1
$\begingroup$

Let us consider the numerical integral $ \ \int_{-1}^{1}w(x) f(x)dx=\sum_{i=0}^{N} f(x_i)w_i$, where $w_i$ are the weights and $w(x)$ is the weight function.

Legendre polynomials, denoted by $ \{p_n \}$ are a list of orthogonal polynomial supported on $[-1,1]$ with weight $w(x)=1$. Then the explicit expression for $p_0,p_1,p_2$.

Answer:

we have show that

$p_0=1, \ p_1=x , \ p_2=\frac{1}{2}(3x^2-1)$.

How to show this using $w(x)=1$?

If $w(x)=1$, then we have

$ \int_{-1}^{1} f(x)dx=\sum_{i=0}^{N} f(x_i) w_i$.

Now how to proceed?

help me

$\endgroup$
  • $\begingroup$ The Legendre weight function $w(x)=1$ is for the definition of the orthogonality. The $w_i$ are not (directly) related to $w(x)$. See en.wikipedia.org/wiki/Gauss-Legendre_quadrature $\endgroup$ – gammatester Nov 25 '18 at 21:16
  • $\begingroup$ I have make correction of my question. I need to find first three legendre polynomials with the information $ w(x)=1$ $\endgroup$ – user612508 Nov 25 '18 at 21:21
0
$\begingroup$

(Up to now I see no correction to the question, here is the computation of the first Legendre polynomials)

This is a simple exercise in integration and using the orthogonality relations.

Let's write $(f,g)=\int_{-1}^1 f(x)g(x) dx$. Then you have to compute the polynomials $p_n$ of degree $n$ with $(p_n, p_m) = \frac{2}{2n+1}\delta_{nm}.$

Firts, with $p_0=a$ you have $$2=(p_0,p_0) = 2a \Longrightarrow a = 1.$$

With $p_1(x) =ax + b$ you get $$0 =(p_1, p_0) = 2b \Longrightarrow b = 0$$ $$\frac{2}{3}=(p_1, p_1) = \frac{2}{3}a^2 \Longrightarrow a = 1.$$

So $p_1(x) = x.$ Can you continue with the Ansatz $p_2(x) = ax^2+bx+c?$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy