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October 15th, 2009, 01:30 AM
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| | Interpolation #2 Im working on higher degree interpolation:
I'm asked to explain why
L0(x) + L1(x) + L2(x) + L3(x) = 1 for n=3
Not sure why this is...any advice? Thanks.
Last edited by mr fantastic; October 18th, 2009 at 12:10 AM.
Reason: Changed post title
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October 15th, 2009, 06:51 AM
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| | Quote:
Originally Posted by jzellt Im working on higher degree interpolation:
I'm asked to explain why
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. | 
October 17th, 2009, 11:38 PM
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| | 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... | 
October 18th, 2009, 12:10 AM
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| | 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.
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