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Formula of nth term of a pattern

  1. Aug 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the formula for nth term of:
    .... , -450 , -270 , -180 , -90 , 90 , 180 , 270 , 450 , ...



    2. Relevant equations
    Arithmetic and geometric series
    Recursive

    3. The attempt at a solution
    Usually when finding the formula of Un, n starts from 1. But since the pattern goes infinite in both ways, how can we set which one is the first term?

    As for recursive, it usually involves the preceding term, Un-1. Again, how can we determine the preceding term for the above case?

    Please give me idea how to start solving this. Thank you very much
     
  2. jcsd
  3. Aug 27, 2014 #2
    For the general formula (in terms of n), you're right that the sequence going both ways complicates things. The answer is that you have to arbitrarily designate a term to be ##a_{1}##, or perhaps ##a_{0}##. You would write the whole formula based around that starting term, in a form like this:

    ##a_{n} = ....##, where ##a_{0} = 3##

    For the recursive definition, you'll just use a term and the preceding term. Since it's recursive, you don't need to know the particular values for either of those terms -- the recursive form just means you describe a term in terms of its preceding term. So it works with any two (adjacent) terms in the sequence, and so it isn't necessary to designate a starting term. The form for that would look like:

    ##a_{n} = a_{n-1}....##

    So ##a_{n}## can refer to any term in the sequence, because this rule works for any pair of terms in the sequence, no matter how long it is and whether it goes infinitely in just one direction or two.
     
    Last edited: Aug 27, 2014
  4. Aug 27, 2014 #3

    Ray Vickson

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    Just fix ##n = 0## anywhere you want; for example, you can take ##U_0 = -450.## All the ##U_n## for ##n < 0## are the ones you do not see to the left of what you have written and all the ##U_n## for ##n > 7## are the ones you don't see to the right of what you have written. Now just start taking differences and see what you get---there is a nice pattern.
     
  5. Aug 28, 2014 #4
    I still can't find the pattern. Do we have to make separate formula for n < 0 and n > 0?

    I tried to set U0 = 0, at the middle of the positive and negative terms. What I got is 270 = 180 + 90 ; 450 = 270 + 180. Maybe it is related to Un = Un-1 + Un-2, but I don't know about 90 and 180.

    Thanks
     
  6. Aug 30, 2014 #5

    SammyS

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    The pattern is evident in the sequence as it's given, before you assign any starting point.

    Hint: Make a difference table with 1st & 2nd differences.
     
  7. Aug 31, 2014 #6

    Ray Vickson

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    For the sequence
    [tex] U = \{U_i,i=0,1,2, \ldots \} = \{-450 , -270 , -180 , -90 , 90 , 180 , 270 , 450 , \ldots\} [/tex]
    the sequence of first differences is
    [tex] dU = \{U_{i+1}-U_{i}, i = 0,1,2,\ldots \} = \{ -270-(-450),-180-(-270), \ldots \}
    = \{ 180,90,90,180,90,90,\ldots \} \\
    = 90 + 90 \{1,0,0,1,0,0,1,0,0,\ldots \} [/tex]
    If you can find a nice formula for the sequence ##\{1,0,0,1,0,0,1,0,0,\ldots \}## you are almost finished. If you know about complex variables and the three (complex) roots of unity, you can finish up with a finite sum of some geometric series, to get a fairly nice formula for the nth term of your original sequence.
     
  8. Aug 31, 2014 #7

    PeroK

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    Alternatively, you should be able to do something with n (mod3) to formulate the sequence of 1, 0, 0, 1, 0, 0 ...
     
  9. Sep 2, 2014 #8

    andrewkirk

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    If we are to be rigorous, we need to take care with our definitions in defining the pattern. Recursive definitions mostly use the Recursion Theorem, which is only defined on the positive integers ##Z_+##. For more general well-ordered sets we can use the Theorem of Transfinite Recursive Definition, but that's not needed here.

    We can't just use the Recursion Theorem here in a single step, because the base set is the full set of integers, which is not well-ordered. The following trick gets around this problem.

    Define ##a_1=90, b_1=-90##
    Define rules to give ##a_{n+1}## and ##b_{n+1}## in terms of ##a_n## and ##b_n## as follows, for ##n\in Z_+##:

    ##a_{n+1}=a_n+180## if ##n+1## is divisible by 3, else ##a_{n+1}=a_n+90##
    ##b_{n+1}=b_n-180## if ##n+1## is divisible by 3, else ##b_{n+1}=b_n-90##

    Then the recursion theorem tells us that these definitions prescribe unique sequences ##a_n## and ##b_n##.

    Now we can just define ##c:Z\to Z## by ##c(n)=a_{n+1}## if ##n\geq 0##, otherwise ##c(n)=b_{-n}##.
     
    Last edited: Sep 2, 2014
  10. Sep 14, 2014 #9
    Sorry for really late reply

    I think I get the idea. I'll try it first. Thanks a lot for the help
     
  11. Sep 25, 2014 #10

    Simon Bridge

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    I'm always puzzled by these things - i.e. notice how the sequence is symmetrical about the -90,90 terms?
    Why not exploit that? Put ##a_1=90##, then ##a_n = (n/|n|)a_{|n|}## ... the sequence for ##a_{|n|}## should be easy ... though there are only four terms so you are spoiled for choice. {90, 180, 270, 450...} the differences than go {1, 1, 2, ...} in units of ##a_1##, which is suggestive.

    What approach you choose seems to depend on what assumptions you make about the pattern in question.
     
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