For a given real number a and a positive integer n, prove that:
(1) there exists exactly one sequence of real numbers x0,x1,…,xn,xn+1, such that {x0=xn+1=0,21(xi+1+xi−1)=xi+xi3−a3,i=1,2,…,n;
(2) the sequence x0,x1,…,xn,xn+1 in (1) satisfies ∣xi∣≤∣a∣,i=0,1,…,n+1.
Solution
(1) Proof of existence: From xi+1=2xi+2xi3−2a3−xi−1, i=1,2,…,n and x0=0, we get that xi is a polynomial of x1 with degree 3i−1 and real coefficients, for 1≤i≤n+1. Specifically, xn+1 is a polynomial of x1 with degree 3n and real coefficients. As 3n is an odd number, there exists a real number x1 such that xn+1=0. Then from this x1 and x0=0 we can calculate xi. The sequence x0,x1,…,xn,xn+1 obtained in this way satisfies the required condition.
Proof of uniqueness: Suppose there are two sequences w0,w1,…,wn,wn+1 and v0,v1,…,vn,vn+1, both of which satisfy Condition (1). Then 21(wi+1+wi−1)=wi+wi3−a3 and 21(vi+1+vi−1)=vi+vi3−a3. Thus, 21(wi+1−vi+1+wi−1−vi−1)=(wi−vi)(1+wi2+wivi+vi2). Suppose ∣wi0−vi0∣ is the greatest. Then ∣wi0−vi0∣≤∣wi0−vi0∣(1+wi02+wi0vi0+vi02)≤21∣wi0+1−vi0+1∣+21∣wi0−1−vi0−1∣≤∣wi0−vi0∣. So, either ∣wi0−vi0∣=0 or (1+wi02+wi0vi0+vi02)=0, that is, either ∣wi0−vi0∣=0 or wi02+vi02+(wi0+vi0)2=0. However it must be ∣wi0−vi0∣=0. Since ∣wi0−vi0∣ is the greatest. Then ∣wi−vi∣=0 for every i=1,2,…,n. This completed the proof of (1).
(2) Suppose that ∣xi0∣ is the greatest. Then we have ∣xi0∣+∣xi0∣3=∣xi0∣(1+∣xi0∣2)=21(xi0+1+xi0−1)+a3≤21∣xi0+1∣+21∣xi0−1∣+∣a∣3 ≤∣xi0∣+∣a∣3. That means ∣xi0∣≤∣a∣. So ∣xi∣≤∣a∣, i=0,1,…,n+1.
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