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Finding equation of a curve through 5 points

  1. Aug 13, 2014 #1

    erg

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    I'm trying to work out the equation of a curve through 5 very specific points... these points are
    (47.5, 46.3) (45,48.1) (43, 38.1) (40, 32.8) (37, 26.5)

    Thought of having general equation ax^4+bx^3+cx^2+dx+e and subbing all the points in and solving but that would take way too long... If there is any simper way to do this or if you could post any link to a program capable of solving it quickly that'd be great.
     
    Last edited: Aug 13, 2014
  2. jcsd
  3. Aug 13, 2014 #2
    Did you ever had the courage to draw the points on a grath ?
     
  4. Aug 13, 2014 #3
    Hi,

    Does the problem specify if the function is a polynomial, exponential, logarithmic, trigonometric, hyperbolic, square root?

    You can't simply assume the function is a polynomial.
     
  5. Aug 13, 2014 #4

    HallsofIvy

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    Staff Emeritus
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    The point is that there exist an infinite number of functions, whether polynomials, logarithms, etc. that will pass through any given 5 points. There exist exactly one fourth order polynomial through 5 points, if that's what you want. If you don't like solving 5 linear equations for the 5 polynomials, try using the "Legendre polynomial formula": The polynomial passing through the five points [itex](x_0, y_0), (x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4)[/itex] is given by
    [tex]\frac{(x- x_1)(x- x_2)(x- x_3)(x- x_4)}{x_0- x_1)(x_0- x_2)(x_0- x_3)(x_0- x_4)}+ \frac{(x- x_0)(x- x_2)(x- x_3)(x- x_4)}{x_1- x_0)(x_1- x_2)(x_1- x_3)(x_1- x_4)}+ \frac{(x- x_0)(x- x_1)(x- x_3)(x- x_4)}{x_2- x_0)(x_2- x_1)(x_2- x_3)(x_2- x_4)}+ \frac{(x- x_0)(x- x_1)(x- x_2)(x- x_4)}{x_3- x_0)(x_3- x_1)(x_3- x_2)(x_0- x_4)}+ \frac{(x- x_0)(x- x_1)(x- x_2)(x- x_3)}{x_4- x_0)(x_4- x_1)(x_4- x_2)(x_4- x_3)}[/tex]

    That is, each fraction has, in the numerator, x minus all x-values except one and, in the denominator, that x-value minus each of the others.
     
  6. Aug 13, 2014 #5

    disregardthat

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    Interpolating polynomial

    Doubt it can be done any easier than Halls formula by hand.
     
  7. Aug 13, 2014 #6
    Just looking at the graph makes me think at a simple problem of linear regression with an outlier. Are you sure that there is no bug in the data ?
     

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