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The idea about Pythagoras stairs?

  1. Nov 15, 2008 #1
    Hello everybody

    I have looked everywhere for some good guides, so the question is:


    You probably know the Fibonacci sequence (1,1,2,3,5,8…) but there is a lesser-known sequence called (Pythagoras stairs) generated by a similar recursion formula.

    This is a sequence of pairs (Xn,Yn) usually arranged as follows:
    1 2
    2 3
    5 7
    12 17

    And so on……. It begins with (x1,y1)=(1,1) and the recursion formula is

    X{n+1} = X{n} + Y{n}
    Y{n+1} = X{n}+ X{n+1}

    Prove by induction that always

    Y^2 = 2 X^2 ± 1

    Pythagoras used this equation to generate rational approximations to (2)^1/2


    Thanks in advance for usefully discuss to solve for this interesting question
     
    Last edited: Nov 15, 2008
  2. jcsd
  3. Nov 15, 2008 #2
    Hi again

    Hello everybody

    I have looked everywhere for some good guides, so the question is:

    You probably know the Fibonacci sequence (1,1,2,3,5,8…) but there is a lesser-known sequence called (Pythagoras stairs) generated by a similar recursion formula.

    This is a sequence of pairs (Xn,Yn) usually arranged as follows:
    1 2
    2 3
    5 7
    12 17

    And so on……. It begins with (x1,y1)=(1,1) and the recursion formula is

    [tex]X\underline{n+1}[/tex]=[tex]X\underline{n}[/tex]+[tex]Y\underline{n}[/tex]

    [tex]Y\underline{n+1}[/tex]= [tex]X\underline{n}[/tex]+[tex]X\underline{n+1}[/tex]


    Prove by induction that always

    [tex]Y\acute{2}[/tex]=2[tex]X\acute{2}[/tex] [tex]\pm1[/tex]

    Pythagoras used this equation to generate rational approximations to [tex]\sqrt{2}[/tex]


    Moreover, if you would like please open the enclosed to find the corrections/clarifications of the question.

    Thanks
     
    Last edited: Nov 25, 2008
  4. Nov 16, 2008 #3
    No comments or solution until now!!!

     
  5. Nov 16, 2008 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    A lot of people will not open a Word file because they are notorious for carrying viruses.
     
  6. Nov 16, 2008 #5
    Dear HallsofIvy

    Thanks for your reply, in fact I already wrote the problem in webpage and at the same time wrote in the word file so there two options


     
  7. Nov 17, 2008 #6
    Hi again

    Hello everybody

    I have looked everywhere for some good guides, so the question is:

    You probably know the Fibonacci sequence (1,1,2,3,5,8…) but there is a lesser-known sequence called (Pythagoras stairs) generated by a similar recursion formula.

    This is a sequence of pairs (Xn,Yn) usually arranged as follows:
    1 2
    2 3
    5 7
    12 17

    And so on……. It begins with (x1,y1)=(1,1) and the recursion formula is

    [tex]X\underline{n+1}[/tex]=[tex]X\underline{n}[/tex]+[tex]Y\underline{n}[/tex]

    [tex]Y\underline{n+1}[/tex]= [tex]X\underline{n}[/tex]+[tex]X\underline{n+1}[/tex]


    Prove by induction that always

    [tex]Y\acute{2}[/tex]=2[tex]X\acute{2}[/tex] [tex]\pm1[/tex]

    Pythagoras used this equation to generate rational approximations to [tex]\sqrt{2}[/tex]


    Moreover, if you would like please open the enclosed to find the question as pdf form.

    Thanks
     
    Last edited: Nov 25, 2008
  8. Nov 17, 2008 #7
    Okay here is how you do it...The method of mathematical induction for proofs follows the induction theorem which requires that for any relation dependent on variable N ,then if N=1,N=2 are TRUE and assuming N= k is TRUE then N=k+1 should also be TRUE...So if you test the Pythagoras' stairs for N=1 and N=2 then you find that it is TRUE...then assume that N=k is true such that y^2(k+1)=2x^2(k+1)+- 1...so using the above you should be able to prove that y^2(k+2)=2x^2(k+2)+-1...you can do this by taking the left hand side..You should be knowing that y(k+2)=2x(k+1)+y(k+1)..square this to have y^2(k+2) =4x^2(k+1)+4x(k+1)y(k+1)+y^2(k+1)...but from previously assuming N=k is true we had y^2(k+1)=2x^2(k+1)+-1..substitute in the previous equation...also keep in mind x(k+2)=x(k+1)+y(k+1)...use this [square it] in the above equations to get the right hand side of y^2(k+2)=2x^2(k+2)+-1 then you have proven using the induction theorem.
     
    Last edited: Nov 17, 2008
  9. Nov 24, 2008 #8
    some wrongs here in the question

    YES I AGREE WITH YOU there are mistake in the question
     
    Last edited by a moderator: Nov 26, 2008
  10. Nov 24, 2008 #9
    Re: some wrongs here in the question

    The above answer is not correct and there are some mistakes in the question, its easy to find the mistakes in the question.

    However the hint for the right question should use these formulae

    Y(n + 1) X(n + 1) = (2X(n+1) + Y(n+3) (X(n+4) + Y(n+5))

     
    Last edited by a moderator: Nov 26, 2008
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