Orthogonality Relationship for Legendre Polynomials

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Homework Statement
I am trying to orthogonalize {1,x,x²}
Relevant Equations
Just the inner product of functions space.
Suppose p = a + bx + cx².
I am trying to orthogonalize the basis {1,x,x²}
I finished finding {1,x,x²-(1/3)}, but this seems different from the second legendre polynomial.
1601451062834.png

What is the problem here? I thought could be the a problem about orthonormalization, but check and is not.
 
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Your basis is orthogonal but not orthonormal. You need to compute the normalization for your third basis function, i.e. $$\int_{-1}^1 dx (x^2 - 1/3)^2$$.
 
eys_physics said:
Your basis is orthogonal but not orthonormal. You need to compute the normalization for your third basis function, i.e. $$\int_{-1}^1 dx (x^2 - 1/3)^2$$.
Yeh, but as i said, i already do it. The integrate gives 8/45, taking the square and dividing by the module not get yet.
 
Vendo seu perfil, acho que daria pra te responder em portugues :D
 
Legendre polynomials are orthogonal but not orthonormal over the interval ##[-1,1]##. Thus, you shouldn't expect your orthonormal basis to be identical to the Legendre polynomials.

NB. If you are trying to construct and orthonormal set ##\{p_0,p_1,p_2\}## of polynomials over the interval ##[-1,1]## from the set ##\{1,x,x^2\}## of monomials. Then the first two are ##p_0(x) = \pm\frac{1}{\sqrt{2}}## and ##p_1(x) = \pm\sqrt{\frac{3}{2}}x## and not ##p_0(x) = 1## and ##p_1(x) = x##. As you can see, you have choice to make regarding the signs of ##p_0(x)## and ##p_1(x)##.
 
The Legendre polynomials are usually normalized such that ##P_n(1) = 1##.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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