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There is a Dirac pulse train following the scheme of the Shah function (or $\delta$-cumb function) with its Fourier series of the form:

$$\varsigma(t,T)=\sum_{n=-\infty}^{\infty}\delta (t-nT)=\frac{1}{T}\sum_{n=-\infty}^{\infty}\exp\left(\frac{i 2\pi n t}{T}\right)$$

here $n\in \Bbb Z$; $t$ time and $T(\in \Bbb N)$ the period between the pulses, which occur exactly at the levels of the integers.

The sum of $\varsigma(t,T)$ over $T$ in a given range $[1,T_{\max}]$ and $n$ in a range $[1,T]$ (instead) is accordingly:

$$S(t,T_{\max})=\sum_{T=1}^{T_{\max}}\varsigma(t,T)=\sum_{T=1}^{T_{\max}}\frac{1}{T}\sum_{n=1}^{T}\exp\left(\frac{i 2\pi n t}{T}\right)$$

And the zeros (at the level of some integers) of this function with respect to $t$, hence:

$$\sum_{T=1}^{T_{\max}}\frac{1}{T}\sum_{n=1}^{T}\exp\left(\frac{i 2\pi n t_0}{T}\right)=0$$

The Question is to identify the number of such zeros $\mathcal Z$ in a range of $t\in[T_{\max},T_{\max}^2]$ by applying Rice's formula, while assuming that the frequencies $1/T$ are rationally independent?

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    $\begingroup$ Why in the definition of $S$ the interior sum is $\sum_{n=1}^T$ and not $\sum_{n=-\infty}^\infty$? And which frequencies are rationally independent? The variable $T$ was assumed to be integer. $\endgroup$ – Andrew Jul 20 '13 at 19:11
  • $\begingroup$ The boundaries are indeed $[1,T]$ although generally Shah defined over infinities. $T$ is an integer, yes and its inverse is meant by frequencies. $\endgroup$ – al-Hwarizmi Jul 20 '13 at 19:55
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    $\begingroup$ Are $t,t_0$ integers or is $t$ (time) continuous? Could you explain the sentence: "And the zeros (at the level of some integers) of this function with respect to t, hence:" $\endgroup$ – Graham Hesketh Jul 24 '13 at 21:40
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    $\begingroup$ For $t,t_0,T,T_{max}\in \mathbb{N}$, $$\frac{1}{T}\sum_{n=1}^{T}\exp\left(\frac{i 2\pi n t}{T}\right)=\cases{1&$T|t$\cr 0 &otherwise\cr}$$ but then how can $$\sum_{T=1}^{T_{\max}}\frac{1}{T}\sum_{n=1}^{T}\exp\left(\frac{i 2\pi n t_0}{T}\right)=0.$$ If $(T=1)|t_0$? If the sum began at $T=2$ I'd be tempted to say that $\mathcal{Z}$ was the number of primes in the interval... $\endgroup$ – Graham Hesketh Jul 26 '13 at 14:17
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    $\begingroup$ One other thing I don't get, in your second equation you are equating a Dirac delta comb ($\infty$ at multiples of $T$) with what is effectively a Kronecker Delta comb (1 at multiples of $T$), how does that work? This was also the issue Andrew was raising I believe.. $\endgroup$ – Graham Hesketh Jul 26 '13 at 16:20
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This is not a complete answer in that it does not apply Rice's formula although it does derive the formula for the counting function $\mathcal{Z}(T_{max})$.

For $t,t_0,T,T_{max}\in \mathbb{N}$:

$$\frac{1}{T}\sum_{n=1}^{T}\exp\left(\frac{i 2\pi n t}{T}\right)=\cases{1&$T|t$\cr 0 &otherwise\cr}$$ so:

$$S(t,T_{\max})=\sum_{T=1}^{T_{\max}}\frac{1}{T}\sum_{n=1}^{T}\exp\left(\frac{i 2\pi n t}{T}\right)$$ counts the number of divisors of $t$ that are $\le T_{max}$ and $t_0$ is defined to be a zero if it satisfies: $$\sum_{T=1}^{T_{\max}}\frac{1}{T}\sum_{n=1}^{T}\exp\left(\frac{i 2\pi n t_0}{T}\right)=\sum_{T=1}^{T_{\max}}\cases{1&$T|t_0$\cr 0 &otherwise\cr}=0 \tag{1}$$ which holds iff $t_0$ is not divisible by any integer $1\le T\le T_{max}$. However $t_0$ is always divisible by $1$ so the sum in $(1)$ must always evaluate to something $\ge1$ for any $T_{max}$. In which case we are forced to say

there are no integers $t_0,\,T_{max}$ such that $(1)$ holds and thus: $$\mathcal{Z}(T_{max})=0, \,\forall T_{max}\in \mathbb{N}.\tag{2}$$

Alternatively, if we chose to exclude the trivial case in which $1$ divides all integers by beginning the outer sum in $(1)$ at $T=2$ we may find a less trivial outcome. Define $t_0$ with $2\le T_{max}\le t_0\le T_{max}^2$ to be a solution to the equation: $$\sum_{T=2}^{T_{\max}}\frac{1}{T}\sum_{n=1}^{T}\exp\left(\frac{i 2\pi n t_0}{T}\right)=0$$ then $\mathcal{Z}$ counts the number of integers $t_0 \in \left[T_{max},T_{max}^2\right]$ such that $T\nmid t_0 \forall\, 2\le T \le T_{max}$, in which case: $$\mathcal{Z}=\pi(T_{max}^2)-\pi(T_{max}) \tag{3}$$ where $\pi(x)$ is the prime counting function.

Proof

If: $$\exists t_0 \,\, \text{such that} \,\,\,T\nmid t_0 \,\forall\, 2\le T \le T_{max},$$ but $t_0$ is not prime, then: $$\exists k>T_{max}\,\, \text{such that} \,\,k|t_0,$$ and: $$\exists n=\dfrac{t_0}{k}>T_{max},n\in \mathbb{N},$$ which implies: $$t_0=nk>T_{max}^2,$$ from which we conclude that $t_0 \in \left[T_{max},T_{max}^2\right]$ is not a solution when $t_0$ is not prime. It is clear that if $t_0$ $\left(2\le T_{max}\le t_0\le T_{max}^2\right)$ is prime, it is not divisible by $2\le T \le T_{max}$ and is therefore a solution, unless $T_{max}$ is itself prime. From this we conclude that all $t_0$ are primes such that $T_{max}<t_0\le T_{max}^2$ and $(3)$ follows.

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  • $\begingroup$ Appreciate your insights, this is of course interesting but unfortunately not a response to my question. The question was not a mathematical reformulation of the sieve of Erathostenes that is resulting from here in case of starting $T=2$ and applying the roots of unity method with the number theoretical function (for counting solutions). This has been studied extensively elsewhere in literature. The problem to me is to solve a signal processing problem of superposition of pulse trains in a discrete (integer) acting system and whether Rice's formula provides a method of approximation. $\endgroup$ – al-Hwarizmi Jul 26 '13 at 18:52
  • $\begingroup$ I wonder, whether your answer would not have a good chance on the following question raised ten days ago. There is a bounty on: math.stackexchange.com/questions/447229/… $\endgroup$ – al-Hwarizmi Jul 28 '13 at 10:11
  • $\begingroup$ I am migrating this to dsp.SE. If you get an account there, you will get reputation there. $\endgroup$ – robjohn Jul 28 '13 at 16:47

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