Logarithms, square roots, value of pi and matrix inverses of Pascals triangle.

In calculating logarithms, square roots and value of \pi/4 we will begin by considering the Pascal triangle:
{A=}\left(\begin{array}{ccccccc}1&0&0&0&0&0&0\\ 1&1&0&0&0&0&0\\ 1&2&1&0&0&0&0\\ 1&3&3&1&0&0&0\\ 1&4&6&4&1&0&0\\ 1&5&10&10&5&1&0\\ 1&6&15&20&15&6&1\end{array}\right)

Logarithms:
To calculate the natural logarithm of {x} consider the following triangle:
{B_1=}\left(\begin{array}{ccccccc}1&0&0&0&0&0&0\\ 1/(1-x)&1&0&0&0&0&0\\ 1/(1-x)&1/(1-x)&1&0&0&0&0\\ 1/(1-x)&1/(1-x)&1/(1-x)&1&0&0&0\\ 1/(1-x)&1/(1-x)&1/(1-x)&1/(1-x)&1&0&0\\ 1/(1-x)&1/(1-x)&1/(1-x)&1/(1-x)&1/(1-x)&1&0\\ 1/(1-x)&1/(1-x)&1/(1-x)&1/(1-x)&1/(1-x)&1/(1-x)&1\end{array}\right)

Multiply elementwise the Pascal triangle, matrix {A_1}, with matrix {B_1} so that we get:

{C_1=}\left(\begin{array}{ccccccc}1&0&0&0&0&0&0\\ 1/(1-x)&1&0&0&0&0&0\\ 1/(1-x)&2/(1-x)&1&0&0&0&0\\ 1/(1-x)&3/(1-x)&3/(1-x)&1&0&0&0\\ 1/(1-x)&4/(1-x)&6/(1-x)&4/(1-x)&1&0&0\\ 1/(1-x)&5/(1-x)&10/(1-x)&10/(1-x)&5/(1-x)&1&0\\ 1/(1-x)&6/(1-x)&15/(1-x)&20/(1-x)&15/(1-x)&6/(1-x)&1\end{array}\right)

Calculate the matrix inverse of matrix {C_1}. Then the first column will be a sequence a(n)
with a property such that \frac{n*a(n)}{a(n+1)} converges to the natural logarithm of {x}, (n=1,2,3...).

Example: For x=2 we get a(n)= 1, 1, 3, 13, 75, 541, 4683, 47293, 545835, 7087261, 102247563
and 10*7087261/102247563 = 0.693147180 which is approximately equal to ln(2)

Square roots:
To calculate the square root of {x} we consider this triangle:
{B_2=}\left(\begin{array}{ccccccc}1&0&0&0&0&0&0\\ x&1&0&0&0&0&0\\ 1&x&1&0&0&0&0\\ 0&1&x&1&0&0&0\\ 0&0&1&x&1&0&0\\ 0&0&0&1&x&1&0\\ 0&0&0&0&1&x&1\end{array}\right)

and this triangle:
{C_2=}\left(\begin{array}{ccccccc}1&0&0&0&0&0&0\\ \frac{(x-1)*x}{2}&1&0&0&0&0&0\\ \frac{(x-1)*x}{2}&\frac{(x-1)*x}{2}&1&0&0&0&0\\ 0&\frac{(x-1)*x}{2}&\frac{(x-1)*x}{2}&1&0&0&0\\ 0&0&\frac{(x-1)*x}{2}&\frac{(x-1)*x}{2}&1&0&0\\ 0&0&0&\frac{(x-1)*x}{2}&\frac{(x-1)*x}{2}&1&0\\ 0&0&0&0&\frac{(x-1)*x}{2}&\frac{(x-1)*x}{2}&1\end{array}\right)

Multiply elementwise the Pascal triangle {A} with matrix {B_2} and divide elementwise with matrix {C_2}. We should then get:

{D_2=}\left(\begin{array}{ccccccc}1&0&0&0&0&0&0\\ \frac{1*2}{(x-1)}&1&0&0&0&0&0\\ \frac{1*2}{(x-1)*x}&\frac{2*2}{(x-1)}&1&0&0&0&0\\ 0&\frac{3*2}{(x-1)*x}&\frac{3*2}{(x-1)}&1&0&0&0\\ 0&0&\frac{10*2}{(x-1)*x}&\frac{4*2}{(x-1)}&1&0&0\\ 0&0&0&\frac{15*2}{(x-1)*x}&\frac{5*2}{(x-1)}&1&0\\ 0&0&0&0&\frac{21*2}{(x-1)*x}&\frac{6*2}{(x-1)}&1\end{array}\right)

we calculate the matrix inverse of matrix {D_2}. Then the first column will be a sequence a(n)
with a property such that x + \frac{n*a(n)}{a(n+1)} converges to the square root of {x}, (n=1,2,3...).

Example: For x=2 we get a(n)= 1, -2, 7, -36, 246, -2100, 21510, -257040, 3510360, -53933040, 920694600
and 2 + 10*(-53933040/920694600) = 1.414213573 which is approximately equal to \sqrt{2}

Value of \pi/4

Consider the following table T(n,k) defined by:
n>=k: if (n-k) modulo 4 = 1 or if (n-k) modulo 4 = 2 then -1 else 1

{B_3}\left(\begin{array}{ccccccc}1&0&0&0&0&0&0\\ -1&1&0&0&0&0&0\\ -1&-1&1&0&0&0&0\\ 1&-1&-1&1&0&0&0\\ 1&1&-1&-1&1&0&0\\ -1&1&1&-1&-1&1&0\\ -1&-1&1&1&-1&-1&1\end{array}\right)

Multiply matrix {B_3} elementwise with the Pascal triangle matrix {A}. We then get:

{C_3}\left(\begin{array}{ccccccc}1&0&0&0&0&0&0\\ -1&1&0&0&0&0&0\\ -1&-2&1&0&0&0&0\\ 1&-3&-3&1&0&0&0\\ 1&4&-6&-4&1&0&0\\ -1&5&10&-10&-5&1&0\\ -1&-6&15&20&-15&-6&1\end{array}\right)

Calculate the matrix inverse of {C_3} so that we get:

{D_3}\left(\begin{array}{ccccccc}1&0&0&0&0&0&0\\ 1&1&0&0&0&0&0\\ 3&2&1&0&0&0&0\\ 11&9&3&1&0&0&0\\ 57&44&18&4&1&0&0\\ 361&285&110&30&5&1&0\\ 2763&2166&855&220&45&6&1\end{array}\right)

The first column then has the property that \frac{n*a(n)}{a(n+1)} converges to \pi/4, (n=1,2,3...).

Example: a(n)= 1, 1, 3, 11, 57, 361, 2763
and 6*361/2763 = 0.783930510 which is approximately equal to \pi/4.

Keywords: Matrix inversion.

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2 Responses to Logarithms, square roots, value of pi and matrix inverses of Pascals triangle.

  1. Larry Kealey says:

    Very cool. I just created 3DListPlot of this in Mathematica, wrapped in a Manipulate, for visualization…just scanning the internet for interesting links to attach to my demo – for their (Mathworld.com) site.

    • Mats Granvik says:

      Thanks,
      Another number triangle with the same properties although somewhat slower convergence is the partial column sums of the Mahonian numbers defined by: T(n,1)=1, k>1: T(n,k) = \sum\limits_{i=1}^{k-1} T(n-i,k-1)

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