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Algebra Difficulty 6.8 National olympiad Prove it China

For a given real number aa and a positive integer nn, prove that:

(1) there exists exactly one sequence of real numbers x0,x1,,xn,xn+1x_0, x_1, \dots, x_n, x_{n+1}, such that
{x0=xn+1=0,12(xi+1+xi1)=xi+xi3a3,i=1,2,,n; \begin{cases} x_0 = x_{n+1} = 0, \\ \frac{1}{2}(x_{i+1} + x_{i-1}) = x_i + x_i^3 - a^3, \quad i = 1, 2, \dots, n; \end{cases}

(2) the sequence x0,x1,,xn,xn+1x_0, x_1, \dots, x_n, x_{n+1} in (1) satisfies xia,i=0,1,,n+1|x_i| \le |a|, i = 0, 1, \dots, n+1.

Solution

(1) Proof of existence: From xi+1=2xi+2xi32a3xi1x_{i+1} = 2x_i + 2x_i^3 - 2a^3 - x_{i-1}, i=1,2,,ni = 1, 2, \dots, n and x0=0x_0 = 0, we get that xix_i is a polynomial of x1x_1 with degree 3i13^{i-1} and real coefficients, for 1in+11 \le i \le n+1. Specifically, xn+1x_{n+1} is a polynomial of x1x_1 with degree 3n3^n and real coefficients. As 3n3^n is an odd number, there exists a real number x1x_1 such that xn+1=0x_{n+1} = 0. Then from this x1x_1 and x0=0x_0 = 0 we can calculate xix_i. The sequence x0,x1,,xn,xn+1x_0, x_1, \dots, x_n, x_{n+1} obtained in this way satisfies the required condition.

Proof of uniqueness: Suppose there are two sequences w0,w1,,wn,wn+1w_0, w_1, \dots, w_n, w_{n+1} and v0,v1,,vn,vn+1v_0, v_1, \dots, v_n, v_{n+1}, both of which satisfy Condition (1). Then 12(wi+1+wi1)=wi+wi3a3\frac{1}{2}(w_{i+1} + w_{i-1}) = w_i + w_i^3 - a^3 and 12(vi+1+vi1)=vi+vi3a3\frac{1}{2}(v_{i+1} + v_{i-1}) = v_i + v_i^3 - a^3. Thus,
12(wi+1vi+1+wi1vi1)=(wivi)(1+wi2+wivi+vi2). \frac{1}{2}(w_{i+1} - v_{i+1} + w_{i-1} - v_{i-1}) = (w_i - v_i)(1 + w_i^2 + w_i v_i + v_i^2).
Suppose wi0vi0|w_{i_0} - v_{i_0}| is the greatest. Then
wi0vi0wi0vi0(1+wi02+wi0vi0+vi02)12wi0+1vi0+1+12wi01vi01wi0vi0. \begin{aligned} |w_{i_0} - v_{i_0}| &\le |w_{i_0} - v_{i_0}| (1 + w_{i_0}^2 + w_{i_0} v_{i_0} + v_{i_0}^2) \\ &\le \frac{1}{2} |w_{i_0+1} - v_{i_0+1}| + \frac{1}{2} |w_{i_0-1} - v_{i_0-1}| \\ &\le |w_{i_0} - v_{i_0}|. \end{aligned}
So, either wi0vi0=0|w_{i_0} - v_{i_0}| = 0 or (1+wi02+wi0vi0+vi02)=0(1+w_{i_0}^2+w_{i_0}v_{i_0}+v_{i_0}^2) = 0, that is, either wi0vi0=0|w_{i_0} - v_{i_0}| = 0 or wi02+vi02+(wi0+vi0)2=0w_{i_0}^2+v_{i_0}^2+(w_{i_0}+v_{i_0})^2 = 0. However it must be wi0vi0=0|w_{i_0} - v_{i_0}| = 0. Since wi0vi0|w_{i_0} - v_{i_0}| is the greatest. Then wivi=0|w_i - v_i| = 0 for every i=1,2,,ni = 1, 2, \dots, n. This completed the proof of (1).

(2) Suppose that xi0|x_{i_0}| is the greatest. Then we have
xi0+xi03=xi0(1+xi02)=12(xi0+1+xi01)+a312xi0+1+12xi01+a3 \begin{aligned} |x_{i_0}| + |x_{i_0}|^3 &= |x_{i_0}| (1 + |x_{i_0}|^2) \\ &= \left| \frac{1}{2}(x_{i_0+1} + x_{i_0-1}) + a^3 \right| \\ &\le \frac{1}{2} |x_{i_0+1}| + \frac{1}{2} |x_{i_0-1}| + |a|^3 \end{aligned}
xi0+a3. \leq |x_{i_0}| + |a|^3.
That means xi0a|x_{i_0}| \leq |a|. So xia|x_i| \leq |a|, i=0,1,,n+1i = 0, 1, \dots, n+1.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.