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

[Dustinsfl]

Space of continuous functions.

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

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

Taking a polynomial of degree 3.

$$x^3+ax^2+bx+c$$

Need to check $$\gamma, x+\alpha, x^2+\beta x+ \lambda$$

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

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

What do I do with that?

 

[Deveno]

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:

$$\int_{-1}^1(ax+b)c\ dx = 0$$

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.