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As you could see in Wikipedia the Fabius function is a known "example of an infinitely differentiable function that is nowhere analytic".

In the following answer where it is explained how is nowhere analytic its shown that (also it is Wikipedia): $$F'(x)=\begin{cases}2F(2x)&x\in[0,1/2]\\ 2F(2(1-x))& x\in[1/2,1]\end{cases}$$

And here is where something I don't understand: the Fabius function showed a behavior of flat function near zero $x=0$ and at the top near $x=1$, so since its derivative matches a scaled version of the function, this means that this "flat top" will be also in its derivative $F'(x)$, implying that there is a neighborhood near $x=\frac12$ where the function have constant slope, so in this neighborhood the function should be analytical since is going to be fitted perfectly by polynomial of order 1 (its just a straight line!).

So in order to be nowhere analytical near $x=\frac12$, it will imply that the Fabius function cannot be a flat function near $x=0$ neither $x=1$, but from the plots through polynomial approximations it really looks like is flat at these points. So I don't know what is happening: It is possible to prove that the Fabius function is not a flat function?

These doubt rises in this other question's answers by @Artes from where I am taking the following plot of $F'(x)$:

Derivative of the Fabius function

Even if I plot a polynomial approximation of $F(x+1)$ obtained as is shown in this answer (considering $m=3$) with function that do have a flat top like $g(x)=\left(1+\exp\left(\frac{1-2|x|}{x^2-|x|}\right)\right)^{-1}$ as you could see here is hard to believe it is not a flat function.

Polynomial vs flat function

Maybe I have misunderstood what means to be nowhere analytical.

Added later

Using what is stated in the mentioned answer that the Fabius function fulfill: $$ \frac{d^n}{dx^n}F(x)\biggr|_{x=\frac12} = 0,\quad\forall n>1,\,n\in\mathbb{Z}$$ I should have that the Taylor Series at $x=\frac{1}{2}$ will be: $$T(x) = \sum\limits_{n=0}^{\infty}\frac{F^{(n)}(a)}{n!}(x-a)^n \Biggr|_{a=\frac12} = F\left(\frac12\right)+F'\left(\frac12\right)\left(x-\frac12\right)$$ and from the Wikipedia I know that $ F\left(\frac12\right) = \frac12$ and $F(1)=1$, and also from the formulas shown at the beginning one can see that: $$\lim\limits_{x\to \frac12^{\pm}}F'(x) = 2F(1)=2 \Rightarrow F'\left(\frac12\right) = 2$$ With this is possible to find the Taylor series at $x=1/2$ as: $$T(x) = 2x -\frac12$$

So for me the Fabius function do "looks analytical" at $x = 1/2$.

Since I cannot plot exactly the Fabius function, if I compare it with a polynomial approximation (as shown here for $m=3$) which should look analytical as is polynomial, but also with another approximation: $$q(x) = \left(1+\exp\left(\frac{1-2|x-1|}{(x-1)^2-|x-1|}\right)\right)^{-1}$$ which has flat top but it don't have a "straight-line slope" near $x=\frac12$ (also fulfill $F(0)=q(0^+)=0$, $F(1)=q(1)=1$, and $F(1/2)=q(1/2)=1/2$), so I could think as the real Fabius function being something "close" to both of them.

Seeing them plotted against $T(x)$ make hard to believe it is not analytical at $x = 1/2$ since the straight line is quite a good approximation to both functions (and the fact it have a defined Taylor series at $x=\frac12$):

Taylor series

Hope you could explain in an intuitive way How it is nowhere analytical when it have a defined Taylor series expansion at $x=\frac12$.

Added after the question was answered

I have realized later that the following function: $$q(x)=\frac{1}{1+\exp\left(\frac{1-2|x|}{x^2-|x|}\right)}$$ Matches "almost perfectly" the comstruction of a "smooth" transition curve show in Wikipedia Non-analytic smooth function:

  1. Define $$f(x)=\begin{cases}e^{-\frac{1}{x}},\quad x>0\\ 0,\quad x\leq 0\end{cases}$$
  2. Define $$p(x)=\frac{f(x)}{f(x)+f(1-x)}$$
  3. Then for some real valued constants $a<b<c<d$ one could built a smooth bump function as: $$r(x)=p\left(\frac{x-a}{b-a}\right)\cdot p\left(\frac{d-x}{d-c}\right)$$ which is non-zero in $[a,\ d]$ such is flat on $[b,\ c]$

One can "experimentally" see that for $[a,\ d]\equiv [-1,\ 1]$, by chosing $b\to 0,\ c\to 0$ then $q(x)$ matches $r(x)$ on $(-1,\ 1)\setminus\{0\}$.

Then, by playing with $q(x)$ I realized two interesting things about the function: $$q(x,a)=\frac{1}{1+\exp\left(\frac{a(1-2|x|)}{x^2-|x|}\right)}$$

  • The function $q(x,4)$ fits "really good" the function $r(x)$ if I change the definition of $f(x)$ to $$f(x)=\begin{cases}e^{-\frac{1}{x^2}},\quad x>0\\ 0,\quad x\leq 0\end{cases}$$
  • The function $q(x,\frac{\sqrt{3}}{2})$ "looks like" a perfect smooth transition function with straight-line edges, fulfilling having a derivative with flat tops - but unfortunately it don't fulfill solving a DDE like $y'(x)=k\left(y(2x+1)-y(2x-1)\right)$ for some constant $k$.

You could see them on Desmos: flat slope

Jointly with what was noted by comments and answers, this kind of discard the flat slope on an open interval of the Fabius function.

Joako
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    “Approximated by polynomials” and “analytic” are not always the same – Тyma Gaidash Jun 01 '23 at 00:33
  • @TymaGaidash The intuition I have about an analytical function in some interval is that are functions that could fitted perfectly by a power series in the mentioned interval. In that sense, if a straight line suit the function in a non-zero measure interval, it should be analytical. I think the converse should be true, proving that the Fabius function is not a flat function, but I don't know how to do it. – Joako Jun 01 '23 at 00:41
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    The function does not have constant slope in a neighbourhood of $x=0$ or $x=1$. – s.harp Jun 01 '23 at 07:14
  • @s.harp Thanks for commenting. Do you have any reference where is proved the Fabius function is not a flat function near $x=0$ and $x=1$? – Joako Jun 01 '23 at 07:53
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    You are confusing two notions. Functions that are flat at certain points in the sense that you mention do not need to have constant slope in some neighbourhood of those points. The Fabius function is flat at the points $x=0$, $x=1$, this follows from the relation between $F'$ and $F$. – s.harp Jun 01 '23 at 08:17
  • @s.harp What I understand is if is flat at a point (where there is no Taylor Series), the only match is a constant value. Now given it have at least two points where is flat, since $F'(x)$ is a scaled version of $F(2x)$, then $F'(x)$ have also some points where is flat, let call them $x_0$, then this should imply that at $x_0$ the derivative matches a constant value, which should be a point where $F(x_0)$ behave as a straight line with slope $F'(x_0)$. I am thinking in the points at the edges of smooth bump functions, like $e^{-\log^2(1-x^2)}$ where the function also match a constant. – Joako Jun 01 '23 at 08:32
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    Look at $x\mapsto e^{-1/x^2}$, this is flat at $x=0$ but note that there is no neighbourhood around $0$ in which the above function is constant. This kind of an implication ("flat $\implies$ constant in some neighbourhood") is simply false, there is no connection. – s.harp Jun 01 '23 at 09:18
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    @s.harp Maybe I have the wrong intuition, but from what I have done in WolframAlpha I believe that it was the case, that since for matching a constant all derivatives becomes zero, that implies some neighbourhood becomes flat: at least numerically is what that software says: $$\lim_{x\to 0.01^{\pm}}e^{-1/x^2}=0$$ – Joako Jun 01 '23 at 15:16
  • @s.harp also from the formulas one could see that $$\lim\limits_{x\to\frac12^{\pm}}F(x)=2$$ and from your answer in the original question $F''(1/2)=0$ so at least in this point looks "feasible" the match by a 1st order polynomial – Joako Jun 01 '23 at 15:54
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    $e^{-1/0.01^2}=\frac1{e^{10^4}}$, which is minuscule but not zero. In fact $e^{-1/x^2}$ is never zero if $-1/x^2\in\Bbb R$ (ie if $x\neq0$). All derivatives being zero does not imply that the function is constant in some neighbourhood - that implication only follows if the function is analytic at that point. – s.harp Jun 01 '23 at 16:32
  • @s.harp I think I don't understand what to be "nowhere analytical" means: I have updated the question with the Taylor series expansion of $F(1/2)$... if the expansion do exist at some point, I don't get what mathematicians want to mean with "nowhere": surely is different to what intuition tells – Joako Jun 01 '23 at 22:09

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A function $f(x)$ is real-analytic$^{[1]}$ at a given point if it is infinitely differentiable at that point (and so, it has the Taylor series at that point) and the Taylor series converges exactly to $f(x)$ in some open interval (however small) containing that point.

How does it apply to the Fabius function? The Fabius function is infinitely differentiable everywhere. For $x<0$, the Fabius function is exactly $0$, so its Taylor series is also $0$, so it coincides with the function in any open interval that is small enough to not include any point $x\ge0$, so the Fabius function is real-analytic there. At $x=0$, its Taylor series is also $0$, but any open interval containing $x=0$ necessarily contains some point $0<x<1$, where the Fabius function is not $0$, so its Taylor series does not converge to the Fabius function in any open interval. So, the Fabius function is not real-analytic at $x=0$. Now, what about $x>0$? At dyadic rational points, the Taylor series of the Fabius function has only a finite number of non-zero coefficients, so it's just a polynomial and has an infinite radius of convergence. But the Fabius function is not a polynomial, so, although the Taylor series converges, it doesn't converge to the Fabius function, and hence, it's not real-analytic there. At all the other points (that are not dyadic rationals), the Taylor series has an infinite number of terms, but they grow too fast and its radius of convergence is $0$, so it doesn't converge in any open interval.

So, the Fabius function is real-analytic for all $x<0$, and is not real-analytic for all $x\ge0$.

Vladimir Reshetnikov
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  • Thank you very much for the explanation. For what I read it look is something more about How is definined to be nowhere: since looks like to be analytic requires to be a power series in an open interval, then having a taylor expansion in one non-zero measure point is not considered to be somewhere XD – Joako Jun 02 '23 at 00:07
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    Well, every infinitely differentiable real-valued function has the Taylor series whose radius of convergence is at least $0$, and its value at that point (the $0$-order term of the Taylor series) necessarily agrees with the value of the function. The condition of being real-analytic has to add something else to that. It requires that the radius of convergence is $>0$ and the result of the summation of the Taylor series exactly agrees with the function on at least some open interval around that point, however small. – Vladimir Reshetnikov Jun 02 '23 at 17:36
  • Didn't thought before asking that if a function is differentiable then obviously has a tangent line, Lol. It makes sense now. Thank you very much. – Joako Jun 02 '23 at 19:26
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    @Joako: FYI, one can consider "has a Taylor series with a positive radius of convergence at a specified point" separate from "real-analytic at a specified point", somewhat analogous to how for continuity one can consider "has a limit at a specified point" separate from "limit equals functional value at a specified point". For details and references, see my 9 May 2002 and 19 May 2002 sci.math posts. – Dave L. Renfro Jun 06 '23 at 16:09
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    Regarding my previous comment, see also Andrés E. Caicedo's answer to the MSE question Is it possible for a function to be smooth everywhere, analytic nowhere, yet Taylor series at any point converges in a nonzero radius? – Dave L. Renfro Jun 06 '23 at 16:15
  • @DaveL.Renfro It is interesting indeed: I had a mistaken intuition about being analytic: the answer above makes me understand that just having a Taylor series' expansion at a point going always feasible if the function is differentiable, since at least has a tangent line at that point. Now since Taylor series fails to exists, it would be interesting to now how the fammily of approximations through piecewise tangent lines could be done (and how are called), as example the polynomials shown in this answer I believe kind of doing that. – Joako Jun 06 '23 at 17:03
  • @DaveL.Renfro Also, proving if trully there are a non-zero meassure interval where the function $q(x,a)$ fits exactly the lines $$L^\pm =\pm 1.7(x\pm 1)-0.35$$ at $x=\frac12$... which visualy I found to be near $a=0.85$ but I don't know how to formaly prove it. – Joako Jun 06 '23 at 17:08