The Method of Finite Differences, Extended -- and Shortened
Date: 12/22/2010 at 10:43:12 From: Abby Subject: Write algrebraic formula for a series I have read all your info about the method of finite differences, but the series I am working with doesn't seem to fit any of the patterns, so I am stuck. The series is 1 16 63 160 I have listed all the differences, hoping to get to a common factor, but no luck. I have looked at the diagonals of the differences to see a pattern, and still no luck. It doesn't seem to be a polynomial. I can't seem to guess it or see any pattern. How can I write an expression for the next term?
Date: 12/22/2010 at 12:20:57 From: Doctor Greenie Subject: Re: Write algrebraic formula for a series Hi, Abby -- The series is (or at least can be) described by a polynomial; and the method of finite differences can lead you to that polynomial. Here is the beginning of the work on the problem using the method of finite differences: 1 16 63 160 15 47 97 32 50 18 With just these numbers, there is no row with a common difference. But you can MAKE a row with a common difference simply by adding another "18" to the row of 3rd differences. Then if you want, you can work backwards to find what the next term in the original sequence would be. That can be useful to check the formula you end up with: if the formula you end up with produces that number for the 5th term in the sequence, then your formula must be right. 1 16 63 160 325 15 47 97 165 32 50 68 18 18 At this point, you could continue with the usual method of finite differences to find the polynomial formula for the sequence. But sometimes the formula for the given sequence is relatively simple and you can find it without going through the entire normal method of finite differences. This example turns out to be a case like that. To use this "shortcut" in the method of finite differences, you need to know that if the n-th row of differences is constant, then that constant difference is (n!) times the leading coefficient of the polynomial. In your example, the constant 3rd difference is 18, so 18 = (3!)a where "a" is the leading coefficient 18 = 6a 3 = a So the polynomial for your sequence is of the form t(n) = 3n^3 + ..... Because the polynomial for this particular example turns out to be relatively simple, we can determine the rest of it by comparing each term of the sequence with the value of the leading 3n^3 term: n t(n) 3n^3 t(n) - n^3 --------------------------- 1 1 3 -2 2 16 24 -8 3 63 81 -18 4 160 192 -32 The differences in the last column represent the "rest" of the polynomial, which should be of degree 2 (or less). Examination of the numbers in the last column leads rather easily to the final determination of the polynomial that defines the sequence: n = 1: -2 = -2(1) n = 2: -8 = -2(4) n = 3: -18 = -2(9) n = 4: -32 = -2(16) We see that the remaining part of the polynomial is simply -2(n^2) And so the polynomial that defines this sequence is t(n) = 3n^3 - 2n^2 We can verify that this formula produces the given first four numbers of the sequence, as well as the fifth number we predicted: t(n) = 3n^3 - 2n^2 = n^2(3n - 2) t(1) = 1^2(3(1) - 2) = 1(1) = 1 t(2) = 2^2(3(2) - 2) = 4(4) = 16 t(3) = 3^2(3(3) - 2) = 9(7) = 63 t(4) = 4^2(3(4) - 2) = 16(10) = 160 t(5) = 5^2(3(5) - 2) = 25(13) = 325 Of course, if you continue with the usual method of finite differences, starting from the point above where we found (created) a constant 3rd difference of 18, you should end up with this same polynomial formula. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
Date: 12/22/2010 at 21:44:08 From: Abby Subject: Thank you (Write algrebraic formula for a series) I learned more from your email than I've learned all year in my math class so far. Thanks for being so complete and for showing me multiple ways to work it out. And thanks for being so fast. I was up for hours last night trying to figure it out. I can sleep well tonight!!
Date: 12/23/2010 at 18:08:11 From: Doctor Greenie Subject: Re: Thank you (Write algrebraic formula for a series) Hi, Abby -- I'm glad I could help out. It sounds as if you are a young student who really enjoys learning. It is especially rewarding for me when I get that feeling from somebody I help here at Dr. Math. Thanks for your good work; and for your note of thanks. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
Date: 12/25/2010 at 00:18:53 From: Abby Subject: Write algrebraic formula for a series Dr. Greenie, I have one more sequence I am working on. Can you check my thinking? The series is 4, 8, 14, 24 I have read all the stuff on Dr. Math, and it seems like a recursive series. I am not sure I know how to write it down on the computer, so let me write it out: 2 times the last term - 2^(n - 2) So for example, the 2nd term is 2*4 minus 0 And the third term would be 2*8 minus 2 I am surprised my teacher would give a problem with a recursive series (that's pretty hard core), so I was wondering if you saw a simpler solution. Looking forward to your view. Thx, Abby
Date: 12/25/2010 at 02:56:41 From: Doctor Greenie Subject: Re: Write algrebraic formula for a series Hi, Abby -- Your recursive sequence formula doesn't work for going from the first term (4) to the second one (8): 2*4 - 2^0 = 8 - 1 = 7 You can get a polynomial formula for this sequence using exactly the same strategy as for the earlier problem. But in this case the formula is not as "nice"; and you probably have to use the formal method of finite differences (instead of a shortcut) to find the formula for this sequence. From the given sequence, we can see 4 8 14 24 4 6 10 2 4 2 We can simply assume that the row of 3rd differences is going to be a constant "2"; using the method of finite differences will again give us a polynomial formula of degree 3 that produces the given sequence. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
Date: 12/26/2010 at 21:09:18 From: Abby Subject: Thank you (Write algrebraic formula for a series) OK Dr. G -- I worked it out. The fractional coefficient was a bit tricky but I was able to figure it out. I forgot that anything to the power of 0 is 1; that's the mistake I made in my first try. Thanks again -- I really appreciate your help!!
Date: 12/26/2010 at 23:31:11 From: Doctor Greenie Subject: Re: Thank you (Write algrebraic formula for a series) Nice work, Abby -- It's been a pleasure to work with you on these problems. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
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