# Existence of solutions to a nonlinear algebraic equation

How can we prove that equation (1) has solutions for every $p$. I mean, is there an analytic method that can be used to show that there exist solutions for every $p$ for this nonlinear equation:

$$\left\{ \begin{array}{ccc} x_1^2 x_2 \cdots x_{p-1} x_p+x_1 &=& 1 \, ,\\ &&\\ x_1 x_2^2 x_3 \cdots x_{p-1} x_p+x_1 x_2 &=& 1 \, ,\\ &&\\ x_1 x_2 x_3^2 x_4 \cdots x_{p-1} x_p+x_1 x_2 x_3&=& 1\, , \\ &&\\ \vdots & \vdots & \vdots \\ &&\\ x_1 x_2 x_3 x_4 \cdots x_{p-1} x_p^2+x_1 x_2 \cdots x_p&=& 1\, . \end{array} \right.$$

For example, let ($p=3$). Then we have:

$$\left\{ \begin{array}{ccc} x_1^2 x_2 x_3+x_1 &=& 1 \, ,\\ &&\\ x_1 x_2^2 x_3 +x_1 x_2 &=& 1 \, ,\\ &&\\ x_1 x_2 x_3^2 +x_1 x_2 x_3&=& 1\, . \end{array} \right.$$

So is there an analytic method for existence of solution for equation (2). For this example I used MAPLE software and got these solutions:

\begin{equation*} \left\{ \begin{array}{ccc} a &=& 0.658418845314095780 \, ,\\ &&\\ b &=& 0.849466898144101812 \, ,\\ &&\\ c&=& 0.927561975482924960 \, . \end{array} \right. \end{equation*}

• Existence of solutions! – Fernando Muro Nov 29 '15 at 20:26

Note that no $x_k$ can vanish. If you put $u :=x_1x_2\dots x_p$ the system gives inductively, for $k=1,\dots,p$:

$$\frac{1}{x_1 x_2\dots x_k}=1+u+\dots+u^k .$$

In particular $u$ solves $$\frac{1}{u}=1+u+\dots+u^p .$$ Incidentally, $u\neq 1$, so we can express the solutions to the systems simply as $$x_k=\frac{u^{k}-1}{u^{k+1}-1},$$ in terms of the solutions $u\neq 1$ to the equation

$$u^{p+2}-2u+1=0.$$

Also note that, by elementary arguments, if $p$ is even the latter equation has a unique real solution (besides $u=1$), which is positive, while if $p$ is odd it has a positive and a negative solution (besides $u=1$). So the system has one or two real solutions according whether $p$ is even or odd, and in any case a unique solution if $x_k$ are assumed to be real and positive.

To complete the analytic solution, we may solve $u^{p+2}-2u+1=0$ by series using the Lagrange inversion formula to invert $u-u^{p+2}/2$ at $1/2$: for the solution $0<u_p<1$ one gets

$$u_p=\frac{1}{2}\sum_{k=0}^\infty {pk+2k\choose k} \frac{2^{-pk-2k}}{pk + k+1}.$$

For instance, $$u_3=\frac{1}{2}\sum_{k=0}^\infty {5k\choose k} \frac{32^{-k}}{4k+1}=0.5187900635...$$ that corresponds to the solutions $x_1=a, x_2=b, x_3=c$ you got by MAPLE.

• Dear Professor P.Majer Thank you very much for your creative and complete answer . It was amazing for me your attitude to this nonlinear equation. I have two questions , First It is possible to explain how do you get this formula for $u_p$ from Lagrange inversion formula and second would you tell me why $0< u_p<1$? I mean is there an elementary way to show that for every $p$, The equation $u^{p+2}-2u+1=0$, has a positive real solution between zero and one. Thank you again for your favor to answer to my question. – Amin235 Nov 29 '15 at 17:03
• You're welcome! 1) The series for the solution is a very standard application of the Lagrange inversion formula to $f(u):=u(1- u^{p+1} /2 )$. Use the binomial series to expand $f^{-m}=u^{-m}(1- u^{p+1} /2 )^{-m}$, then find the residue at $0$ extracting the coefficient of $u^{-1}$. 2) Yes, write the equation as $u^p=2u-1$ and compare the values of both LHS and RHS at $0$, and their derivatives at $u=1$. Also, by convexity of $u^p$ there are at most $2$ positive solutions. – Pietro Majer Nov 29 '15 at 18:02
• That $u_p$ can't be larger than $1$ also follows immediately from the first equation for it, $1/u=1+u+\dots+u^p$. – Pietro Majer Nov 29 '15 at 19:03
• I would appreciate your help in advance. – Amin235 Nov 29 '15 at 19:06
• The latter equation for $u$ also comes from a generalization of the Golden Ratio ($p=1$): dividing a segment into $p+1$ sub-intervals that are in a geometric progression of $p+1$ terms, whose next term is the segment itself. Also, $u_p$ is the limit of the ratio of two consecutive terms $F_m/F_{m+1}$ of a linear recursive sequence $F_m$ where each term is the sum of the preceding $p+1$ terms e.g. starting with $1,1,\dots,1.$ emis.de/journals/JIS/VOL18/Szczyrba/sz3.pdf For instance, 0.5187900635… is the reciprocal of the "Tetranacci constant" oeis.org/A000288 – Pietro Majer Nov 29 '15 at 22:56