Linear Algebra Lagrange polynomials & Basis

  • #1
162
0
http://img100.imageshack.us/img100/9016/linalggp1.jpg [Broken]
for (a): does that mean i must compute l0(t), l1(t) and l2(t), and i wasn't sure how to do this with the lagrange polynomial formula given, so i found one online and did it, i'm not sure if this is correct, but my l0(t) looks like this:
= (t-a1)(t-a2)/(ao-a1)(ao-a2)
= (t-2)(t-3)/(1-2)(1-3)
=1/2(t-2)(t-3)

for (b) how do i go about proving this?

and also (c) what does it mean to deduce, i have looked up the definition but i still don't understand.
 
Last edited by a moderator:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,847
965
Yes, your polynomial for (a) is correct. I'm not sure why you say you "weren't sure how to do it with the given polynomial", it looks like you have done exactly that.
With i= 0, the fraction is [itex](t- a_1)(t-a_2)/(a_0-a1)(a_0-a2)= (t- 2)(t-3)/(1-2)(1-3)= (t-2)(t-3)/(-1)(-2)= -1/2 (t- 2)(t- 3)[/itex].
With i= 1, the fraction is [itex](t- a_0)(t-a_2)/(a_1-a_0)(a_1-a_2)= (t-1)(t-3)/(2-1)(2-3)= (t-1)(t-3)/(1)(-1)= -1(t-1)(t-3)[/itex].
With i= 2, the fraction is [itex](t- a_0)(t-a_1)/(a_2-a_0)(a_2-a_1)= (t-1)(t-2)/(3-1)(3-2)= (t-1)(t-2)/(2)(1)= 1/2(t-1)(t-2)[/itex]
Multiply those out and add.

To prove that a set is a "basis", you just prove that any function can be written as a "linear combination" of those (the sum of a number times each of those, as in (c)) in only one way. An easy way to do that is to note that there exist a unique nth order polynomial through any given n+1 points. Do you see that "lj(t)" is an l-1 degree polynomial? Given any n degree polynomial, what do you get if you evaluate that function at a1, a2, etc.?

Once you have proved (b), (c) follows easily. Saying that the {li} forms a basis means that any such function can be written as a linear combination. In fact, you really need to show (c) in order to prove (b)!
 

Related Threads on Linear Algebra Lagrange polynomials & Basis

Replies
5
Views
908
  • Last Post
Replies
4
Views
860
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
9
Views
10K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
3K
Replies
17
Views
2K
  • Last Post
Replies
3
Views
2K
Top