Weird Shit with Numbers

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?