Solution

Chebyshev's equal-ripple criterion says that a polynomial

q

of degree at most

m

is a best uniform approximation to a continuous function

F

if and only if the error

F - q

attains alternating extrema of maximum magnitude at

m + 2

ordered points.

Since

T_{n}

has leading coefficient

2^{n - 1}

2^{1 - n} T_{n}

is monic and alternates between

\pm 2^{1 - n}

t_{j} = cos (jπ / n)

. Therefore

q_{*} (t) = t^{n} - 2^{1 - n} T_{n} (t), d e g q < n min ∥ t^{n} - q ∥_{\infty} = 2^{1 - n} .

At each

t_{j}

, both

T_{n} - f

and

T_{n} + f

have sign

(- 1)^{j}

, because

∣ f (t_{j}) ∣ < 1

. Each has a root in every interval between consecutive extrema, hence all its at most

n

roots lie in

(- 1, 1)

. For

t > 1

both polynomials remain positive, giving

∣ f (t) ∣ < T_{n} (t)

. For

t < - 1

both have sign

(- 1)^{n}

, giving

∣ f (t) ∣ < ∣ T_{n} (t) ∣

. On

[- 1, 1]

the hypothesis gives

∣ f ∣ < 1

. Thus

∣ f (t) ∣ < max {1, ∣ T_{n} (t) ∣} (t \in R) .

Solved by gpt-5.6-sol high.

 Ancestors (10)