\delta_{0j}-1)^{\delta_{i0}+j-1}\,\dfrac{(i-1)!(j-1)!}{(v-u)^{i+j}},\\[4pt] \end{align}

$\color{brown}{\textbf{Derivatives in the given point.}}$

Note that \begin{align} &p(x) = 0,\quad r(x) = 0,\quad q(y) = 0,\quad s(y) = 0,\quad w(x,y) = 0, \quad f(x,v) = f(u,y) = 0,\\[4pt] &p_i(x) = (\delta_{i0}-1)^{i-1}\,\dfrac{(i-1)!}{x^i},\\ &q_j(y) = (\delta_{0j}-1)^{j-1}\,\dfrac{(j-1)!}{y^j},\\ &r_i(x) = (\delta_{i0}-1)\,\dfrac{(i-1)!}{(1-x)^i},\\ &s_j(y) = (\delta_{0j}-1)\,\dfrac{(j-1)!}{(1-y)^j},\\ &w_{ij}(x,y) = (\delta_{i0}\delta_{0j}-1)^{\delta_{i0}+j-1}\,\dfrac{(i-1)!(j-1)!}{(y-x)^{i+j}}\\[4pt] &=\begin{cases} 0,\quad\text{if}\quad i=0\wedge j=0\\ (-1)^j\,\dfrac{(j-1)!}{(y-x)^j},\quad\text{if}\quad i=0 \wedge j>0\\ -\dfrac{(i-1)!}{(y-x)^i},\quad\text{if}\quad i>0 \wedge j=0\\ (-1)^{j-1}\,\dfrac{(i-1)!(j-1)!}{(y-x)^{i+j}},\quad\text{if}\quad i>0 \wedge j>0,\\ \end{cases}\\[4pt] &f_{mn}(x,y) = \sum\limits_{i=1}^m\sum\limits_{j=1}^n \binom mi \binom nj \begin{vmatrix} a_{ij}(x,y) & b_{ij}(x,y) \\ p_{m-i}(x)s_{n-j}(y) & r_{m-i}(x)q_{n-j}(y) \end{vmatrix},\\ &a_{ij}(x,y) = w_{ij}(x,y)-\delta_{0j}p_i(x)-\delta_{i0}s_j(y),\\ & b_{ij}(x,y) = w_{ij}(x,y)-\delta_{0j}r_i(x)-\delta_{i0}q_j(y),\\ \end{align} Taking in account zero values of $p,q,r,s,w$ in the given point $\dbinom uv = \dbinom xy,$ unzero derivatives can start from $i+j=3.$

$\color{brown}{\textbf{2D Taylor series.}}$

The previous results allow to present the issue equality in the form of $f_l(u,v)=0,$ where $$f_l(u,v) = \sum\limits_{k=3}^l \dfrac1{k!}\sum\limits_{m=1}^{k-1}\binom km f_{m,k-m}(x,y)(u-x)^m(v-y)^{k-m}.$$ is $2D$ Taylor series of $l-$th order.

The calculated values of the derivatives are in the table below.

\begin{vmatrix} & i=0 & i = 1 & i=2 & i = 3 \\ p_i(x) & 0 & \dfrac1x & -\dfrac1{x^2} & \dfrac2{x^3} \\ q_i(y) & 0 & \dfrac1y & -\dfrac1{y^2} & \dfrac2{y^3} \\ r_i(x) & 0 & -\dfrac1{1-x} & -\dfrac1{(1-x)^2} & -\dfrac2{(1-x)^3}\\ s_i(y) & 0 & -\dfrac1{1-y} & -\dfrac1{(1-y)^2} & -\dfrac2{(1-y)^3} \\ w_{i0}(x,y) & 0 & -\dfrac1{y-x} & -\dfrac1{(y-x)^2} & -\dfrac2{(y-x)^3} \\ w_{i1}(x,y) & \dfrac1{y-x} & \dfrac1{(y-x)^2} & \dfrac2{(y-x)^3} & \dfrac2{(y-x)^4} \\ w_{i2}(x,y) & -\dfrac1{(y-x)^2} & -\dfrac2{(y-x)^3} & -\dfrac6{(y-x)^4} & -\dfrac{24}{(y-x)^5} \\ \end{vmatrix}

$\color{brown}{\textbf{The third order model.}}$

One can get \begin{align} &a_{01} = w_{01}(x,y)-p_0(x)-s_1(y) = \dfrac1{y-x} + \dfrac1{1-y}=\dfrac{1-x}{(y-x)(1-y)},\\ &b_{01} = w_{01}(x,y)-r_0(x)-q_1(y) = \dfrac1{y-x}-\dfrac1y = -\dfrac x{y(y-x)},\\ &a_{10} = w_{01}(x,y)-p_1(x)-s_0(y) = -\dfrac1{y-x} - \dfrac1x = -\dfrac{y}{x(y-x)},\\ &b_{10} = w_{01}(x,y)-r_1(x)-q_0(y) = \dfrac1{y-x}+\dfrac1{1-x} = \dfrac{1+y-2x}{(1-x)(y-x)},\\ &f_{12} = a_{01}r_1(x)q_1(y) - b_{01}p_1(x)s_1(y) = -\dfrac{a_{01}}{(1-x)y} + \dfrac{b_{01}}{x(1-y)} = -\dfrac2{(y-x)y(1-y)},\\ &f_{21} = a_{10}r_1(x)q_1(y) - b_{10}p_1(x)s_1(y) = -\dfrac{a_{10}}{(1-x)y} + \dfrac{b_{10}}{x(1-y)}\\ &= \dfrac1{x(y-x)(1-x)} + \dfrac{1+y-2x}{x(1-x)(y-x))(1-y)} =\dfrac2{x(y-x)(1-y)},\\ &f_3(u,v) = \dfrac12(f_{12}(v-y)+f_{21}(u-x))(u-x)(v-y) =\left(\dfrac{u-x}{x}-\dfrac{v-y}{y}\right)\dfrac{(u-x)(v-y)}{(y-x)(1-y)},\\ &f_3(u,v)=0\rightarrow (u=x)\vee(v=y)\vee(uy=vx). \end{align}

The order of model can be increased.

Yuri Negometyanov
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  • Impressive! Can you explain the overall intuition behind this method? You take the 2d Taylor approximation of $u$ near $x$ and $v$ near $y$, then argue that each term of the expansion is zero? – Thomas Ahle Mar 20 '19 at 16:42
  • @ThomasAhle Thank you for the comment. I've understood only that there are possible models of order 2 and higher, because $f(x,y)=0.$ There is new idea for me too, and I'm researching it just now. At first, I'm trying to get something better than $(u-x)(v-y)=0.$ – Yuri Negometyanov Mar 20 '19 at 16:59
  • @ThomasAhle I think, there is my finish. Good luck! – Yuri Negometyanov Mar 21 '19 at 04:06
  • Thank you for the extra details! I still wonder if you could explain the intuition a bit behind your calculations? It also seems like the solutions implied in the end don't include the Hyperbolic solution. $w - r = q$ I think it is in your notation. Perhaps it is clear enough from your determinant equation that $w - r -q=0$ and $w - p -s=0$ is a solution, if achievable. – Thomas Ahle Mar 21 '19 at 14:14
  • @ThomasAhle Thank you for the comments and for the interesting task. 2) The Hyperbolic solution is satisfied near $(x,y).$ Also, the third order Taylor series takes in account three cases, and you have pointed how to obtain the fourth solution from the discriminant form of the equation. Besides, the perspective approach is researching of the high-order models, using obtaned results. 1) Successfully founded discriminant form provides easy differentiation. The 2D series was choosen due to the implicit form, which allows to compare solutions with the Hyperbolic one. – Yuri Negometyanov Mar 21 '19 at 16:57