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The content of this question is a modified form of the calculation that was the precursor to this question, answered by joriki.

In short it is this Mathematica program:

Clear[x, nn, A1, A2, A3, A4, A5]
x = 2
nn = 12;
A1 = Table[
   Table[(n - k)*(n - k + 1 + 4/(x - 1)) + 1, {k, 1, nn}], {n, 1, nn}];
MatrixForm[A1]
A2 = Table[Table[Binomial[n - 1, k - 1], {k, 1, nn}], {n, 1, nn}];
MatrixForm[A2]
A3 = Inverse[A2];
MatrixForm[A3]
A4 = A1*A3;
MatrixForm[A4]
A5 = Inverse[A4];
MatrixForm[A5]
N[(A5[[nn, 1]] + A5[[nn, 2]])/(A5[[nn, 1]] - A5[[nn, 2]]), 30]
N[Sqrt[x], 30]

and the question is whether the second last expression tends to the square root of x.

Written in latex:

The program starts with the number $x$ to calculate the square root of, for example:

$$x=2$$

First consider the table: $$A_1(n,k)=(n-k) \left(-k+n+\frac{4}{x-1}+1\right)+1$$

$$A_1=\left( \begin{array}{cccccccc} 1 & -3 & -5 & -5 & -3 & 1 & 7 & 15 \\ 7 & 1 & -3 & -5 & -5 & -3 & 1 & 7 \\ 15 & 7 & 1 & -3 & -5 & -5 & -3 & 1 \\ 25 & 15 & 7 & 1 & -3 & -5 & -5 & -3 \\ 37 & 25 & 15 & 7 & 1 & -3 & -5 & -5 \\ 51 & 37 & 25 & 15 & 7 & 1 & -3 & -5 \\ 67 & 51 & 37 & 25 & 15 & 7 & 1 & -3 \\ 85 & 67 & 51 & 37 & 25 & 15 & 7 & 1 \end{array} \right)$$

and thereafter the inverse of the Pascal triangle:

$$A_3=\left( \begin{array}{cccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & -2 & 1 & 0 & 0 & 0 & 0 & 0 \\ -1 & 3 & -3 & 1 & 0 & 0 & 0 & 0 \\ 1 & -4 & 6 & -4 & 1 & 0 & 0 & 0 \\ -1 & 5 & -10 & 10 & -5 & 1 & 0 & 0 \\ 1 & -6 & 15 & -20 & 15 & -6 & 1 & 0 \\ -1 & 7 & -21 & 35 & -35 & 21 & -7 & 1 \end{array} \right)$$

Multiply elementwise $A_1$ with $A_3$ and calculate the matrix inverse of the resulting matrix to get matrix $A_5$:

$$A_5=\left( \begin{array}{cccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 7 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 83 & 14 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1453 & 249 & 21 & 1 & 0 & 0 & 0 & 0 \\ 33877 & 5812 & 498 & 28 & 1 & 0 & 0 & 0 \\ 987251 & 169385 & 14530 & 830 & 35 & 1 & 0 & 0 \\ 34524727 & 5923506 & 508155 & 29060 & 1245 & 42 & 1 & 0 \\ 1408573993 & 241673089 & 20732271 & 1185695 & 50855 & 1743 & 49 & 1 \end{array} \right)$$

Then the ratio $(A_5(n, 1) + A_5(n, 2))/(A_5(n, 1) - A_5(n, 2))$ should tend to square root of 2

$$(1408573993+241673089)/(1408573993-241673089) = 1.41421356033159778921552707958$$

$$\sqrt 2 = 1.41421356237309504880168872421$$

Leaving the $x$ unevaluated in the program one gets the approximate form for $\sqrt x$:

$$\sqrt x = \frac{\frac{48650112}{(x-1)^2}+\frac{203064960}{(x-1)^3}+\frac{445992960}{(x-1)^4}+\frac{534159360}{(x-1)^5}+\frac{330301440}{(x-1)^6}+\frac{82575360}{(x-1)^7}+\frac{5324788}{x-1}+178102}{\frac{17435712}{(x-1)^2}+\frac{99738240}{(x-1)^3}+\frac{276971520}{(x-1)^4}+\frac{399974400}{(x-1)^5}+\frac{289013760}{(x-1)^6}+\frac{82575360}{(x-1)^7}+\frac{1178212}{x-1}+13700}$$

So the question again, does this last ratio tend to $\sqrt x$?

share|improve this question
    
Does what tend to $\sqrt{x}$ as what tends to what? I suspect the latter two "what"s were supposed to be nn and $\infty$, and that for the first "what" was actually supposed to be the penultimate line of the program, rather than the last ratio you wrote. –  Hurkyl Jan 1 '13 at 16:10
    
I might to have been clear when asking. The Mathematica program is correct however. –  Mats Granvik Jan 1 '13 at 17:12
1  
The comment above should have been: "I might not have been clear when asking". –  Mats Granvik Jan 1 '13 at 17:19

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