A polynomial f(x) of degree 2000 is given. It's known that f(x2-1) has exactly 3400 real roots while f(1-x2) has exactly 2700 real roots. Prove that there exist two real roots of f(x) such that the difference between them is less than 0.002.
Suppose f(x) has a real root at c so that f(c)=0
Then the nature some of the real roots of f(x^2-1) can be determined by x^2-1=c namely:
none if c<-1, one if c=-1, two if c>-1
and the nature of some of the real roots of f(1-x^2) can be determined by 1-x^2=c namely
none if c>1, one if c=1, two if c<1
Since f(x^1-1) and f(1-x^2) both have an even number of roots, f(x) cannot have roots at 1 or -1. Also there are 3400/2=1700 unique roots less than 1 and 2700/2=1850 unique roots greater than -1.
As f(x) can only have 2000 roots total, the only possible distribution of these roots is
150 below -1
1550 between -1 and 1
300 above 1
The maximum number of roots that can fit between -1 and 1 and still be spaced 0.002 apart is 1001. We have more than this, so there must be some whose distance is less.
(2/1549 = 0.00129 is the true widest spacing)
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Posted by Jer
on 2021-04-17 17:04:49 |
Possible solution
