mmmmmmaths time

Formerly @positively-knotted. 20s || Maths || Linguistics || "overpowering woke agenda meets book of revelation"

[neatlymanifolded]

Hey! I used to be @positively-knotted on here, posting things about doing a maths phd (4-manifolds) and occasionally other things (which were usually better).

My account got nuked a few weeks ago and I've basically given up on getting it back, so new account I guess. Sad times.

Time to start over I guess :/

[ultramanzer0]

is prev a warrior of light or an agent of darkness

warrior of light

agent of darkness

Remaining time: 11 hours 29 minutes

[algebraic-dumbass]

i'd like to test something

what are you?

I am not a math person and I don't do rock climbing/bouldering

I am a math person and I don't do rock climbing/bouldering

I am not a math person and I do rock climbing/bouldering

I am a math person and I do rock climbing/bouldering

Voting ended onMay 31

("a math person" is a large term. If you enjoy math in any capacity, it doesn't matter if you are highschool student, a grad student, a hobbyist, a tenure-track professor if they still make those, or anything, as long as you enjoy math, you count!)

[algebraic-dumbass]

Results are in: 23,3% of math people do rock climbing, while 19,8% of non-math people do rock climbing.

63,5% of rock climbers are math people, while only 59,5% of all respondents are math people.

therefore, according to the My™ Ass™ Institute of Statistics, this is a significant correlation between being a math person and a rock climber

[grfn-btbtas]

Not sure why but for some reason when I was 18 or 19 I was convinced that becoming an Old English scholar would have been somewhat akin to becoming a genuine wizard, would put me in touch with the ability to manipulate the fabric of reality, as opposed to, you know, being a course of study in a dead language

[neatlymanifolded]

This is the plot of BBC Merlin

[compositionalotterfolk]

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 f∘f∘...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 (⊙ _ ⊙ ) 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 (ᵕ • ᴗ •) 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??

[weirdshitwithnumbers]

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 you’re missing important details.

[compositionalotterfolk]

I left out the details on the first point because I thought it might be unnecessary, but the proof does seem incomplete without it:

Let’s consider a point “p” in A a very small distance “r” 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 “moves.” The real question is “by how much?” 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+f’(x_f)r. This is because our first order Taylor expansion about x_f is: x_f-(x-x_f)f’(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)[f’(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 didn’t 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 it’s all about deciding what versions of our function we are considering. I should probably prove that a valid analytic function exists; but that’ll take me more time. I might post it later.

If this doesn’t seem like enough, or if anyone wants to take a stab at the proof just lmk!

[neatlymanifolded]

Lol Claude just cited me at me* *me at myself? myself at me? none of these sound right

[possiblytheworstrapperofalltime]

88

in the club, i throw shapes like a knight

then I move like a bishop, give the dj a fright

[spectracled-eider]

The age old task for a mathematician to come up with good notation

[depsidase]

[neatlymanifolded]

It's crazy that we can't classify surfaces up to (non-orientable) cobordism in non-orientable 4-manifolds yet. We couldn't even do this in orientable manifolds until this year!

The oriented case was done in the 1950s!!! This cannot be that difficult. What's going on

[max1461]

she overfull on my \hbox until i badness 10000000

[neatlymanifolded]

The experience of being a mother seems very needlessly stressful

[neatlymanifolded]

love it how people's apparent types can vary from "blonde" or "tall" to "bisexuals who used to think they were lesbians and have medically-curious diseases"

[neatlymanifolded]

The Dalabon word for policeman is nguhyirru, lit. "(someone with) angry intestines"

[sqrt-73]

you've heard of optimal square packing, but packing cubes in cubes also exists.

is the optimal for 12 cubes. (image from https://erich-friedman.github.io/packing/cubincub/ )

ok this is even more cursed than squares. i love it and hate it at the same time. my eyes bleed out with hearts to this.

[locally-normal]

Me trying to convince my wife this really is the best way to load a moving truck

[p-adict]

l infinity sphere packing

[locally-normal]

This one weird trick will make it so much easier to pack your infinity spheres!!

[neatlymanifolded]

Fuck I haven't actually posted about maths on here for ages. Been busy with uhhh otherthings

maybe ill do something about the state of unknotting surfaces in S⁴, bc there's been a lot of cool developments there recently that I should be more on top of

[anxeious]