JEEOlympiad
Let m(a) = max over x in [-1, 1] of |x^3 - a·x|. As a ranges over all…
Question
Let m(a) = max over x in [-1, 1] of |x^3 - a·x|. As a ranges over all real numbers, m(a) attains its smallest possible value M. Report M as a decimal.
✓ Verified answer: 0.25checked by our engine — not a guess
Step-by-step solution
We minimize the sup-norm on [-1,1] of x^3 - a x over the choice of a.
The Chebyshev polynomial T3(x) = 4x^3 - 3x satisfies |T3(x)| <= 1 on [-1,1] and is the monic-leading minimax polynomial (after scaling): x^3 - (3/4)x = (1/4)(4x^3 - 3x) = (1/4)T3(x), whose maximum absolute value on [-1,1] is 1/4.
By the minimax property of Chebyshev polynomials, no other choice of a makes the leading-coefficient-1 cubic x^3 - a x smaller in sup-norm.
Hence the optimal a = 3/4 and M = 1/4 = 0.25.
Final answer0.25
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