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How to use finite differences to determine an equation?   X | Y -3  | 0 -2  | 15 -1...

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oyechalphut | Student, Grade 10 | (Level 1) Honors

Posted March 5, 2013 at 6:24 PM via web

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How to use finite differences to determine an equation?

  X | Y

-3  | 0

-2  | 15

-1  | 16

0   | 9

1   | 0

2   | -5

3   | 0

1 Answer | Add Yours

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted March 6, 2013 at 3:21 AM (Answer #1)

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If the data given is generated by a polynomial, then we can use the method of finite differences to find the degree of the polynomial. Since the x values all differ by one, consider the y-values:

0      15      16      9      0      -5      0    Now take the difference
   \  /     \  /     \  /    \  /    \  /    \  /
    15       1       -7     -9     -5      5
        \  /     \  /     \  /    \  /    \  /
        -14     -8      -2      4      10
              \ /      \/      \/      \/
               6       6       6       6

Since the third order differences are the same, the generating polynomial is a cubic.

The general cubic is `y=ax^3+bx^2+cx+d` Substituting known x,y pairs we create a system of linear equations; here use (0,9),(1,0),(2,-5) and (3,0):

0a+0b+0c+d=9 ==> d=9

 a+ b+ c+ 9=0  Using x=1,y=0 and d=9
8a+4b+2c+9=-5 Using x=2,y=-5 and d=9
27a+9b+3c+9=0 Using x=3,y=0,d=9

We have three equations and three unknowns. Solve the system (there are a number of methods: inverse matrices, Cramer's rule, Gaussian elimination, Gauss-Jordan row reduction, linear combinations, substitution, or technology come to mind.) Or you could see that you have three zeros so the function factors as

 

`y=(x+3)(x-3)(x-1)`  ** if -3 is a zero, x+3 is a factor.

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The function we seek is `y=x^3-x^2-9x+9` or `y=(x-3)(x+3)(x-1)`

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A formula that you will not have learned in class. Start at x=0. The y-value is 9. Now go along the diagonal of the finite differences from 9 -- you get 9,-9,4,6. According to a formula from Newton, the formula for the nth number of the sequence we are studying is given by:

`a+bn+(cn(n-1))/2+(dn(n-1)(n-2))/(2*3)+...` where a is the y-value for x=0, and b,c,d,... are the numbers along that diagonal.

Substituting a=9,b=-9,c=4 and d=6 (going further the finite differences are zero and thus do not contribute anything) we get:

`9-9n+(4n(n-1))/2+(6n(n-1)(n-2))/6`

`=9-9n+2n^2-2n+n^3-3n^2+2n`

`=n^3-n^2-9n+9` as we got above.

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