Last week, I wrote in Primes
matrix about a
possibility to represent primes. The negative part of this was, that
there isn’t a nice way to generate \(X_{\left(i\right)}^{n\times n}\).
Today, I want to add a few lines, how you can approximate it.
Let’s look again at
$$
X_{\left(1\right)}^{n\times n} :=
\begin{pmatrix}
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{pmatrix}
= \left(x_{\left(1\right),kj}\right)_{k=1,\dots,n \ , j=1,\dots, n} \ \delta_{kj}
$$
$$
X_{\left(2\right)}^{n\times n} :=
\begin{pmatrix}
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{pmatrix}
= \left(x_{\left(2\right),kj}\right)_{k=1,\dots,n \ , j=1,\dots, n} \ \delta_{kj}
$$
Instead of the definition from my last post, now, we will define
\(x_{\left(i\right),kj}\) in the following way
$$
x_{\left(i\right),kj} := \lim_{m \rightarrow \infty} \cos^{2m}\left(2\pi\frac{k - x_{i}}{2x_{i} + 1}\right) \delta_{kj}
$$
and
$$
\delta_{kj} := \left\{
\begin{array}{l@{\quad \quad}l}
1 & \mathrm{if} \ k = j \\
0 & \mathrm{else}
\end{array}
\right.
$$
Explanation: Like in our old definition, we always get an ‘1’ at all
places which presents a divisible number for a given \(x_{i}\). Now, at
the other positions, we get an value \(v\) between \(0 < |v| < 1\), instead
of ‘0’ in our old definition. To get our ’0’s, like before, we use
\(\lim_{m \rightarrow \infty}\). The factor two ensures a positive
approxiation value.
With this, we also get
$$
\overline{x_{\left(i\right),kj}} := \left(1 - x_{\left(i\right),kj}\right)\delta_{kj} \\
= \lim_{m \rightarrow \infty} \left(1 - \cos^{2m}\left(2\pi\frac{k- x_{i}}{2x_{i} + 1}\right)\right)\delta_{kj}
$$
An alternative way also to write this, is given by
$$
\overline{x_{\left(i\right),kj}} := \lim_{m \rightarrow \infty} \sin^{\frac{2}{m}}\left(2\pi\frac{k - x_{i}}{2x_{i} + 1}\right) \delta_{kj}
$$
Explanation: We have a similiar situation like for our
\(x_{\left(i\right),kj}\), but in this case now, we have to use \(2/m\) to
get an approximation for ‘1’.
We see that we can use \(\cos\) and also \(\sin\) depending on which values
we are interested in. Of course, don’t forget the different cases for
\(\lim\) with \(2m\) respectively \(2/m\).
So, we can put this together to
$$
\eta_{\left(i\right),kj} := \lim_{m \rightarrow \infty} \exp\left(I 2\pi\frac{k-x_{i}}{2x_{i} + 1} \epsilon\left(m\right)\right) \delta_{kj}
$$
\(I\) is the Imaginary unit, with \(I^{2} = -1\) and
$$
\epsilon\left(m\right) := \left\{
\begin{array}{l@{\quad \quad}l}
2m & \mathrm{for \ real \ part \ Re\left(\eta\right)} \ \\
\frac{2}{m} & \mathrm{for \ imaginary \ part \ Im\left(\eta\right)}
\end{array}
\right.
$$
Ok. What do we need, next? We need
$$
\exp\left(Iz_{1}\right) \cdot \exp\left(Iz_{2}\right) = \exp\left(I\left(z_{1} + z_{2}\right)\right) = \cos\left(z_{1} + z_{2}\right) + I\sin\left(z_{1} + z_{2}\right)
$$
so let’s look at
$$
z_{1} + z_{2} = 2\pi \epsilon\left(m\right)\left(\frac{k - x_{1}}{2x_{1} + 1} + \frac{k - x_{2}}{2x_{2} + 1}\right) \\
= 2\pi \epsilon\left(m\right) \frac{\left(k - x_{1}\right)\left(2x_{2} + 1\right) + \left(k - x_{2}\right)\left(2x_{1} + 1\right)}{\left(2x_{1} + 1\right)\left(2x_{2} + 1\right)}
$$
In https://github.com/Samdney/primescalc we already
discussed a similiar situation, which showed us that we don’t have any
problems if \(2x_{1} + 1\) and \(2x_{2} + 1\) are primes. For more
information about this, please read the mentioned paper.
So we can do our final step
$$
x_{\left(a,\dots,b\right),kj} = \lim_{m \rightarrow \infty} \left(\prod_{i=a}^{b} \exp\left(I 2\pi\frac{k-x_{i}}{2x_{i} + 1} \epsilon\left(m\right)\right)\right) \delta_{kj} \\
= \lim_{m \rightarrow \infty} \exp\left(\sum_{i = a}^{b} I2\pi\frac{k - x_{i}}{2x_{i} + 1}\epsilon\left(m\right)\right) \delta_{kj}
$$
Now we look at the special case of
\(x_{i} = a, a+1, a+2, \dots, b-2, b-1,b\) for the given sum
$$
\sum_{i = a}^{b} I2\pi\frac{k - x_{i}}{2x_{i} + 1}\epsilon\left(m\right) = \\
= I2\pi\frac{1}{4}\left(2k\psi^{\left(0\right)}\left(b + \frac{3}{2}\right) + \psi^{\left(0\right)}\left(b + \frac{3}{2}\right) - 2k\psi^{\left(0\right)}\left(\left(a - 1\right) + \frac{3}{2}\right) - \psi^{\left(0\right)}\left(\left(a-1\right) + \frac{3}{2}\right)\right)\epsilon\left(m\right)
$$
since
$$
\sum_{x=1}^{n} \frac{k - x}{2x + 1}\\
= \frac{1}{4}\left(2k\psi^{\left(0\right)} \left(n + \frac{3}{2}\right) -
2k\psi^{\left(0\right)}\left(\frac{3}{2}\right) - 2k + \psi^{\left(0\right)}\left(n + \frac{3}{2}\right) - \psi^{\left(0\right)}\left(\frac{3}{2}\right)\right)
$$
\(\psi^{\left(n\right)}\left(x\right)\) is the n-the derivate of the
digamma function.
Finally, a small comment about permitted ranges.
Since our functions work with fix period \((k - x_{i})/(2x_{i} + 1)\), we
have to ignore the range \([1,x_{i}]\) for each \(x_{i}\)-equation, else we
will also receive invalid results.