Define the nonlinear function mapping

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Homework Statement

Let P2 be the vector space of real polynomials of degree less or equal than 2. Define the (nonlinear) function E : P2 to R as

E(p)=integral from 0 to 1 of ((2/pi)*cos((pi*x)/2)-p(x))^2 dx

where p=p(x) is a polynomial in P2. Find the point of minimun for E, i.e. find the polynomial q exists in P2 such that

E(q) is less than or equal to E(p) for all p exists in P2

Homework Equations

-Both cos(x) and q(x) belong to the vector space C([0; 1]) of continuous functions on the interval [0; 1].

-The mapping (u, v) to the integral between 0 and one of (u(x)v(x)) dx defines a scalar product on C([0; 1]).

-The squared length of a vector u according to this scalar product would be
tha magnitude of u squared = (u, u) = the integral between 0 and one of (u(x))^2 dx

The Attempt at a Solution

TIP: Try to understand it geometrically (i.e. make a sketch with lines and points in R^2). Compare with the following: in the usual linear systems, how do you minimize |Ax - b| when b is not in R(A)?

integral from 0 to 1 of ((2/pi)*cos((pi*x)/2)-p(x))^2 dx = the magnitude squared of (2/pi)*cos((pi*x)/2)-p(x)

I'm really quite lost as to what this question in asking. Any help would be greatly apreciated : )

 

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The prof issued a clarification on this question:

P_1 is like Ax, and cos... is like a vector b outside the range of A. The equation Ax + b cannot be solved but the distance between Ax and b can be minimized with an orthogonal projection.

I have been working on this problem and this is what I have:

Process:
-find a basis for P1 on the interval [0,1]
-use Gram-Schmid procedure to make this basis orthonormal
-compute the orthogonal projection of f=2/pi*cos(pi*x/2) on P_1, using the orthonormal basis and the scalar product
-minimize the integral with the information found above

Basis for P1 ( a subset of the vector space P(containing all polynomials)):
P1 = a+bx
Basis of P1 is [1,x]

GramSchmit:
orthogonal basis: [1, x - (<x,1>/<1,1>)*1] = [1,x-1/2]
normalized: [1,(1/12)(x-1/2)]

Projection:
if you have a space W spanned by an orthogonal set {x, y} and you want to project a vector v on it orthogonally, then you just compute the sum <v, x> x + <v, y> y so...
if v = 2/pi*cos(pi*x/2) and {x,y} = [1,(1/12)(x-1/2)] then:
<2/pi*cos(pi*x/2), 1> * 1 + < 2/pi*cos(pi*x/2), (1/12)*(x-1/2)> * (1/12)(x-1/2)
-> (4/pi^2) + (pi-4)/(6pi^3)

I'm not sure if that's right but any help wuold be greatly appreciated.