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Orthogonal set

Amer

Active member
Mar 1, 2012
275
Let [tex]\{ \phi_0,\phi_1,...,\phi_n\}[/tex] othogonal set of polynomials on [a,b] n>0, with a weight function w(x) prove that

[tex]\int_{a}^b w(x)\phi_n Q_k (x) \; dx = 0 [/tex]

for any polynomail [tex]Q_k(x) [/tex] of degree k<n ?

My work :

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

but if we take x

[tex]\int_{-1}^{1} x(x^3 ) \; dx \neq 0 [/tex]
 
Last edited:

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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

Active member
Mar 1, 2012
275
thanks, I edited my post
can you check it again ?
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Let [tex]\{ \phi_0,\phi_1,...,\phi_n\}[/tex] othogonal set of polynomials on [a,b] n>0, with a weight function w(x) prove that

[tex]\int_{a}^b w(x)\phi_n Q_k (x) \; dx = 0 [/tex]

for any polynomail [tex]Q_k(x) [/tex] of degree k<n ?

My work :

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

but if we take x

[tex]\int_{-1}^{1} x(x^3 ) \; dx \neq 0 [/tex]
Then why do you assert that they are orthogonal. In particular, what is your definition of "orthogonal"?
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Let [tex]\{ \phi_0,\phi_1,...,\phi_n\}[/tex] othogonal set of polynomials on [a,b] n>0, with a weight function w(x) prove that

[tex]\int_{a}^b w(x)\phi_n Q_k (x) \; dx = 0 [/tex]

for any polynomail [tex]Q_k(x) [/tex] of degree k<n ?

My work :

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

but if we take x

[tex]\int_{-1}^{1} x(x^3 ) \; dx \neq 0 [/tex]
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

Active member
Mar 1, 2012
275
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 ([tex]\phi_k[/tex] is of order k )and i solved it