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Inner product integration 
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#1
Nov311, 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. [tex]x^3+ax^2+bx+c[/tex] 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}(x^3+ax^2+bx+c)(x+\beta)dx[/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\beta10a\alpha\beta=0[/tex] What do I do with that? 


#2
Nov311, 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 p_{0}(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 <p_{1}(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 p_{1}(x) = ax. again there is no reason not to choose a = 1. for n = 2, we need <p_{2},c> = 0, and <p_{2},ax+b> = 0 if p_{2}(x) = rx^{2}+sx+u, this means r = 3u, from the first inner product, and s = 0 from the second. so p_{2}(x) = u(3x^{2}  1). again, any nonzero choice will do, although one might be inclined to choose u such that <p_{2}(x),p_{2}(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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