Equation of a Sequence with Constant Third DifferencesDate: 05/26/98 at 12:20:43 From: Sohail Malik Subject: the method of difference Dear Dr. Math, I am fully capable of using the method of difference when the sequence has a second difference. But I have a sequence that has a third difference: 1 4 10 20 35 56 .... Please could you tell me the method of difference when there is a third difference? I would be eternally grateful if you could help me with this problem. Yours faithfully, Sohail Malik Date: 05/26/98 at 16:03:51 From: Doctor Anthony Subject: Re: the method of difference Given your sequence, you can find out the following information: n = 1 2 3 4 5 6 ... f(n) = 1 4 10 20 35 56 ... 1st difference D 3 6 10 15 21 ... 2nd difference D^2 3 4 5 6 ... 3rd difference D^3 1 1 1 ... With constant 3rd differences, the expression for f(n) will be a cubic polynomial in n: f(n) = an^3 + bn^2 + cn + d Substituting in for n, you find: f(1) = a + b + c + d = 1 f(2) = 8a + 4b + 2c + d = 4 f(3) = 27a + 9b + 3c + d = 10 f(4) = 64a + 16b + 4c + d = 20 And you have 4 equations to solve for the 4 unknowns a, b, c, and d. There is another way, depending on the Gregory-Newton formula, which in this case would be: f(n) = f(1) + C(n-1,1)D(1) + C(n-1,2)D^2(1) + C(n-1,3)D^3(1) = 1 + (n - 1)(3) + (n - 1)(n - 2)/2 (3) + (n - 1)(n - 2)(n - 3)/6 (1) = 1 + 3n - 3 + (3/2)(n^2 - 3n + 2) + (1/6)(n^3 - 6n^2 + 11n - 6) = (1/6)n^3 + (1/2)n^2 + (1/3)n We can check this with n = 1 and n = 2: f(1) = 1 and f(2) = 4 Thus the equation is giving the correct values. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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