0

Suppose a germ of a function at a point (from one direction) or at infinity has finite, infinite, oscillating and infinitesimal parts.

Is it possible to separate it into all four, so that

  • The infinite part was monotonously growing (by absolute value and without changing sign), with all its derivatives also were growing monotonously.

  • The finite part was a constant equal to the regularized value of the integral of the derivative of the function at the point.

  • The oscillating part was neither growing, nor decreasing by absolute value, but have no limit and changing sign, also having convergent Fourier series.

  • The infinitesimal part heaving limit $0$?

Is there an algorithm?

Anixx
  • 10,161
  • Could you be more explicit? When you say "separate", do you mean write as a sum? What does "the regularized value of the integral of the derivative of the function at the point" mean? Is the original function supposed to be differentiable? If so, do you simply mean the limit of the function at the point? Is the "oscillating part" supposed to be periodic? Any (relevant) restriction on the initial function? – nombre Jan 05 '21 at 20:26
  • @nombre 1. Yes, writing it as a sum. 2. https://math.stackexchange.com/questions/3893672/how-do-you-regularize-infinite-integrals/3952388#3952388 3. Yes, let it be differentiable 4. Not limit but a germ, a more general concept. If the function has a limit, the finite part is equal to it. 5. No 6. Well, we can consider only analytic functions for simplicity. – Anixx Jan 05 '21 at 20:31
  • I am not sure I understand what the regularized value means since it is not defined in the link you give. But in any case, how would you decompose the function $x\mapsto \sin(x^2)$ on $\mathbb{R}$? – nombre Jan 05 '21 at 20:54
  • @nombre well, the finite part is zero (Cesaro regularization of integral of cosine). Infinite part is zero. Infinitesimal part is zero. There is only oscillating part. – Anixx Jan 05 '21 at 20:57
  • So this function has convergent Fourier series? (I don't know much about Fourier series, I'm just asking) If that's the case, then I suggest first trying to prove that any real analytic function can be written as a sum of four such functions. – nombre Jan 05 '21 at 21:18
  • @nombre I cannot. And even the outlines for those parts that I gave are loose. – Anixx Jan 05 '21 at 21:24

2 Answers2

1

This is not an anwser, but the OP might find it interesting.

I don't think there is an algorithmic way to do this. For now I am not even sure that such a decomposition exists even for analytic functions. However this is difficult to tell because I don't understand the third category (I take the second one to be "constant function"), because for me, Fourier series regard periodic functions. But perhaps you meant convergent (or existing) Fourier transform?

But if one is willing to consider only certain types of functions as explained below, one may have a simpler (but more restrictive) decomposition.

Let $\mathcal{R}$ denote the set of germs $[f]$ at $+\infty$ of real-valued partial functions $f$ such that for all $n \in \mathbb{N}$, there is $a_n \in \mathbb{R}$ such that $f$ is defined and $n$-times differentiable on $(a_n,+\infty)$. Under pointwise sum and product, this is a ring. Under pointwise derivation, it is a differential ring. It is equipped with the partial order $[f]<[g]$ iff $f(x)<g(x)$ for sufficiently large $x \in \mathbb{R}$.

A Hardy field is a differential subfield $\mathcal{H}$ of $\mathcal{R}$ which contains the set of germs of constant functions. The structure $(\mathcal{H},<)$ is then an ordered field. So in $\mathcal{H}$, any element is either in category $1$, or a sum of one unique element of category $2$ and one unique element of category $4$.

Hardy fields include the fields generated by real constants and the identity function, the exponential function, the logarithm, all functions on neighborhoods of $+\infty$ which are first order definable in the language of exponential fields, or in any o-minimal structure on $\mathbb{R}$. It is conjectured that every Hardy field embeds into a Hardy field where differential polynomials have the intermediate value property. So one can get quite big while satisfying an even stronger sort of separation than that you propose.

By a theorem of Aschenbrenner, van den Dries and van der Hoeven, every Hardy field embeds as a differential field in a field of generalized transseries (surreal numbers). Every such transseries $f$ can be written as $f=f_{\succ}+f_=+f_{\prec}$ where $f_{\succ},f_=$ and $f_{\prec}$ correspond to categories $1$, $2$ and $4$ respectively. The difference here with the general case of Hardy fields is that the function $f\mapsto f_{\succ}$ is an additive morphism (so are the other two).

However the embedding is not unique and not at all constructive in general. In fact, assuming a subfield $\mathcal{H}'$ of $\mathcal{H}$ is embedded as a differential field of transseries, and $\varphi \in \mathcal{H} \setminus \mathcal{H}'$ is d-transcendant over $\mathcal{H}'$, meaning that any differential equation $g_n {\varphi}^{(n)}+...+g_1 \varphi'+ g_0 \varphi=0$ where $g_0,...,g_n \in \mathcal{H}'$ must satisfy $g_0=...=g_n=0$, then there are many transseries $f$ that $\varphi$ could be identified with, and in some cases one can arrange that $f_=$ for instance be any real number.

nombre
  • 5,295
  • 17
  • 28
  • Surreal numbers are well-ordered. The category 3 is not. So, I think we should speak about games here when considering category 3. – Anixx Jan 07 '21 at 19:06
  • I am still unsure what the third category is. Are you talking about Fourier transforms? The surreal numbers are (linearly) ordered, whereas well-orderedness refers to a specific type of linear order. I don't think games would be relevant to this type of question. – nombre Jan 07 '21 at 20:34
  • The third catgory is a germ that 1) Has zero mean value at infinity 2) Has alternating signs 3) The sum of mean values of positive part and negative part is zero at infinity – Anixx Jan 07 '21 at 22:19
  • @Anixx Okay. I have no intuition as to whether the decomposition exists in the first place. I think the existence of a natural decomposition is highly unlikely but I may be wrong. – nombre Jan 10 '21 at 17:48
0

I do not know exactly how to separate the oscillating part, but if your function's germ at infinity satisfies the Hardy field requirements, it already has no oscillating parts.

Now, we can separate the remaining three.

First, canonically embeed the Hardy field into the surreal numbers.

Second, represent the germ in question as the Conway normal form.

Now, all terms with negative powers compose the infinitesimal part, the real term compose the real part and all terms with positive powers compose the purely-infinite part.

The purely-infinite surreal numbers are denoted as $\mathbb J$. They are also known as "stars" or "divisible integers".

For instance, the germ of $\ln x\in \mathbb J$ (because its Conway form is $\omega^{1/\omega}$, and $1/\omega>0$), while $\ln(x+1)$ does not belong to this set.

Gonshor calls stars purely infinite numbers, and shows that any number is uniquely the sum of a purely infinite number, a real number and an infinitesimal number.

https://www.m-a.org.uk/resources/4H-Jim-Simons-Meet-the-surreal-numbers.pdf

Anixx
  • 10,161