Complete the Array Series

AntonVrba

[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


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

Edit-2
OOPS - redface shame etc

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]

BicycleTree

A[9, 11] = .061803? (Numbering from [0,0]; from [1,1] that would be A[10,12] = .061803

AntonVrba

BicycleTree - sorry try again

BicycleTree

Perhaps:
A(9, 11) = 1/18

AntonVrba

Sorry - wrong again

BicycleTree

Maybe
A(9, 11) = .06867

<<<GUILLE>>>

[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????

marcus

my first guess:

A(3,2) = 0.80473...

marcus

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)

AntonVrba

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, each induvidual row or column also forms its own logical arithmic series .

BicycleTree

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.

AntonVrba

OOPS - redface shame etc

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]