# Minimum absolute value of polynomial

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.

• 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. Jun 14, 2021 at 23:41
• Or maybe, the major idea could be the application of Chebyschev's polynomials and their involved inequalities?! Jun 14, 2021 at 23:58
• 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. Jun 15, 2021 at 0:11

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 $$x_2=1$$).