Method of Finite DifferencesDate: 10/12/2000 at 14:36:30 From: Victoria Gillett Subject: Turning a series into an equation I was given the equation 3n^2 - 4n - 2. By plugging in 1, 2, 3, 4, 5 for n, I worked out the series: -3, 2, 13, 30, 53. If I were given only the series, how would I work through the series in order to find it the equation? Date: 10/12/2000 at 17:56:01 From: Doctor Greenie Subject: Re: Turning a series into an equation Hi, Victoria. Thanks for sending your question to us here at Dr. Math. The most elementary method I know of to find the equation from the series is called the method of "finite differences." The key to this method is the fact that the equation is a polynomial of degree k if and only if the k-th row of differences generated by the series is constant. That probably doesn't mean anything to you if you haven't seen the method before; I'll show you the method for your series in a few moments. I found an in-depth demonstration of WHY the method of finite differences works in "Finite Differences and Polynomials," from the Dr. Math archives, at the following URL: http://mathforum.org/dr.math/problems/harris.6.17.99.html Often you can find answers to the questions you have in the Dr. Math archives. It takes some practice to be able to find the information you are looking for (and of course sometimes the help you need is not in the archives). I found the above URL by going to the Dr. Math main page, clicking on the "Search the Archives" link, and searching on the phrase "finite difference." I had to click on the button that told the search to look for that exact phrase, since it would have found many unrelated URLs if it looked for the words "finite" and "difference" separately. Of course, in your case, you probably didn't know the term "finite differences," so you would not have been able to perform the same search. Anyway, always be sure to take a minute or two to TRY to find answers to your questions in the archives before you send us your questions. In my search, I did not find any place in the archives where an example of the use of the method of finite differences is provided, so I will demonstrate the method using your example. So, suppose I have the sequence: -3, 2, 13, 30, 53 and I need to find the polynomial expression that generates this sequence. I start by writing out the terms of the sequence, with a bit of space between the entries: -3 2 13 30 53 Beneath this row I write a "first row of differences" - the numbers showing the differences between the successive numbers in the first row: -3 2 13 30 53 5 11 17 23 It is absolutely essential that you understand what is being done here. The first entry in this "first row of differences" is 5 because the difference between the first and second numbers in the original row is 5; the second entry in this row is 11 because the difference between the second and third numbers in the original row is 11; and so on. Each entry in each successive row of differences is written between the two entries in the preceding row to help the reader see where it came from. I see that the numbers in my first row of differences are not all the same, so I next write a "second row of differences," showing the differences between the successive entries in the first row of differences: -3 2 13 30 53 5 11 17 23 6 6 6 Here the differences are all the same; my method of finite differences tells me that, since this occurs in my second row of differences, the polynomial that generates the sequence will be of degree 2 - that is, it is a polynomial of the form: f(n) = an^2 + bn + c ..........................[1] To find the values of a, b, and c for your example (we know we should end up with a = 3, b = -4, and c = -2, because we know the generating polynomial was 3n^2-4n-2), we can substitute values of n = 1, n = 2, and n = 3 into the polynomial [1] above. For example, if n = 1, we know the polynomial generates the first number in the sequence, which is -3, so we get the equation: f(1) = a*(1^2) + b*1 + c = -3 or a + b + c = -3 ................................[2] Evaluating f(2) and f(3) similarly (these are the 2nd and 3rd numbers in the sequence), we find 4a + 2b + c = 2 ...............................[3] and 9a + 3b + c = 13 ..............................[4] Equations [2], [3], and [4] are a system of 3 equations in 3 unknowns that can be solved; the solution is (as it must be): (a,b,c) = (3,-4,-2) Write back if you need help solving systems of 3 equations in 3 unknowns - or better yet, try finding help on that in our archives. Note that I needed 3 equations because I had 3 unknowns. I could have used any of the numbers in the sequence to build these three equations. For instance, I could have used the 5th number in the sequence to get the equation f(5) = 25a + 5b + c = 53 However, I chose to use the first three numbers in the sequence so that the coefficients in my system of 3 equations and 3 unknowns would be as small as possible. If your original sequence had been generated by a polynomial of degree 4, that is: f(n) = an^4 + bn^3 + cn^2 + dn + e then you would have found that it was not until the fourth row of differences that the differences were constant; then you would have 5 unknowns (a, b, c, d, and e) so you would need 5 equations in those variables to solve for a, b, c, d, and e; to get those equations you would use the first 5 terms of your sequence. I remember having fun as a young math student (probably between about 7th and 9th grades) playing with this method and discovering how nicely it works. If you have the time and interest, make up another polynomial of degree 2, generate the sequence, and then go through the method again to see if you can get back to your original polynomial by starting from the numbers in the sequence. Then maybe you can get really bold and try the same thing with an original polynomial of degree 3. Have fun! - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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