Evaluating Arithmetic Functions In The Complex Plane Caleb Fellow Posts: 51 Threads: 6 Joined: Feb 2023 02/19/2023, 07:03 AM I'm interested if others have any ideas on how to extend certain arithmetic functions to the complex plane. The particular case I am interseted in the Louiville function $\lambda(n) = (-1)^{\Omega(n)}$. Some more information on that function can be found here.  My interest in this particular function is the following. We have the very nice identity  $\sum \frac{\lambda(n)q^n}{1-q^n} = \sum q^{n^2}$ Since the Louiville function is basically the characteristic function for the perfect squares. Using a residue theorem approach, we should have that $\sum q^{n^2}$ can be continued beyond its natural boundary as the function $\sum \frac{\lambda(n)q^n}{1-q^n} + \sum \frac{\lambda( \frac{2 \pi i n}{\ln(q)})}{1-e^{ -\frac{4 \pi^2}{\ln(q)}}}$  These extra terms require looking at the Louiville function at imaginary values, so I'm curious about any approach to this.  Second, I'd also like to see what $\ln{\lambda(z)}$ looks like on the complex plane. In particular, I think it would just be cool to see how prime factorization functions extends to the whole complex plane (for instance, if we extend the indiciator function of the primes to the complex plane, are there are 'prime' complex numbers (i.e. numbers where f(z) = 0)?) Caleb Fellow Posts: 51 Threads: 6 Joined: Feb 2023 02/19/2023, 08:59 AM Hmm, this might be non-sense, but I'll write it anyway.  The goal will be to construct some functions made up of lambda that allow us to compute $\lambda(n)$ while only evaulating it on the positive integers. One very useful identity we can use is that $g(n) = \sum_{n=1}^\infty \frac{\lambda(n)}{n^s} = \frac{\zeta(2s)}{\zeta(s)}$ Now, let's consider the two functions   $\sum_{n=0}^\infty \frac{g(-n)}{z^{n+1}} \qquad \sum_{k=1}^\infty \frac{\lambda(k)}{z-k}$ These are linked in the following way. We have that $\star_1 \quad \sum_{n=0}^\infty \frac{g(-n)}{z^{n+1}} = \sum_{n=0}^\infty \frac{1}{z^{n+1}} \sum_{k=1}^\infty \lambda(k)k^n =$  $\sum_{k=1}^\infty \frac{\lambda(k)}{z} \sum_{n=0}^\infty (\frac{k}{z})^n=$ $\star_2 \quad \sum_{k=1}^\infty \lambda(k) \frac{1}{z-k}$ The stars indicate lines where I replaced a divergent series with an analytical continuation. In these lines, we have to either add or subtract the residues that were introduced. The usefulness of this approach is that we introduce a residue at $k=z$, and its value is some multiple of $\lambda(z)$. Therefore, we should be able to compute the value of $\lambda(z)$.  So, going back, let's go back and carefully keep track of the residues introduced. At $\star_1$ we replace an analytical continuation with a divergent series. Therefore, we create some residues, so we need to add them back in to get equality. Thus, we should we should have that  $\sum \frac{g(-n)}{z^{n+1}}+\sum \text{Res}\left(\frac{\zeta(-2s)}{\zeta(-s)}\frac{1}{z^{s+1}(e^{2 \pi i s}-1)}, s = -\rho \right) = \sum \frac{1}{z^{n+1}} \sum \lambda(k) k^n$ In other words, we have to sum over the residues from the non-trivial zeroes of the zeta function (notice that the trivial zeroes get canceled out since we are looking at $\frac{\zeta(2s)}{\zeta(s)}$.) Next, we look at $\star_2$. In this case, we are replacing a divergent series by an analytical continuation, so we need to subtract the residue it generates. However, that residue it generates is equal to  $2 \pi i \frac{\lambda(z)}{e^{2 \pi i z} - 1}$ But this involves $\lambda(z)$! Therefore, we should be able to compute $\lambda(z)$ by evaluating these series. In particular, we now have  $\sum \frac{\lambda(k)}{z-k} + 2 \pi i \frac{\lambda(z)}{e^{2 \pi i z} - 1} = \sum_{n=0}^\infty \frac{g(-n)}{z^{n+1}} +\sum \text{Res}\left(\frac{\zeta(-2s)}{\zeta(-s)}\frac{1}{z^{s+1}(e^{2 \pi i s}-1)}, s = -\rho \right)$ Which implies that  $\lambda(z) = \frac{e^{2 \pi i z} -1}{2 \pi i}\left(\sum_{n=0}^\infty \frac{g(-n)}{z^{n+1}}- \sum_{k=1}^\infty \frac{\lambda(k)}{z-k} +\sum \text{Res}\left(\frac{\zeta(-2s)}{\zeta(-s)}\frac{1}{z^{s+1}(e^{2 \pi i s}-1)}, s = -\rho \right)\right)$ There's a much easier way to pick up all those zeta-zero residues however, we just do a contour integral at some point that picks them up. In particular, it just has to behind $z = -\frac{1}{2}$ assuming the RH, but behind $z=-1$ will work no matter what. JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 02/19/2023, 09:13 AM I believe you are confused, Caleb. To ask for an analytic continuation of $\lambda(n)$--is very unnatural. To ask for an analytic continuation, you must specify some additional condition, to what the continuation looks like. In this case we are very unnatural. Let's say we write: $\lambda(n) =(-1)^{\Omega(n)}\\$ Where $\Omega(n)$--counts how many prime factors $n$ has. When we write: $\Omega(s)\\$ For $\Re(s) > 1$; where $\Omega(s) \Big{|}_{\mathbb{N}} = \Omega(n)$. We have to make an additional distinction claim on $\Omega(s)$. Let's say that $\Omega(s)$ is in our Fractional Calculus space. So let's say: $\Omega(s+1) = \frac{d^{s}}{dw^s}\Big{|}_{w=0} \sum_{n=0}^\infty \Omega(n+1) \frac{(-w)^n}{n!}\\$ In order to call this an interpolation, you have to show that: \begin{align} f(w) &= \sum_{n=0}^\infty \Omega(n+1) \frac{(-w)^n}{n!}\\ \Omega(s+1) &= \frac{1}{\Gamma(1-s)} \int_0^\infty f(w) w^{-s}\,dw\\ \int_0^\infty |f(w)|\,dw &< \infty\\ \end{align} Now this would perfectly solve your problem. BUT, AND THIS IS A HUGE FUCKING BUT! It's on you to show that: $\int_0^\infty |f(w)|\,dw < \infty$ In order to say that: $\lambda(s)\\$ Is holomorphic. Or it's on you to say that: $\lambda(s+1) = \frac{d^s}{dw^s}\Big{|}_{w=0} \sum_{n=0}^\infty \lambda(n+1) \frac{(-w)^n}{n!}\\$ And test that: $\int_0^\infty \left| \sum_{n=0}^\infty \lambda(n+1) \frac{(-w)^n}{n!}\right| \,dw < \infty\\$ BUT YOU HAVE TO PROVE THIS. AND IT IS SUPER NONTRIVIAL!!!!!!! Let's instead look at your problem at face value: $\sum_{n=0}^\infty \frac{\lambda(n) z^n}{1-z^n} = \sum_{n=0}^\infty z^{n^{2}}\\$ Setting $\lambda(0) = 1$ and $\Omega(0) = 0$. The question, and area of research you are broaching on, Caleb. Is an old avenue of research. I am not that well versed in it; but it's referred to as: THE CIRCLE METHOD. Or, Ramanujan's and Hardy's CIRCLE METHOD. What happens is close to what you are writing. But it gets very harder. $\sum_{n=0}^\infty z^{n^2} = \sum_{n=0}^M \frac{\lambda(n)z^n}{1-z^n} + \sum_{j=0}^M H_j(z)\\$ Where as $M\to \infty$, the sum $H_j(z)$ turns into asymptotic series. Which allow for estimates on growth. END OF STORY. Your quest to analytically continue this sum is futile. The function: $F(z) = \sum_{n=0}^\infty z^{n^2}\\$ Has a dense amount of singularities on $|z| =1$. This function is reducible into: $F(e^{2\pi i x}) = \sum_{n=0}^\infty e^{2\pi i x n^2}\\$ Which is holomorphic for $\Im(x) > 0$. Where a natural boundary appears along $x \in \mathbb{R}$. There is no analytic continuation. But for; $\Im(x) < 0$ we have: \begin{align} F(e^{2\pi i x}) &= \overline{\sum_{n=0}^\infty e^{2\pi i \overline{x}n^2}}\\ &= \sum_{n=0}^\infty e^{-2\pi i x n^2}\\ \end{align} And this finds $F(z)$ for $|z| >1$. But each are not holomorphic on the unit circle $|z| = 1$. Where upon we have two different functions; one holomorphic for $|z| < 1$--the other for $|z| > 1$. Where each complements the other; and both blow up as $|z|\to 1$. In conclusion; we have shown asymptotics on $\lambda(n)$ and $\Omega(n)$... Please state them! Regards, James Caleb Fellow Posts: 51 Threads: 6 Joined: Feb 2023 02/19/2023, 09:30 AM (02/19/2023, 09:13 AM)JmsNxn Wrote: I believe you are confused, Caleb. To ask for an analytic continuation of $\lambda(n)$--is very unnatural. To ask for an analytic continuation, you must specify some additional condition, to what the continuation looks like. In this case we are very unnatural. Let's say we write: $\lambda(n) =(-1)^{\Omega(n)}\\$ Where $\Omega(n)$--counts how many prime factors $n$ has. When we write: $\Omega(s)\\$ For $\Re(s) > 1$; where $\Omega(s) \Big{|}_{\mathbb{N}} = \Omega(n)$. We have to make an additional distinction claim on $\Omega(s)$. Let's say that $\Omega(s)$ is in our Fractional Calculus space. So let's say: $\Omega(s+1) = \frac{d^{s}}{dw^s}\Big{|}_{w=0} \sum_{n=0}^\infty \Omega(n+1) \frac{(-w)^n}{n!}\\$ In order to call this an interpolation, you have to show that: \begin{align} f(w) &= \sum_{n=0}^\infty \Omega(n+1) \frac{(-w)^n}{n!}\\ \Omega(s+1) &= \frac{1}{\Gamma(1-s)} \int_0^\infty f(w) w^{-s}\,dw\\ \int_0^\infty |f(w)|\,dw &< \infty\\ \end{align} Now this would perfectly solve your problem. BUT, AND THIS IS A HUGE FUCKING BUT! It's on you to show that: $\int_0^\infty |f(w)|\,dw < \infty$ In order to say that: $\lambda(s)\\$ Is holomorphic. Or it's on you to say that: $\lambda(s+1) = \frac{d^s}{dw^s}\Big{|}_{w=0} \sum_{n=0}^\infty \lambda(n+1) \frac{(-w)^n}{n!}\\$ And test that: $\int_0^\infty \left| \sum_{n=0}^\infty \lambda(n+1) \frac{(-w)^n}{n!}\right| \,dw < \infty\\$ BUT YOU HAVE TO PROVE THIS. AND IT IS SUPER NONTRIVIAL!!!!!!! Let's instead look at your problem at face value: $\sum_{n=0}^\infty \frac{\lambda(n) z^n}{1-z^n} = \sum_{n=0}^\infty z^{n^{2}}\\$ Setting $\lambda(0) = 1$ and $\Omega(0) = 0$. The question, and area of research you are broaching on, Caleb. Is an old avenue of research. I am not that well versed in it; but it's referred to as: THE CIRCLE METHOD. Or, Ramanujan's and Hardy's CIRCLE METHOD. What happens is close to what you are writing. But it gets very harder. $\sum_{n=0}^\infty z^{n^2} = \sum_{n=0}^M \frac{\lambda(n)z^n}{1-z^n} + \sum_{j=0}^M H_j(z)\\$ Where as $M\to \infty$, the sum $H_j(z)$ turns into asymptotic series. Which allow for estimates on growth. END OF STORY. Your quest to analytically continue this sum is futile. The function: $F(z) = \sum_{n=0}^\infty z^{n^2}\\$ Has a dense amount of singularities on $|z| =1$. This function is reducible into: $F(e^{2\pi i x}) = \sum_{n=0}^\infty e^{2\pi i x n^2}\\$ Which is holomorphic for $\Im(x) > 0$. Where a natural boundary appears along $x \in \mathbb{R}$. There is no analytic continuation. But for; $\Im(x) < 0$ we have: \begin{align} F(e^{2\pi i x}) &= \overline{\sum_{n=0}^\infty e^{2\pi i \overline{x}n^2}}\\ &= \sum_{n=0}^\infty e^{-2\pi i x n^2}\\ \end{align} And this finds $F(z)$ for $|z| >1$. But each are not holomorphic on the unit circle $|z| = 1$. Where upon we have two different functions; one holomorphic for $|z| < 1$--the other for $|z| > 1$. Where each complements the other; and both blow up as $|z|\to 1$. In conclusion; we have shown asymptotics on $\lambda(n)$ and $\Omega(n)$... Please state them! Regards, JamesYou are correct in stating that these are not *analytic* continuations-- this is a generalized analytical continuation. There is a particular case where analytical continuation makes sense beyond natural boundaries, which is the case of series like  $\sum \frac{1}{1-e^{i \theta_n}}$ Where $\theta_n$ creates a natural boundary by being dense on the unit circle. This case, we have a very natural candidate for the analytical continuation, which is the limit obtained by looking at the sequence of partial sums. Poincare studied this particular case in detail, and showed some interesting ways in which the sum inside the natural boundary and outside the natural boundary are connected. I believe (Weirestrass?) also studied generalized analytical continuation in detail, and was looking for more general ways to connect functions besides analytical continuation.  Therefore, the idea would be that we can naturally continue the series $\sum \frac{\lambda(k)}{1-x^k}$ using generalized analytical continuation, and then in the processing of linking that series to the theta function series, we pick up some new residues that depend on $\lambda(z)$ at imaginary values. I'm hoping there is some canonical way to extend $\lambda(n)$ to the compelx plane. Of course its not unique, but I was hoping this type of idea had been considered before to at least give some suggestions on the right approach. I will look in more detail the fractional calculus approach you mention. Hopefully it will reveal some representation of the $\lambda$ function on the complex plane and I can compare it to the result I obtained JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 02/19/2023, 10:08 AM (02/19/2023, 09:30 AM)Caleb Wrote: (02/19/2023, 09:13 AM)JmsNxn Wrote: I believe you are confused, Caleb. To ask for an analytic continuation of $\lambda(n)$--is very unnatural. To ask for an analytic continuation, you must specify some additional condition, to what the continuation looks like. In this case we are very unnatural. Let's say we write: $\lambda(n) =(-1)^{\Omega(n)}\\$ Where $\Omega(n)$--counts how many prime factors $n$ has. When we write: $\Omega(s)\\$ For $\Re(s) > 1$; where $\Omega(s) \Big{|}_{\mathbb{N}} = \Omega(n)$. We have to make an additional distinction claim on $\Omega(s)$. Let's say that $\Omega(s)$ is in our Fractional Calculus space. So let's say: $\Omega(s+1) = \frac{d^{s}}{dw^s}\Big{|}_{w=0} \sum_{n=0}^\infty \Omega(n+1) \frac{(-w)^n}{n!}\\$ In order to call this an interpolation, you have to show that: \begin{align} f(w) &= \sum_{n=0}^\infty \Omega(n+1) \frac{(-w)^n}{n!}\\ \Omega(s+1) &= \frac{1}{\Gamma(1-s)} \int_0^\infty f(w) w^{-s}\,dw\\ \int_0^\infty |f(w)|\,dw &< \infty\\ \end{align} Now this would perfectly solve your problem. BUT, AND THIS IS A HUGE FUCKING BUT! It's on you to show that: $\int_0^\infty |f(w)|\,dw < \infty$ In order to say that: $\lambda(s)\\$ Is holomorphic. Or it's on you to say that: $\lambda(s+1) = \frac{d^s}{dw^s}\Big{|}_{w=0} \sum_{n=0}^\infty \lambda(n+1) \frac{(-w)^n}{n!}\\$ And test that: $\int_0^\infty \left| \sum_{n=0}^\infty \lambda(n+1) \frac{(-w)^n}{n!}\right| \,dw < \infty\\$ BUT YOU HAVE TO PROVE THIS. AND IT IS SUPER NONTRIVIAL!!!!!!! Let's instead look at your problem at face value: $\sum_{n=0}^\infty \frac{\lambda(n) z^n}{1-z^n} = \sum_{n=0}^\infty z^{n^{2}}\\$ Setting $\lambda(0) = 1$ and $\Omega(0) = 0$. The question, and area of research you are broaching on, Caleb. Is an old avenue of research. I am not that well versed in it; but it's referred to as: THE CIRCLE METHOD. Or, Ramanujan's and Hardy's CIRCLE METHOD. What happens is close to what you are writing. But it gets very harder. $\sum_{n=0}^\infty z^{n^2} = \sum_{n=0}^M \frac{\lambda(n)z^n}{1-z^n} + \sum_{j=0}^M H_j(z)\\$ Where as $M\to \infty$, the sum $H_j(z)$ turns into asymptotic series. Which allow for estimates on growth. END OF STORY. Your quest to analytically continue this sum is futile. The function: $F(z) = \sum_{n=0}^\infty z^{n^2}\\$ Has a dense amount of singularities on $|z| =1$. This function is reducible into: $F(e^{2\pi i x}) = \sum_{n=0}^\infty e^{2\pi i x n^2}\\$ Which is holomorphic for $\Im(x) > 0$. Where a natural boundary appears along $x \in \mathbb{R}$. There is no analytic continuation. But for; $\Im(x) < 0$ we have: \begin{align} F(e^{2\pi i x}) &= \overline{\sum_{n=0}^\infty e^{2\pi i \overline{x}n^2}}\\ &= \sum_{n=0}^\infty e^{-2\pi i x n^2}\\ \end{align} And this finds $F(z)$ for $|z| >1$. But each are not holomorphic on the unit circle $|z| = 1$. Where upon we have two different functions; one holomorphic for $|z| < 1$--the other for $|z| > 1$. Where each complements the other; and both blow up as $|z|\to 1$. In conclusion; we have shown asymptotics on $\lambda(n)$ and $\Omega(n)$... Please state them! Regards, JamesYou are correct in stating that these are not *analytic* continuations-- this is a generalized analytical continuation. There is a particular case where analytical continuation makes sense beyond natural boundaries, which is the case of series like  $\sum \frac{1}{1-e^{i \theta_n}}$ Where $\theta_n$ creates a natural boundary by being dense on the unit circle. This case, we have a very natural candidate for the analytical continuation, which is the limit obtained by looking at the sequence of partial sums. Poincare studied this particular case in detail, and showed some interesting ways in which the sum inside the natural boundary and outside the natural boundary are connected. I believe (Weirestrass?) also studied generalized analytical continuation in detail, and was looking for more general ways to connect functions besides analytical continuation.  Therefore, the idea would be that we can naturally continue the series $\sum \frac{\lambda(k)}{1-x^k}$ using generalized analytical continuation, and then in the processing of linking that series to the theta function series, we pick up some new residues that depend on $\lambda(z)$ at imaginary values. I'm hoping there is some canonical way to extend $\lambda(n)$ to the compelx plane. Of course its not unique, but I was hoping this type of idea had been considered before to at least give some suggestions on the right approach. I will look in more detail the fractional calculus approach you mention. Hopefully it will reveal some representation of the $\lambda$ function on the complex plane and I can compare it to the result I obtained Ramanujan and Hardy LITTLE CIRCLE METHOD! The statement you are trying to make is: $F(t+ e^{2\pi i \theta}) = \sum_{n=0}^\infty a_n t^n\\$ Where $a_n = O(n!)$ and this is an asymptotic series! Please! READ ABOUT LITTLE CIRCLE METHOD! And the different ways of doing this. JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 02/19/2023, 11:58 AM (This post was last modified: 02/19/2023, 12:59 PM by JmsNxn.) I am going to explain The Little Circle Method; with the pivotal example. I'm not going to do the rigor, or any of the hard math. I'm just going to give the story board; of how this behaves. This is a very deep subject. I am not a master, or someone who is exceptionally good at these types of questions. But they appear, and I know a fair amount of literature. We will call the function $\#n$ as the number of partitions of the number $n \in \mathbb{N}$. In which: \begin{align} \# 0 &= 1\\ \# 1 &= 1\\ \# 2 &= 2\\ \# 3 &= 3\\ \# 4 &= 5\\ &\vdots \\ \#n &= \#\{A \in \{\mathbb{N}\cup \{0\}\}^n\,|\,\sum_{j=1}^n A[j] = n\}\\ \end{align} The function $\#n$ is known as the partition function; or the additive partition function. The following result; low and behold; dates to Euler: $\sum_{n=0}^\infty \frac{x^n}{n!} \#n = \prod_{j=1}^\infty \frac{1}{1-x^j}\\$ Euler proved his own results about this; which culminate in The Pentagonal Number Theorem. Which honestly, is such a bananas kind of result. You're missing out if you haven't seen it. We won't need that; we just need the above expression. The goal that Ramanujan, and later, Hardy, set out; was to construct an analytic expression for $\# n$. The truth is, no analytic expression exists which satisfied any of their constraints. Largely because; they weren't integrable in the manner they hoped. Instead, the best they could do, was find an asymptotic solution. Begin by calling: $F(z) = \prod_{j=1}^\infty \frac{1}{1-z^j}\\$ At every rational point $z^k = 1$, we have a pole. And therefore, for $|z| = 1$ we have a dense amount of singularities. There is no beating this. There is no making it better. Additionally, we have the Euler formula (And For Fuck's Sake, this is why you study Euler!), which is written: $F(z) = \sum_{n=0}^\infty \#n \frac{z^n}{n!}\\$ You are trying to guess something similar that Ramanujan was trying to guess. He wanted to write $\# n \approx P(n)$, and he wanted arbitrary accuracy. (Please recall, I'm just giving a rough history overview, and relating your problem in broad motions). [Enter LITTLE CIRCLE] Let us take $\gamma_\rho(t) : [0,2\pi] \to \rho e^{2 \pi i t}$. Where $0 < \rho < 1$. Then, we have a formula for $\#n$ which is written as: $\#n = \frac{1}{2\pi i} \int_{\gamma_\rho} \frac{F(z)}{z^{n+1}}\,dz\\$ Where now, this result holds by Cauchy's Integral theorem, and there's really no magic involved. $\frac{1}{2\pi i} \int_{\gamma_\rho} \frac{\sum_{n=0}^\infty c_n z^n}{z^{n+1}}\,dz = c_n\\$ Which is just Residue magic... The idea of the Little Circle method, is to let $\rho \to 1$. But as we let $\rho \to 1$, we produce singularities at all the values $z^k = 1$. And we want to avoid these values. So we make cuts in the perfect circle of $\gamma_\rho$. We make "little circles", in the contour $\gamma_\rho$. What this eventually does (I'm not going to explain it, because it takes 40 pages of hard math to get here, I'm just giving you an idea), is that we get: $\#n = \frac{1}{2\pi i}\int_{\tau_\rho} \frac{F(z)}{z^{n+1}}\,dz + \sum_{k=1}^M H_k(n)\\$ The arc $\tau_\rho$, is a broken circle, with small circles deleted. And the sum is up to $M$ terms, where $M = M(\rho)$. And as $\rho \to 1$, the small circles get smaller, and $M\to \infty$. Where finally we end up with Ramanujan's first REAL HANDS DOWN UNBELIEVABLE RESULT, that $\#n$ can be approximated with an exponential series. Because $H_k(n)$ devolves to an $O(e^{a_k\sqrt{n}})$  kind of expression. You are trying to do something similar, Caleb. Whether you see it or not. You are. And I wish you luck. I'm not trying to discourage you. I just want you to know that, studying the asymptotic boundary values of: $G(z)= \sum_{n=0}^\infty z^{n^2}\\$ And trying to guess the asymptotics of $\lambda(n)$ from this--requires appreciating history of these results. Nothing but love, Caleb.    Just giving my two cents. I'm excited for everything you post. You're going places if you keep working like this! I mean that sincerely! I might seem like a no-one, but I've met a lot of "famous mathematicians"--and I can smell it on you. Keep hitting the books and doing the raw demonstrations you are doing Sincerely, James PS: The first German mathematician to run through this; discovered a faster Exponential square root series than Ramanujan-Hardy did. He also pretty much gave the "reduce to its elements" Little Circle, analysis. So, if you want the more modern, better explanation, study Hans Rademacher. Dude just stumbled upon a better Little Circle analysis. And the Germans do it better tommy1729 Ultimate Fellow Posts: 1,924 Threads: 415 Joined: Feb 2009 02/20/2023, 12:16 AM (02/19/2023, 08:59 AM)Caleb Wrote: Hmm, this might be non-sense, but I'll write it anyway.  The goal will be to construct some functions made up of lambda that allow us to compute $\lambda(n)$ while only evaulating it on the positive integers. One very useful identity we can use is that $g(n) = \sum_{n=1}^\infty \frac{\lambda(n)}{n^s} = \frac{\zeta(2s)}{\zeta(s)}$ Now, let's consider the two functions   $\sum_{n=0}^\infty \frac{g(-n)}{z^{n+1}} \qquad \sum_{k=1}^\infty \frac{\lambda(k)}{z-k}$ These are linked in the following way. We have that $\star_1 \quad \sum_{n=0}^\infty \frac{g(-n)}{z^{n+1}} = \sum_{n=0}^\infty \frac{1}{z^{n+1}} \sum_{k=1}^\infty \lambda(k)k^n =$  $\sum_{k=1}^\infty \frac{\lambda(k)}{z} \sum_{n=0}^\infty (\frac{k}{z})^n=$ $\star_2 \quad \sum_{k=1}^\infty \lambda(k) \frac{1}{z-k}$ ... Im sorry Caleb but as stated that does not make sense. First you define  $g(n) = \sum_{n=1}^\infty \frac{\lambda(n)}{n^s} = \frac{\zeta(2s)}{\zeta(s)}$ This is a function of s !! n is your index of summation. you even get g(n) = zeta(2s)/zeta(s) the n magically dissapeared. the things that follow from it are even worse. you cannot simple toy with these variables n , s , z etc like you do. I wasted time looking for a typo or so, but it just does not make sense. *** Ok so you have a natural boundary at the unit circle. The most logical way to get out is a reflexion formula to get the interior biholomorphic to the exterior of the unit circle. We also want it to be almost continu near the unit circle. So we get f(z) = f( conjugate 1/z )    or f(z)  =  conjugate f ( 1/z ) where conjugate is the complex conjugate ofcourse. ( this is somewhat analogue of automorphic functions (or q-series) such as the 2 functional equations satisfied by the dedekind eta function ) This way the unit circle works a bit like a mirror. This is the most mathematical and physical and logical idea I can see. However why would that be interesting ? Notice the riemann mapping theorem strongly limits the ways a circle maps to itself , or the way a function maps to the circle. If you want non-analytic ( not even anti-analytic ) reflection formulas you need more motivation ( because it seems random ) and you would loose analyticity outside the unit circle ! Which you actually dont want right ? Not trying to be rude. regards tommy1729 « Next Oldest | Next Newest »

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