Table of Differences

• The finite differences of a function are found by setting up a table matching x and '(x) coordinates. The values for x should be evenly spaced, though the values for '(x) will not be unless the function is linear. Subtract each value of '(x) from the next consecutive value for '(x) to get an additional line of values. Repeat this last step to find the differences in that line of values. Repeat until you reach a line of values that are identical. If the function is quadratic, the process should be done twice. Here is an example:
  
  X___'(x)___diff(1)____diff(2)
  
  0____2_____8________6
  
  1___10____14________6
  
  2___24____20________6
  
  3___44____26________6
  
  4___70____32
  
  5___102
  
  Since only two repetitions of finding the differences are necessary, the function has two degrees, of the form '² (x)=ax²+bx+c.
  
  

Solving through Systems

• There are two common methods of solving the equation if points on the curve are known. The easiest is probably to plug values of x and '(x) into the function to get at least three equations. Then manipulate the equations to find values for the coefficients a, b and c. Here's an example taken from the values listed above:
  
  2=a*0+b*0+c, which simplifies to c=2,
  
  10=a*1²+b*1+c, which simplifies to a+b=8 after substituting 2 in for c, and
  
  24=a*2²+b*2+c, which simplifies to 2a+b=11.
  
  By subtracting the second simplified equation (a+b=8) from the third one (2a+b=11), we get a=3. Substitute this value into the other equation to get b=5. The quadratic function is '(x)=3x²+5x+2.
  
  

Taylor's Method

• If for some reason it is not practical to solve with systems of equations, the quadratic function can be solved through Taylor's method, which uses the top row of differences from the finite differences chart. Taylor's method can be used for polynomial functions of any degree, but for dealing with a quadratic equation the following simplified version is adequate:
  
  '(x)= '(0)+'´(0)x+'´´(0)x(x-1)/2, where '´ is the first value from the first row of differences between values of '(x), and '´´ is from the second row of differences.
  
  Substituting in for the values of '(0) yields '(x)=2+8x+6(x²-x)/2, which simplifies to '(x)=3x²+5x+2.
  
  

Extensions and Limitations

• The finite differences can be used to solve equations of any degree, using either systems of equations or Taylor's method. They will not be helpful, though, if the equation is not polynomial, and if the values of '(x) are not integers, some inaccuracies may creep into the equation through rounding or truncating errors.