[math-fun] DGF --> OGF?
Can every suitably nice function A(s) be "inverted" to give an arithmetic function a(n) such that A(s) = sum_{n >= 1} a(n)/n^s? (E.g. by Perron's formula?) In other words, given ordinary generating functions A_i(x) = sum_{n >= 0} a^i_n x^n, I can combine them into a Dirichlet generating function A(s) = prod_{i >= 1} A_i(p_i^{-s}). My question is whether given a suitably nice but otherwise arbitrary function of s and think of it as a Dirichlet generating function, can I decompose it into a collection of ordinary generating functions? If so, what constitutes "suitably nice"? Merely convergence for Re(s) > sigma? -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com
On May 8, 2019, at 12:00 PM, Mike Stay <metaweta@gmail.com> wrote: My question is whether given a suitably nice but otherwise arbitrary function of s and think of it as a Dirichlet generating function, can I decompose it into a collection of ordinary generating functions?
An OGF seems to imply a smooth function over a finite region of agreement, where there is some hope to make all precision by including more and more terms. If the function has poles, perhaps you could assume it differentiably finite, and build the poles into the defining DE. But this is just the usual local analysis. I can’t tell what you are hoping for on a global scale? Before asking a question in such broad generality, can you give one or even a few minimal working example(s) showing us what you are aiming for? If you only want equivalence on one point, the question is much easier, and increasingly well studied. A few days ago I gave a nice example for Pi^2 on a thread about unusual integrals. —Brad
On Wed, May 8, 2019 at 12:01 PM Mike Stay <metaweta@gmail.com> wrote:
In other words, given ordinary generating functions A_i(x) = sum_{n >= 0} a^i_n x^n, I can combine them into a Dirichlet generating function A(s) = prod_{i >= 1} A_i(p_i^{-s}).
But If you are talking about generalizing the prime product formula for Riemann Zeta (?), it is pretty easy to over-constrain the coefficients. Assume only that F = Sum a_n/n^s. a_1 = 1. F_i = Sum a_{i,m}/p_i^(m*s). a_i,0 = 1. If the product over F_i equals F then "a_{i,m}=a_{p_i^m}", and coefficients a_n to terms with more than one prime are no longer free. It does not help to relax initial values, for "b_{i,m}/b_{i,0} = b_{p_i^m}/b_{0}", reduces to the form "a_i,m=a_{p_i^m}", and again the coefficients are mostly over-determined. Unless I am mistaken, this is a valid argument saying that your suggested product formulas generally do not exist. --Brad
On Wed, May 8, 2019 at 5:18 PM Brad Klee <bradklee@gmail.com> wrote:
On Wed, May 8, 2019 at 12:01 PM Mike Stay <metaweta@gmail.com> wrote:
In other words, given ordinary generating functions A_i(x) = sum_{n >= 0} a^i_n x^n, I can combine them into a Dirichlet generating function A(s) = prod_{i >= 1} A_i(p_i^{-s}).
<snip>
Unless I am mistaken, this is a valid argument saying that your suggested product formulas generally do not exist.
I'm just talking about https://en.wikipedia.org/wiki/Bell_series . I agree that an arbitrary function interpreted as a Dirichlet series will not give rise to a multiplicative function in general. Are there results on when it does? -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com
Of course, but that's a condition on a, not directly on A. It's like saying "exp(z) equals its conjugate when exp(z) is real" rather than "exp z equals its conjugate when Im(z) is an integer multiple of pi". On Wed, May 8, 2019 at 8:32 PM <bradklee@gmail.com> wrote:
a_{p1^m1 p2^m2 ... p^m} = a_{p1^m1} a_{p2^m2} ... a_{p^m}
?? —Brad
On May 8, 2019, at 8:31 PM, Mike Stay <metaweta@gmail.com> wrote: Are there results on when it does?
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-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com
Okay, thanks for the clarification. I wasn't following the thread closely, and it seemed likely that people are well aware of what I wrote, but I sent it to one person just in case. Jerry Shurman On 5/9/19, Mike Stay <metaweta@gmail.com> wrote:
Of course, but that's a condition on a, not directly on A. It's like saying "exp(z) equals its conjugate when exp(z) is real" rather than "exp z equals its conjugate when Im(z) is an integer multiple of pi".
On Wed, May 8, 2019 at 8:32 PM <bradklee@gmail.com> wrote:
a_{p1^m1 p2^m2 ... p^m} = a_{p1^m1} a_{p2^m2} ... a_{p^m}
?? —Brad
On May 8, 2019, at 8:31 PM, Mike Stay <metaweta@gmail.com> wrote: Are there results on when it does?
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-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com
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Do not rudely accuse me of writing tautologies! Compare with: * A function A(x) admits a hypergeometric form whenever a_0=1, a_{n+1}=poly1(n)/poly2(n)*a_{n}. * The function A(s) admits a product form whenever each a_n obeys the previously stated identity. In both of these statements a deciding criterion is given for dividing the set of all functions into has and has not classes with regard to some apparently arbitrary property. That is distinctly different from the tautological form “A is A”! We also have a lot of cases where “B is not A”. Honestly, if you are hoping to get a better response, you might want to do some self-critique and try to improve on communicating what you want to hear. As far as I can tell, I answered your question, and then you attacked me with a fallacious argument, meaning to imply that I am some sort of idiot. Wrong! Cheers, Brad
On May 9, 2019, at 6:39 AM, Mike Stay <metaweta@gmail.com> wrote:
Of course, but that's a condition on a, not directly on A. It's like saying "exp(z) equals its conjugate when exp(z) is real" rather than "exp z equals its conjugate when Im(z) is an integer multiple of pi".
On Wed, May 8, 2019 at 8:32 PM <bradklee@gmail.com> wrote:
a_{p1^m1 p2^m2 ... p^m} = a_{p1^m1} a_{p2^m2} ... a_{p^m}
?? —Brad
On Thu, May 9, 2019 at 8:57 AM <bradklee@gmail.com> wrote:
Do not rudely accuse me of writing tautologies!
<snip>
improve on communicating what you want to hear. As far as I can tell, I answered your question, and then you attacked me with a fallacious argument, meaning to imply that I am some sort of idiot.
I never attacked anyone. (But my wife will readily agree with you that I am not always clear in my communications.) I was just hoping for some characterization of Dirichlet generating functions whose corresponding series were multiplicative without referring to the series itself. If that's not possible, that's fine. -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com
participants (4)
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Brad Klee -
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Jerry Shurman -
Mike Stay