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.
Let S = {r_1 < r_2 < .. < r_N} be the set of distinct real roots of f(x). So card(S) <= 2000.
The roots of f(y^2 - 1) are the solutions to y^2 - 1 = r_k, i.e.
y_k = +- (1 + r_k)^(1/2). So that A = the number or distinct real roots of f(y^2 - 1) is 2 times the number of r_k in S satisfying r_k >= -1.
The roots of f(1 - z^2) are the solutions to 1 - z^2 = r_k, i.e.
z_k = +- (1 - r_k)^(1/2). So that B = the number or distinct real roots of f(1 - z^2) is 2 times the number of r_k in S satisfying 1 >= r_k.
Let C be the elements in S satisfying 1 >= r_k >= -1 then
2* card(S) =< A + B - 2C
2(2000) =< 3400 + 2700 - 2C
1050 =< C
So the minimum separation between root to f(x) = 0 in the range [-1,+1] is no more than 2/1050 < .002
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Posted by FrankM
on 2021-04-18 11:40:51 |
Solution
