1
$\begingroup$

How to show that any polynomial of degree $\le n$ can be written as a linear combination of the $L_j(x)$ for $j = 0,1,2,...,n$

$P_n(x)=a_0L_0(x)+a_1L_1(x)+a_2L_2(x)+\dots+a_nL_n(x)$

ANSWER :: the legendre polynomials of degree ≤ n, are linearly independent, and thus form a basis for all polynomials of degree ≤ n. therefore, every polynomial of degree ≤ n can be written as a linear combination

thanks. :-)

$\endgroup$
  • $\begingroup$ Sorry I'm confused, are you asking for a proof of why this is so? $\endgroup$ – Aldon Nov 8 '15 at 15:59
  • $\begingroup$ I cannot explain how to proof it. Do you have any suggestion? Thank you. $\endgroup$ – Bob Nov 8 '15 at 16:12
  • $\begingroup$ If I remember correctly it has something to do with the functions being orthogonal, therefore you can represent a function $f(x)$ as a sum of Legendre polynomials. I just don't remember how to do it mathematically. $\endgroup$ – Aldon Nov 8 '15 at 16:30
  • $\begingroup$ See math.stackexchange.com/questions/512295/… $\endgroup$ – Maestro13 Nov 20 '15 at 21:24
1
$\begingroup$

A little mixup in notation there. $P_n(x)$ is normally denoting the $n^{th}$ Legendre polynomial, and $L_n(x)$ the $n^{th}$ Laguerre polynomial.

Both $\{ P_n| n \in \mathbb{N} \}$ and $\{ L_n| n \in \mathbb{N} \}$ form sets of orthogonal functions, which means that when taking an inner product of two of its members which are different, then the result is zero. Here the inner product is some integral of the product of the functions.

See for example Legendre Polynomial Orthogonality and Size and Legendre polynomials, Laguerre polynomials: Basic concept.

Any function on the relevant interval ($[-1,1]$ for Legendre polynomials, $[0,\infty]$ for Laguerre polynomials) can then be expanded into a series with terms $a_nP_n(x)$ or $a_nL_n(x)$ instead of the 'normal' $a_nx^n$.

In particular, any polynomial $P(x)$ can be expanded on $[-1,1]$ as $\sum_{j=0}^{\infty} a_jP_j(x)$ with appropriate coefficients $a_j$.

See Using Legendre polynomial to approximate any polynomial.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.