Roos β They/Them β Eighth Year Math Student β Main is @bubbliterally β Looking to become the world's leading expert on the empty set β Icon and banner are by @tamberella
Anya is live and ready to show you everything. Watch her strip, dance, and perform exclusive shows just for you. Interact in real-time and make your fantasies come true.
β Live Streamingβ Interactive Chatβ Private Showsβ HD Quality
Anya is LIVE right now
FREE
Free to watch β’ No registration required β’ HD streaming
Somehow I never realized that in any topological ring containing the rationals, the exponential map must take values in units whenever it converges. Pretty neat!
@lilith-hazel-mathematics said: #Can we get a proof? I couldn't think of one off the top of my head. I assume we need to use Exp(-x) somehow but couldn't figure it out
Yeah that's it. Because x β¦ -x is a homeomorphism, exp(x) converges if and only if exp(-x) does. And because exp(x) and exp(-x) are multiplicative inverses as formal power series, necessarily they will be as elements of the ring as well.
After thinking about it for a while, I'm not really convinced by any part of this argument. At the same time, the claims involved do seem plausible, and I couldn't find any counter examples after testing a few topological rings I'm familiar with.
Regarding my skepticism, I can't imagine how the fact that x β¦ -x is a homeomorphism implies exp(-x) will converge. Namely, exp(-x) is doing something quite different from exp(x), where all and only the odd-indexed terms are negated. That said, if we strengthen our assumption to unconditional convergence, or something a tad stronger like "all subseries converge unconditionally", then I think it follows that exp(-x) converges, so it's not a big deal.
Even assuming they both converge, I tried to carry out the usual argument showing that exp(x)*exp(-x)=1, but I get stuck near the end trying to show that some error terms tend to zero. If we know that the product series will also converge unconditionally, then it should follow easily, but I don't know of any proof to that effect. This is related to an apparently open problem I posted on Math Stack Exchange a few years back.
Maybe there's something I'm missing, but it seems like we may need a fair bit of additional structure on the topological ring to guarantee the claim holds. Of course, if we know exp(a)*exp(b)=exp(a+b) in general, then it becomes trivial, but this seems like a nontrivial assumption.
You know what, that's really fair! I only thought very briefly about that fact that because negation is a homeomorphism, all topological things you can say about x are equivalent to ones you can say about -x, but actually thinking about it in this case it's not obvious at all! The same holds for the product of the series. Good catch!
I suppose a more conservative statement (and one still covering the thing that inspired the thought, namely that it works for the dual numbers) would be that it holds for normed algebras over the reals, as the factorial denominator will always dominate the polynomial terms. I'll admit I haven't thought about that too hard either, though, so maybe I'm missing something obvious!
That matches the types of rings I was testing actually, yeah. One of the examples I tried, where the claim also holds, was the ring of arithmetical functions, using pointwise addition and dirichlet convolution for multiplication, under the pointwise limit topology. I don't think it can be given a norm, but you can impose an extremely similar structure which often serves the same purposes. Incidentally the Dual numbers are obtainable as a quotient of that space.
Somehow I never realized that in any topological ring containing the rationals, the exponential map must take values in units whenever it converges. Pretty neat!
@lilith-hazel-mathematics said: #Can we get a proof? I couldn't think of one off the top of my head. I assume we need to use Exp(-x) somehow but couldn't figure it out
Yeah that's it. Because x β¦ -x is a homeomorphism, exp(x) converges if and only if exp(-x) does. And because exp(x) and exp(-x) are multiplicative inverses as formal power series, necessarily they will be as elements of the ring as well.
After thinking about it for a while, I'm not really convinced by any part of this argument. At the same time, the claims involved do seem plausible, and I couldn't find any counter examples after testing a few topological rings I'm familiar with.
Regarding my skepticism, I can't imagine how the fact that x β¦ -x is a homeomorphism implies exp(-x) will converge. Namely, exp(-x) is doing something quite different from exp(x), where all and only the odd-indexed terms are negated. That said, if we strengthen our assumption to unconditional convergence, or something a tad stronger like "all subseries converge unconditionally", then I think it follows that exp(-x) converges, so it's not a big deal.
Even assuming they both converge, I tried to carry out the usual argument showing that exp(x)*exp(-x)=1, but I get stuck near the end trying to show that some error terms tend to zero. If we know that the product series will also converge unconditionally, then it should follow easily, but I don't know of any proof to that effect. This is related to an apparently open problem I posted on Math Stack Exchange a few years back.
Maybe there's something I'm missing, but it seems like we may need a fair bit of additional structure on the topological ring to guarantee the claim holds. Of course, if we know exp(a)*exp(b)=exp(a+b) in general, then it becomes trivial, but this seems like a nontrivial assumption.
You know what, that's really fair! I only thought very briefly about that fact that because negation is a homeomorphism, all topological things you can say about x are equivalent to ones you can say about -x, but actually thinking about it in this case it's not obvious at all! The same holds for the product of the series. Good catch!
I suppose a more conservative statement (and one still covering the thing that inspired the thought, namely that it works for the dual numbers) would be that it holds for normed algebras over the reals, as the factorial denominator will always dominate the polynomial terms. I'll admit I haven't thought about that too hard either, though, so maybe I'm missing something obvious!
Have you ever wondered what a "half composition" of a function would be?
If we say a function "f" is composed with itself "k" times, we can intuitively interpret that as fβfβ...k times (ie, the kth composition of f), so long as k is an integer; but what about when k isn't an integer?
We could interpret the "a/b th composition of f," as some function "g", such that, when g is composed "b" times with itself, it is equal to f composed "a" times with itself. While this interpretation is sound it still begs a few questions: is there an intuitive way to think of "real" compositions, what about complex?
These questions are interesting, however, I feel they are dwarfed by one you could be considering right now:
If we have some smooth function f, can we find functions that are these "exotic composites" of f? Can we do so for any smooth function?
Recommended preliminaries:
A good understanding of taylor expansions is very necessary.
Familiarity with ODEs would be helpful, but they are not used directly.
I had initially planned to include alt text for formulas but after much trial and error, I have given up. If anyone finds this interesting but was unable to engage with the content properly due to the lack of alt text I would be (genuinely) happy to converse about the subject via dm.
We will be breaking this into two steps:
Determining the a first order approximation of the desired function.
Finding higher order derivatives of such a function function.
We will be using the following notation to denote multiple compositions, as well as preserving the following homomorphism:
Theorem 1:
given a smooth function f:A->A, that fixes for some value x_f in A (ie, f(x_f)=x_f), then:
proof: If we repeatedly apply chain rule we can form the following product:
Clearly, this product can't be made to work for all values of s naturally, as that would inherently require some non-integer composition, but what is well understood is extending the product of a constant.
If we choose to evaluate this at x=x_f a peculiarity arises. Compositions of fixed points do not change, that is f(f(x_f))=f(x_f)=x_f. This allows the product to be simplified as follows:
we can then allow s to vary beyond integer values. QED
Now we could generate an approximation of our desired function nearby the fix point, however, this is fairly pointless.
one could notice that by repeatedly differentiating the initial product, you can actually continue to get higher order approximations, by applying the definition of the geometric series whenever nessesary. However, this endeavor is Extremely computationally intense; and no obvious pattern immediately discern itself (at least for me, I would invite anyone with blatant disregard for their own mental well being to try to find one directly).
Instead, we will be constructing an alternate method of improving our approximation.
Theorem 2:
For a given smooth function f:A->A, with some fix point x_f, and for some integer m>0, the following holds:
where B_{m,k} denote bell polynomial coefficients. (Just to be clear, the "mathcal B" defined for shorthand, it is not related to the statement directly.)
proof: Fa di Bruno's theorem is a generalization of the chain rule to higher order derivatives. It states the following:
where B_{m,k} are bell polynomial coefficients with the derivatives of g as "inputs".
However, rather than let this define the mth derivative of a composition of different functions, we instead use it to take the mth derivative of the s+1th composition of f with itself:
Note that the final term of this sum contains the mth derivative of the sth composition of f, making this expression recursively defined. We rectify this by subtracting this problem term from both sides (also, we begin to evaluate at the fix point, as it allows some slight clean up of the expression).
While this does get all unknowns on the same side of the expression, this is still unsolved.
To begin to solve this we will attempt to factor the "left side." There is a linear operator T_s called the "forward operator" defined so that T_sF(s)=F(s+1). We will simply take it for granted that this operator commutes with differential operators, but there are many proofs of this fact. This allows us to modify the above to a more neat form (the "right side" is unchanged):
our motivation for this simplification will be apparent momentarily. First, we will be applying the inverse of T_s to both sides (ie, the "backwards operator" T_s^-1F(s)=F(s-1)). We can directly apply it to the right by replacing all s with s-1.
(Here we use a slight abuse of notation, as the "1" on the left should in reality be the identity operator "I," however it is not of consequence.)
Note that the expression on the "left" consists of a linear operator L=(1-[f(x_f)]^mT_s^-1) next to an expression. Now, if we where to naively attempt to invert this operator we would have the reciprocal expression in the form 1/(1-L). This is where we use the definition of the geometric series.
We say this series approaches the desired operator so long as it converges (the convergence condition is noted below it). If this is unfamiliar, I would recommend really trying to grasp it (it's a rather interesting concept; try to apply this to (1-L)g(s) and see what happens I would also recommend replacing the main in the sum with L^l).
If we apply this inverse operator to both sides we can reach a formula explicitly for the derivative, assuming it converges.
Since m is finite we can exchange these sums without issue. I'm going to avoid specifics, but we can arrange to reach the final formula. introducing the shorthand.
Substitute the expression on the left side into the convergence condition found before to reach the above convergence condition. note that for any chosen m this expression only references derivatives less than m, which allows us to recursively apply this without issue.
To reach the expression in the beginning simply replace m with m+1 in this formula. QED
This formula allows us to find all derivatives of the desired function so long as the expression converges and the function fixes. The convergence condition is a little rough to determine, however from what I've found it is sufficient that |f'(x_f)|<1 (that proof was more arduous than the one above, so I have omitted it, more for my sake than yours); which is rather restrictive, however, if this condition is not fulfilled, you may simply invert the function on a nearby interval and the derivative must necessarily have a reciprocal magnitude by the inverse function theorem; leaving out only the case where |f'(x_f)|=1
Using theorems 1 and 2, we can obviously create a taylor expansion. Letting the following hold (yes I'm restating the short hand again; sue me):
"Most" differentiable functions defined on f:C->C, have at least one fix point, so as long as you aren't "unlucky" this should work.
A few notes I think are important:
is this the best way to do this?
Probably not. I was looking at math overflow for information, and I heard vague talk about a professor that had solutions to this very problem that worked for |f'(x_f)|=1, but I couldn't find anything substantial at the time. From the discussion I saw I was getting the vibe it was never published so idk.
I couldn't even begin to consider how my formula fairs in terms of convergence speed.
2. Is this even right?
Uhhh, I think; but idk I guess. I don't see much why it isn't correct, and like, I did just go over the proof, so if you think I'm wrong just like... idk what to say (β _ β ) sry. The formula is disastrously hard to test so I just haven't tested it; if anyone wants to test it feel free. Just know if it isn't working you're actually wrong and just need more terms (α΅ β’ α΄ β’) yeah, yeah...
If you do find out I'm wrong I actually would like to know.
3. How long did it take to find this solution?
... I don't wanna talk about that... then why did I bring it up??
One issue here I can see is that you assume that if you have a formula for f*s(x_f) and its derivatives at integer values of s that can be extended to the reals, that it does extend to the reals. Unless you prove continuity with respect to s the non integer values could be unrelated to the integer values. Assuming you do (you probably can), you need to prove that f*s(x_f) is analytic as a function of s (you cannot prove this by the formula for integer values being analytic when extended to reals) in order to say that the analytic continuation of the integer values is equal to the true function.
You probably can do so, but youβre missing important details.
I left out the details on the first point because I thought it might be unnecessary, but the proof does seem incomplete without it:
Letβs consider a point βpβ in A a very small distance βrβ away from the fix point. Now, if we apply f to p and x_f, x_f clearly does not move, however f(p) is not p, ie it βmoves.β The real question is βby how much?β That is totally dependent on the distance r. As we let the distance r grow very small, it is clear that this point p is moved closer to x_f+fβ(x_f)r. This is because our first order Taylor expansion about x_f is: x_f-(x-x_f)fβ(x_f). This kinda feel like where not actually getting anywhere, but the point is if we substitute this approximation into itself repeatedly we at first can say we seem to get x_f-(x-x_f)[fβ(x_f)]^s. Now why is this any better evidence? For linear functions, composition can already be extended, and in exactly this fashion (you can check from here that the homomorphism is held, which does mean that the expression is valid; but uniqueness is uncertain).
The proof that I placed up there is essentially the veiled version of this logic: composing an approximation with certain properties with itself that are certainly preserved.
Tbh, I thought the second point didnβt have much merit, however, I double checked and, it is very important to consider. The entire formulation does not explicitly reference the differentiability of f*s(x) with respect to s anywhere, and almost none of the intermediate steps actually require analyticity in terms of s, only that f*s(x) exists. That is, all but commuting the forward operator and the differentiation operator.
The quickest proof that the forward operator and derivative commute uses the identity e^{partial_s}=T_s, which clearly commutes with other partial derivatives, unless the chosen function is not smooth.
But I do think that this is the only point of issue, as nothing else obviously requires f*s(x) be analytic in terms of s, it all more acts as a proof that it must be (which can be see via and inductive argument).
I think this might be due to a uniqueness issue. If we assume f*s(x) must be analytic, the formulation continues without issue and leads to a result without obvious contradiction. This result should have all the desired properties regardless of the fact that it was constrained. It is possible that there is an f*s(x) that is not analytic, but those answers are not considered.
I think the point you were trying to make was slightly different, but itβs all about deciding what versions of our function we are considering. I should probably prove that a valid analytic function exists; but thatβll take me more time. I might post it later.
If this doesnβt seem like enough, or if anyone wants to take a stab at the proof just lmk!
Anya is live and ready to show you everything. Watch her strip, dance, and perform exclusive shows just for you. Interact in real-time and make your fantasies come true.
β Live Streamingβ Interactive Chatβ Private Showsβ HD Quality
Anya is LIVE right now
FREE
Free to watch β’ No registration required β’ HD streaming
So you can define smooth functions on arbitrary subsets X β ββΏ by saying that f is smooth if it extends to a smooth function (in the normal sense of having however many continuous partial derivatives) on an open neighbourhood of X. This definition then agrees with the usual definition if X is an embedded manifold or whatever. However! What if you consider Cβ° smoothness, i.e. continuity? We already know what it means for functions on arbitrary subsets of ββΏ to be continuous, so the question becomes whether any continuous function on X into a Euclidean space can be extended to a continuous function on an open neighbourhood of X. That doesn't seem so obvious to me!
I'm pretty sure that isn't true: take the characteristic function on [Ο, inf). Restricted to Q, this is continuous, but I don't think it can be extended to a continuous map on R.
Right! That works. It has jump continuities within every positive distance from 0. So this notion of Cβ° smoothness is strictly stronger than continuity.
Somehow I never realized that in any topological ring containing the rationals, the exponential map must take values in units whenever it converges. Pretty neat!
@lilith-hazel-mathematics said: #Can we get a proof? I couldn't think of one off the top of my head. I assume we need to use Exp(-x) somehow but couldn't figure it out
Yeah that's it. Because x β¦ -x is a homeomorphism, exp(x) converges if and only if exp(-x) does. And because exp(x) and exp(-x) are multiplicative inverses as formal power series, necessarily they will be as elements of the ring as well.
I promised some maths so here comes. The overarching theme for these posts is a framework called deformation/rigidity (you have to give it to us, we have pretty cool names) introduced by Sorin Popa in the early 2000's. One crucial aspect here is the notion of an s-malleable deformation. I'm trying to keep things intuitive for now so I won't include yet the actual definition but some intuition below:
Naive definition: One starts with an action of a group G on a von Neumann algebra, say (A, Ο). The commutative counterpart would be an action of a group by (potentially discontinuous) Borel automorphisms of [0,1] which preserve the Lebesgue measure - this is in fact a much more general example than one might expect. Then an s-malleable deformation of this action consist of an action of G on a larger algebra (π, Ο) which contains two copies of A, sitting 'orthogonal' to each other inside π and which can be 'deformed' into one-another by a continuous path of automorphisms that commute with the G-action. Furthermore we ask that this path is symmetric in time.
There will ideally be a number of posts explaining why anyone would care and what this is useful for but what I want to draw attention to is this idea of 'giving yourself space' to work by putting your object inside a larger object. It's really ubiquitous in maths I feel but somehow I get excited every time I run into it even if I've seen it so many times already (maybe because it inevitably goes wrong whenever I try to use it myself).
So really I just though I'd draw attention to this kind of phenomena:
The Lebesgue integral and DCT. At the end of the day, every time I want to compute an integral I do it by reducing to a Riemann integral so clearly the point of the Lebesgue integral is not that it's computable, but that by developing this fancier language you can prove things which are true for the Riemann integral itself but which you couldn't prove before - e.g. without secretly using the Lebesgue measure or Fatou's Lemma try proving that if a sequence of Riemann integrable functions on [0,1] is unformly bounded and converges pointwise to 0 then the integrals converge to 0. You need 0 knowledge about the Lebesgue measure to state it but good luck proving it without the language of Lebesgue (*I am not saying it's impossible but rather non-trivial I think)
Very often in number theory/diophantine equations. Just one example which I've seen recently: say a is a positive integer and consider n=a^3-3a+1. Prove all prime factors of n are either 1 (mod 9) or -1 (mod 9). Clearly you'll want to work in π½_p but really you'll have to move to a degree-2 extension of π½_p in order to write a=t+t^(-1) which will generally not have roots in the base field. (It's a cute exercise to finish the problem for here)
Furstenberg's topological proof for the infinity of primes *but also* his proof of Van der Waerden's Theorem (any way you partition the naturals into finitely many subsets one of them contains arithmetic progressions of arbitrarily long lengths) using the Poincare/Birkhoff multiple recurrence theorem from ergodic theory.
Bernstein's p^Ξ±q^Ξ² theorem using rep theory but also a bit of Galois theory.
Even the basic Cauchy's Theorem from group theory is just a really cute counting argument, I wasn't aware combinatorics was invited to this party??
I'll leave you with two more examples. They both can be done with func-anal tools but you wouldn't expect it. It's quite fun thinking about them:
Let G be a discrete group acting on a set I. Prove that the following are equivalent: 1. all orbits are infinite and 2. whenever A and B are finite subsets of I there is some g in G such that gA is disjoint from B. As an aside, if you endow I with the counting measure this is saying that the action is ergodic iff it is weakly mixing but this is rather irrelevant for the proof.
(This one you can also do with martingales I think but also there is a functional-analytic proof) Suppose on each lattice point (m,n) of the plane you write a real number a_{m,n}. You know that the bi-infinite family (a_{m,n})_{m,n} is bounded and that every number is the (arithmetic) mean of its four neighbours. Show that in fact all the a_{m,n} must be equal.
Anya is live and ready to show you everything. Watch her strip, dance, and perform exclusive shows just for you. Interact in real-time and make your fantasies come true.
β Live Streamingβ Interactive Chatβ Private Showsβ HD Quality
Anya is LIVE right now
FREE
Free to watch β’ No registration required β’ HD streaming
Somehow I never realized that in any topological ring containing the rationals, the exponential map must take values in units whenever it converges. Pretty neat!
So I've been learning some cool maths recently and every time I stumble upon a new gem I feel the need to share it with people.
I've attempted writing notes before but it gets a bit tedious and time consuming, and most of the time I give up on writing the notes before they get to a point where someone else could actually use them.
Maybe making some posts here is the way to go? Unclear if anyone will ever read them but at least I can get the thoughts out of my head.
Unfortunately my research includes words like von Neumann algebras which I cannot always avoid (sometimes I really wish I was a topologist so I could talk about shapes instead) but then again the philosophy is that von Neumann algebras are a model for non-commutative probability/ergodic theory so potentially stay tuned if you might be interestead in what an s-malleable deformation of an action is?
Anya is live and ready to show you everything. Watch her strip, dance, and perform exclusive shows just for you. Interact in real-time and make your fantasies come true.
β Live Streamingβ Interactive Chatβ Private Showsβ HD Quality
Anya is LIVE right now
FREE
Free to watch β’ No registration required β’ HD streaming
