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Dirichlet series inversion and prime number coutnign function

  1. Aug 7, 2004 #1
    Hello, i have developed a formula for inverting a dirichlet series ,and i apply to solve Phi(n),Mu(n) and d(n) arithmetical function,also i give a formula to obtain the Pi(n) function by means of a triple integral, the paper is in .doc format but if someone want to see it in .pdf can go to www.gohtm.com htere you can convert any word file to .pdf,i have submittedto several journals,now i want you to give me your opinion.
     

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  2. jcsd
  3. Aug 7, 2004 #2

    shmoe

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    Hi, a few things

    1. What is your [tex]\delta[/tex]?
    2. At one point you are assuming that a(x) is defined for every x, then you go on to use this "fact" to show you've generalized these a(n)'s to non-integer values of x.
    3. You should learn LaTeX
    4. I don't know what journals you've sent this to, but are you sure it meets their formatting standards? You should try reading a few articles in these journals to get a feel for their standards- things like citing "www.mathworld.com" are generally not done in journals.
    5. You should use [tex]\zeta[/tex] for the Riemann Zeta function, not [tex]\xi[/tex].
    6. it's not clear to me how you've come up with eq. (3). You say you divide something be (21), but there is no (21) and I'm not sure exactly what you're trying to divide in the first place.
    7. in your integrals you should seperate your "dx" term better. It doesn't have to go at the end, but it should be clear you aren't trying to do some kind of Stieltjes integral (maybe you are?)

    That's all for now. You'll notice I didn't comment much on the math-it's really hard to understand what you're trying to do. You really need to work on clarity. Seriously, take the time to learn latex and proper formatting. You'll find people will take you far more seriously and spend the time to try to understand your work if you can explain yourself clearly.
     
  4. Aug 9, 2004 #3
    I tried to send it manuscripted to several teachers at my university but until now i got no reply, i have observed the conditions to pubish and they accept word or .pdf format.

    I think that the most important contribution is to give a formula to invert a dirichlet series by means of the formula
    M^-1[f(4-s)]
    a(n)=---------------- and its application tonumber theory
    M^-1[R(4-s)]

    Also i give a formula for the prime number counting function Pi(x)/x^4 in the form of a triple integral with limits (c-i8,c+i8), (0,8) , (c-i8,c+i8) of F(n,q,x,s= where

    F(n,q,x,s)=(n^-s+2)(x^-q)LnR(4n-nq)/(4-q)R(4-s)
     
  5. Aug 9, 2004 #4

    shmoe

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    Hi, there's more to the format than just what type of file it is. Legibility is key. Think about the professors point of view- you've already got a busy schedule and someone (probably several people infact) sends you an unsolicited manuscript. When you get some free time, you take a look at it. If it's incomprehensible or it takes straining to understand, they'll probably just toss it in the "ignore" pile.

    Something as simple as notation could make me quickly pass judgement on a manuscript. For heavens sake, why do you use [tex]\xi[/tex] for the riemann zeta function and not [tex]\zeta[/tex]? It doesn't affect the math, sure, but every single paper or text book I've ever read uses [tex]\zeta[/tex], it is the riemann zeta function after all. It's industry standard that makes sense, you don't arbitrarily change it and expect people to take the time to read it. ([tex]\xi[/tex] is actually pretty standard notation for a function intimitely related to zeta, which just makes matters worse).

    That aside, you haven't defined some terms, like your [tex]\delta[/tex] function I asked you about. This just makes the read frustrating, I'm not going to spend my precious time trying to sort out what the heck you mean when I have no guarantee that you have any idea what you're talking about (no offence). I'd definitely be more willing to give it a thorough read if it was more clear

    You mentioned you were a physicist, are you a student at your university or faculty? Either way, I highly suggest you take or audit the next intro to analytic number theory course they offer. That or beg your closest number theorest to offer something as a reading course for you. You've been asking many very basic questions about Dirichlet series and [tex]\zeta[/tex] (which is fine, I'm happy to oblige) that you should know if you expect to be able to even judge a piece of work for it's validity, let alone produce something usefull. Learn to walk before you try to run.

    I do hope you take the above as constructive criticism. I'm always happy to encourage someone to study number theory, but they should do it in as thorough a manner as possible.
     
  6. Aug 10, 2004 #5
    Thanks again,i changed the format in the Riemann function by [tex]\zeta[/tex] and define the [tex]\delta[/tex]
    as the fourier transform of y=1 is that enough?..do i hvae to enumerate the prperties of the delta in my work?...
     
  7. Aug 11, 2004 #6

    shmoe

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    Hi, that would have helped. I had thought the Dirac delta function was one of the possibilities, but there was at least one another option.

    You didn't respond to my a(x) objection before. When you come up with this integral:

    [tex]\int\limits_{0}^{\infty}a(x)x^{s-4}w(x)dx[/tex]

    You are assuming that your coefficients a(n) somehow extend to all positive real numbers in a sensible way. This isn't the case for most arithmetic functions.

    Assuming this is valid anyways, when you apply the Mellin inversion to your equation (2), you get
    [tex]a(x)x^{3}w(x)=\frac{1}{2\pi i}\int\limits_{c-i\infty}^{c+i\infty}\frac{g(4-s)}{x^s}ds[/tex]
    This sum [tex]w(x)[/tex] of delta functions isn't a function in the normal sense, it's the sum of a bunch of linear functionals (or point masses if you like). You certainly can't divide by [tex]w(x)[/tex], and expressing it as an inverse Mellin transform doesn't hide this unpleasantness. The RHS of this expression isn't a function either, it's a pointmass kind of deallie like your delta functions. I'm quite confidant that the actual integral involved will diverge if you try to evaluate it at whatever x you like, if you were trying to pick off [tex]a(n)[/tex] like you hoped to do. You can try to equate them in a linear functional kind of way to conclude that [tex]a(n)=a(n)[/tex], which doesn't help at all.
     
  8. Aug 11, 2004 #7
    I,m assuming that every arithmetical function can be defined for any x,just as happens for the prime number counting function , for example Pi(sqrt(2)=Pi([sqrt(2)]) where [x] is the integer part of x,the same would happen to other arithmetical functions,for example mu(sqrt(2))=mu([sqrt(2)]) and the same for the others.

    in fact when you try to obtain a(r) for r integer you get that in the sum of the delta functions only d(x-r) will give a non-zero value so setting x=r we have d(0)a(r)/d(0) the delta function diverges as n with n tending to infinity but d(0)/d(/0)=1 as n/n=1 then you have a(r) for r being an integer.

    for non-integers values you are right,all the delta functions would be 0 so we got an expression 0/0 indetermined,a possible solution to that would be taking c-iT,c+iT with T a big numnet and dividing the two Mellin transforms

    for the case w(x)mu(x)=M^-1[1/R(4-s)] you can check is correct by expanding 1/R(4-s) and doing the Mellin inverse transform term by term.
     
  9. Aug 11, 2004 #8

    shmoe

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    This isn't all what you appeared to be doing, as you state "Iwith these expressions we can generalize the value of arithmetical function a(x) to non-integer real or complex values." It looks to me like you think you have some way of naturally extending these a(n) to all real numbers. Taking a([x]) is a very artificial way of extending it.

    No, you can't do this. Both the integrals are going to be divergent. You can't take their ratio over a finite interval and end up with a sensible answer. Compare with the following "cancellation":

    [tex]2\lim_{n\rightarrow \infty} n=\lim_{n\rightarrow \infty} 2n=\infty=\lim_{n\rightarrow \infty} n[/tex]

    therefore

    [tex]2=\frac{\lim_{n\rightarrow \infty} n}{\lim_{n\rightarrow \infty} n}=\lim_{n\rightarrow \infty} \frac{n}{n}=1[/tex]

    This is essentially the same as what you propose.

    You're missing my point. These delta functions cannot be treated like real functions. You're hoping that when you put in, say x=n, that you'll have [tex]\mu(n)[/tex] on the left and something sensible real number on the right. You can try integrating term by term like a normal function. You'll get an infinite sum of integrals, every one of them oscillating and not converging because of it, except the one involving the [tex]\mu(n)[/tex] term which is the integral of a constant function over an infinite interval. It does not produce a real number. It produces one of these delta functions. It's not something that could be evaluated in any way to give you [tex]\mu(n)[/tex]. You can't go from having a delta function being a point mass to being a normal well-behaved honest to goodness real function like you're trying to do.
     
  10. Aug 12, 2004 #9
    so all my work is bad?.. :[ it,s a pity i tried hard to obtain a formula for the prime number counting function and now it seems is false...i feel sad.
     
  11. Aug 20, 2004 #10
    A last post does the Laplace Transform of Sum(1<n<8)d(x-n)mu(x) exist?.
     
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