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For given equidistant values $u_{-1}, u_{0}, u_{1},$ and $u_{2}$, a value is interpolated by Lagrange's formula. Show that it may be written in the form

$$u_{x} = yu_{0} + xu_{1} + \frac{y\left (y^{2} -1 \right )}{3!}\Delta ^{2}u_{-1} + \frac{x\left (x^{2} -1 \right )}{3!}\Delta ^{2}u_{0},$$ where $x+y=1$

The point where I am facing a doubt is:

a) Are the $x's$ equidistant (i.e. $-1, 0, 1, 2$) or the $y's$ equidistant or both (which seems improbable).

b) Even in either cases, I am not able to proceed.

OGC
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AJ_
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  • The nodes are equidistant, and then the interpolant $u_x$ (which is defined at the very least for $x \in [-1,2]$, not just discrete values) is supposed to be given by the formula on the right hand side, where $y$ is defined to be $1-x$. – Ian Sep 26 '18 at 03:44
  • The why does it say "equidistant value u-1 etc...". Also, I am still unable to get the answer. – AJ_ Sep 26 '18 at 03:48
  • It's kind of awkward phrasing IMO, I'd say "equidistant nodes with corresponding values" or something like this. Anyway, one way to do it is to just plug in $x=-1,0,1,2$ and check that you get $u_{-1},u_0,u_1,u_2$. A way to do it without assuming the result in the first place is to look at the polynomials $x,(1-x),(1-x)((1-x)^2-1)$ and $x(x^2-1)$ and treat these as your basis elements. Then evaluate them at the four nodes and set up the linear system. I don't really see a nice way to go from Lagrange or even Newton to this form, though. – Ian Sep 26 '18 at 11:08

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