How to express $2x^3-x^2-3x+2$ as a linear combination of Legendre polynomials

Question

I have used the formula

\begin{align}p_0(x)&=1\\ p_1(x)&=x\\ p_2(x)&=\frac12(3x^2−1)\\ p_3(x)&=\frac12(5x^3−3x) \end{align}

$$2x^3-x^2-3x+2=Ap_3(x)+Bp_2(x)+Cp_1(x)+Dp_0(x)$$

EDIT-

\begin{align}&=\frac A2(5x^3−3x) + \frac B2(3x^2−1) + Cx +D \end{align}

but don't know how to continue after this.

Answer 1

Continuing from your steps, we compare the coefficients of each term on the L.H.S and R.H.S.

$$2x^3-x^2-3x+2=\dfrac A2(5x^3−3x) + \dfrac B2(3x^2−1) + Cx +D$$

So, $$ 2x^3-x^2-3x+2=\dfrac{5}{2}Ax^3+\dfrac{3}{2}Bx^2+(C-\dfrac{3}{2}A)x+(D-\dfrac{B}{2})$$

• For the $x^3$ term, we have $\dfrac{5}{2}A=2 \Rightarrow A=\dfrac{4}{5}$

• For the $x^2$ term, we have $\dfrac{3}{2}B=-1 \Rightarrow B=\dfrac{-2}{3}$

• For the $x$ term, we have $C-\dfrac{3}{2}A=-3 \Rightarrow C-\dfrac{3}{2}\cdot\dfrac{4}{5}=-3 \Rightarrow C-\dfrac{6}{5}=-3 \Rightarrow C=\dfrac{6}{5}-3\Rightarrow C=\dfrac{-9}{5}$

• For the constant term, we have $D-\dfrac{1}{2}B=2 \Rightarrow D-\dfrac{1}{2} \cdot \dfrac{(-2)}{3}=2 \Rightarrow D=2-\dfrac{1}{3} \Rightarrow D=\dfrac{5}{3}$

Answer 2

First get the $x^3$ term right. That will set A. That is $4/5$. Then do the quadratic to set B, and so on.

Edit: the coefficient of $x^3$ on both sides are 2 and $5/2*A $. Set these equal to solve for A.

Answer 3

In matrix form,

$$\begin{bmatrix} p_0\\ p_1\\ p_2\\ p_3\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ -\frac{1}{2} & 0 & \frac{3}{2} & 0\\ 0 & -\frac{3}{2} & 0 & \frac{5}{2}\end{bmatrix} \begin{bmatrix} 1\\ x\\ x^2\\ x^3\end{bmatrix}$$

We want to find a weight vector $\mathrm{w}$ such that

$$2 x^3 - x^2 - 3 x + 2 = \begin{bmatrix} w_0\\ w_1\\ w_2\\ w_3\end{bmatrix}^T \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ -\frac{1}{2} & 0 & \frac{3}{2} & 0\\ 0 & -\frac{3}{2} & 0 & \frac{5}{2}\end{bmatrix} \begin{bmatrix} 1\\ x\\ x^2\\ x^3\end{bmatrix}$$

and, thus, we have to solve the linear system

$$\begin{bmatrix} 2\\ -3\\ -1\\ 2\end{bmatrix}^T = \begin{bmatrix} w_0\\ w_1\\ w_2\\ w_3\end{bmatrix}^T \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ -\frac{1}{2} & 0 & \frac{3}{2} & 0\\ 0 & -\frac{3}{2} & 0 & \frac{5}{2}\end{bmatrix}$$

or, in a more standard form,

$$\begin{bmatrix} 1 & 0 & -\frac{1}{2} & 0\\ 0 & 1 & 0 & -\frac{3}{2}\\ 0 & 0 & \frac{3}{2} & 0\\ 0 & 0 & 0 & \frac{5}{2}\end{bmatrix} \begin{bmatrix} w_0\\ w_1\\ w_2\\ w_3\end{bmatrix} = \begin{bmatrix} 2\\ -3\\ -1\\ 2\end{bmatrix}$$

The solution is

$$\frac{5}{3} \, p_0 (x) - \frac{9}{5} \, p_1 (x) - \frac{2}{3} \, p_2 (x) + \frac{4}{5} \, p_3 (x) = 2 x^3 - x^2 - 3 x + 2$$

Answer 4

Another approach. Through the orthogonality of Legendre polynomials, if $$ f(x) = \sum_{k=0}^{3}c_k\cdot P_k(x)\tag{1} $$ we have: $$\forall k\in\{0,1,2,3\},\qquad c_k = \frac{2k+1}{2}\int_{-1}^{1}f(x)\,P_k(x)\,dx. \tag{2}$$