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Math Help - lagragian interpolation

  1. #1
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    lagragian interpolation

    Let Q(t) be a quadratic polynomial such that Q(0) = 0,
    Q(1) = 2, and Q(3) = 6. Use Lagrangian interpolation to
    obtain Q(2).


    im not understanding these things at all.

    any help?
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  2. #2
    Senior Member
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    Quote Originally Posted by p00ndawg View Post
    Let Q(t) be a quadratic polynomial such that Q(0) = 0,
    Q(1) = 2, and Q(3) = 6. Use Lagrangian interpolation to
    obtain Q(2).


    im not understanding these things at all.

    any help?
    Basically, they are giving you three points on the polynomial:

    Q({\color{red}0}) = 0
    Q({\color{blue}1}) = 2
    Q({\color{magenta}3}) = 6

    You need to determine the quadratic (i.e. degree 2) polynomial that will go through all three of these points. The formula looks like this:

    Q(t) = 0 \cdot \frac{t-{\color{blue}1}}{{\color{red}0}-{\color{blue}1}} \cdot \frac{t-{\color{magenta}3}}{{\color{red}0}-{\color{magenta}3}} + 2 \cdot \frac{t-{\color{red}0}}{{\color{blue}1}-{\color{red}0}} \cdot \frac{t-{\color{magenta}3}}{{\color{blue}1}-{\color{magenta}3}} + 6 \cdot \frac{t-{\color{red}0}}{{\color{magenta}3}-{\color{red}0}} \cdot \frac{t-{\color{blue}1}}{{\color{magenta}3}-{\color{blue}1}}

    I've color-coded the numbers to make it easier to see where each number comes from. For example, look at just the second term:

    2 \cdot \frac{t-{\color{red}0}}{{\color{blue}1}-{\color{red}0}} \cdot \frac{t-{\color{magenta}3}}{{\color{blue}1}-{\color{magenta}3}}

    In the front we multiply by the function value at t=1, which is 2. Then there is one fraction corresponding to each point given OTHER than t=1. Hopefully you can see the general format by studying what I am doing here. It's a bit hard to explain this online so perhaps you will get it just by studying the pattern in my work.

    Once you have written out the terms like I did above, you can simplify it to get to a simple polynomial form, although it's not completely necessary. In the problem, it asks you to solve for Q(2) which just means you take the polynomial Q(t) we just created and then plug in t=2.
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  3. #3
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    ahh i get it now, thank you for that.
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