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Consider a finite sequence $x_i \in (0,1)$ for $i=1,\ldots, n$ and define $y_i=\dfrac{\Pi_{j=1}^n x_j }{x_i}$.

I solved this system for $x$ in terms of $y$ and got $$x_i=\dfrac{\left(\Pi_{j=1}^n y_j \right)^\frac{1}{n-1}}{y_i}.$$

Now pick some $m<n$. Is there a simple way of solving $(x_1,\ldots x_m, y_{m+1},\ldots, y_n)$ in terms of $(y_1,\ldots y_m, x_{m+1},\ldots, x_n)$?

If we take logs, this is a linear system. My only idea on how to solve is to write the system for $n=2,3$ and solve it by brute-force to see if some pattern emerges and if so, make a conjecture and prove it... But perhaps someone can easily see some clever trick I am missing. This is not homework. It is for some lemma I need in my paper.

Sergio Parreiras
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2 Answers2

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Define $x_i'=x_i$ and $y_i'=\dfrac{y_i}{x_{m+1}\ldots x_n}$, for $i\leq m$. Notice that $y_i'=\dfrac{\prod_{i=1}^m x_i'}{x_i'}$.

Now use your second formula to write $x_i=x_i'(i\leq m)$ as a function of $y_i'$. Notice that $y_i'$ depends only on $y_1,\ldots,y_m,x_{m+1},\ldots ,x_n$.

Next, for $j>m$, notice that $y_j=\dfrac{\prod_{i=1}^m x_i'\prod_{i=m+1}^n x_i}{x_{j}}$. Remind that we know how to write $x_i'$ depending on $y_1,\ldots,y_m,x_{m+1},\ldots ,x_n.$ So you can write $y_j(j>m)$ depending on $y_1,\ldots,y_m,x_{m+1},\ldots ,x_n.$

Daniel
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Rewriting the equations in the linearized form, you can use a block decomposition

$$\left[ \begin{array}{ccc} A & B\\ C& D \end{array} \right] \left[ \begin{array}{ccc} x_0\\ x_1\end{array} \right]=\left[ \begin{array}{ccc} y_0\\ y_1\end{array} \right]$$

and solve for the vector sections that you want

$$x_0=A^{-1}(y_0-Bx_1),\\y_1=Cx_0+Dx_1=CA^{-1}y_0+(D-CA^{-1}B)x_1.$$

Letting $O_{mk}$ be a rectangular matrix of all ones, we have

$$A=O_{mm}-I_m,$$ and $$A^{-1}=\frac1{m-1}O_{mm}-I_m.$$

Given the simple structure of the blocks, a closed formula is possible. Let us denote $X_1$ and $Y_0$ the sum of the components of $x_1$ and $y_0$. Then,

$$A^{-1}y_0=\frac{Y_0}{m-1}O_{m1}-y_0$$ $$A^{-1}Bx_1=(\frac1{m-1}O_{mm}-I_m)O_{m,n-m}x_1=(\frac1{m-1}O_{mm}-I_m)O_{m1}X_1=\frac{X_1}{m-1}O_{m1}$$ and $$x_0=\frac{Y_0-X_1}{m-1}O_{m1}-y_0.$$

Next,

$$y_1=O_{n-m,m}x_0+(O_{n-m,n-m}-I_{n-m})x_1\\ =(m\frac{Y_0-X_1}{m-1}-Y_0)O_{n-m,1}+X_1O_{n-m,1}-x_1\\ =\frac{Y_0-X_1}{m-1}O_{n-m,1}-x_1.$$