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Fitting a Function given 3 points

  1. May 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Using a simple "trick," you can find a polynomial function whose graphi goes through any given set of points. For example, the graph of the following function goes through the points (1, 3), (2, 5), and (3, -1).

    [tex]f(x) = 3\;\frac{(x-2)(x-3)}{(1-2)(1-3)} \;+\; 5\frac{(x-1)(x-3)}{(2-1)(2-3)} \;+\;(-1) \frac{(x-1)(x-2)}{(3-1)(3-2)}[/tex]

    Find a polynomial function g that satisfies g(-2) = 1, g(0) = 3, g(1) = -2 and g(4) = 3

    2. Relevant equations

    (Above)

    3. The attempt at a solution

    [tex]g(x) = (1)\frac{(x-1)(x-4)}{(-2-1)(-2-4)} \; + \;(-2)\frac{(x+2)(x-4)}{(1+2)(1-4)} \;+ \;3\frac{(x+2)(x-1)}{(4+1)(4-1)}[/tex]

    I cannot figure out how to produce g(0) = 3.
     
  2. jcsd
  3. May 16, 2010 #2

    D H

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    Why are you making your polynomial a quadratic? Perhaps it should be something else -- like a cubic.
     
  4. May 16, 2010 #3
    Hint: each "term" should have more factors in parenthesis
     
  5. May 17, 2010 #4

    HallsofIvy

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    As both DH and Chaz say, you need four terms, not three, and you need three factors in numerator and denominator of each term.

    A polynomial passing through f(a)= t, f(b)= u, f(c)= v, and f(d)= w must look like
    [tex]t\frac{(x- b)(x- c)(x- d)}{(a- b)(a- c)(a- d)}+[/tex][tex] u\frac{(x- a)(x- c)(x- d)}{(b- a)(b- c)(b- d)}+[/tex][tex] v\frac{(x- a)(x- b)(x- d)}{(c- a)(c- b)(c- d)}[/tex][tex]+ w\frac{(x- a)(x- b)(x- c)}{(d- a)(d- b)(d- c)}[/tex]

    Look at what happens at, say, x= c. Every fraction except the third has a factor of x- c in the numerator and so will be 0 at x= c. The third fraction will have, at x= c, (c- a)(c- b)(c- d) in both numerator and denominator and so will be 1. The entire polynomal will be f(c)= 0+ 0+ v(1)+ 0= v.

    This is the "Lagrange interpolation formula"- to write a polynomial that passes through n given points, you need a sum of n fractions, each having n-1 factors in numerator and denominator.
     
  6. May 19, 2010 #5
    I have never heard of this formula, but it is so clever! I love it :smile:

    As for helping OP, all I can say is, look for the pattern, and, once you've got it, be meticulous in your bookkeeping.
     
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