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Finite Difference (Interpolating Polynomial)

  1. Jan 21, 2013 #1
    1. The problem statement, all variables and given/known data
    http://puu.sh/1QFsA [Broken]

    2. Relevant equations



    3. The attempt at a solution
    I'm actually not sure how to do this question. How do i find Δx^2. I kind of understand the question but I don't know how to prove it. I know that Δy becomes dy when the width becomes infinitesimally small (Ʃ (infinity)). And y'' is basically the second derivative... but I have no clue how to go about verifying this...

    This is my attempt:
    http://puu.sh/1QFqE [Broken]

    I, myself, think it looks more like gibberish. Could someone nudge me to the right direction, please? Thanks!
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 21, 2013 #2

    LCKurtz

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    You have calculated ##y_0,\,y_1,\,y_2##. The first differences are ##y_1-y_0, y_2-y_1##, as you have calculated. Simplify them both by adding the fractions. Then take their difference and simplify it similarly. That will give you ##\Delta^2y_0##. Then divide that by ##\Delta^2x## and take the limit as ##\Delta x\rightarrow 0##. Just push it through and determine whether it gives the second derivative of ##\frac 1 x## at ##x=x_0## or not.
     
  4. Jan 21, 2013 #3
    But the question asks for http://puu.sh/1QKd3 [Broken], and you are saying to divide http://puu.sh/1QKeC [Broken] by http://puu.sh/1QKdV [Broken]. The square is associated with x, rather than delta. Also, when I take the second difference of x, I get zero...
     
    Last edited by a moderator: May 6, 2017
  5. Jan 21, 2013 #4

    Ray Vickson

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    You don't need ##\Delta x^2## (not ##\Delta^2 x##); that is just "notation", the same as ##dx^2## is merely a notation in differentiation. We have ##\Delta y / \Delta x## = first divided-difference, and so ##\Delta^2 y / \Delta x^2## = second divided difference = divided difference of the divided differences.
     
    Last edited by a moderator: May 6, 2017
  6. Jan 21, 2013 #5

    LCKurtz

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    Yes, I meant ##\Delta x^2 = (\Delta x)^2## as stated in the problem. That's why I almost (and should have) suggested you use ##h## instead of ##\Delta x## when working that problem. Using ##h^2## there would have been no typo and no misunderstanding.
     
    Last edited by a moderator: May 6, 2017
  7. Jan 21, 2013 #6
    http://puu.sh/1QMxk [Broken]

    I'm still very confused about the "Then divide that by Δ^2x and take the limit as Δx→0."

    I got it simplified using CAS, and replaced Δx by h like you suggested. However, how do I convert that into an derivative. If I just take the limit as h->0, then it would be 0 and the derivative that I need is 2/x^2.
     
    Last edited by a moderator: May 6, 2017
  8. Jan 21, 2013 #7

    LCKurtz

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    Your steps look OK. Weren't you asked to divide by ##(\Delta x)^2=h^2## before you let ##h\rightarrow 0##? And hopefully you will get the second derivative.
     
    Last edited by a moderator: May 6, 2017
  9. Jan 21, 2013 #8
    Thanks a lot! It makes sense now. The question was so confusing but it is actually a very simple question!!

    Thanks again everyone for your help.
     
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