# Minimizing a integral

1. Sep 26, 2011

### saxen

1. The problem statement, all variables and given/known data

Determine the polynomial of the form P(x)=x^3+ax^2+bx+c
that minimizes

$\int[P(x)]^2$

2. Relevant equations

3. The attempt at a solution

My first thought was that I should find a second degree polynomial that minimizes x^3. That didn't work at all! So now I'm stuck and have no ideas what to do. Anyone got a hint for me?

Thanks!

edit: Latex and I dosen't seem to get along. Its integral(P(x)^2)

Last edited by a moderator: Sep 26, 2011
2. Sep 26, 2011

### Staff: Mentor

Should the integral be a definite integral? Otherwise I don't understand how you can minimize it.

3. Sep 28, 2011

### saxen

Sorry! Missed that the integral is between 0 and 1

4. Sep 28, 2011

### SammyS

Staff Emeritus
Three parameters to work with ...

It seems certain that the zeros of the polynomial should all be on the interval [0,1].

If you assume there's one zero with multiplicity 3, then it's easy to find the minimum. However, that's not the most general case, so it doesn't prove that it's the absolute minimum.

5. Sep 28, 2011

### hotvette

Unless I'm missing something, it seems to me the indefinite integral evalutes to a 7th order polynomial in x and the definite integral evaluates to a non-linear function of a,b,c, say G(a,b,c). The task would then be to find a,b,c to minimize G.

Last edited: Sep 28, 2011
6. Sep 28, 2011

### Ray Vickson

In fact, the integral will be a quadratic in a, b, etc.

RGV

7. Sep 28, 2011

### susskind_leon

So it comes down to a (simple) minimization of G(a,b,c). You should check how this works in case you don't know. The gradient with respect to a,b,c must vanish and the Hessian must be positive definite. Therefor, you might need to look at the eigenvalues of H. Anyhow, since G(a,b,c) is really simple, it shouldn't be too hard really.

8. Sep 28, 2011

### SammyS

Staff Emeritus
So, OP (saxen) needs to minimize $\displaystyle \int_0^1(x^3+ax^2+bx+c)^2\,dx\,.$

If a, b, and c are all real, then there is no need to use the absolute value .

I'm pretty sure the polynomial, P(x) = x3+ax2+bx+c must have three real zeros, one at x = 1/2, the other two placed equal distance from and on opposite sides of x=1/2 . Let one of those zeros be at x = t, then the other is at 1-t.

Then P(x) = (x-t)(x+t-1)(x-1/2).

(P(x))2 is quartic in t. The above definite integral is also quartic in t.

It's derivative (W.R.T. t) is thus cubic in t with critical points symmetric about t=1/2.

9. Sep 28, 2011

### lurflurf

You are going to want to use Shifted Legendre polynomials.
The idea being that you will write
P(x)=x^3+ax^2+bx+c=(1/20) P3*+A P2*+B P1*+C P0*
Integral[P(x)]=1/20^2/7+A^2/5+B^2/3+C^2
Which is easy to minimize.

Last edited: Sep 29, 2011
10. Sep 28, 2011

### lurflurf

That is true. It might be more clearly seen as P(x)=(x-.5)(x-.5+t)(x-.5-t).

11. Sep 29, 2011

### saxen

Sorry but I don't really get it. We seem to be thinking the same, rewriting with orthonormal basis, but after that?

Last edited by a moderator: May 5, 2017
12. Sep 29, 2011

### lurflurf

The advantage of an orthogonal basis is that norms (as your problem is to minimize a norm) have a simple form.
||A u+B v+C w+D x||^2=|A|^2 ||u||^2+|B|^2 ||v||^2+|C|^2 ||w||^2+|D|^2 ||x||^2
we desire to minimize
||P(x)||^2=||x^3+ax^2+bx+c||^2=||(1/20) P3*+A P2*+B P1*+C P0*||^2
=|1/20|^2 ||P3*||^2+|A|^2 ||P2*||^2+|B|^2 ||P1*||^2+|C|^2 ||P0*||^2
=|1/20|^2 (1/7)+|A|^2 (1/5)+|B|^2 (1/3)+|C|^2 (1)
Which is easy to minimize by taking A,B,C=0