Writing [itex]x^3[/itex] in Legendre base

[Msilva]

Hello friends. I need help to write the function x^3 as a somatory using the Legendre polinomials as base. Something like:
f(x)=\sum^{\infty}_{n=0}c_{n}P_{n}(x)

Basically is to find the terms c_{n}.
But, the problem is that Legendre polinomials does't form a orthonormal base: \langle P_{m}|P_{n}\rangle=\delta_{mn}\frac{2}{2n+1}, and I don't know how exactly to use this information.

May I use c_n=\frac{2n+1}{2}\int_{-1}^{-1}P_n(x)x^3\,dx? Is that right?

 

[Chandra Prayaga]

You can do this problem without doing all the integrals. I am writing the first three Legendre polynomials below:

P₀ = 1
P₁ = x
P₂ = (3x² - 1)/2
P₃ = (5x³ - 3x)/2

You want to write:

x³ = a₀P₀ + a₁P₁ + a₂P₂ + a₃P₃

I took only up to P₃ because I have x³ on the left.

Now just look at the Legendre polynomials listed above, and keep adjusting the coefficients a₀ .. etc. until you get what you want. For example, since I want x³, and there is a coefficient of 5/2 in front of x³ in P³, I can choose a₃ = 2/5, so that a₃ P₃ yields x³. But this also gives me an extra term in x. So now adjust the coefficient a₁ so that a₁ P₁ cancels out that extra term in x, and you are left with exactly what you want. The rest of the coefficients a₀, a₂, and then a₄, a₅... are all zeroes.[/SUB][/SUB]

If you are fond of doing integrals, you can also use the result that you wrote for each c_n. It is correct.

 

[Msilva]

Actualy x^3 is an aleatory function that I wrote here just to ilustrate my question, but you helped me a lot.

With these integrals the coefficients are exactly those expected by your arguments. This is a hint that the logic of the integral is right.

Thank you for your suport, Prayaga.