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Here's the question:

Let $x_0=-2, x_1=0, x_2=1, x_3=4$ and let $L_j$ be the Lagrange polynomial for $j=0,1,2,3$. Show that $L_0, L_1, L_2, L_3$ form a basis for $\Pi_3$.

So I've calculated $L_0,...,L_3$. I'll supply $L_3$ and $L_0$ just to see if I got them right. I am, on the other hand, completely lost on what I should be doing next.

$L_0= x(x-1)(x-4)/(-36)$

$L_3= x(x+2)(x-1)/(24)$

user314580
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2 Answers2

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First of all, as far as I can see (if I haven't miscalculated as well ^^), $L_3$ should be $$L_3 = \frac{x(x+2)(x-1)}{(4-(-2)) \cdot 4 \cdot (4-1)} = \frac{x(x+2)(x-1)}{72}$$

Now to the question why they form a basis. The Lagrange polynomials are defined in such a way that you have $$L_j(x_i) = \delta_{ij}, i,j=0,\dots,3$$ where $\delta_{ij}$ is the Kronecker delta. We know that $\dim(\Pi_3) = 4$ (because $\{1,x,x^2,\dots,x^3\}$ is a basis). Thus it suffices to show that $L_0, \dots, L_3$ are linearly independent. So let $\lambda_0, \dots, \lambda_3$ so that $$\phi := \sum_{j=0}^{3}{\lambda_j L_j} = 0$$

We now need to show that $\lambda_j = 0\ \ \ \forall j \in \{0,\dots,3\}$. But this follows easily from $$0 = \phi(x_i) = \sum_{j=0}^{3}{\lambda_j L_j}(x_i) = \sum_{j=0}^{3}{\lambda_{j}\delta_{ij}} = \lambda_i$$

johnnycrab
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  • Yes, only $L_3$ was incorrect. Thanks for the guidance sir! – user314580 Mar 12 '16 at 13:20
  • @user314580 you're welcome. But bubba is right. Upvoting / accepting answer is a nice way of approving efforts. – johnnycrab Mar 12 '16 at 13:42
  • Must I register to do so? – user314580 Mar 12 '16 at 14:08
  • @user314580 No, simply click the "up" or "down" array over/beneath the score of a post. If the post is an answer to a question you asked, you can also accept one answer (note: you can upvote multiple answers, but "accept" only one). This can be done by clicking the checkmark below the score number. – johnnycrab Mar 12 '16 at 15:18
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I assume that $\Pi_3$ means the space of all polynomials of degree 3.

This space has dimension $4$, so, if we can show that $L_0$, $L_1$, $L_2$, $L_3$ span it, then we're done.

So, given any cubic polynomial $a_0 + a_1x + a_2x^2 + a_3x3$, we have to show that there exist numbers $\lambda_0$, $\lambda_1$, $\lambda_2$, $\lambda_3$ such that $$ a_0 + a_1x + a_2x^2 + a_3x^3 = \lambda_0 L_0(x) + \lambda_1 L_1(x) + \lambda_2 L_2(x) + \lambda_3 L_3(x) $$ Equating coefficients of powers of $x$, you'll get a system of four linear equations for $\lambda_0$, $\lambda_1$, $\lambda_2$, $\lambda_3$. Use your favorite technique to show that this linear system has a solution (e.g. non-zero determinant).

bubba
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