How to Find a Basis for W in P4[x] Using the Gram-Schmidt Process?

[Bachelier]

I need some direction with respect to this problem please:

Define the inner product on P₄[x] over $$\Re$$ as follows

<f,g> = $$\int_{0}^{1}\f(x)g(x) dx$$

let W be the subspace of P₄[x] consisting of the poly. ) and all polynomials with degree 0, that is W =R

Find a basis for $$W^{\perpendicular}$$per

 

[HallsofIvy]

I assume you meant
$$<f, g>= \int_0^1 f(x)g(x)dx$$.

$P_4[x]$ is the 5 dimensional space of all polynomials, with real coefficients, of degree 4 or less. A basis consists of $\{1, x, x^2, x^3, x^4\}$.
$$< 1, 1>= \int_0^1 1 dx= 1$$
so "1" is already a unit vector. Use "Gram-Schmidt" to extend that to an orthonormal basis for $P_4[x]$, then drop the "1" vector.

 

[Bachelier]

I assume you meant
$$<f, g>= \int_0^1 f(x)g(x)dx$$.

$P_4[x]$ is the 5 dimensional space of all polynomials, with real coefficients, of degree 4 or less. A basis consists of $\{1, x, x^2, x^3, x^4\}$.
$$< 1, 1>= \int_0^1 1 dx= 1$$
so "1" is already a unit vector. Use "Gram-Schmidt" to extend that to an orthonormal basis for $P_4[x]$, then drop the "1" vector.

cool I'll try it.

 

[Bachelier]

I did that. Actually this was the path I followed before posting my question:
Here's the deal:

I use V₁ = 1

V₂ = X - (Projection of X onto 1)

I get V₂ = X - 1/2

in their answer in this book, they gave 1 - X/2 which is not even orthogonal to X - 1/2.

I don't know how they got these answers:
$\{1-X/2 ; X/2 - X^2/3 ; X^2/3 - 1/4X^3\}$

Also we know Dim V = Dim W + Dim W^per

Dim V = 5, Dim W = 1, so we should have four polynomials in the basis, not 3!

 

[HallsofIvy]

Oh, blast! I suspect then, that they are using a different convention than I am and are using "$P_4[x]$" to mean the four dimensional space of polynomials of degree 3 or less: $\{p| p(x)= a+ bx+ cx^2+ dx^3\}$.

Use Gram-Schmidt to construct an orthonormal basis from $\{1, x, x^2, x^3\}$ and drop "1".