# Interpolation #2

• Oct 14th 2009, 11:30 PM
jzellt
Interpolation #2
Im working on higher degree interpolation:

L0(x) + L1(x) + L2(x) + L3(x) = 1 for n=3

Not sure why this is...any advice? Thanks.
• Oct 15th 2009, 04:51 AM
HallsofIvy
Quote:

Originally Posted by jzellt
Im working on higher degree interpolation:

L0(x) + L1(x) + L2(x) + L3(x) = 1 for n=3

Not sure why this is...any advice? Thanks.

I have no idea what you are doing. What Data are you interpolating? Is this a third degree interpolation? What are L0, L1, L2, and L3?

Just offhand it looks to me like this is saying that you are giving "equal weight" to the four calculations but other than that I can't say without more information.
• Oct 17th 2009, 09:38 PM
jzellt
Ok...sorry.

The hint given is to use Pn(x) = y0L0(x) + y1L1(x) + ... + ynLn(x).

(This is the interpolating polynomial of degree >= n)

So, Im supposed to use the above hint with suitable choices for {y0, y1, y2, y3) to show:

L0(x) + L1(x) + L2(x) + L3(x) = 1. Any advice...
• Oct 17th 2009, 10:10 PM
mr fantastic
Quote:

Originally Posted by jzellt
Ok...sorry.

The hint given is to use Pn(x) = y0L0(x) + y1L1(x) + ... + ynLn(x).

(This is the interpolating polynomial of degree >= n)

So, Im supposed to use the above hint with suitable choices for {y0, y1, y2, y3) to show:

L0(x) + L1(x) + L2(x) + L3(x) = 1. Any advice...

This still makes no sense.