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How can I proceed to prove that there are constants $α_0, α_1, ..., α_n$ such that $x^n = α_0P_0(x) + α_1P_1(x) + ... + α_nP_n(x)$ where $P_n$ is legendre polynomial.

I guess that this has to do with orthogonality property and the fact that $P_n$'s are linearly independent, but I have no clue where to start. Can anybody help?

idriskameni
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user635977
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  • for every $x$ ? that I don't know but if $|x| \leq 1$ then the Legendre polynomials is an orthogonal basis of the space $L^2([-1,1])$ which is a Hilbert space and it just happens that $x^n \in L^2([-1,1]) $ therefore it can be written as a Legendre polynomials series – the_firehawk Jan 18 '19 at 13:59

2 Answers2

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Any function on $[-1,1]$ whether continuous or not can be expanded in a series of Legendre polynomials. That is the same as saying that $\{P_n\}$ form a basis for this kind of expansions. Now $\{P_n\}$ is also an orthogonal basis, meaning that $\int_{-1}^{1} P_n(x)P_m(x)dx = 0$ if $n \neq m$. This helps us determining the values $\alpha_j$: suppose $f(x) = \sum_{j=0}^{\infty} \alpha_j P_j(x)$, then

$$\int_{-1}^{1} P_n(x)f(x)dx = \int_{-1}^{1} P_n(x)\sum_{j=0}^{\infty} \alpha_j P_j(x) dx = \sum_{j=0}^{\infty} \alpha_j \int_{-1}^{1} P_n(x)P_j(x)dx = \\\alpha_n \int_{-1}^{1} P_n^2(x)dx = \frac{2\alpha_n}{2n+1}$$

Hence

$$\alpha_j = \frac{2j+1}{2} \int_{-1}^{1} P_j(x)f(x)dx$$

Now plug in $x^n$ for $f(x)$ and evaluate the integrals - I leave that part to you. You also need to show that $\alpha_j = 0$ for $j > n$.

Maestro13
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While Maestro’s answer works, he doesn’t show that you produce $x^n$, only that you approximate arbitrarily closely it by orthogonal polynomials. Rather, notice that the first n legendre polynomials are linearly independent and contained in the sub space of n degree polynomials. Since the dimensions are equal, the legendre polynomials span the subpace of up-to-n dimensional polynomials, of which $x^n$ is. Orthogonality doesn’t actually matter, only linear independence.

Alex
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