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New recursive series identity

  1. Oct 17, 2007 #1
    I derived the following identity after considering my thread "Recursive series equality" but the result is so clean and neat that I post it as a new topic.

    Let [tex]S_{0} = 0 \quad S_{1} = 1 \quad S_{n} = b*S_{n-1} - S_{n-2}[/tex]

    Then [tex]S_{n}*(S_{n+b} -S_{b-2}) = (S_{n+1}+1)*(S_{n+b-1} -S_{b-1})[/tex]

    Has anyone seen this before?
     
  2. jcsd
  3. Oct 18, 2007 #2
    Ramsey, not to rain on your party, but it *might* turn out to be a little of a tautology. Since 0=S0 and 1=S1, it can be rewritten as

    [tex](S_{n} + S_0)*(S_{n+b} -S_{b-2}) = (S_{n+1}+S_1)*(S_{n+b-1} -S_{b-1})[/tex]

    which just means that the LHS formula has the same value when the indexes displace. Now, I don't know if this is common or not, the fact of having this kind of (quadratic) invariants along the sequence.

    Edit: Doh, this note is probably rubbish, since two terms are fixed while the other two vary with n. It was a hasty comment.
     
    Last edited: Oct 18, 2007
  4. Oct 18, 2007 #3
    Now I can generalize this identity such that [tex]S_{0}[/tex] and [tex]S_{1}[/tex] can each be any integer!

    [tex](S_{n} - S_{0})*(S_{n+b} - S_{b-2}) = (S_{n+1} - S_{0} + 1)*(S_{n+b-1}-S_{b-1})[/tex]

    Edit. sorry for misinformation, but this works only for b=3 if [tex]S_{0} <> 0 [/tex]
     
    Last edited: Oct 19, 2007
  5. Oct 19, 2007 #4
    Allow me some more deranging. Note that your last equation comes out by replacing each term of the sequence, say S_i, by a constant displacement, S_i + c. (In your case this constant is minus S0, thus bringing the generic sequence back to a "canonical" sequence where S0=0.). The places in the equation where two terms are subtracting cancel the constant out, so it only shows up where terms are by themselves.

    I can but wonder what happens to recurrence sequences when a constant is added to each term. Your last edit suggests that for b=3 (and for -1 as the coefficient of S_n-2) the original equation in Post#1 is invariant to addition of a constant.

    Deranging end. : )
     
  6. Oct 19, 2007 #5
    I don't understand what you are saying
     
  7. Oct 19, 2007 #6
    Define a new sequence Rn by the rule: Rn = Sn + c, where c is a constant.

    Then R0 = S0 + c, R1 = S1 + c, and the new recurrence rule for R is given by
    Rn = Sn + c = b . Sn-1 - Sn-2 + c = b(Rn-1 - c) - (Rn-2 - c) + c
    Rn = b . Rn-1 - Rn-2 - c(b - 2)​

    which, when b=2, becomes the same rule that defines Sn (why did you say b=3?).

    So you can displace by any constant a sequence defined by the rule 2 . Xn-1 - Xn-2, and it will be transformed in another sequence with the same rule.

    If c = -S0, R0 becomes 0, and you can apply your first identity (on post#1) to R:

    Rn (Rn+b - Rb-2) = (Rn+1 + 1)(Rn+b-1 - Rb-1)​

    and substituting Rn = Sn + c = Sn - S0, you get the last identity (on post #3).

    But you will obtain a similar effect for any constant c, not only for c = -S0 (as long as b=2). (Edit: And as long as R1 is not arbitrary, but displaced by the same constant as the others.)
     
    Last edited: Oct 19, 2007
  8. Oct 19, 2007 #7
    Fact is that my identity in the third post doesn't work for b = 2 so you need to consider this further.
     
  9. Oct 19, 2007 #8
    My new identity in fact only works when [tex]S_{1} = (b-1)*S_{0} +1[/tex] and it works for all b. Dodo noted that you can add or subtract a constant to the series when b = 2 but if [tex] S_1 <> S_0 + 1[/tex] it doesn't work for b = 2 which I noted in the last post.
     
    Last edited: Oct 20, 2007
  10. Oct 20, 2007 #9
    It is a neat equation, anyway. How come the b coefficient does not appear multiplying anything, but as an index displacement? Funny.

    It seems to stem from the conjectures

    [tex]
    \begin{align*}
    S_{n-1} \ S_{b-1} \ +\ S_{n+b-1} &= S_n \ S_b \\
    S_{n-1} \ S_{n+b-1} \ +\ S_{b-1} &= S_n \ S_{n+b-2}
    \end{align*}
    [/tex]
     
    Last edited: Oct 20, 2007
  11. Oct 20, 2007 #10
    Edit
    should have been
    Then [tex]S_{n}*(S_{n+a} -S_{a-2}) = (S_{n+1}+1)*(S_{n+a-1} -S_{a-1})[/tex]
     
  12. Oct 20, 2007 #11
    Actually I derived a proof for the case of S(0) = 0 based upon your proof in the previous thread. Now that I have my new conjecture for S(0) not equal to 0, I will work on a proof for the general case of S(1) = bS(0) +1. Your conjectures seem to be off base but I excuse that in view of the confusion I created with the dual use of the b variable.
    If you start with series [tex]S[/tex] with for d =0 as in my previous thread and consider what happens when you subtract S(n) from each term of the series you get a new series with [tex]d = (b-2)*S(n)[/tex] but the index n becomes the zero index in the equations of my last post. Play with the equation that you proved in the last thread and you get something like S(a) - bS(a-1) wich is the same as S(a-2).
     
  13. Oct 21, 2007 #12

    Gib Z

    User Avatar
    Homework Helper

    I have also found a remarkable identity:

    [tex] a+6 = 8[/tex], but it only works when a=2 !!

    :D:D:D
     
  14. Oct 21, 2007 #13
    But I can prove your identity, can you prove mine?
     
  15. Oct 21, 2007 #14
    OK this identity works for all S(0), S(1) and b:

    [tex](S_{n} - S_{0})*(S_{n+a} - S_{a-2}) = (S_{n+1} + S_{1} -bS_{0})*(S_{n+a-1} - S_{a-1})[/tex]
     
  16. Oct 22, 2007 #15
    Changing the index (other than n and a) by adding 1 and noting that [tex]S_{2} - bS_{1} = -S_{0} [/tex] we have the following rewrite of the identity

    [tex](S_{n} -S_{1}) \cdot (S_{n+a} - S{a-1}) = (S_{n+1} - S_{0}) \cdot (S_{n+a-1} - S_{a})[/tex]

    Now since the the bijections of recursive series [tex]R_{n} = bR_{n-1} + R_{n-2}[/tex] (one bijection S1 consists of every even term and the other bijection S2 consists of every odd term) each have the recursive formula [tex]S_{n} = (b^2+2)S_{n-1} - S_{n-1}[/tex] we have the following two recursive formulas

    [tex](R_{2n} - R_{2}) \cdot (R_{2n+2a} -R_{2a-2}) = (R_{2n+2} - R_{0}) \cdot (R_{2n+2a -2) - R_{2a}) [/tex]

    [tex](R_{2n+1} - R_{3}) \cdot (R_{2n+1 +2a} - R_{2a-1}) = (R_{2n+3} - R_{1}) \cdot (R_{2n+2a-1} -R_{2a+1})[/tex]

    This has some implications for Fibonacci type series and possibly situations where
    [tex] a^{2} + b^{2} = c^{2}[/tex]
     
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