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Complete the Array Series

  1. May 1, 2005 #1
    [tex]
    A[r,c] = \left(\begin{array}{cccccc}
    \frac{1}{2}\multsp \left(1+{\sqrt{5}}\right)&-&-&-&-&....\\
    \frac{1}{4}\multsp \left(2+{\sqrt{8}}\right)&\frac{1}{4}\multsp
    \left(1+{\sqrt{5}}\right)&-&-&-&....\\
    \frac{1}{6}\multsp \left(3+{\sqrt{13}}\right)&-&\frac{1}{6}\multsp
    \left(1+{\sqrt{5}}\right)&-&-&.... \\
    \frac{1}{8}\multsp \left(4+{\sqrt{20}}\right)&-&-&\frac{1}{8}\multsp
    \left(1+{\sqrt{5}}\right)&-&.... \\
    \frac{1}{10}\multsp
    \left(5+{\sqrt{29}}\right)&-&-&-&\frac{1}{10}\multsp \left(1+{\sqrt{5}}\right)&....\\
    ....&....&....&....&....&....
    \end{array}\right)
    [/tex]

    How often have we solved linear series, how about a two dimensional series array

    Can you complete the matrix?

    Once you have it, instead of white just give the answer as a decimal number for say A[9,11] or any other pair

    The answer is suprisingly simple :biggrin:


    Edit-1
    :biggrin: To note is that each induvidual row or column also forms its own logical arithmic series :biggrin:

    Edit-2
    :surprised OOPS - redface shame etc :redface:

    It is not the solution I had in mind - so you are correct. The solution I had in mind has a further property

    [tex]\frac {A[r,c]}{A[n.r,n.c]} = n [/tex]
     
    Last edited: May 2, 2005
  2. jcsd
  3. May 1, 2005 #2
    A[9, 11] = .061803? (Numbering from [0,0]; from [1,1] that would be A[10,12] = .061803
     
  4. May 1, 2005 #3
    BicycleTree - sorry try again
     
  5. May 1, 2005 #4
    Perhaps:
    A(9, 11) = 1/18
     
  6. May 1, 2005 #5
    Sorry - wrong again
     
  7. May 1, 2005 #6
    Maybe
    A(9, 11) = .06867
     
    Last edited: May 1, 2005
  8. May 1, 2005 #7
    [tex]A[r,c] = \left(\begin{array}{cccccc} \frac{1}{2}\multsp \left(1+{\sqrt{5}}\right)&-&-&-&-&....\\ \frac{1}{4}\multsp \left(2+{\sqrt{8}}\right)&\frac{1}{4}\multsp \left(1+{\sqrt{5}}\right)&-&-&-&....\\ \frac{1}{6}\multsp \left(3+{\sqrt{13}}\right)&-&\frac{1}{6}\multsp \left(1+{\sqrt{5}}\right)&-&-&.... \\ \frac{1}{8}\multsp \left(4+{\sqrt{20}}\right)&-&-&\frac{1}{8}\multsp \left(1+{\sqrt{5}}\right)&-&.... \\ \frac{1}{10}\multsp \left(5+{\sqrt{29}}\right)&-&-&-&\frac{1}{10}\multsp \left(1+{\sqrt{5}}\right)&....\\....&....&....&....&....&.... \end{array}\right)[/tex]

    Is it,
    A(11, 13)=O.6867????
     
  9. May 1, 2005 #8

    marcus

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    my first guess:

    A(3,2) = 0.80473...
     
  10. May 1, 2005 #9

    marcus

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    Ahh! I see that my proposal is the same as bicycle tree gave in post #6, just two posts back.

    I would also say, like bicycle tree, that

    A(9,11) = 0.0.06867...

    so he and I must be using the same formula for the general term A(r,c)
     
  11. May 1, 2005 #10
    Folks, none of the proposed answers match my series - either there are two solutions which would surprise me. So lets check marcus' or bicycletree's solution.

    For the answers to be correct, :biggrin: each induvidual row or column also forms its own logical arithmic series :biggrin: .
     
  12. May 2, 2005 #11
    The formula I got is:
    A(r, c) = 1/(2*r)*(r-c+1+sqrt((r-c+2)^2-2(r-c)+1))
    That matches the values you've given.
     
  13. May 2, 2005 #12
    :surprised OOPS - redface shame etc :redface:

    It is not the solution I had in mind - so you are correct. The solution I had in mind has a further property

    [tex]\frac {A[r,c]}{A[n.r,n.c]} = n [/tex]
     
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