The Mobius function Blog

Useless relationship between the Riemann zeta and prime zeta function.

Posted on May 12, 2012 by Mats Granvik

Mathematica 8:

(*program start*) Clear[a, b, epsilon, n, k]; a = {0, 1, 1, 2, 1, 0, 1, 3, 2, 0, 1, 0}; b = {0, 0, 0, 2, 0, 1, 0, 3, 2, 1, 0, 0}; epsilon = 1/10000; MatrixForm[ Table[Table[ Sqrt[(Im[Log[n - k - 1]] - Im[Log[n - k]])/Pi*a[[n]]*b[[k]]], {n, 1, 12}], {k, 1, 12}]] (*program end*)

Posted in Uncategorized | Comments Off

Triangle of numbers A131689 in the oeis.

Posted on May 8, 2012 by Mats Granvik

Mathematica 8:

Clear[nn, X, A] nn = 12; A = Range[nn + 1]*0; X = Transpose[Table[Table[Binomial[n, k], {n, 0, nn}], {k, 0, nn}]] - IdentityMatrix[nn + 1]; Do[A[[i + 1]] = MatrixPower[X, i][[All, 1]], {i, 0, nn}] MatrixForm[Transpose[A]] Total[A]

Matrix A is triangle A131689 in the oeis:
A131689

Which outputs:
{1, 1, 3, 13, 75, 541, 4683, 47293, 545835, 7087261, 102247563, 1622632573, 28091567595}

Posted in Uncategorized | Comments Off

Matrix formulation of the Fourier transform of a sequence – example

Posted on May 7, 2012 by Mats Granvik

Taken from an email conversation with Gary W. Adamson on February 14 2012.
”
Just one thing, (the number theoretic) Schramm triangle “B” or “chaotic set” (?) is:

cos(2*Pi*1/1),
cos(2*Pi*1/2), cos(2*Pi*2/2),
cos(2*Pi*1/3), cos(2*Pi*2/3), cos(2*Pi*3/3),
cos(2*Pi*1/4), cos(2*Pi*2/4), cos(2*Pi*3/4), cos(2*Pi*4/4),
cos(2*Pi*1/5), cos(2*Pi*2/5), cos(2*Pi*3/5), cos(2*Pi*4/5), cos(2*Pi*5/5),
cos(2*Pi*1/6), cos(2*Pi*2/6), cos(2*Pi*3/6), cos(2*Pi*4/6), cos(2*Pi*5/6), cos(2*Pi*6/6),
cos(2*Pi*1/7), cos(2*Pi*2/7), cos(2*Pi*3/7), cos(2*Pi*4/7), cos(2*Pi*5/7), cos(2*Pi*6/7), cos(2*Pi*7/7),
=B

While the (numerical method used in signal processing) Fourier array “F” is a square array:

cos(-2*Pi*(1-1)*(1-1)/4), cos(-2*Pi*(2-1)*(1-1)/4), cos(-2*Pi*(3-1)*(1-1)/4), cos(-2*Pi*(4-1)*(1-1)/4)
cos(-2*Pi*(1-1)*(2-1)/4), cos(-2*Pi*(2-1)*(2-1)/4), cos(-2*Pi*(3-1)*(2-1)/4), cos(-2*Pi*(4-1)*(2-1)/4)
cos(-2*Pi*(1-1)*(3-1)/4), cos(-2*Pi*(2-1)*(3-1)/4), cos(-2*Pi*(3-1)*(3-1)/4), cos(-2*Pi*(4-1)*(3-1)/4)
cos(-2*Pi*(1-1)*(4-1)/4), cos(-2*Pi*(2-1)*(4-1)/4), cos(-2*Pi*(3-1)*(4-1)/4), cos(-2*Pi*(4-1)*(4-1)/4)

which has the general form:

F(n,k) = cos(-2*Pi*(k-1)*(n-1)/N)

where capital “N” is the dimension of the 4*4 matrix or the length of the sequence to be

Fourier transformed, and n=1,2,3… and k=1,2,3…

(The definition in wikipedia “Discrete Fourier Transform”

is exactly the same but with shifted indexes for “n” and “k”.

Also, I would have written a 7*7 array as in the example with the Schramm triangle

but it does not fit on the screen in a mail, so I use a 4*4 array as example instead.)

So to take the Fourier transform of a sequence

x=2,3,1,5

just multiply the matrix “F” with the vector “x”

2*cos(-2*Pi*(1-1)*(1-1)/4), 3*cos(-2*Pi*(2-1)*(1-1)/4), 1*cos(-2*Pi*(3-1)*(1-1)/4), 5*cos(-2*Pi*(4-1)*(1-1)/4)
2*cos(-2*Pi*(1-1)*(2-1)/4), 3*cos(-2*Pi*(2-1)*(2-1)/4), 1*cos(-2*Pi*(3-1)*(2-1)/4), 5*cos(-2*Pi*(4-1)*(2-1)/4)
2*cos(-2*Pi*(1-1)*(3-1)/4), 3*cos(-2*Pi*(2-1)*(3-1)/4), 1*cos(-2*Pi*(3-1)*(3-1)/4), 5*cos(-2*Pi*(4-1)*(3-1)/4)
2*cos(-2*Pi*(1-1)*(4-1)/4), 3*cos(-2*Pi*(2-1)*(4-1)/4), 1*cos(-2*Pi*(3-1)*(4-1)/4), 5*cos(-2*Pi*(4-1)*(4-1)/4)

=F . x

which when multiplied is equal to:

2, 3, 1, 5
2, 0, -1, 0
2, -3, 1, -5
2, 0, -1, 0

Then take row sums and you get a new vector capital “X”:

X=11, 1, -5, 1

We then say that upper case “X” is the Fourier transform of lower case “x”

———————————————————————–

This above is called the matrix formulation of the Discrete Fourier Transform.

The default Fourier (cosine) transform in Mathematica would be:

Re[Fourier[{2, 3, 1, 5}]

Which can be entered into Wolfram Alpha.

But the Mathematica code for the matrix formulation of the
Fourier (cosine) transform I tried to explain in the previous mail, is:

Re[Fourier[{2, 3, 1, 5}, FourierParameters -> {1, -1}]]

where
x = 2, 3, 1, 5

But Wolfram Alpha does not accept “FourierParameters” command
so to get the same result:
X = 11, 1, -5, 1

you have to use the Wolfram Alpha line:

Re[Fourier[{2, 3, 1, 5}]]*Total[{2, 3, 1, 5}]/Re[Fourier[{2, 3, 1, 5}]][[1]]

which gives the answer 11, 1, -5, 1

“Re” stands of course for real part, while “Im” would gives imaginary part.
“

Posted in Uncategorized | Comments Off

Reciprocal of a number raised to s-1 as a sum of reciprocals raised to s

Posted on May 1, 2012 by Mats Granvik

Mathematica 8

Clear[s, n, k] s = 2; Table[Sum[n/(k + n - 1)^s - n/(k + n)^s, {k, 1, Infinity}], {n, 1, 12}]
{1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10, 1/11, 1/12}

$\frac{1}{n^{s-1}} = \sum\limits_{k=1}^{\infty} \frac{n}{(k+n-1)^{s}} - \frac{n}{(k+n)^{s}}$

Posted in Uncategorized | Comments Off

Mertens function Lambert series

Posted on April 5, 2012 by Mats Granvik

Mathematica 8

Clear[k, x, n, nn, h] nn = 90; g1 = Plot[ N[Total[Table[ Accumulate[Table[MoebiusMu[k], {k, 1, nn}]][[n]]*(x^n/(1 - x^n) - x^(n + 1)/(1 - x^(n + 1))), {n, 1, nn}]]] , {x, -5, 5}, PlotRange -> {-5, 5}, AspectRatio -> 1] g2 = Plot[-x^(-1), {x, -5, 5}, PlotRange -> {-5, 5}, AspectRatio -> 1] Show[g1, g2]

M(n)=1,0,-1,-1,-2,-1,-2,-2,-2,-1,-2,-2… = oeis sequence A002321.
A002321

$y = M(n)(\frac{x^{n}}{(1 - x^{n})} - \frac{x^{(n + 1)}}{(1 - x^{(n + 1)})})$

which evaluates to:

$y = x$ if $-1 \le x \le 1$

$y = -\frac{1}{x}$ if $x \textless -1$ or $x \textgreater 1$ .

Singularities at x = -1 and x = 1

Posted in Uncategorized | Comments Off

Expansion of the Dirichlet series for the mobius function

Posted on April 4, 2012 by Mats Granvik

Summing to infinity:

$\sum \limits_{n=1}^{\infty} \frac{\mu(n)}{n^{s}} = 1 - \sum \limits_{a=2}^{\infty} \frac{1}{a^{s}} + \sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \frac{1}{(a \cdot b)^{s}} - \sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} \frac{1}{(a \cdot b \cdot c)^{s}} + \sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} \sum \limits_{d=2}^{\infty} \frac{1}{(a \cdot b \cdot c \cdot d)^{s}} -...$

Mertens function, partial sums of the Mobius function:

$\sum \limits_{k=1}^{n} \mu(k) = 1 - \sum \limits_{a=2}^{a \leq n} 1 + \sum \limits_{a=2}^{a \leq \frac{n}{b}} \sum \limits_{b=2}^{b \leq \frac{n}{a}} 1 - \sum \limits_{a=2}^{a \leq \frac{n}{b \cdot c}} \sum \limits_{b=2}^{b \leq \frac{n}{a \cdot c}} \sum \limits_{c=2}^{c \leq \frac{n}{a \cdot b}} 1 + \sum \limits_{a=2}^{a \leq \frac{n}{b \cdot c \cdot d}} \sum \limits_{b=2}^{b \leq \frac{n}{a \cdot c \cdot d}} \sum \limits_{c=2}^{c \leq \frac{n}{a \cdot b \cdot d}} \sum \limits_{d=2}^{d \leq \frac{n}{a \cdot b \cdot c}} 1 -...$

For the first partial sum:
$a \leq n$

The partial double sum:
$a \cdot b \leq n$

The partial triple sum:
$a \cdot b \cdot c \leq n$

The partial quadruple sum:
$a \cdot b \cdot c \cdot d \leq n$

This notation is not entirely correct since the minimum value for indices in for example the quadruple sum is 2*2*2*2=16 which is too large if n is less than that number

Posted in Uncategorized | Comments Off

Multiples of Van der Pauw constant Pi divided by logarithm of two as solutions to equation

Posted on April 3, 2012 by Mats Granvik

Mathematica 8

Re[s /. Table[ FindRoot[ N[Sum[1/2^s, {s, 2, Infinity}], 20] == 1 + 1/2^s, {s, a*I}], {a, 1, 60*2*Pi/Log[2], 2*Pi/Log[2]}]] Im[s /. Table[ FindRoot[ N[Sum[1/2^s, {s, 2, Infinity}], 20] == 1 + 1/2^s, {s, a*I}], {a, 1, 60*2*Pi/Log[2], 2*Pi/Log[2]}]]/(Pi/Log[2])

Real part:
{1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., \
1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., \
1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., \
1., 1., 1., 1., 1., 1., 1., 1., 1.}

Imaginary part divided by Van der Pauw constant Pi/Log[2] = 4.5323601418271938…

{1., 3., 5., 7., 9., 11., 13., 15., 17., 19., 21., 23., 25., 27., \
29., 31., 33., 35., 37., 39., 41., 43., 45., 47., 49., 51., 53., 55., \
57., 59., 61., 63., 65., 67., 69., 71., 73., 75., 77., 79., 81., 83., \
85., 87., 89., 91., 93., 95., 97., 99., 101., 103., 105., 107., 109., \
111., 113., 115., 117., 119.}

Posted in Uncategorized | Comments Off

Natural numbers, harmonic numbers, logarithms and partial sums of square roots with BBP type formulas

Posted on April 1, 2012 by Mats Granvik

With this Mathematica 8 program:

(*program start*) Clear[s, i, n, j] s = 1; i = 1; j = 0; Sum[1/(n + 0)^s - 1/(n + 1)^s, {n, 1, Infinity, i + j}] Sum[1/(n + 0)^s + 1/(n + 1)^s - 2/(n + 2)^s, {n, 1, Infinity, i + 2*j}] Sum[1/(n + 0)^s + 1/(n + 1)^s + 1/(n + 2)^s - 3/(n + 3)^s, {n, 1, Infinity, i + 3*j}] Sum[1/(n + 0)^s + 1/(n + 1)^s + 1/(n + 2)^s + 1/(n + 3)^s - 4/(n + 4)^s, {n, 1, Infinity, i + 4*j}] Sum[1/(n + 0)^s + 1/(n + 1)^s + 1/(n + 2)^s + 1/(n + 3)^s + 1/(n + 4)^s - 5/(n + 5)^s, {n, 1, Infinity, i + 5*j}] Sum[1/(n + 0)^s + 1/(n + 1)^s + 1/(n + 2)^s + 1/(n + 3)^s + 1/(n + 4)^s + 1/(n + 5)^s - 6/(n + 6)^s, {n, 1, Infinity, i + 6*j}] Sum[1/(n + 0)^s + 1/(n + 1)^s + 1/(n + 2)^s + 1/(n + 3)^s + 1/(n + 4)^s + 1/(n + 5)^s + 1/(n + 6)^s - 7/(n + 7)^s, {n, 1, Infinity, i + 7*j}] (*program end*)

I noticed some probably known facts:

With parameters:
s = 1;
i = 1;
j = 0;
-> 1,2,3,4,5,6,7…

With parameters:
s = 2;
i = 1;
j = 0;
-> 1, 3/2, 11/6, 25/12, 137/60, 49/20, 363/140 …
which starts as the harmonic numbers.

With parameters:
s = 1;
i = 1;
j = 1;
-> Log[2], Log[3], Log[4], Log[5], Log[6], Log[7], Log[8] …
(Jaume Oliver Lafont has commented on this)

With parameters:
s = 1/2;
i = 1;
j = 0;
and after wrapping the answer with N[answer,20] I get these
inaccurate numbers accompanied with error messages for every number:

2.4142135623730931752 = Sqrt[1] + Sqrt[2],
4.1462643699419685951 = Sqrt[1] + Sqrt[2] + Sqrt[3],
6.1462643699419710784 = Sqrt[1] + Sqrt[2] + Sqrt[3] + Sqrt[4],
8.3823323474417526706 = Sqrt[1] + Sqrt[2] + Sqrt[3] + Sqrt[4] + Sqrt[5],
10.831822090224927022 = Sqrt[1] + Sqrt[2] + Sqrt[3] + Sqrt[4] + Sqrt[5] + Sqrt[6],
13.477573401289513240 = Sqrt[1] + Sqrt[2] + Sqrt[3] + Sqrt[4] + Sqrt[5] + Sqrt[6] + Sqrt[7], …

Posted in Uncategorized | Comments Off

Partition numbers with Gary W. Adamsons aerate convolve algorithm

Posted on March 27, 2012 by Mats Granvik

Mathematica 8

Clear[t, A, B] nn = 16; A = Table[Table[If[n >= k, 1, 0], {k, 1, nn}], {n, 1, nn}]; MatrixForm[A]; B = A; Do[ Clear[t, A]; t[n_, k_] := t[n, k] = If[k == 1, If[Mod[n, j] == 1, 1, 0], If[n >= k, t[n - 1, k - 1], 0]]; A = Table[Table[t[n, k], {k, 1, nn}], {n, 1, nn}]; B = B.A; (*Print[MatrixForm[B]]*) , {j, 2, nn}] B[[All, 1]]

{1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176}

Posted in Uncategorized | Comments Off

Dirichlet series converging to zero or logarithm of one

Posted on March 19, 2012 by Mats Granvik

Mathematica 8:

Sum[1/(i + 0) - 1/(i + 1) - 2/(i + 2) - 1/(i + 3) + 1/(i + 4) + 2/(i + 5), {i, 1, Infinity, 6}]

And a variant:

ListLinePlot[ N[Table[Sum[ 1/(i + 0) - 1/(i + 1) - 2/(i + 2) - 1/(i + 3) + 1/(i + 4) + 2/(i + 5), {i, 1, Infinity, j}], {j, 1, 24}]]]

Posted in Uncategorized | Comments Off

	Are the amplitudes o… on Fourier transform of exponenti…
	blogging on Logarithmic square root w…
	afmsjksua@gmail.com on First differences of most nega…
	Nick on The inverse of a triangular ma…
	www.tuinbank-teak.nl on The inverse of a triangular ma…

Recent Comments