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Recursive sequence problem, or something similar

  1. May 8, 2016 #1
    1. The problem statement, all variables and given/known data
    define calculation for the numbers ( a ; b; c) such that the new triplet is as follows.

    (a ; b ; c) --> transforms into --> (b+c ; c+a ; a+b)

    the first terms for a,b,c are : (1 ; 3 ; 5)

    Perform the calculation 2015 times, such that the new resulting numbers of the triplet are (x ; y ; z)

    calculate the subtraction for numbers x-z


    2. Relevant equations


    3. The attempt at a solution

    I would probably not have attempted this problem. This was an old college entrance exam question. But it was supposedly solvable with high school and trade school knowledge.

    Our class at school back in the day only had recursive sequences post calculus in high school mathematics.
    As I recall I did rather mediocre or less than average at that class.

    I was also wondering what would be the "smart way" to solve this problem assuming one only had pen and paper, and perhaps function calculator at hand. (no computer, no graphing calculator with recursive programs, no excel...).

    I tend to do these kind of recursions first by making a table of the numbers and transformations and then try to find some kind of connections.

    Off the top of my head, it looks like the B value (b=c+a)

    It looks like B value grows as follows.

    3*2w

    where w= number of transformations,

    also the beginning number 3 could be regarded as the zeroeth transformation such that
    3*20= 3


    It looks like C grows somewhat as follows.

    C0= 5
    C1=3*2^0+1
    C2=3*2^1+ 3*2^0+ 5
    C3=3*2^2 + 3*2^1 + 3*2^0 +1
    C4=3*2^3 + 3*2^2 + 3*2^1 + 3*2^0 +5
    C5=3*2^4 + 3*2^3 + 3*2^2 + 3*2^1 + 3*2^0 +1

    etc.


    In a similar fashion some sort of pattern can be found for the A value growth.

    I was simply wondering what kind of subtraction there would become for the 2015-th term.

    The B value alone would be staggeringly high. The B value should be according to my estimation something like 3*2^2015

    What would be the "smart way" to solve what would be the 2015-th term for the X value?

    You could say X value, or A value, either way it is the first number in the triplet.

    The question itself asked for the
    subtraction between

    X2015 - Y2015


    Now that I looked again at the growth of the values of A and C... It looks like the B is always like... the arithmetic mean, of the A and C values. Whether this is useful or not... it does not seem obvious to me.

    This arithmetic mean connection holds for all the first 10 terms, whichI did on pen and paper. Hehe... it was laborious but the connection jumped up at me when I looked at the paper.
    Thus it seems likely that the
    [X2015+Z2015 ] / 2=Y2015

    I'm not very good at these recursions it seems. It was one troublesome problem in my math book.

    Obviously had I known perfect and good way to solve the problem easily, I would not have posted a thread about it. I could definitely use some tips or pointers.
     
  2. jcsd
  3. May 8, 2016 #2

    haruspex

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    Yes.
    you posted the question as x-z, not x-y.
    Either way it is quite simple. Did you write out the a, b and c values for the first few iterations? Try four or five of them.
     
  4. May 9, 2016 #3
    original question was X - Y subtraction. That was my mistake sorry about that. I was up late until midnight or something. This problem was nagging at my brain though, so I decided to post the thread!

    X ; Y ; Z are the 2015-th iteration of a; b ; c (as far as I know.)

    (b+c) (a+c) (a+b)​
    I did manually the first nine iterations of a; b; c
    here are the first five iterations however,
    0. 1 ; 3 ; 5​
    1. 8 ; 6 ; 4
    2. 10 ; 12 ; 14
    3. 26; 24 ; 22
    4. 46 ; 48 ; 50
    5. 98 ; 96 ; 94
    It looks like the answer is either of the options below. I wouldn't know which one to choose though.

    X-Y = -2
    or
    X-Y = 2

    I suppose in real life, one would have to some rationale on paper, about how one arrives at the result of the subtraction. I guess it's called mathematical induction in this case. Why would it work in this case, and to what effect upon the subtraction?

    3*2^2015 is a massive number (number for Y, 2015-th iteration), too big for my graphing calculator to show it. But it is indeed still a finite number, which could be computed with enough resources.

    I was wondering also as a sidenote, what kind of clauses or functions, could be writen out for the iteration of X and Z values.

    Y value clause I managed to do on my own effort. But that's only part of the puzzle as it were.
     
    Last edited: May 9, 2016
  5. May 9, 2016 #4

    haruspex

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    How would you write the nth term of the sequence 1, -1, 1, -1, ....?
     
  6. May 9, 2016 #5

    I suppose the connection with X Y Z seems to be whether the number of iterations, n, is uneven or even number.

    I noticed this pattern yesterday already, but I was not feeling very confident about this problem at all...

    For X values and Z values the +2 or -2 depends on whether the iteration number (number of iterations to be performed) is uneven or even number. This number -2 or +2 is added to the Y value of the same iteration in such manner as below.

    It seems as though the resulting subtraction X2015 - Y2015 = 2
    In the 2015-th iteration, it should be so, that the X value is 2 greater than the Y value.


    n= number of iterations, such that n belongs to natural numbers
    Note that the number n, must be essentially the same for all the numbers of the triplet, (let those numbers of the triplet be x,y,z) which we are boosting with this growth.

    x0 y0 z0 = a b c = 1, 3, 5

    For X values
    when n is even
    Xn= 3*2n -2

    when n is uneven
    Xn= 3*2n +2


    For Y values
    for Y value itself, n can be uneven or even natural number.
    Yn= 3*2n

    For Z values
    when n is even
    Zn= 3*2n +2

    when n is uneven
    Zn= 3*2n -2



    choosing n=1


    X1 = 3*21+2 = 8
    Y1 = 3*21= 6
    Z1= 3*21-2 = 4


    choosing n= 2


    X2 = 3*22-2 = 10
    Y2 = 3*22= 12
    Z2= 3*22+2 = 14


    choosing n= 3

    X3 = 3*23+2 = 26
    Y3 = 3*23= 24
    Z3= 3*23-2 = 22
     
    Last edited: May 9, 2016
  7. May 9, 2016 #6

    haruspex

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    That is all correct, but you need to answer my question in post #4. Hint: what is (-1)2?
     
  8. May 10, 2016 #7
    (-1)^2 = 1

    because the even number of times in the exponent, the sign switched to positive. With odd number in the exponent, the sign stays negative, assuming there was a minus within those brackets... like there is indeed currently.

    negative*negative = positive
    however,

    with odd number of exponents (from a base of negative number within brackets)
    (-1)^3
    = (-1)*(-1)*(-1)
    = -1

    with 3 in the exponent, written out it would have been like this essentially
    negative*negative *negative--> comes out essentially as...--> positive*negative = negative



    I've never really thought about that in any more detail.

    If you had, (-2)^10
    This equals 2^10


    If you had = (-2)^11
    this would be essentially= (-2)*[(-2)^10]

    The recursion notation is completely forgotten by me sadly...I think the recursives, were not really focused highly in our high school class. (Or I simply did poorly in that particular area)

    It might be that my function notation usage skills are also little bit rusty because I can't put my finger on it, beyond my attempts thusfar. I can't really say how to write that out better than I already did with the worded description in English.

    I'm in a bit of a hurry so I'll check in on this thread later during the day.
     
  9. May 10, 2016 #8

    haruspex

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    Right, so what is the nth term in -1, +1, -1.....?
     
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