inner-productslegendre polynomials
I have this Legendre polynomial:
$p_k(x) = \frac{1}{2^k k!} \frac{d^k}{dx^k} ((x^2 – 1)^k) $
I have calculated that: $p_0 = 1, p_1 = x, p_2 = \frac{1}{3}(3x^2-1), p_3 = \frac{1}{2}(5x^3-3x) $
Now I have to show that $<p_2,p_1> \; = 0$.
I have then integrated $p_2 $ and $ p_1$ from [-1,1] and get 0 and 0. And <0,0> = 0 is true.
My question is, am I on the right track? Is there 'more' I have to show?
The goal is to approximate a function $f(x)$ over the interval $(-1,1)$ with a sequence of Legendre polynomials $P_{k}(x)$ through order $d$: $$ f(x) = \lambda_{0} P_{0}(x) + \lambda_{1} P_{1}(x) + \dots + \lambda_{d} P_{d}(x) = \sum_{k=0}^{d} \lambda_{k} P_{k}(x) $$ What amplitudes $\lambda$ provide the best fit?
Projection will extract the $\lambda$ values without having to form a linear system. The Legendre functions are an orthogonal basis over the domain: $$ \int_{-1}^{1} P_{j}(x) P_{k}(x) dx = \frac{2(k+1)}{2} \delta_{j}^{k} $$ Orthogonality allows the amplitudes to be computed independently. This will project out each $\lambda$ value; integrate both side of the approximation equation over the domain. $$ \int_{-1}^{1} f(x) P_{j}(x) dx = \int_{-1}^{1} \left( \sum_{k=0}^{d} \lambda_{k} P_{k}(x) \right) P_{j}(x) dx = \int_{-1}^{1} \lambda_{j} P_{j}^{2}(x) dx $$ Therefore $$ \lambda_{j} = \frac{2}{2(j+1)} \int_{-1}^{1} f(x) P_{j}(x) dx $$ The computed fit parameters are $$ \lambda = \left[ \begin{array}{c} \lambda_{0} \\ \lambda_{1} \\ \lambda_{2} \\ \lambda_{3} \end{array} \right] = \left[ \begin{array}{c} \frac{1}{2} \left(e-\frac{1}{e}\right) \\[2pt] \frac{3}{e} \\ \frac{5}{2} \left(e-\frac{7}{e}\right) \\[3pt] \frac{7}{2} \left(\frac{37}{e}-5 e\right) \end{array} \right] $$ The following plot shows the error $$ r(x) = f(x) - \sum_{k=0}^{3}\lambda_{k} P_{k}(x) $$ plotted over the domain.
Here you should take the weight function $r(x)$ to be just the constant function $1$. There is nothing in the problem statement that explicitly tells you this, but normally if you say that two functions are orthogonal without specifying the weight function that means that the weight function is $1$ (or in other words, there is no weight function and the inner product is just the integral of the product of the two functions). If this convention has not been stated previously and you have only seen a definition of orthogonality with respect to a specified weight function, then the omission of the weight function is just an error in the problem statement.
Best Answer
You got the definition of $\langle p_2, p_1 \rangle$ wrong. $\langle p_2, p_1 \rangle$ is not $(\int p_1(x)dx )(\int p_2(x)dx)$.
$\langle p_2, p_1 \rangle=\int_{-1}^{1}p_1(x)p_2(x)dx=\int_{-1}^{1} \frac x 3 (3x^{2}-1)dx= 0 -0=0$.