The Mobius function Blog

Plot of the Riemann zeta function in the complex plane – Wolfram Alpha

Posted on September 26, 2011 by Mats Granvik

A few months ago I got my copy of Mathematica 8. It wasn’t until a few days ago that I learned how to plot the Riemann zeta function in the complex plane. I did not figure this out myself, instead I searched Google and found Jeffrey Stopple’s home page here with some of his Mathematica code:

Show[Graphics[RasterArray[Table[Hue[Mod[3 Pi/2 + Arg[Zeta[sigma + I t]], 2 Pi]/(2 Pi)], {t, -30, 30, .1}, {sigma, -30, 30, .1}]]], AspectRatio -> Automatic]

Wikipedia has perhaps the best plot. It is however possible to plot the argument of the Riemann zeta function in Wolfram Alpha with this short command:

ContourPlot[Arg[Zeta[s + I t]], {s, -30, 30}, {t, -30, 30}]

Link to the contourplot in Wolfram Alpha.

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Sequence from question on math stackexchange

Posted on September 14, 2011 by Mats Granvik

Yesterday I asked a question on the math stackexchange forum about a sequence where the $n$ first terms are the solutions to the polynomials formed from $n$ x $n$ determinants. Here are the 250 first terms as requested in a comment:

{ }, 2, 3, 0, 5, 3/2, 7, 0, 0, 5/3, 11, 0, 13, 7/4, 15/7, 0, 17, 0, 19, 0, 7/3, 11/6, 23, 0, 0, 13/7, 0, 0, 29, 15/11, 31, 0, 33/13, 17/9, 35/11, 0, 37, 19/10, 13/5, 0, 41, 7/5, 43, 0, 0, 23/12, 47, 0, 0, 0, 51/19, 0, 53, 0, 11/3, 0, 19/7, 29/15, 59, 0, 61, 31/16, 0, 0, 65/17, 33/23, 67, 0, 69/25, 35/23, 71, 0, 73, 37/19, 0, 0, 77/17, 13/9, 79, 0, 0, 41/21, 83, 0, 85/21, 43/22, 87/31, 0, 89, 0, 91/19, 0, 31/11, 47/24, 95/23, 0, 97, 0, 0, 0, 101, 51/35, 103, 0, 35/19, 53/27, 107, 0, 109, 11/7, 37/13, 0, 113, 19/13, 115/27, 0, 0, 59/30, 119/23, 0, 0, 61/31, 123/43, 0, 0, 0, 127, 0, 43/15, 65/41, 131, 0, 133/25, 67/34, 0, 0, 137, 69/47, 139, 0, 141/49, 71/36, 143/23, 0, 145/33, 73/37, 0, 0, 149, 0, 151, 0, 0, 77/47, 31/7, 0, 157, 79/40, 159/55, 0, 161/29, 0, 163, 0, 33/17, 83/42, 167, 0, 0, 85/53, 0, 0, 173, 87/59, 0, 0, 177/61, 89/45, 179, 0, 181, 91/55, 61/21, 0, 185/41, 31/21, 187/27, 0, 0, 95/59, 191, 0, 193, 97/49, 65/33, 0, 197, 0, 199, 0, 67/23, 101/51, 29/5, 0, 41/9, 103/52, 0, 0, 209/29, 35/27, 211, 0, 213/73, 107/54, 215/47, 0, 217/37, 109/55, 73/25, 0, 221/29, 37/25, 223, 0, 0, 113/57, 227, 0, 229, 115/71, 77/37, 0, 233, 0, 235/51, 0, 79/27, 119/71, 239, 0, 241, 0, 0, 0, 0, 123/83, 247/31, 0, 249/85, 0

The solutions as they come out from Mathematica are listed by magnitude so they need to be sorted according to the difference between two successive lists of solutions. This list above has been sorted in a Excel spreadsheet using the MATCH(value,range,0) function.

Link to the question:

Are the primes found as a subset in this sequence a_n?

plot-of-32-nd-polynomial

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Binomial series

Posted on July 9, 2011 by Mats Granvik

$(a+x)^n = a^n + {n \choose 1}a^{n-1}x^1 + {n \choose 2}a^{n-2}x^2 + {n \choose 3}a^{n-3}x^3 +...$

Special cases are:
$(a+x)^{1} = a^{1}+x^{1}$
$(a+x)^{2} = a^{2}+2a^{1}x^{1}+x^{2}$
$(a+x)^{3} = a^{3}+3a^{2}x^{1}+3a^{1}x^{2}+x^{3}$
$(a+x)^{4} = a^{4}+4a^{3}x^{1}+4a^{2}x^{2}+4a^{1}x^3+x^{4}$

These expansions below are valid in the interval: $-1<x<1$

$(a+x)^{-1} = 1-x+x^2-x^3+x^4-...$
$(a+x)^{-2} = 1-2x+3x^2-4x^3+5x^4-...$
$(a+x)^{-3} = 1-3x+6x^2-10x^3+15x^4-...$

And these expansions (below) are valid in the interval: $-1<x\leq1$

$(a+x)^{-1/2} = 1-\frac{1}{2}x+\frac{1\cdot3}{2\cdot4}x^{2}-\frac{1\cdot3\cdot5}{2\cdot4\cdot6}x^{3}+...$

$(a+x)^{1/2} = 1+\frac{1}{2}x-\frac{1}{2\cdot4}x^{2}+\frac{1\cdot3}{2\cdot4\cdot6}x^{3}-...$

$(a+x)^{-1/3} = 1-\frac{1}{3}x+\frac{1\cdot4}{3\cdot6}x^{2}-\frac{1\cdot4\cdot7}{3\cdot6\cdot9}x^{3}+...$

$(a+x)^{1/3} = 1+\frac{1}{3}x-\frac{2}{3\cdot6}x^{2}+\frac{2\cdot5}{3\cdot6\cdot9}x^{3}-...$

All of these can be found in Schaum's outlines, Mathematical Handbook of Formulas and Tables, Third Edition, Murray R. Spiegel PhD, Seymour Lipschutz PhD, John Liu PhD, page 138.

These expansions also work for triangular matrices with ones in the main diagonal by replacing $a$ with the identity matrix $I$ and replacing $x$ with $X$ which stands for the part of the lower triangular matrix below the main diagonal.

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Mobius function as signs of eigenvalues of a real symmetric matrix

Posted on June 30, 2011 by Mats Granvik

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

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Ordinary generating functions for coefficients of some Dirichlet series

Posted on April 18, 2011 by Mats Granvik

The Dirichlet series $a_n$ corresponding to a power of the Riemann zeta function $\zeta(s)^m = \sum\limits_{n=1}^{\infty}\frac{a_n}{n^s}$ has the ordinary generating function:

$\sum \limits_{n=1}^{\infty} a_nx^n = x + {m \choose 1}\sum \limits_{a=2}^{\infty} x^{a} + {m \choose 2}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} x^{ab} + {m \choose 3}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} x^{abc} + {m \choose 4}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} \sum \limits_{d=2}^{\infty} x^{abcd} +...$

Examples:
$\sum \limits_{n=1}^{\infty} 1(n)x^n = x + \sum \limits_{a=2}^{\infty} x^{a}$

$\sum \limits_{n=1}^{\infty} \tau(n)x^n = x + 2\sum \limits_{a=2}^{\infty} x^{a} + \sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} x^{ab}$

$\sum \limits_{n=1}^{\infty} \tau_3(n)x^n = x + 3\sum \limits_{a=2}^{\infty} x^{a} + 3\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} x^{ab} + \sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} x^{abc}$

$\sum \limits_{n=1}^{\infty} \tau_4(n)x^n = x + 4\sum \limits_{a=2}^{\infty} x^{a} + 6\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} x^{ab} + 4\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} x^{abc} + \sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} \sum \limits_{d=2}^{\infty} x^{abcd}$

$\sum \limits_{n=1}^{\infty} \mu(n)x^n = x - \sum \limits_{a=2}^{\infty} x^{a} + \sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} x^{ab} - \sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} x^{abc} + \sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} \sum \limits_{d=2}^{\infty} x^{abcd} -...$

$\sum \limits_{n=1}^{\infty} \mu_2(n)x^n = x - 2\sum \limits_{a=2}^{\infty} x^{a} + 3\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} x^{ab} - 4\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} x^{abc} + 5\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} \sum \limits_{d=2}^{\infty} x^{abcd} -...$

$\sum \limits_{n=1}^{\infty} \mu_3(n)x^n = x - 3\sum \limits_{a=2}^{\infty} x^{a} + 6\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} x^{ab} - 10\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} x^{abc} + 15\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} \sum \limits_{d=2}^{\infty} x^{abcd} -...$

$\sum \limits_{n=1}^{\infty} \mu_{-1/2}(n)x^n = x - \frac{1}{2}\sum \limits_{a=2}^{\infty} x^{a} + \frac{1*3}{2*4}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} x^{ab} - \frac{1*3*5}{2*4*6}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} x^{abc} +...$

$\sum \limits_{n=1}^{\infty} \mu_{1/2}(n)x^n = x + \frac{1}{2}\sum \limits_{a=2}^{\infty} x^{a} - \frac{1}{2*4}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} x^{ab} + \frac{1*3}{2*4*6}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} x^{abc} +...$

$\sum \limits_{n=1}^{\infty} \mu_{-1/3}(n)x^n = x - \frac{1}{3}\sum \limits_{a=2}^{\infty} x^{a} + \frac{1*4}{3*6}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} x^{ab} - \frac{1*4*7}{3*6*9}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} x^{abc} +...$

$\sum \limits_{n=1}^{\infty} \mu_{1/3}(n)x^n = x + \frac{1}{3}\sum \limits_{a=2}^{\infty} x^{a} - \frac{2}{3*6}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} x^{ab} + \frac{2*5}{3*6*9}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} x^{abc} +...$

$\sum \limits_{n=1}^{\infty} D(n)x^n = x + \sum \limits_{a=2}^{\infty} x^{a} + \frac{1}{2!}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} x^{ab} + \frac{1}{3!}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} x^{abc} + \frac{1}{4!}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} \sum \limits_{d=2}^{\infty} x^{abcd} +...$

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

Posted on March 19, 2011 by Mats Granvik

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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Catalan numbers as a convergent of iterative matrix inversion, INVERT transform.

Posted on March 19, 2011 by Mats Granvik

I have here left out the indices of the elements for easier typing but hopefully this can still be understood.

${A=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ x&1&0&0&0&0&0&0\\ x&x&1&0&0&0&0&0\\ x&x&x&1&0&0&0&0\\ x&x&x&x&1&0&0&0\\ x&x&x&x&x&1&0&0\\ x&x&x&x&x&x&1&0\\ x&x&x&x&x&x&x&1\end{array}\right)$

Shift down the elements in matrix ${A}$ one step to get matrix ${B}$ .

${B=}\left(\begin{array}{cccccccc}0&0&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ x&1&0&0&0&0&0&0\\ x&x&1&0&0&0&0&0\\ x&x&x&1&0&0&0&0\\ x&x&x&x&1&0&0&0\\ x&x&x&x&x&1&0&0\\ x&x&x&x&x&x&1&0\end{array}\right)$

Add a diagonal of ones to the main diagonal in matrix ${B}$ to get matrix ${C}$ .

${C=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&1&0&0&0&0&0&0\\ x&1&1&0&0&0&0&0\\ x&x&1&1&0&0&0&0\\ x&x&x&1&1&0&0&0\\ x&x&x&x&1&1&0&0\\ x&x&x&x&x&1&1&0\\ x&x&x&x&x&x&1&1\end{array}\right)$

Change the sign of all the elements below the main diagonal in matrix ${C}$ so that you get matrix ${D}$ .

${D=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ -1&1&0&0&0&0&0&0\\ -x&-1&1&0&0&0&0&0\\ -x&-x&-1&1&0&0&0&0\\ -x&-x&-x&-1&1&0&0&0\\ -x&-x&-x&-x&-1&1&0&0\\ -x&-x&-x&-x&-x&-1&1&0\\ -x&-x&-x&-x&-x&-x&-1&1\end{array}\right)$

Calculate the matrix inverse of matrix ${D}$ to get matrix ${E}$

Iterate by replacing matrix ${A}$ with matrix ${E}$ .

Matrix ${E}$ should then converge to the Catalan numbers in all columns.

Keywords: Catalan numbers, INVERT transform, matrix inverse.

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The ordinary generating function for the Mobius function

Posted on March 8, 2011 by Mats Granvik

$\sum \limits_{n=1}^{\infty} \mu(n)x^n = x - \frac{x^{2}}{1-x} + \sum \limits_{a=2}^{\infty} \frac{x^{2a}}{1-x^{a}} - \sum \limits_{b=2}^{\infty} \sum \limits_{a=2}^{\infty} \frac{x^{2ab}}{1-x^{ab}} + \sum \limits_{c=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{a=2}^{\infty} \frac{x^{2abc}}{1-x^{abc}} - \sum \limits_{d=2}^{\infty} \sum \limits_{c=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{a=2}^{\infty} \frac{x^{2abcd}}{1-x^{abcd}} + ...$

This expression above is basically what I posted to the seqfan mailing list:
seqfan-2011-March-5

And here:
seqfan-2011-March-8

And here:
Wikipedia Möbius function

Link to Wikipedia revision history:
Wikipedia, Möbius function, difference between revisions 1

Wikipedia, Möbius function, difference between revisions 2

Using the six first terms (on the right hand side) I get 14 accurate decimals, of 14 possible in a spreadsheet, in the interval -0.5 to +0.5. The technique for inverting triangular matrices as a binomial series, which is the basis for this formula, was learned together with Gary W. Adamson.

This expression above simplifies to:

$\sum \limits_{n=1}^{\infty} \mu(n)x^n = x - \sum \limits_{a=2}^{\infty} x^{a} + \sum \limits_{b=2}^{\infty} \sum \limits_{a=2}^{\infty} x^{ab} - \sum \limits_{c=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{a=2}^{\infty} x^{abc} + \sum \limits_{d=2}^{\infty} \sum \limits_{c=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{a=2}^{\infty} x^{abcd} - ...$

mats.granvik(AT)abo.fi

Keywords: Mobius function, power series, binomial series.

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Motzkin numbers as a convergent of iterative matrix inversion

Posted on February 28, 2011 by Mats Granvik

I have here left out the indices of the elements for easier typing but hopefully this can still be understood.

${A=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ x&1&0&0&0&0&0&0\\ x&x&1&0&0&0&0&0\\ x&x&x&1&0&0&0&0\\ x&x&x&x&1&0&0&0\\ x&x&x&x&x&1&0&0\\ x&x&x&x&x&x&1&0\\ x&x&x&x&x&x&x&1\end{array}\right)$

Shift down the elements in matrix ${A}$ two steps to get matrix ${B}$ .

${B=}\left(\begin{array}{cccccccc}0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ x&1&0&0&0&0&0&0\\ x&x&1&0&0&0&0&0\\ x&x&x&1&0&0&0&0\\ x&x&x&x&1&0&0&0\\ x&x&x&x&x&1&0&0\end{array}\right)$

Add two diagonals of ones to the main diagonal (where row index = column index) and to the diagonal below (where row index = column index + 1) in matrix ${B}$ to get matrix ${C}$ .

${C=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&1&0&0&0&0&0&0\\ 1&1&1&0&0&0&0&0\\ x&1&1&1&0&0&0&0\\ x&x&1&1&1&0&0&0\\ x&x&x&1&1&1&0&0\\ x&x&x&x&1&1&1&0\\ x&x&x&x&x&1&1&1\end{array}\right)$

Change the sign of all the elements below the main diagonal in matrix ${C}$ so that you get matrix ${D}$ .

${D=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ -1&1&0&0&0&0&0&0\\ -1&-1&1&0&0&0&0&0\\ -x&-1&-1&1&0&0&0&0\\ -x&-x&-1&-1&1&0&0&0\\ -x&-x&-x&-1&-1&1&0&0\\ -x&-x&-x&-x&-1&-1&1&0\\ -x&-x&-x&-x&-x&-1&-1&1\end{array}\right)$

Calculate the matrix inverse of matrix ${D}$ to get matrix ${E}$

Iterate by replacing matrix ${A}$ with matrix ${E}$ .

Matrix ${E}$ should then converge to the Motzkin numbers in all columns.

Keywords: Matrix inversion, Motzkin numbers, iteration.

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Changed to new theme.

Posted on February 26, 2011 by Mats Granvik

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