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Mathematical Paradox

by ramsey2879
Tags: mathematical, paradox
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ramsey2879
#1
Sep3-10, 07:16 PM
P: 894
I discovered a mathamatical truth that seems to have no clear logical foundation. Could anyone here explain the math behind the following?

A.Given the following recursive series:

S(1) = 3
S(2) = 2
S(n) = 6*S(n-1) - S(n-2) - 4

Define a new series as follows
N(i) = (S(i) + S(i+1) + 1)/2

B.Now note that for all i
1) S(i)*S(i+1) = N(i)*(N(i) + 1)/2
2) (N(i) + N(i +1) + 1)/4 = S(i+1)
3) N(i)*N(i+1)/4 = S(i+1)*(S(i+1) + 1)/2

C. A even more amazing paradox is that you can rewrite line 2 of part A as S(2) = 12, and make no other changes except to redo the math. Still B1, B2, and B3 again hold true!

What factors lurk in the background of this paradox?
Could yet another number be substituted for the numbers 2 or 12 in line 2 of part A?
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Hurkyl
#2
Sep3-10, 07:43 PM
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You know there's a straightforward procedure to solve recurrence relations of that type, right?

Second order linear homogeneous recurrences like that are of the form
S(n) = A rn + B sn
unless they degenerate; I forget how to treat them when they degenerate.


Paradox?
ramsey2879
#3
Sep3-10, 08:08 PM
P: 894
Quote Quote by Hurkyl View Post
You know there's a straightforward procedure to solve recurrence relations of that type, right?

Second order linear homogeneous recurrences like that are of the form
S(n) = A rn + B sn
unless they degenerate; I forget how to treat them when they degenerate.


Paradox?
I saw that before but I lost the book that it was in. Is the a web site that explains how to do so. Don't worry about the -4 in the recursive relation of my amended post. By subtracting 1 from each S(i) term you get a series that is truely S(n) = 6*S(n-1)-S(n-2)! So I would subtract 1 from each term solve to get the series S'(n) then write [tex]S(n) = Ar^{n} + Bs^{n} + 1[/tex] where A and B are derived from the series S'(n).


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