Fitting a Function given 3 points

AI Thread Summary
To find a polynomial function g that passes through the points g(-2) = 1, g(0) = 3, g(1) = -2, and g(4) = 3, the Lagrange interpolation formula is recommended. This formula requires constructing a polynomial as a sum of fractions, each containing factors that correspond to the other points, ensuring that the polynomial equals the desired value at each specified point. The discussion emphasizes the need for four terms in the polynomial, each with three factors in the numerator and denominator. Participants highlight the importance of careful bookkeeping and recognizing the pattern in the interpolation process. The Lagrange interpolation method is praised for its cleverness and effectiveness in fitting polynomials through given points.
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Homework Statement



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).

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)}

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

Homework Equations



(Above)

The Attempt at a Solution



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)}

I cannot figure out how to produce g(0) = 3.
 
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Why are you making your polynomial a quadratic? Perhaps it should be something else -- like a cubic.
 
Hint: each "term" should have more factors in parenthesis
 
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
t\frac{(x- b)(x- c)(x- d)}{(a- b)(a- c)(a- d)}+u\frac{(x- a)(x- c)(x- d)}{(b- a)(b- c)(b- d)}+v\frac{(x- a)(x- b)(x- d)}{(c- a)(c- b)(c- d)}+ w\frac{(x- a)(x- b)(x- c)}{(d- a)(d- b)(d- c)}

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.
 
HallsofIvy said:
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.

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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