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Write an equation based on data

by musicgold
Tags: based, data, equation, write
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Oct23-13, 08:58 PM
P: 180

I am trying to understand what is the best way to solve problems like the one below. I have a few data points and I need to which function best describes that data.

(x, y) = (-3, -32) (-2, 16) (-1, 8) (0,4) (1,2) (2, 1)

note that the y parameter in the first data point is -32 and not 32.

Is there a process to solve such problems or do I have to guess the function?

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Oct23-13, 11:56 PM
P: 746
put it on a graph and see what you get using excel
Oct24-13, 07:45 AM
Sci Advisor
PF Gold
P: 39,682
There are, of course, an infinite number of functions whose graphs go through those points. The unique fifth degree polynomial that fits is the Lagrange polynomial:
[tex]-32\frac{(x+ 2)(x+ 1)(x)(x- 1)(x- 2)}{(-3+2)(-3+ 1)(-3)(-3- 1)(-3-2)}+ 16\frac{(x+ 3)(x+ 1)(x)(x- 1)(x- 2)}{(-2+3)(-2+ 1)(-2)(-2- 1)(-2-2)}+ 8\frac{(x+ 3)(x+ 2)(x)(x- 1)(x- 2)}{(-1+3)(-1+ 2)(-3)(-1-1)(-1-2)}+ 4\frac{(x+ 3)(x+ 2)(x+1)(x- 1)(x- 2)}{(0+3)(0+2)(-0+ 1)(0- 1)(0-2)}+ 2\frac{(x+3)(x+ 2)(x+ 1)(x)(x- 2)}{(1+ 3)(1+2)(1+ 1)(1)(1-2)}+ \frac{(x+ 3)(x+ 2)(x+ 1)(x)(x- 1)}{(1+ 3)(1+2)(-3+ 1)(-3)(-3-2)}[/tex]

Oct25-13, 03:13 PM
P: 28
Write an equation based on data

There are several methods for example newton's forward difference interpolation which though quite difficult to do and limited in the fact that you have to have a constant difference between the terms it is very effective, because if the data points are from an arithmetic series it will give you an expression of the series. Above is an example of Lagrange's method which is far easier to do, less limited as the data points can be anything and returns a polynomial that works.
Oct26-13, 10:16 PM
P: 6
1. As you've emphasized that the first data is in fact not an error it may not be helpful to point out that the last five data points are in the function: y = (2^(2-x)) , but if a piecewise function is an option than it's worth a thought.
2. Otherwise you can use excel as Woopydalan said. However adding a linear trend line will do you no good. If I remember correctly there is an option for fitting a polynomial of higher degrees. It may go up to a sixth degree polynomial.
3. The Lagrange polynomial method that HallsofIvy posted will definitely actually forget what I said and just copy and paste that into wolfram alpha

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