# Orthogonal set

#### Amer

##### Active member
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

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

Last edited:

#### Deveno

##### Well-known member
MHB Math Scholar
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

#### Amer

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

#### HallsofIvy

##### Well-known member
MHB Math Helper
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"?

• Amer

#### CaptainBlack

##### Well-known member
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

#### Amer

##### Active member
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