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Inner product integration

by Dustinsfl
Tags: integration, product
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Nov3-11, 03:23 PM
P: 629
Space of continuous functions.

Inner product [tex]<f,g>=\int_{-1}^{1}f(x)g(x)dx[/tex].

Find a monic polynomial orthogonal to all polynomials of lower degrees.

Taking a polynomial of degree 3.


Need to check [tex]\gamma, x+\alpha, x^2+\beta x+ \lambda[/tex]

[tex]\int_{-1}^{1}(\gamma x^3+\gamma a x^2 +\gamma bx + \gamma c)dx[/tex]
[tex]=\frac{\gamma x^4}{4}+\frac{\gamma a x^3}{3}+\frac{\gamma b x^2}{2}+\gamma c x|_{-1}^{1}[/tex]
[tex]=\frac{2\gamma a}{3}+2\gamma c=0\Rightarrow c=-\frac{a\gamma}{3}[/tex]

[tex]\int_{-1}^{1}\left(x^4+ax^3+bx^2-\frac{a\alpha x}{3}+\beta x^3 +\alpha\beta x^2+b\beta x-\frac{a\alpha\beta}{3}\right)dx=6+10b+10a\beta-10a\alpha\beta=0[/tex]

What do I do with that?
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Nov3-11, 04:51 PM
Sci Advisor
P: 906
well, are you given the degree you polynomial is supposed to be, or are you suppose to find a formula for any n (degree)?

for n = 0, we can choose p0(x) = 1 (we don't have any polynomials of lesser degree, so any constant will do. i like 1, don't you?).

for n = 1, the only requirement is that <p1(x),c> = 0 for any constant polynomial k(x) = c, that is:

[tex]\int_{-1}^1(ax+b)c\ dx = 0[/tex]

or: 2b = 0, so b = 0, thus p1(x) = ax. again there is no reason not to choose a = 1.

for n = 2, we need <p2,c> = 0, and <p2,ax+b> = 0

if p2(x) = rx2+sx+u, this means r = -3u, from the first inner product, and s = 0 from the second.

so p2(x) = u(3x2 - 1). again, any non-zero choice will do, although one might be inclined to choose u such that <p2(x),p2(x)> = 1.

now, for n = 3:

you may as well assume that γ ≠ 0, since it is arbitrary, which gives:

c = -a/3, not c = -aγ/3 (just divide by γ).

in your second inner product, you start with x+β, instead of x+α, and somehow wind up with something with α's and β's. huh? pick a variable for the constant term of your generic linear polynomial, and stick with it.

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