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: $T(n,1)=1, k>1: T(n,k) = \sum\limits_{i=1}^{k-1} T(n-i,k-1)$