0
$\begingroup$

I've defined three general methods for derivation of formulas for prime counting functions where each prime counting function is represented by an infinite series of Fourier series, and illustrated application of the first method to six different prime counting functions on the following website.

Illustration of Fourier Series for Prime Counting Functions

In the distributional framework for prime counting functions the first-order derivatives are represented by Dirac delta distributions. In the case of $\psi'(x)$, there is a Dirac delta distribution at each prime-power value of $x$ with weight $Log(p)$ where $x$ is of the form $x=p^n$. Note the real portion of the Fourier transform of a Dirac delta function is a Cosine term.

(1) $\quad \operatorname{FourierTransform}(\delta(x-a),x,y)=e^{-2\,i\,\pi\,a\,y}=\cos(2\,\pi\,a\,y)-i\,\sin(2\,\pi\,a\,y)\,,\quad a\in\mathbb{R}$

In the Fourier series framework for prime counting functions, the first-order derivatives are represented by infinite series of Fourier series which converge to the Dirac delta distributions in the distributional framework. These Fourier series consist of Cosine terms, and note the Fourier transform of a Cosine function is a pair of Dirac delta distributions.

(2) $\quad \operatorname{FourierTransform}(\cos(2\,\pi\,b\,x),x,y)=\dfrac{\delta(b-y)}{2}+\dfrac{\delta(b+y)}{2}$

The Fourier transforms in (1) and (2) above both assume the Fourier parameters $\{0,\,-2\,\pi\}$.

Question 1: What is the relationship between the Cosine terms in the distributional framework and the Dirac delta distributions in the Fourier series framework? For example, do the Cosine terms in the distributional framework converge to the Dirac delta distributions in the Fourier series framework, analogous to the way the Cosine terms in the Fourier series framework converge to the Dirac delta distributions in the distributional framework?

Question 2: If the analogous relationship in question 2 is valid, is this convergence in any way sensitive to the Fourier parameters used for the two Fourier transforms? For example, will the convergence only apply when using the same set or a specific set of Fourier parameters for both Fourier transforms? Or will the convergence perhaps be faster if the same Fourier parameters are used for both Fourier transforms versus using different Fourier parameters for the two Fourier transforms?

1/1/2018 Update: I believe the discrepancies between the Fourier transforms of the distributional and Fourier series representations of prime-counting functions result from the idealization of the Fourier transform of $sin$ and $cos$ functions as Dirac delta ($\delta$) functions. Please see the answer I posted below which I believe provides a fair amount of insight into the theory and value of Fourier series representations of non-periodic functions. Happy New Year!

Direct link to answer I posted below

3/26/2018 Update:

The following link defines and illustrates a general method for derivation of a Fourier series representation for $f(x)=\sum\limits_{n=1}^x a(n)$ related to the Dirichlet series $F(s)=\sum\limits_{n=1}^\infty\frac{a(n)}{n^s}$.

Derivation of Fourier Series Representation for $f(x)=\sum\limits_{n=1}^x a(n)$

$\endgroup$
  • 2
    $\begingroup$ Please don't vandalize your question, Steve. $\endgroup$ – Gerry Myerson Oct 30 '16 at 5:44
  • $\begingroup$ You cannot delete your question because it has received two (upvoted) answers. Other people have spent effort on the question, and their effort has been deemed useful by users. It would not be fair to them to let the question poser delete the question single-handedly. As for the chat, chat messages can - except by moderators - only be deleted for a short time (a few minutes, I don't remember the exact limit). I've deleted the link to the chat room, that's about as good as deleting the whole conversation. $\endgroup$ – Daniel Fischer Oct 30 '16 at 12:30
  • $\begingroup$ @DanielFischer I didn't create my comments in chat, they were moved to chat, but thanks for at least deleting the link to chat. The answers below were both in response to my original pair of questions, and don't make sense in the context of the additional questions which I've since added, so I'll edit my question back to my original pair of questions. $\endgroup$ – Steven Clark Oct 30 '16 at 15:59
3
$\begingroup$

Your question means nothing. Work on $$\text{I}\Pi(x) = \sum_{k=-\infty}^\infty \delta(x-k) = 1+2 \sum_{n=1}^\infty \cos(2\pi n x)$$ where the Fourier series on the right converges only is in the sense of distributions, that is, for every $\varphi \in C_c^\infty$ with (*) compact support $[a,b]$ : $$\langle \text{I}\Pi, \varphi \rangle = \int_{-\infty}^\infty \text{I}\Pi(x) \varphi(x) \, dx = \sum_{k \in \mathbb{Z} \cap [a,b]} \varphi(k)$$ $$=\lim_{N \to \infty} \langle 1+2 \sum_{n=1}^N \cos(2\pi n x), \varphi \rangle = \lim_{N \to \infty} \int_{-\infty}^\infty (1+2 \sum_{n=1}^N \cos(2\pi n x)) \varphi(x) \, dx$$

So what I mean is : there is no Fourier series for the Dirac delta, there is only a Fourier series for the Dirac comb $\text{I}\Pi(x)$.

And read a course on : the Fourier series and the Fourier transform, on the distributions, on some complex analysis and the Laplace/Mellin transform.


(*) Since $\text{I}\Pi(x)$ is tempered distribution of order $1$, you can extend $\langle \text{I}\Pi,\varphi \rangle$ to any $\varphi(x)$ continuous (at $x \in\mathbb{Z}$) and with compact support, or decreasing fast enough at $x \to \infty$.

For example, it is perfectly true that $\langle \text{I}\Pi(x) ,x^{-s}\Lambda(\lfloor x+1/2 \rfloor) \rangle = \sum_{n=1}^\infty n^{-s}\Lambda(n) = \frac{-\zeta'(s)}{\zeta(s)}$ for $Re(s) > 1$, but it doesn't mean that $$\frac{-\zeta'(s)}{\zeta(s)} = \lim_{N\to \infty} \int_{1/2}^\infty x^{-s} \Lambda(\lfloor x+1/2 \rfloor)(1+2\sum_{n=1}^N \cos(2\pi n x)) \, dx$$ and this is all the point of the theory of distributions (and the theory of the Riemann zeta function) to study those kind of things.

$\endgroup$
  • $\begingroup$ Arguing that my formulas for $\psi[x]$, $\psi'[x]$, $UnitStep[x-1]$, and $UnitStep'[x-1]=\delta[x-1]$ can not converge because a single Fourier series does not converge is like arguing that von Mangoldt's formula can not converge because a single zeta zero contribution does not converge. $\endgroup$ – Steven Clark Jan 31 '17 at 4:12
  • $\begingroup$ Would you also argue that von Mangoldt's formula for $\psi[x]$ does not converge to a set of distributions at prime powers, and it's first-order derivative does not converge (up to a very simple term) to a set of Dirac delta functions at prime powers? The corresponding convergences of my formulas for $\psi[x]$ and it's first-order derivative are just as valid, and likewise the convergence of my formula for $UnitStep[x-1]$ to a single distribution at $x=1$ and the convergence of it's first-order derivative to the single Dirac delta function $\delta(x-1)$ are just as valid. $\endgroup$ – Steven Clark Jan 31 '17 at 15:46
  • $\begingroup$ @SteveC. When I say "read some courses" I mean proving many theorems. You can't understand what is in the books about $\zeta(s)$ without proving first some theorems : about the Fourier series, the Fourier/Laplace/Mellin transform, the holomorphic functions. Did you try reading a course on the (convergence of) Fourier series ? There are many difficult problems in this field, and it teaches you a fundamental fact in analysis : many obvious formulas are completely false, because it doesn't converge as we expected. $\endgroup$ – reuns Feb 1 '17 at 0:05
  • $\begingroup$ If you accept that the Fourier series for the Dirac comb converges to the Dirac comb, then it is but a small additional step to accept that my formulas for the prime counting functions converge to the prime counting functions. $\endgroup$ – Steven Clark Feb 12 '17 at 19:35
  • $\begingroup$ Riemann's formula for $J[x]$ and von Mangoldt's formula for $\psi[x]$ are conditionally convergent, and likewise my formulas for prime counting functions are conditionally convergent. The conditional convergence of my formulas seems fairly elementary to me, and it seems very unlikely to me that I'm the first one ever to investigate the conditional convergence requirements for sums of Fourier series. Was this topic not covered in any of the books you've read or courses you've taken in Fourier analysis? $\endgroup$ – Steven Clark Feb 12 '17 at 19:35
1
$\begingroup$

Indeed, this is not an answer, except to some degree to the meta-question aspects: apart from the literal difficulties in terminology, etc, as others have raised, the apparent question implicitly assumes a number of things that are not entirely wrong, but really are partly wrong. Of course, this degrades the sense of the sequel.

In particular, it is unwise to apparently distinguish "conventional" X from the supposedly novel version of X, when, in fact, the "conventional" one is factually misrepresented, and at the same time the "novel" thing is standard, too. It'd be a bit analogous to saying that, although the sun has not shone in the past, new technology has made the sun shine ... and will clean your carpets, too. I realize this is parody, but it is not entirely different from the unfortunate dislocated sense of the "question". Perhaps understandable in some ways, but not good to be so aggressive on a basis of minimal information.

$\endgroup$
  • $\begingroup$ Continuing some of @user1952009's indications... your questions can perhaps be given mathematical sense in various possible ways, but a fundamental obstacle is determining exactly what you mean by what you say. E.g., in your comment here, what does "evaluation parameters can be chosen..." mean? I can't guess it. Also, probably you didn't mean that $\delta_{1}$ is a "sum of cosine terms", if the latter create an even (generalized) function, since $\delta_1$ is not even. But, sure, it is possible to take Fourier transforms of tempered distributions... [cont'd] $\endgroup$ – paul garrett Oct 28 '16 at 22:02
  • $\begingroup$ [cont'd] ... But, indeed, non-periodic functions simply do not have "Fourier series", since they in no way descend to the circle. But maybe that's not what you really intend... By the way, a certain version of what you may be getting after would be Guinand's version of (von Mangoldt's version of) Riemann's so-called explicit formula. (Also considered by A. Weil a few years after Guinand.) I think that is the number-theoretic content of this situation. Still, yes, it is true that there is some motivation to look at "modern analysis" here... and elsewhere... which is not well documented. $\endgroup$ – paul garrett Oct 28 '16 at 22:05
  • $\begingroup$ @SteveC., a problem that I have is the "derived form a sieve which recovers..." An infinite sum of Fourier series is not necessarily a problem. Sawtooth functions are fine. But this is just delaying "paying the piper", to my mind. E.g., convergence in some topology? $\endgroup$ – paul garrett Oct 28 '16 at 23:44
1
$\begingroup$

I realize this answer is rather lengthy and contains a considerable number of formulas, but I think a significant amount of background information is necessary to understand my answer below as the topic of Fourier series representations of prime-counting functions and functions such as $\theta(x-1)$, $\delta(x-1)$, and $\delta'(x-1)$ seems to be unfamiliar territory for most everyone but myself. Also, I think the Fourier series representations of $\theta(x-1)$, $\delta(x-1)$, and $\delta'(x-1)$ deserve some elaboration for several reasons. First, these Fourier series representations can be used with various convolution integrals to derive new formulas for a variety of functions providing new insights into functions and their relationships. Second, these Fourier series representations are closely related to the Riemann zeta function $\zeta(s)$ which is illustrated by their Mellin transforms and various other derived formulas. Finally, the Fourier series representations of $\theta(x-1)$, $\delta(x-1)$, and $\delta'(x-1)$ can be generalized to representations of $\theta(x-a)$, $\delta(x-a)$, and $\delta'(x-a)$ which can in turn be used to derive Fourier series representations of prime-counting functions and their first and second order-derivatives.

Quite some time ago I noticed what appeared to be discrepancies between the Fourier transforms of the distributional and Fourier series representations of prime-counting functions which was the motivation for this question. I was initially trying to analyze these discrepancies in the context of the Fourier series representations of the second Chebyshev function $\psi(x)$ and it's first and second-order derivatives, but I decided to shift my focus to the considerably simpler context of the Fourier series representations of $U(x)$ and it's first and second-order derivatives defined below where $\theta(x)$ is the Heaviside step function (also referred to as the Unit step function) and $\delta(x)$ is the Dirac delta function.

(1) $\quad U(x)=-1+\theta(x+1)+\theta(x-1)$

(2) $\quad U'(x)=\delta(x+1)+\delta(x-1)$

(3) $\quad U''(x)=\delta'(x+1)+\delta'(x-1)$

The Fourier series representation of $U(x)$ is derived as follows.

(4) $\quad U(x)=\sum\limits_{n=1}^\infty\mu(n)\,Floor[\frac{x}{n}]$

(5) $\quad U(x)=\sum\limits_{n=1}^\infty\mu(n)\left(\frac{x}{n}-SawtoothWave(\frac{x}{n})\right)$

(6) $\quad U(x)=\sum\limits_{n=1}^\infty\mu(n)\,\left(\frac{x}{n}-\left(\frac{1}{2}-\frac{1}{\pi}\sum\limits_{k=1}^\infty\frac{\sin\left(\frac{2\,\pi\,k\,x}{n}\right)}{k}\right)\right)$

When evaluated at finite limits, the Fourier series representation defined in formula (6) above is conditionally convergent and must be evaluated as illustrated in formula (7) below. In formula (7) and all formulas derived from it below, all $SawtoothWave$ functions under evaluation must be evaluated to the same frequency versus the same harmonic which is one of several conditional convergence requirements. The parameter $f$ controls the evaluation frequency and is assumed to be a positive integer.

(7) $\quad U(x)=\sum\limits_{n=1}^N\mu(n)\,\left(\frac{x}{n}-\left(\frac{1}{2}-\frac{1}{\pi}\sum\limits_{k=1}^{f\,n}\frac{\sin\left(\frac{2\,\pi\,k\,x}{n}\right)}{k}\right)\right)$

Formula (7) above and all functions derived from it below converge under the conditions specified in (8) below where $\mathcal{M}(N)$ is the Mertens function. When $\mathcal{M}(N)$ evaluates to zero, $U(0)$ evaluates to zero since $U(0)=-\frac{1}{2}\sum\limits_{n=1}^N\mu(n)=-\frac{1}{2}\mathcal{M}(N)$ which is the net offset of all $SawtoothWave$ functions under evaluation.

(8) $\quad\mathcal{M}(N)=0\,\land\,N\to\infty\,\land\,f\to\infty$

Assuming the conditions stated in (8) above the $\frac{1}{2}$ term in formula (7) above can be ignored and formula (7) above can be simplified to (9) below.

(9) $\quad U(x)=\sum\limits_{n=1}^N\mu(n)\,\left(\frac{x}{n}+\frac{1}{\pi}\sum\limits_{k=1}^{f\,n}\frac{\sin\left(\frac{2\,\pi\,k\,x}{n}\right)}{k}\right)$

The Fourier series representation of $U'(x)$ is defined in (10) below.

(10) $\quad U'(x)=\sum\limits_{n=1}^N\frac{\mu(n)}{n}\,\left(1+2\sum\limits_{k=1}^{f\,n}\cos\left(\frac{2\,\pi\,k\,x}{n}\right)\right)$

Assuming the conditions stated in (8) above, since $\sum\limits_{n=1}^\infty\frac{\mu(n)}{n}=\frac{1}{\zeta(1)}=0$ formulas (9) and (10) above can be simplified to (11) and (12) below.

(11) $\quad U(x)=\frac{1}{\pi}\sum\limits_{n=1}^N\mu(n)\sum\limits_{k=1}^{f\,n}\frac{\sin\left(\frac{2\,\pi\,k\,x}{n}\right)}{k}$

(12) $\quad U'(x)=2\sum\limits_{n=1}^N\frac{\mu(n)}{n}\,\sum\limits_{k=1}^{f\,n}\cos\left(\frac{2\,\pi\,k\,x}{n}\right)$

The Fourier series representation of $U''(x)$ is defined in (13) below.

(13) $\quad U''(x)=-4\,\pi\sum\limits_{n=1}^N\frac{\mu(n)}{n^2}\sum\limits_{k=1}^{f\,n}k\,\sin\left(\frac{2\,\pi\,k\,x}{n}\right)$

The following three plots illustrate the Fourier series representations of $U(x)$, $U'(x)$, and $U''(x)$ defined in (9), (12), and (13) above where all three functions are evaluated at $N=101$ and $f=4$. Note $\mathcal{M}(101)=0$ consistent with (8) above. The red discrete portions of the three plots below illustrate the evaluation of the Fourier series representations of $U(x)$, $U'(x)$, and $U''(x)$ at integer values of $x$. Formula (9) is illustrated for $U(x)$ instead of formula (11) because formula (9) converges much better at modest evaluation limits than formula (11).


The following plot illustrates the Fourier series representation of $U(x)$ defined in (9) above in blue and the distributional representation of $U(x)$ defined in (1) above in orange. Under the conditions specified in (8) above, the Fourier series representation of $U(x)$ converges to the distributional representation of $U(x)$ consistent with the Heaviside step function half-maximum convention (i.e. converges to $-\frac{1}{2}$ at $x=-1$ and to $\frac{1}{2}$ at $x=1$).

Evaluation of Formula (9) for Base Function

Figure (1): Illustration of Formula (9) for $U(x)$


The following plot illustrates the Fourier series representation of $U'(x)$ defined in (12) above. The first-order derivative $U'(x)$ exhibits a strict convergence at integer values of $x$. $U'(x)$ defined in (12) above always evaluates exactly to $2\,f$ at $|x|=1$ and exactly to zero at $|x|\ne 1$ for $x\in \mathbb{Z}$ and $0<|x|\le N$. The convergence of $U'(x)$ to $0$ at $x=0$ requires $\mathcal{M}(N)=0$.

Evaluation of Formula (12) for First-Order Derivative

Figure (2): Illustration of Formula (12) for $U'(x)$


The following plot illustrates the Fourier series representation of $U''(x)$ defined in (13) above.

Evaluation of Formula (13) for Second-Order Derivative

Figure (3): Illustration of Formula (13) for $U''(x)$


Formulas (9) and (11) above can be integrated in the right-half plane resulting in formulas (14) and (15) below. Note that under the conditions specified in (8) above $f(x+1)$ represents the linear function $x$ for $x\ge 0$. I believe one can continue to integrate in this manner to derive a formula for $x^m$ for $x\ge 0$ and $m\in\mathbb{Z}^+$.

(14) $\quad f(x)=\int_0^x U(t)\,dt=\frac{1}{2}\sum\limits_{n=1}^N\mu(n)\left(\frac{x^2}{n}+\frac{n}{\pi^2}\sum\limits_{k=1}^{f\,n}\frac{1-\cos\left(\frac{2\,\pi\,k\,x}{n}\right)}{k^2}\right)$

(15) $\quad f(x)=\int_0^x U(t)\,dt=\frac{1}{2\,\pi^2}\sum\limits_{n=1}^N\mu(n)\, n\sum\limits_{k=1}^{f\,n}\frac{1-\cos\left(\frac{2\,\pi\,k\,x}{n}\right)}{k^2}$

The Laplace transforms of the distributional representations of $U(x)$, $U'(x)$, and $U''(x)$ defined in (1), (2), and (3) above are as follows.

(16) $\quad\mathcal{L}_x[U(x)](y)=\frac{e^{-y}}{y}\,,\quad\Re(y)>0$

(17) $\quad\mathcal{L}_x[U'(x)](y)=e^{-y}$

(18) $\quad\mathcal{L}_x[U''(x)](y)=y\,e^{-y}$

The Laplace transforms of the Fourier series representations of $U(x)$, $U'(x)$, and $U''(x)$ defined in (11), (12), and (13) above are as follows. Under the conditions stated in (8) above, the Laplace transforms of the Fourier series representations defined below converge to the Laplace transforms of the distributional representations for $Re(y)>0$.

(19) $\quad\mathcal{L}_x[U(x)](y)=\frac{e^{-y}}{y}=2\sum\limits_{n=1}^N\mu(n)\,n\sum\limits_{k=1}^{f\,n}\frac{1}{4\pi^2\,k^2+n^2\,y^2}\,,\quad\Re(y)>0$

(20) $\quad\mathcal{L}_x[U'(x)](y)=e^{-y}=2\,y\sum\limits_{n=1}^N\mu(n)\,n\sum\limits_{k=1}^{f\,n}\frac{1}{4\pi^2\,k^2+n^2\,y^2}\,,\quad\Re(y)>0$

(21) $\quad\mathcal{L}_x[U''(x)](y)=y\,e^{-y}=-8\,\pi ^2\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\,n}\frac{k^2}{4\pi^2\,k^2+n^2\,y^2}\,,\quad\Re(y)>0$

The Laplace transforms defined in (19), (20), and (21) above are illustrated at the following link.

Laplace Transforms of $U[x]$, $U'[x]$, and $U''[x]$

The Laplace transforms of the Fourier series representations of $U(x)$, $U'(x)$, and $U''(x)$ defined in (11), (12), and (13) above can also be written as (22), (23), and (24) below which, under the conditions stated in (8) above, also converge to the Laplace transforms of the distributional representations for $Re(y)>0$.

(22) $\quad\mathcal{L}_x[U(x)](y)=\frac{e^{-y}}{y}=-\frac{8\,\pi ^2}{y^2}\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\,n}\frac{k^2}{4\pi^2\,k^2+n^2\,y^2}\,,\quad\Re(y)>0$

(23) $\quad\mathcal{L}_x[U'(x)](y)=e^{-y}=-\frac{8\,\pi ^2}{y}\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\,n}\frac{k^2}{4\pi^2\,k^2+n^2\,y^2}\,,\quad\Re(y)>0$

(24) $\quad\mathcal{L}_x[U''(x)](y)=y\,e^{-y}=2\,y^2\sum\limits_{n=1}^N\mu(n)\,n\sum\limits_{k=1}^{f\,n}\frac{1}{4\pi^2\,k^2+n^2\,y^2}\,,\quad\Re(y)>0$

The distributional representations of $U(x)$, $U'(x)$, and $U''(x)$ defined in (1), (2), and (3) above all have special relationships between their Fourier transforms and inverse Fourier transforms which are illustrated below where I believe the sign is related to whether each is an odd or even function. The Fourier transforms below and all subsequent Fourier transforms assume the Fourier parameters $\{0,\,-2\pi\}$.

(25) $\quad\mathcal{FT}_x[U(x)](z)=-\mathcal{FT}_x^{-1}[U(x)](z)=-\frac{i\,\cos(2\,\pi\,z)}{\pi\,z}$

(26) $\quad\mathcal{FT}_x[U'(x)](z)=\mathcal{FT}_x^{-1}[U'(x)](z)=2\,\cos(2\,\pi\,z)$

(27) $\quad\mathcal{FT}_x[U''(x)](z)=-\mathcal{FT}_x^{-1}[U''(x)](z)=4\,i\,\pi\,z\,\cos(2\,\pi\,z)$

The $sin$ and $cos$ terms associated with the Fourier series representations of $U(x)$, $U'(x)$, and $U''(x)$ also have special relationships between their Fourier transforms and inverse Fourier transforms which are illustrated below.

(28) $\quad\mathcal{FT}_x\left[\sin\left(\frac{2\,\pi\,k\,x)}{n}\right)\right](z)=-\mathcal{FT}_x^{-1}\left[\sin\left(\frac{2\,\pi\,k\,x)}{n}\right)\right](z)=\frac{1}{2}i\,\delta\left(\frac{k}{n}+z\right)-\frac{1}{2}i\,\delta\left(\frac{k}{n}-z\right)$

(29) $\quad\mathcal{FT}_x\left[\cos\left(\frac{2\,\pi\,k\,x)}{n}\right)\right](z)=\mathcal{FT}_x^{-1}\left[\cos\left(\frac{2\,\pi\,k\,x)}{n}\right)\right](z)=\frac{\delta \left(\frac{k}{n}+z\right)}{2}+\frac{\delta \left(\frac{k}{n}-z\right)}{2}$

Term-wise integration of the Fourier series representation of $U(x)$, $U'(x)$, and $U''(x)$ defined in (11), (12), and (13) above implies the following Fourier transforms for the Fourier series representation of each of these functions which are obviously inconsistent with the corresponding Fourier transforms of the distributional representation defined in (25), (26), and (27) above.

(30) $\quad\mathcal{FT}_x[U(x)](z)=\frac{i}{2\,\pi}\sum\limits_{n=1}^N\mu(n)\sum\limits_{k=1}^{f\,n}\frac{\delta\left(\frac{k}{n}+z\right)-\delta\left(\frac{k}{n}-z\right)}{k}$

(31) $\quad\mathcal{FT}_x[U'(x)](z)=\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\,n}\left(\delta\left(\frac{k}{n}+z\right)+\delta\left(\frac{k}{n}-z\right)\right)$

(32) $\quad\mathcal{FT}_x[U''(x)](z)=-2\,i\,\pi\sum\limits_{n=1}^N\frac{\mu (n)}{n^2}\sum\limits_{k=1}^{f\,n}k\left(\delta\left(\frac{k}{n}+z\right)-\delta\left(\frac{k}{n}-z\right)\right)$

I believe the reason for the seeming discrepancies between the Fourier transforms of the distributional and Fourier series representations for $U(x)$, $U'(x)$, and $U''(x)$ is the idealization of the Fourier transforms of $sin$ and $cos$ functions as Dirac delta ($\delta$) functions. Note this discrepancy exists even for the Fourier transforms of the distributional and Fourier series representations of the Dirac comb. In the case of the distributional and Fourier series representations of the Dirac comb each representation transforms into the other representation which I suppose makes the discrepancy a bit easier to accept, but the case being explored here is considerably different.

The bilateral Laplace transforms of the Fourier series representations of $U(x)$, $U'(x)$, and $U''(x)$ don't converge for $Re(y)=0$, and consequently they cannot be evaluated at $y=2\,\pi\,i\, z$ to obtain their corresponding Fourier transforms. However, the contribution of the right and left-half planes can be evaluated at $y=2\,\pi\,i\,z+\epsilon$ and $y=2\,i\,\pi\,z-\epsilon$ respectively and summed together to approximate their corresponding Fourier transforms for $z\in\mathbb{R}$ as defined in the formulas below.

(33) $\quad\mathcal{FT}_x[U(x)](z)\approx 2\sum\limits_{n=1}^N\mu(n)\,n\sum\limits_{k=1}^{f\,n}\left(\frac{1}{(2\,\pi\,k)^2+n^2(2\,\pi\,i\,z+\epsilon)^2}-\frac{1}{(2\,\pi\,k)^2+n^2(2\,i\,\pi\,z-\epsilon)^2}\right)\,,\,z\in\mathbb{R}_{\ne 0}$

(34) $\quad\mathcal{FT}_x[U'(x)](z)\approx 2\sum\limits_{n=1}^N\mu(n)\,n\sum\limits_{k=1}^{f\,n}\left(\frac{2\,i\,\pi\,z+\epsilon}{(2\,\pi\,k)^2+n^2(2\,i\,\pi\,z+\epsilon)^2}-\frac{2\,\pi\,i\,z-\epsilon}{(2\,\pi\,k)^2+n^2(2\,\pi\,i\,z-\epsilon)^2}\right)\,,\,z\in\mathbb{R}$

(35) $\quad\mathcal{FT}_x[U''(x)](z)\approx -8\,\pi^2\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\,n}\left(\frac{k^2}{(2\,\pi\,k)^2+n^2(2\,i\,\pi\,z+\epsilon)^2}-\frac{k^2}{(2\,\pi\,k)^2+n^2(2\,\pi\,i\,z-\epsilon)^2}\right)\,,\,z\in\mathbb{R}$

The following three plots illustrate the Fourier transforms of the distributional representations of $U(x)$, $U'(x)$, and $U''(x)$ defined in (25), (26), and (27) above in blue and the approximations to the Fourier transforms of the Fourier series representations of $U(x)$, $U'(x)$, and $U''(x)$ defined in (33), (34), and (35) above in orange. All three plots use the same evaluation limits $N=101$, $f=4$, and $\epsilon=0.1$. The second plot below corresponds to the evaluation of the real part of the Fourier transform of $U'(x)$, and the first and third plots below correspond to the evaluations of the imaginary parts of the Fourier transforms of $U(x)$ and $U''(x)$ respectively.


Evaluation of Imaginary Part of Formula (33)

Figure (4): Illustration of $\Im$ Part of Formula (33) for $\mathcal{FT}_x[U(x)](z)=-\frac{i\,\cos(2\,\pi\,z)}{\pi\,z}$


Evaluation of Real Part of Formula (34)

Figure (5): Illustration of $\Re$ Part of Formula (34) for $\mathcal{FT}_x[U'(x)](z)=2\,\cos(2\,\pi\,z)$


Evaluation of Imaginary Part of Formula (35)

Figure (6): Illustration of $\Im$ Part of Formula (35) for $\mathcal{FT}_x[U''(x)](z)=4\,i\,\pi\,z\,\cos(2\,\pi\,z)$


The Mellin transforms of the distributional representations of $U(x)$, $U'(x)$, and $U''(x)$ defined in (1), (2), and (3) above are as follows.

(36) $\quad\mathcal{M}_x[U(x)](s)=-\frac{1}{s},\quad\Re(s)<0$

(37) $\quad\mathcal{M}_x[U'(x)](s)=1$

(38) $\quad\mathcal{M}_x[U''(x)](s)=1-s$

The Mellin transforms of the $sin$ and $cos$ terms associated with the Fourier series representations of $U(x)$, $U'(x)$, and $U''(x)$ are as follows.

(39) $\quad\mathcal{MT}_x\left[\sin\left(\frac{2\,\pi\,k\,x)}{n}\right)\right](s)=(2\pi )^{-s}\,\sin\left(\frac{\pi\,s}{2}\right)\,\Gamma(s)\,\left(\frac{k}{n}\right)^{-s}$

(40) $\quad\mathcal{MT}_x\left[\cos\left(\frac{2\,\pi\,k\,x)}{n}\right)\right](s)=(2\,\pi)^{-s}\,\cos\left(\frac{\pi\,s}{2}\right)\,\Gamma(s)\,\left(\frac{k}{n}\right)^{-s}$

Term-wise integration of the Fourier series representation of $U(x)$, $U'(x)$, and $U''(x)$ defined in (11), (12), and (13) above implies the following Mellin transforms for the Fourier series representation of each of these functions. These transforms can be shown to be equivalent to the Mellin transforms of the distributional representations defined in (36), (37), and (38) above via analytic continuation and the definition of the Riemann zeta functional equation.

(41) $\quad\mathcal{M}_x[U(x)](s)=\frac{1}{\pi}\,(2\,\pi)^{-s}\,\sin\left(\frac{\pi\,s}{2}\right)\,\Gamma(s)\sum\limits_{n=1}^\infty\mu(n)\,\left(\frac{1}{n}\right)^{-s}\sum\limits_{k=1}^\infty k^{-1-s}$

(42) $\quad\mathcal{M}_x[U'(x)](s)=2\,(2\,\pi)^{-s}\,\cos\left(\frac{\pi\,s}{2}\right)\,\Gamma(s)\sum\limits_{n=1}^\infty\mu(n)\,\left(\frac{1}{n}\right)^{1-s}\sum\limits_{k=1}^\infty k^{-s}$

(43) $\quad\mathcal{M}_x[U''(x)](s)=-4\,\pi\,(2\,\pi)^{-s}\,\sin\left(\frac{\pi\,s}{2}\right)\,\Gamma(s)\sum\limits_{n=1}^\infty\mu(n)\,\left(\frac{1}{n}\right)^{2-s}\sum\limits_{k=1}^\infty k^{1-s}$

Formulas (44), (45), and (46) below illustrate the analytic continuations of formulas (41), (42), and (43) above are equivalent to the Mellin transforms of the distributional representations defined in (36), (37), and (38) above per the definition of the Riemann zeta functional equation. These results are achieved by replacing the sums over $n$ and $k$ in formulas (41), (42), and (43) above with their corresponding zeta functions and then simplifying using the definition of the Riemann zeta functional equation.

(44) $\quad\mathcal{M}_x[U(x)](s)=\frac{1}{\pi}\,(2\,\pi)^{-s}\,\sin\left(\frac{\pi\,s}{2}\right)\,\Gamma(s)\,\frac{\zeta(1+s)}{\zeta(-s)}=-\frac{1}{s}$

(45) $\quad\mathcal{M}_x[U'(x)](s)=2\,(2\,\pi)^{-s}\,\cos\left(\frac{\pi\,s}{2}\right)\,\Gamma(s)\,\frac{\zeta(s)}{\zeta(1-s)}=1$

(46) $\quad\mathcal{M}_x[U''(x)](s)=-2\,(2\,\pi)^{1-s}\,\sin\left(\frac{\pi\,s}{2}\right)\,\Gamma(s)\,\frac{\zeta(s-1)}{\zeta(2-s)}=1-s$

Note the Fourier series representation of $U'(x)$ represents $\delta(x-1)$ in the right-half plane and the Mellin transform $\mathcal{M}_x[\delta(x-1)](s)=1$. I'm currently investigating convergence of formulas derived from convolutions such as (47) to (49) below where the Fourier series representation of $U'(x)$ is substituted for $\delta(x-1)$ in the convolutions.

(47) $\quad g(y)=\delta(x-1)\,*_\mathcal{M_1}\,g(x)=\int_0^\infty\delta(x-1)\,g\left(\frac{y}{x}\right)\,\frac{dx}{x}$

(48) $\quad g(y)=\delta(x-1)\,*_\mathcal{M_2}\,g(x)=\int_0^\infty\delta(x-1)\,g(y\,x)\,\,dx$

(49) $\quad g(y)=\delta(x-1)\,*_\mathcal{M_3}\,g(x)=\int_0^\infty\delta(x-1)\,g(y+1-x)\,\,dx$

It's also possible to derive formulas for derivatives of functions using the Fourier series representations of the derivatives of $\delta(x-1)$ via relationships such as the following.

(50) $\quad g^{(n)}(y)=(-y)^{-n}\left(\delta^{(n)}(x-1)\,*_{\mathcal{M}_2}\,g(x)\right)=(-y)^{-n}\int_0^\infty\delta^{(n)}(x-1)\,g(y\,x)\,dx$

(51) $\quad g^{(n)}(y)=\delta^{(n)}(x-1)\,*_{\mathcal{M}_3}\,g(x)=\int_0^\infty\delta^{(n)}(x-1)\,g(y+1-x)\,dx$

An initial formula derived via a convolution such as those described above may be used to derive additional formulas for related functions via differentiation, integration, Laplace normal/inverse transforms, Mellin normal/inverse transforms, Hankel transforms, and in rare cases Fourier normal/inverse transforms.

I've derived a considerable number of formulas for a variety of functions using the techniques described above, but these formulas vary widely in complexity and a few of them don't seem to converge. The following formulas are mostly toward the simpler end of the spectrum and all seem to exhibit observational evidence of convergence under the general conditions specified in (8) above and the specific conditions specified for each formula below. The $E_m(y)$ function below (typically referred to as $E_n(y)$) is one of the more difficult functions with respect to determining convergence conditions as it's a function of two complex variables ($m$ and $y$).


(52) $\quad e^{-y}=2\,y\sum\limits_{n=1}^N\mu(n)\,n\sum\limits_{k=1}^{f\,n}\frac{1}{4\,\pi^2\,k^2+n^2\,y^2}\,,\quad\Re[y]>0$

(53) $\quad e^{-y}=-\frac{8\,\pi^2}{y}\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\,n}\frac{k^2}{4\,\pi\,^2\,k^2+n^2\,y^2}\,,\quad\Re[y]>0$

(54) $\quad e^{-y}=-\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\,n}\log\left(\left(\frac{2\,\pi\,k}{n\,y}\right)^2+1\right),\quad\Re[y]>0$


(55) $\quad\log(y)=\frac{1}{\pi }\sum\limits_{n=1}^N\mu(n)\sum\limits_{k=1}^{f\, n}\frac{\pi\,\sin^2\left(\frac{k\,\pi\,(y-1)}{n}\right)-\sin\left(\frac{2\,k\,\pi\,(y-1)}{n}\right)\,Ci\left(\frac{2\,k\,\pi\,(y-1)}{n}\right)+\cos\left(\frac{2\,k\,\pi\,(y-1)}{n}\right)\,Si\left(\frac{2\,k\,\pi\,(y-1)}{n}\right)}{k},\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad y>1$


(56) $\quad\cos(y)=2\,e\sum\limits_{n=1}^N\mu(n)\,n\sum\limits_{k=1}^{f\,n}\frac{n^2\,\left(y^2+1\right)+4\,\pi^2\,k^2}{\left(n^2\,\left(y^2-1\right)-4\,\pi^2\,k^2\right)^2+4\, n^4\,y^2},\quad y\in\mathbb{R}$

(57) $\quad\sin(y)=4\sum\limits_{n=1}^N\mu(n)\sum\limits_{k=1}^{f\,n}\frac{\pi\,k\,\sin (y+1)\,\sin \left(\frac{2\,\pi^2\,k}{n}\right)-n \cos (y+1) \cos ^2\left(\frac{\pi^2\,k}{n}\right)}{n^2-4\,\pi ^2\,k^2},\quad Re(y)\ne 0\lor y=0$

(58) $\quad sinc(y)=\frac{\sin (y)}{y}=2\,e\sum\limits_{n=1}^N\mu (n)\,n\sum\limits_{k=1}^{f n}\frac{n^2\,\left(y^2+1\right)-4\,\pi^2\,k^2}{\left(n^2\,\left(y^2-1\right)-4\,\pi^2\,k^2\right)^2+4\,n^4\, y^2},\quad y\in\mathbb{R}$


(59) $\quad sech(y)=\frac{\pi}{y}\sum\limits_{n=1}^{N}\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\,n} sech\left(\frac{k\,\pi^2}{n\,y}\right),\quad\Re(y)>0$


(60) $\quad erf(y)=\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\,n}\Gamma\left(0,\frac{k^2\,\pi^2}{n^2\,y^2}\right)\,,\quad\Re(y)>0$

(61) $\quad erfc(y)=1-erf(y)=2\,\pi^{-3/2}\sum\limits_{n=1}^N\mu(n)\sum\limits_{k=1}^{f\,n}\frac{F\left(\frac{k\,\pi}{n\,y}\right)}{k}\,,\quad\Re(y)>0$

(62) $\quad F(y)=\frac{1}{2}\pi^{3/2}\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\,n} erfc\left(\frac{\pi\, k}{n\,y}\right)\,,\quad y\in\mathbb{R}\quad\text{(Dawson function)}$


(63) $\quad K_0(y)=\pi\sum\limits_{n=1}^N\mu(n)\sum\limits_{k=1}^{f\,n}\frac{1}{\sqrt{4\,\pi^2\,k^2+n^2\,y^2}}\,,\quad\Re(y)>0$


(64) $\quad Si(y)=-\pi\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\,n}\left( \begin{array}{cc} \{ & \begin{array}{cc} \log\left(\frac{2\,k\,\pi}{n\,y}\right) & 2\,k\,\pi<n\,y \\ 0 & \text{True} \\ \end{array} \\ \end{array} \right),\quad y>0$

(65) $\quad Ei(y)=-\frac{1}{2\,\pi}\sum\limits_{n=1}^N\mu(n)\sum\limits_{k=1}^{f\,n} \frac{\pi+2\,\cot^{-1}\left(\frac{2\,\pi\,k}{n\,y}\right)}{k},\quad y<0$


(66) $\quad E_1(y)=\frac{1}{\pi}\sum\limits_{n=1}^N\mu(n)\sum\limits_{k=1}^{f\,n}\frac{\tan ^{-1}\left(\frac{2\,\pi\,k}{n\,y}\right)}{k},\quad Re[y]>0$

(67) $\quad E_m(y)=\frac{2}{m\,y}\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\,n}\,_2F_1\left(1,\frac{m}{2};\frac{m+2}{2};-\left(\frac{2\,\pi\,k}{n\,y}\right)^2\right),\\$ $\qquad\qquad\qquad\qquad\Re(y)>0\,\land\,\Re(m)\geq 0\,\land\,m\neq 0$


(68) $\quad\frac{1}{\Gamma[s]}=e\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\,n}\left(\left(1+\frac{2\,i\,\pi\,k}{n}\right)^{-s}+\left(1-\frac{2\,i\,\pi\,k}{n}\right)^{-s}\right),\quad\Re[s]>0$

(69) $\quad\Gamma(0,\,s)=e\sum\limits _{n=1}^N\mu(n)\sum\limits_{k=1}^{f\,n}\left(\frac{\log\left(1+\frac{n+2\,i\,\pi\,k}{n\,s}\right)}{n+2\,i\,\pi\,k}+\frac{\log \left(1+\frac{n-2\,i\,\pi\,k}{n\,s}\right)}{n-2\,i\,\pi\,k}\right),\quad s>0$

(70) $\quad\Gamma[s]=e\,\pi\,\csc\,(\pi\,s)\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\,n}\left(\left(\frac{n}{n+2\,i\,\pi\,k}\right)^{1-s}+\left(\frac{n}{n-2\,i\,\pi\,k}\right)^{1-s}\right),\quad-1<\Re(s)<1$


Most of the formulas above are illustrated at the following link. I typically use $f=4$ for evaluation plots as most of the formulas are much more sensitive to the magnitude of $N$ than they are to the magnitude of $f$. I'd appreciate feedback anyone might have with respect to convergence of these formulas.

Illustration of Formulas Derived From Fourier Series Representation of $U'(x)$ and $U''(x)$

The Fourier series representation of $U(x)$ is an example of what I refer to as a method 1 Fourier series. Similar Fourier series can be derived for the log-step staircase function $T(x)$ and prime-counting functions such as the base prime-counting function $\pi(x)$, Riemann's prime-power counting function $\Pi(x)$, the first Chebyshev function $\vartheta(x)$, and the second Chebyshev function $\psi(x)$.

Assuming $a\in\mathbb{R}$, the $U(x)$ function can be shifted by subtraction of parameters as illustrated in (71) below. Formula (71) below evaluates to $\theta(x-a)$ for $x\ge a-1$. This representations of $\theta(x-a)$ is the basis for what I refer to as the method 2 Fourier series representation of prime-counting functions. Since both $\theta$ functions move in the same direction when shifting $U(x)$ via subtraction of parameters, the lower evaluation bound must be restricted to $x\ge a-1$ when evaluating $U(x-(a-1))$ to avoid contribution of the $\theta$ function originally in the left-half plane.

(71) $\quad U(x-(a-1))=-1+\theta(x-(a-1)+1)+\theta(x-(a-1)-1)\\$ $\qquad\qquad\qquad\qquad\qquad=-1+\theta(x-(a-2))+\theta(x-a)$

Assuming $a\in\mathbb{R}$ and $a>1$, the $U(x)$ function can also be shifted by division of parameters as illustrated in (72) below. Formula (72) below evaluates to $\theta(x-a)$ for $x\ge 0$. This representation of $\theta(x-a)$ is the basis for what I refer to as the method 3 Fourier series representation of prime-counting functions. Note that in this case the two $\theta$ functions move in opposite directions (since $a>1$), and the $\theta$ function in the left-half plane can be ignored when evaluating formulas for $x\ge 0$ which is the primary interest with respect to evaluation of formulas related to prime-counting functions.

(72) $\quad U\left(\frac{x}{a}\right)=-1+\theta\left(\frac{x}{a}+1\right)+\theta\left(\frac{x}{a}-1\right)=-1+\theta(x+a)+\theta(x-a)$

Method 3 has the advantage over method 2 that it's not necessary to restrict the lower evaluation bound when evaluating $\theta(x-a)$, but has the disadvantage that conditional convergence becomes a bit more complicated. Additional terms must be evaluated in order to obtain convergence, and this leads to longer evaluation times.

The following three plots illustrate method 1, method 2, and method 3 Fourier series representations related to the first-order derivative $\psi'(x)$ of the second Chebyshev function. The method 1 representation in the first plot is for $\psi'(x)-1$, whereas the method 2 and method 3 representations illustrated in the second and third plots are both for $\psi'(x)$. All three plots are evaluated at $f=4$. The orange curve illustrated in the plots is the reference function $2\,f\log(x)$ and the horizontal dashed grid-lines are at $2\,f\log(2)$ and $2\,f\log(3)$. The red discrete portions of the plots illustrate all three Fourier series representations exhibit the same strict convergence to exactly to $2\,f$ times the step size of $\psi(x)$ at positive integer values of $x$.


Method 1 Fourier Series Representation of $\psi'(x)-1$

Figure (7): Illustration of Method 1 Fourier Series Representation of $\psi(x)-1$


Method 2 Fourier Series Representation of $\psi'(x)$

Figure (8): Illustration of Method 2 Fourier Series Representation of $\psi(x)$


Method 3 Fourier Series Representation of $\psi'(x)$

Figure (9): Illustration of Method 3 Fourier Series Representation of $\psi(x)$


Personally I find the method 2 and method 3 Fourier series representations of prime-counting functions somewhat artificial compared with the method 1 Fourier series representation. The method 2 Fourier series representation seems particularly artificial to me since it requires a different lower evaluation bound be used to evaluate each different instance of $\theta(x-a)$.

3/26/2018 Update:

The following link defines and illustrates a general method for derivation of a Fourier series representation for $f(x)=\sum\limits_{n=1}^x a(n)$ related to the Dirichlet series $F(s)=\sum\limits_{n=1}^\infty\frac{a(n)}{n^s}$.

Derivation of Fourier Series Representation for $f(x)=\sum\limits_{n=1}^x a(n)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.