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Derivation of Lagrange Family of Interpolation functions

  1. Mar 21, 2013 #1
    Folks,

    I am puzzled how the linear interpolation functions (see attached) were determined based on the following equation below

    ##\displaystyle \psi_i=\frac{(\xi-\xi_1)(\xi-\xi_2)....(\xi-\xi_{i-1})(\xi-\xi_{i+1})...(\xi-\xi_n)}{(\xi_i-\xi_1)(\xi_i-\xi_2)...(\xi_i-\xi_{i-1})(\xi_i-\xi_{i+1})...(\xi_i-\xi_n)}##

    What do the dots represent above?

    and

    [itex]\psi_i(\xi_j)= 1[/itex] if ##i=j## and ##0## if ##i\ne j##

    For example how is ##\psi_1(\xi)## determined?

    thanks
     

    Attached Files:

  2. jcsd
  3. Mar 21, 2013 #2

    mathman

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    The dots mean fill in all the like terms for indices between the given. In this case, 3 to i-2 and i+2 to n-1.

    For the second question, when j = i, the numerator and denominator are the same. When j ≠ i, one term of the numerator product = 0.
     
  4. Mar 21, 2013 #3
    Sorry, I still dont follow...

    I dont see how ##\psi_1(\xi)=\frac{1}{2}(1-\xi)## is obtained...?
     
  5. Mar 21, 2013 #4

    Mark44

    Staff: Mentor

    The dots (...) are called an ellipsis, and appear as three periods. The ellipsis means "continuing in the same fashion." The first missing factor in both the numerator and denominator would be (x - x3) and the next would be (x - x4), and so on. (I don't see any purpose in writing ##\xi## when plain old x will do just fine.)
     
  6. Mar 21, 2013 #5

    rbj

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    like any other dots like that which you see in a mathematical expression. it means "fill in the blank" with the continuing pattern.

    it's not as good as my Applied Engineering Mathematics book from Kreyszig, but the wikipedia article on Lagrange polynomials should have the answers to your question.
     
  7. Mar 22, 2013 #6
    ##\displaystyle \psi_1=\frac{(\xi-\xi_1)(\xi-\xi_2)(\xi-\xi_{0})(\xi-\xi_{2})(\xi-\xi_1)}{(\xi_1-\xi_1)(\xi_1-\xi_2)(\xi_1-\xi_{0})(\xi_1-\xi_{2})(\xi_1-\xi_1)}##.........?
     
  8. Mar 22, 2013 #7

    mathman

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    ξ1 - ξ1 is NOT a term in the denominator. Look at the definition of ψi, the ith term is left out of the numerator and denominator.

    Where did you get the expression for ψ1(ξ)?
     
  9. Mar 23, 2013 #8
    Its the first equation shown in the picture in first post...
     
  10. Mar 23, 2013 #9

    mathman

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    The pictures are slightly criptic. The important thing is that the definition of ψi depends on n. The particular item you ask about is for n = 2.
    In that case:
    ψ1(ξ) = (ξ-ξ2)/(ξ12), where ξ1= -1 and ξ2=+1.
     
  11. Mar 25, 2013 #10
    OK, thanks
     
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