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Applied Linear Algebra

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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

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)?

P_1 is like Ax, and cos... is like a vector b outside the range of A, you cannot solve the equation, but you can minimize the distance between Ax and b. the way of doing this is with an orthogonal projection.

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



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 wanna 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)
 

Answers and Replies

  • #2
lanedance
Homework Helper
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haven't checked your working, however your reasoning is sound

one point though, shouldn't you find an orthonormal basis for P2 to complete the problem, this amounts to finding one more coeffienct for the extra basis vector?
 
  • #3
lanedance
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also you should probably define what you inner product is (though its reasonably obvious)
 
  • #4
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Here sorry the question was modified and i forgot to change it...

Let P1 be the vector space of real polynomials of degree less or equal than 1. Defne the (nonlinear) function E : P1 -> R as
E(p) = blah
(where p = p(x) is a polynomial in P1). Find the point of minimum for E, i.e., find the polynomial q exists in P1 such that
E(q) <= E(p) for all p exists in P1.

thats why i only defined p1
 
  • #5
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But im not sure how to find the minimum for E
 
  • #6
lanedance
Homework Helper
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E will be minimised when p(x) is the projection of f(x) = (2/pi)*cos((pi*x)/2) onto P2, to show this consider experessing f(x) as an infinite sum of polynomials in Pinf, and look at the orthogonlity of the basis vectors
 

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