Let's say we have some polynomial of degree 2: $P(x) = 2x^2 - 7x + 6$. How can I find polynomial $Q(x)$ of degree at most 1 such that maximum value of $|P(x) - Q(x)|$ in range $x\in[0,1]$ is minimum.

Is there any trick for this kind of problems?

I have tried to set $Q(x) = ax + b$, then I can see that $b$ just shifts the range of $P(x) - Q(x)$. Since values of $x$ are positive, it kind of make sense to set $a=7$, but I am not sure.

  • $\begingroup$ I guess the Hilbert matrix would play a critical role in this problem since you have the desired coefficients for it. But, I am not sure how the procedure for determining polynomial $Q$ could be translated to the solution of the Hilbert matrix. $\endgroup$
    – Snowflake
    Jun 14, 2021 at 23:41
  • $\begingroup$ Or maybe, the major idea could be the application of Chebyschev's polynomials and their involved inequalities?! $\endgroup$
    – Snowflake
    Jun 14, 2021 at 23:58
  • 3
    $\begingroup$ Hints: * show that if you can solve it for one quadratic polynomial then you can solve it for any quadratic polynomial; * solve the problem instead for $P(x) = x(1-x)$—the symmetry with respect to $[0,1]$ should make this simpler. $\endgroup$ Jun 15, 2021 at 0:11

1 Answer 1


Your problem is a straightforward application of Chebyshev's Equioscillation Theorem.

Since $P$ is convex, all you have to do is:

  1. Draw the secant through $(0,P(0))$ and $(1,P(1))$ i.e. $y=-5x+6$
  2. Calculate maximum $y$-distance between the secant and $P$ i.e. $\max\{-5 x + 6 - (2 x^2 - 7 x + 6)\} = 1/2$ at $x = 1/2$
  3. Bring the secant halfway down: $y=-5x+6-\frac{1}{4}=-5x+\frac{23}{4}$
  4. Use Equioscillation Theorem to prove that $Q(x)=-5x+\frac{23}{4}$ is the linear approximation you are looking for ($x_0=0, x_1=1/2$ and $x_2=1$).

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