Proving Orthogonal Polynomials: A Weighted Integral

[Amer]

Let $$\{ \phi_0,\phi_1,...,\phi_n\}$$ othogonal set of polynomials on [a,b] n>0, with a weight function w(x) prove that

$$\int_{a}^b w(x)\phi_n Q_k (x) \; dx = 0$$

for any polynomail $$Q_k(x)$$ of degree k<n ?

My work :

I think there is a problem in the question since if we take $$x^2,x^3$$ on the interval [-1,1] they are orthogonal

but if we take x

$$\int_{-1}^{1} x(x^3 ) \; dx \neq 0$$

 

[Deveno]

you haven't defined your weight function w(x), but let's assume it is the constant function 1. clearly $1,x$ are orthogonal, so we can start with a basis:

$B = \{1,x,\dots \}$

now let's look at what our third basis element $ax^2 + bx + c$ might be:

being orthogonal to 1 requires that $\int_{-1}^1ax^2 + bx + c\ dx = 0$. evaluating the integral, we find that:

$\frac{a}{3} + c - (\frac{-a}{3} + (-c)) = \frac{2a}{3} + 2c = 0$, and simplifying we get: $c = \frac{-a}{3}$.

so our second degree polynomial is of the form: $ax^2 + bx - \frac{a}{3}$.

since we must also have our second-degree polynomial orthogonal to x, this means that:

$\int_{-1}^1 ax^3 + bx^2 - \frac{ax}{3}\ dx = 0$, and evaluating THAT interval leads to $b = 0$.

traditionally, these polynomials are "normalized" so that $\phi_k(1) = 1$, doing so for:

$\phi_2(x) = ax^2 - \frac{a}{3}$ leads to: $a = \frac{3}{2}$, so that we get: $\phi_2(x) = \frac{1}{2}(3x^2 - 1)$.

the point is, there is no reason to assume that the "standard" basis: $\{1,x,x^2,x^3,\dots \}$ will be orthogonal with respect to the inner product defined by:

$\langle f,g \rangle = \int_{-1}^1 f(x)g(x)\ dx$ or the "weighted inner product" $\langle f,g \rangle = \int_{-1}^1 w(x)f(x)g(x)\ dx$

if you continue the process i started above (or by using gram-schmidt), you will get:

$\phi_3(x) = \frac{1}{2}(5x^3 - 3x)$ which can be verified to be orthogonal to $\phi_0, \phi_1,\phi_2$.

 

[Amer]

thanks, I edited my post
can you check it again ?

 

[HallsofIvy]

Let $$\{ \phi_0,\phi_1,...,\phi_n\}$$ othogonal set of polynomials on [a,b] n>0, with a weight function w(x) prove that

$$\int_{a}^b w(x)\phi_n Q_k (x) \; dx = 0$$

for any polynomail $$Q_k(x)$$ of degree k<n ?

My work :

I think there is a problem in the question since if we take $$x^2,x^3$$ on the interval [-1,1] they are orthogonal

but if we take x

$$\int_{-1}^{1} x(x^3 ) \; dx \neq 0$$

Then why do you assert that they are orthogonal. In particular, what is your definition of "orthogonal"?

 

[CaptainBlack]

Let $$\{ \phi_0,\phi_1,...,\phi_n\}$$ othogonal set of polynomials on [a,b] n>0, with a weight function w(x) prove that

$$\int_{a}^b w(x)\phi_n Q_k (x) \; dx = 0$$

for any polynomail $$Q_k(x)$$ of degree k<n ?

My work :

I think there is a problem in the question since if we take $$x^2,x^3$$ on the interval [-1,1] they are orthogonal

but if we take x

$$\int_{-1}^{1} x(x^3 ) \; dx \neq 0$$

There is something missing from this, there seems to be an implicit assumption that \( \phi_k(x) \) is of degree \( k \) (or rather that \( \phi_n(x) \) is of degree \(n\) and every degree less than \(n\) is represented by one of the other \(\phi\)s ). If this is not the case then the result can fail.

CB

 

[Amer]

There is something missing from this, there seems to be an implicit assumption that \( \phi_k(x) \) is of degree \( k \) (or rather that \( \phi_n(x) \) is of degree \(n\) and every degree less than \(n\) is represented by one of the other \(\phi\)s ). If this is not the case then the result can fail.

CB

it is true, our instructor fixed the question as what you said ($$\phi_k$$ is of order k )and i solved it