It is well known that we can determine the degree of a polynomial can be found by finding when the differences are the same. i.e. if the second differences are the same, it is a polynomial of the 2nd degree and if the third differences are the same, it is a polynomial of the 3rd degree.

I've noticed that the coefficient of the highest degree in the equation can be determined by looking at the corresponding differences.

I.e. Given:

x   y    D1y  D2y  D3y
1   1    -4   12   12
2   -3   8    24   12
3   5    32   36   
4   37   68   
5   105 

D here stands for Delta.

Right away, you can tell the the equation that matches this will be a polynomial of the third degree.

I've found that the relation between the differences for $\ \Delta\ $ 3y will be related to the coefficient such that:

$\ a\ = \frac{\Delta 3y}{3!} $ where $\ a\ $ is the coefficient for $\ x^3\ $ Here this would be a = 2

What I would like to figure out is how to determine the coefficient for $\ x^2\ $.

I know it is possible to create a matrix of equations and solve, but I am curious about a different approach.

My thought was to consider the equation as if it had no $\ x^3\ $ term. So in order to do that I must change the differences as if $\ x^3\ $ wasn't apart of the equation and this would let me extract the $\ x^2\ $ term.

So I calculated out all of the $\ \Delta y\ $ terms using an equation of just $\ 2x^3\ $

x   y    D1y   D2y   D3y
1   2    14    24    12
2   16   38    36    12
3   54   74    48
4   128  122  
5   250  

I found that if I subtracted these values from the ones in the original equation I would be left with a constant second difference of -12 meaning that $\ b = \frac{-12}{2!}\ = -6 $ where b is the coefficient of $\ x^2\ $

What I can't figure out is a quick way to determine a value of $\ \Delta 2y\ $ for $\ 2x^3\ $ without adding up all of the numbers manually. If I could do this then the process would be much faster and thus this process would be more practical for polynomial equations of very large degrees.


From the first line of values of $x, y, D1y, D2y, D3y$ which are $1, 2, 14, 24, 12$, you can get that the polynomial is

$$ \frac{12}{3!} (x-1)(x-2)(x-3) + \frac{24}{2!} (x-1)(x-2) + \frac{14}{1} (x-1) + 2 =2x^3$$

This is known as the method of differences. It should be easy to see how this generalizes. As to why this works, you should convince yourself

  1. There exists a unique polynomial of degree $n$ in which the first $n-1$ differences are 0, and the last difference is 1.
  2. Show that this polynomial has the form $\frac{1}{n!} \prod_{i=1}^n (x-i)$.
  3. The starting value of $x$, which is 1, tells us that we want the polynomial product to start with $(x-1)$. If your starting value of $x$ was 0 (more common), then the polynomials will be shifted by 1 and hence have the form $x(x-1)(x-2) ... $.
  • $\begingroup$ Nice work, @Calvin, as usual ;-) $\endgroup$ – Namaste Jun 12 '13 at 23:10
  • $\begingroup$ I read the link you posted and from what I understand, the page refers to a method to calculate more values of y given a set of x and y values and while not knowing the original function. It didn't answer my question per se, but it did give me a great idea about how to solve my problem. I'm sort of wondering if you knew that before you directed me there :P. $\endgroup$ – Klik Jun 12 '13 at 23:15
  • $\begingroup$ @TheWeirdNerd This is a common method, and is the discrete version of Maclaurin expansion. Sadly I can't seem to find a good writeup of it anywhere. You can try this. $\endgroup$ – Calvin Lin Jun 12 '13 at 23:24

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