lagrange-interpolation
I tried to find an answer for some time, the answer is simple probably ,say I have a function given by set of points and their values like ${(4,1),(5,0),(6,0)}$,
how do I calculate Lagrange interpolation polynomial for them?
We know that $$w(x) = 1*l_0(x) + 0*l_1(x)+0*l_2(x)$$
But I think we are obviously losing information here calculating only for the first term the $w(x)$. I can't find a clue.
How to proceed?
There is this theorem:
Given two sequences $z_n$ and $w_n$ of complex numbers such that $|z_n| \to \infty$, there exists a holomorphic function $f$ such that $f(z_n) = w_n$ for all $n$.
It is a consequence of the Weierstrass factorization theorem and the Mittag-Leffler theorem.
See this question.
It is difficult to say what's the problem without seeing what you did exactly, but here are some tips
Make sure you're using radians, for the looks of it, that may be the problem
Your solution must pass for all the input points, at least $f(0) = \cos(0) = 1$
Here's just a plot showing your results, just to help you confirm the solution in the book is actually correct
This is the full procedure
\begin{eqnarray} L(x) &=& \cos(0) \frac{(x - 0.6)(x - 0.9)}{(0 - 0.6)(0 - 0.9)} + \cos(0.6)\frac{(x - 0)(x - 0.9)}{(0.6 - 0)(0.6 - 0.9)} + \cos(0.9) \frac{(x - 0)(x - 6)}{(0.9 - 0)(0.9 - 0.6)} \\ &=& \frac{\cos(0)}{0.54}(x - 0.6)(x - 0.9) - \frac{\cos(0.6)}{0.18}x(x - 0.9) + \frac{\cos(0.9)}{0.27} x (x - 0.6) \\ &=& 1.85185 (x - 0.6)(x - 0.9) - 4.5852x(x - 0.9) + 2.30226 x(x - 0.6) \\ &=& -0.431087 x^2 - 0.0324552 x + 1 \end{eqnarray}
Best Answer
If you have points like $(4,1),(5,0),(6,0)$, it is true that only one of the Lagrange polynomials for those points will actually contribute to the final interpolating polynomial. However, all of the Lagrange polynomials in the basis "know" where the other nodes are located. For example here the Lagrange polynomial that is involved is $\frac{(x-5)(x-6)}{2}$, so it "knows" where those other zeros were, in the sense that you'd get a different result out of, say, $(4,1),(6,0),(7,0)$.