Max of the absolute value of a polynomial

• Sick0Fant
In summary, the maximum value of |P_1(x)| for x in [x_0,x_1] is h^2/4, where h = x_1 - x_0. This can be proven using Rolle's Theorem and the fact that the x_i terms are evenly spaced. Additionally, for P_2(x), the maximum value with x in [x_0,x_2] is (2*(3)^(1/2)/9)*h^3, where x_0 = 0, x_1 = h, and x_2 = 2h, or x_0 = -h, x_1 = 0, and x_2 = h, as

Sick0Fant

What I have is this:

Let P_n(x)=(x-x_0)(x-x_1)...(x-x_n), _i are subscripts.
Prove that the maximum value of |P_1(x)| for x in [x_0,x_1] is h^2/4, where h =x_1 - x_0.

All the x_i terms are evenly spaced. That is, x_(i+1)-x_i is the same for all i.

What I noticed is that P_1(x_0)=P_1(x_1)=0. So by Rolle's Theorem, there exists a c in [x_0,x_1] such that P_1'(c) = 0. Since the polynomial is of degree two, there will be at most one of these points. Also, since we are taking the absolute value of the function, that point is guaranteed to be the maximum. Problem is, I do not get that the max is h^2/4.

P_1'(x) = 2x - x_0 - x_1.
=> x = (x_0+x_1)/2.

Is there something I'm missing? Edit: yes there was... I found the maximum x, not the function value that that input returns. ;-) Sorry for wasting your time.

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I've got a follow-up question:

I have to prove that the maximum value of |P_2(x)| with x in [x_0,x_2] is (2*(3)^(1/2)/9)*h^3.

Any ideas?

x_0 = 0
x_1 = h
x_2=2h

or try

x_0=-h
x_1=0
x_2=h

(so P_2 is and odd function)

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Thanks a bunch!