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How can I find three points such that $t_0,t_1,t_2\in [0,1]$ are so that the interpolation of Lagrange basis functions $k_0(t),k_1(t),k_2(t)$ associate with them are orthogonal to each other, that is $\int_0^1k_i(t)k_j(t)dt=0$ for any $i,j\in\{0,1,2\}$?

I know the Lagrange Basis functions are $$l_k(x):=\frac{\prod_{0\le i\le n,\ i\ne k} (x - x_i)}{\prod_{0\le i\le n, \ i\ne k} (x_k - x_i)}, \qquad k=0,\dots,n.$$

and if I let $x_0,...,x_n$ be distinct real numbers and $l_k(x)$ be the Lagrange's basis function then $\delta_n = ∏^n _{k=0}(x-x_k)$ which will lead that $\sum^n_{k=0}x^j_kl_k(x) ≡ x^j$, but how can I apply this to my question in gray? Any help will be greatly appreciated.

user60514
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1 Answers1

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I think the best way to do this problem is just to go ahead and try it.

$k_0(x) = \frac{(x-t_1)(x-t_2)}{(t_0-t_1)(t_0-t_2)}$

$k_1(x) = \frac{(x-t_0)(x-t_2)}{(t_1-t_0)(t_1-t_2)}$

$k_2(x) = \frac{(x-t_0)(x-t_1)}{(t_2-t_0)(t_2-t_1)}$

So,

$\int_0^1k_0(x)k_1(x)dx = \int_0^1\frac{(x-t_1)(x-t_2)}{(t_0-t_1)(t_0-t_2)}\cdot\frac{(x-t_0)(x-t_2)}{(t_1-t_0)(t_1-t_2)}dx = 0$.

We divide out all the fluff and get

$\int_0^1(x-t_0)(x-t_1)(x-t_2)^2dx = 0$.

In a similar fashion, we'll be able to get

$\int_0^1(x-t_0)^2(x-t_1)(x-t_2)dx = 0$, and

$\int_0^1(x-t_0)(x-t_1)^2(x-t_2)dx = 0$.

From here, you can just expand and integrate, it's just a polynomial. You'll 3 equations with 3 unknowns. I'd like to say there's an easier solution somewhere out there, but this should work.

John K.
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  • The process you're going through actually is very similar to how Gaussian-Legendre Quadrature works. If you're interval was [-1,1], you would get the nodes for the 3-point Gaussian-Legendre Quadrature. You can probably just apply a linear transformation.

    link

     – John K. Apr 29 '13 at 01:40
  • So the answers should be

    0.5sqrt(3/5)+0.5, 0.5, -0.5sqrt(3/5)+0.5.

    Which is just the points from the linear transformation from [1,1] -> [0,1] from the nodes x0 = -sqrt(3/5), 0, sqrt(3/5). If you haven't seen Gaussian-Quadrature yet, just do it the way I presented in my original answer. I checked it on my calculator, it does work. This was an interesting problem.

     – John K. Apr 29 '13 at 02:19
  • Thank you very much John! This is great information. – user60514 Apr 29 '13 at 02:43