- #1
AntonVrba
- 92
- 0
[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
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]
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 etcBicycleTree said: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.
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: