+- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Modding out functional relationships; An introduction to congruent integration. (/showthread.php?tid=1338) Modding out functional relationships; An introduction to congruent integration. - JmsNxn - 06/16/2021 Let's be as rigorous as possible in this post. Let's try to be straight forward too. Let's restrict ourselves to the \bullet and \Omega notation. This post is largely in response to Leo W.'s posts and MphLee's functional relationships. Regardless of this; all of this is drawn from my paper on compositional integration. Let $\mathcal{N},\mathcal{N}'$ be topological neighborhoods of $z = 0$. Let's assume that $\phi(s,z) : \mathcal{S} \times \mathcal{N} \to \mathcal{N}'$ and that $\phi(s,0) = 0$. This will make things a bit simpler as we progress. These aren't necessary conditions; but they're well enough to get at the root of Mphlee'stheory. Let $\gamma : [a,b] \to \mathcal{S}$ be a continuously differentiable arc. Additionally, we can consider $\gamma$ as a path along a larger arc; so that we can take a derivative in $b$. Let's clarify some of the language too. If I write, for $\gamma \subset \mathcal{S}$, $ Y_\gamma(z) = \int_\gamma \phi(s,z)\,ds\bullet z\\ Y_\gamma(z)= ze^{\int_\gamma \frac{\partial}{\partial z} \phi(s,0)\,ds} + ...\\ \frac{\partial}{\partial b} Y_\gamma = \phi(\gamma(b), Y_\gamma)\gamma'(b)\\$ This produces a well-behaved compositional integral, such that, $ \forall \gamma\,\,Y_\gamma : \mathcal{N}_\gamma \to \mathcal{N}_\gamma'\\$ These are topologically neighborhoods of 0; so just call this $\mathcal{N}$: "there exists a neighborhood around zero". With that cleared up, we can talk about when $\gamma$ is a Jordan curve. Which basically just means $\gamma(b) = \gamma(a)$ and we don't self intersect on our path. Theorem 1: Let $\mathcal{S}$ be simply connected. For all Jordan curves $\gamma \subset \mathcal{S}$: $ \int_\gamma \phi(s,z)\,ds\bullet z = z\\$ Now, what happens when we start putting poles everywhere? ... I hope you remember our discussion about the residual theorem. $ \int_\gamma f(s) \phi(s,z)\,ds\bullet z \simeq \Omega_j \text{Rsd}(s=\zeta_j,f\phi;z)\bullet z\\$ With which you can visualize with Mphlee's beautiful picture: But, we're going to move $\alpha$ around to whatever you want. And make these statements equivalent. For a more detailed description: I suggest reading this thread Each of these closed contours are equivalent to each other under some mapping $\sigma:[a,b] \to \mathcal{S}$ in which, $ \int_\gamma f(s,z)\,ds\bullet z = \int_\sigma \bullet \int_{\varphi} \bullet \int_{\sigma^{-1}} f(s,z)\,ds\bullet z\\$ We can mod out by this equivalence relation; and we get our desired first object. $ \oint_\gamma f(s,z)\,ds\bullet z = \Omega_j \text{Rsd}(s = \zeta_j,f;z) \bullet z\\$ So, we're going to invent The Congruent Integral. Which is a modded out version of the above formula. In which the fundamental identity is for any two jordan curves $\gamma,\varphi$ (which contain the same singularities), $ \oint_\gamma f(s) \phi(s,z)\,ds\bullet z = \oint_\varphi f(s)\phi(s,z)\,ds\bullet z = \Omega_j \text{Rsd}(s=\zeta_j,f\phi;z)\bullet z\\$ Which is up to conjugation... (I've explained this over and over, I hope you remember). Now from this you can make another "mod out". If $h(s)$ is holomorphic about $s = \zeta$; then, $ \oint_\gamma (f(s)+h(s))\phi(s,z)\,ds\bullet z = \oint _\gamma f(s) \phi(s,z)\,ds\bullet z\\$ Which is proved using a limit. Suppose that $\gamma_\delta = \zeta + \delta e^{ix}$, then, $ \int_{\gamma_\delta} (f(s) + h(s))\phi(s,z)\,ds\bullet z - \int_{\gamma_\delta}f(s) \phi(s,z)\,ds\bullet z = \mathcal{O}(\delta)\\$ which implies the two classes agree in a limit. So if we define a brand new equivalence class: $ \oint_\gamma f(s)\phi(s,z)\,ds\bullet z \equiv \oint_\gamma g(s)\phi(s,z)\,ds\bullet z\\$ If we know that $f,g$ share the same poles and residues at the poles... We get the true congruent integral. In which, for a Jordan curve $\gamma\subset \mathcal{S}$ and arbitrary meromorphic functions $f,g$, $ \oint_\gamma (f(s) + g(s))\phi(s,z)\,ds\bullet z = \oint f(s)\phi(s,z)\,ds\bullet \oint g(s)\phi(s,z)\,ds\bullet z\\$ And we've effectively abelianized a lot of tangential relations to MphLee's work. It borders on what Leo W. talked about; but I believe he has his own descriptors. In the book I, then, start taking infinite compositions; which is just $f = \sum_j f_j$; and creating nested integrations $f = \int g$. Where you can do pretty much everything in Cauchy's analysis; but we're in some weird modded out space. Which nonetheless; can be pulled back to a normal complex analytic scenario... Just with a lot of conjugations. I thought I'd post this, mostly, as a direction of thought for all of MphLee's posts lately; and it's my own interpretation. Again, the paper is at, https://arxiv.org/abs/2003.05280 RE: Modding out functional relationships; An introduction to congruent integration. - MphLee - 06/16/2021 This is a good moment for that post! In the last weeks I want back to reading and studding basic stuff to make order in my brain. In fact I was asking recently on MSE also about a path non-abelian algebra/category so to formalize the integral as a functor. Three days ago I started a second complete read (skipping proof details) of your long paper after you were referencing the congruent integral. I'm at page 27 now, I'm surprised how much helpful your forum posts were to understand better your paper. RE: Modding out functional relationships; An introduction to congruent integration. - JmsNxn - 06/17/2021 I thought I'd write some plain examples using $\phi(s,z) = z$; in which we are reduced into the exponential case. $ \int_\gamma f(s)z \,ds\bullet z = ze^{\displaystyle \int_\gamma f(s)\,ds}\\$ Now when we write, $ \int_\gamma (f(s) + g(s))z\,ds\bullet z = \int_\gamma f(s)z \,ds \bullet \int_\gamma g(s)z\,ds\bullet z\\$ We mean that if, $ F(z) = ze^{\displaystyle \int_\gamma f(s)\,ds}\\$ And, $ G(z) = ze^{\displaystyle \int_\gamma g(s)\,ds}\\$ Then, $ F(G(z)) = ze^{\displaystyle \int_\gamma f(s) + g(s)\,ds}\\$ And the residue theorem is, $ \int_\gamma f(s)z\,ds\bullet z = \Omega_j \text{Rsd}(s = \zeta_j,f(s)z;z)\bullet z = \Omega_j ze^{2 \pi i\text{Res}(s=\zeta_j,f(s))}\,\bullet z = z e^{2 \pi i \sum_j \text{Res}(s=\zeta_j, f(s))}$ The thesis of this paper was that in some modded out space; it works exactly the same. In fact, for any function $\phi = \phi(z)$, we can prove this result using all of the old analysis, $ \int_\gamma (f(s) + g(s))\phi(z)\,ds\bullet z = \int_\gamma f(s)\phi(z) \,ds \bullet \int_\gamma g(s)\phi(z)\,ds\bullet z\\$ Which is exactly what mphlee has been talking about lately; and the content of the YT video he posted. It's just written very strangely here. The benefit of writing it this strangely; is that it generalizes in an algebraic way much better than the typical vector space mumbo jumbo. In which we can now talk about, $ \int_\gamma f(s)\phi(s,z)\,ds\bullet z\\$ And if we conjugate these things; and "hide" the conjugations using $\oint$; we get, $ \oint_\gamma (f(s)+g(s))\phi(s,z)\,ds\bullet z = \oint_\gamma f(s)\phi(s,z) \,ds \bullet \oint_\gamma g(s)\phi(s,z)\,ds\bullet z\\$ This means; explicitly; if $f,g$ are meromorphic and $\gamma$ is a Jordan curve: $ F = \int_\gamma f(s)\phi(s,z) \,ds \bullet z\\ G = \int_\gamma g(s)\phi(s,z)\,ds\bullet z\\ H = \int_\gamma (f(s)+g(s))\phi(s,z)\,ds\bullet z\\$ And there exists functions $a,b,c$ such that, $ a(F(a^{-1}(b(G(b^{-1}(z)))))) = c(H(c^{-1}(z)))\\$ And these functions are always solvable as compositional contour integrations. RE: Modding out functional relationships; An introduction to congruent integration. - JmsNxn - 06/23/2021 I thought I'd add a bit about Taylor Series too. If I write, $ F_k(w,z) = \int_\gamma \frac{(w-\zeta)^k}{(s-\zeta)^{k+1}} f(s)z\,ds\bullet z = z e^{\displaystyle 2 \pi i \frac{f^{(k)}(\zeta)}{k!} (w-\zeta)^k}\\$ Then, $ \Omega_{k=1}^\infty F_k(w,z)\bullet z = z e^{\displaystyle 2 \pi i \sum_{k=1}^\infty\frac{f^{(k)}(\zeta)}{k!} (w-\zeta)^k} = z e^{2\pi i f(w)}= \int_\gamma \frac{f(s)z}{s-w}\,ds\bullet z\\$ Or, $ G_k(w,z) = \int_\gamma \frac{(w-\zeta)^k}{(s-\zeta)^{k+1}} f(s)z^2\,ds\bullet z = \frac{1}{\displaystyle 1/z + 2\pi i\frac{f^{(k)}(\zeta)}{k!}(w-\zeta)^k}\\$ Then, similarly, $ \Omega_{j=1}^\infty G_k(w,z)\bullet z = \int_\gamma \frac{f(s)z^2}{s-w}\,ds\bullet z\\$ This always extends to separable functions, in which, $ \Omega_{k=1}^\infty \int_\gamma \frac{(w-\zeta)^k}{(s-\zeta)^{k+1}} f(s)\phi(z)\,ds\bullet z=\int_\gamma \frac{f(s)\phi(z)}{s-w}\,ds\bullet z$ Using the congruent integral we can show, $ \Omega_{k=1}^\infty \oint_\gamma \frac{(w-\zeta)^k}{(s-\zeta)^{k+1}}\phi(s,z)\,ds\bullet z=\oint_\gamma \frac{\phi(s,z)}{s-w}\,ds\bullet z$