0

I’d like an image and/or a series for a real, nowhere analytic, smooth everywhere function $f(x)$ with a Maclaurin series of $0$ i.e. $f^{(n)}(0)=0$ for $n\in\mathbb{N}$. The easiest way to generate such a function would be to use a smooth everywhere, analytic nowhere function and subtract from it its own Maclaurin series.

The reason for this request is to get a stronger intuition for how smooth functions are more “chaotic” than analytic functions. Such a flat function can be well approximated by the $0$ function precisely at $x=0$, but this approximation quickly deteriorates away from the origin in some sense. Seeing this visually would help my intuition.

Null Simplex
  • 421

1 Answers1

2

The Fabius function is infinitely differentiable, but nowhere analytic. But a plot of this function looks extremely well behaved: see this math.stackexchange question. As pointed in the answer to the above question, at most points, the Taylor series diverges while at dyadic rational points, the Taylor series is a polynomial which obviously converges, but it cannot converge to the Fabius function for the simple reason that that function is not a polynomial. This example illustrates the two ways in which a smooth function can fail to be real analytic: the Taylor series may diverge, or it may converge to the wrong function.

I think this example also shows that the difference between smooth and analytic is not visually discernible from a plot.

However, the divergence of the Taylor series almost everywhere blocks your idea of subtracting this series to get a function with a Taylor series that is zero everywhere. In fact, if $f'$ is zero everywhere, then $f$ is constant by the mean value theorem, and the constant function is real analytic. What functions like $\exp(-1/x^2)$ show is that a non zero function can have all its derivatives equal to zero at one point, but this cannot happen everywhere because of the mean value theorem.

Jayanth R Varma
  • 2,914
  • After thinking about your answer, I realize that if two functions $f$ and $g$ have the same Taylor series at some point $x$, and a third function $h$ has a different Taylor series at $x$, then there exists some $\epsilon>0$ such that $|f(x+\lambda)-g(x+\lambda)|$ is less than both $|f(x+\lambda)-h(x+\lambda)|$ and $|g(x+\lambda)-h(x+\lambda)|$ where $0<|\lambda|<\epsilon$. In other words, two functions with the same taylor series behave extremely similar at that point relative to functions with a different taylor series. Not confident if this works for divergent taylor series. – Null Simplex Mar 18 '25 at 09:45
  • 1
    It would be useful to use Taylor's theorem which is a truncated Taylor series with a remainder term. For the smooth functions that you are considering these always exist and there are no convergence issues. The Peano form of the remainder will perhaps give you what you want. – Jayanth R Varma Mar 18 '25 at 10:42
  • 1
    If the Taylor series of $f$ and $g$ agree to $k$ terms but that of $h$ agrees with $f$ and $g$ only to $k-1$ terms, then using the Peano form of the remainder, $f-g$ can be written as $u(\lambda)\lambda^k$ where $\lim_{\lambda \to 0}u(\lambda)=0$ but $f-h$ will have a term $v(\lambda)\lambda^k$ where $\lim_{\lambda \to 0}v(\lambda)$ will not be zero but will instead be the difference between the $k$'th order derivatives of $f$ and $h$ – Jayanth R Varma Mar 18 '25 at 10:45
  • If $f$ is analytic, $g$ is smooth but nowhere analytic, and they have the same Taylor series at $x$, do truncated Taylor series locally approximate both functions near $x$ equally well at $x$, or is the truncated Taylor series approximation of $f$ more accurate in some sense? – Null Simplex Mar 19 '25 at 02:21
  • 1
    I do not know. But if $f=x+x^2$ and we subtract a flat function to get $g=f-\exp(-1/x^2)$ and we truncate the Taylor series after the linear term, the linear approximation $x$ will approximate $g$ better than $f$ at $x=0$? But $g$ is analytic everywhere except at $0$, and so does not meet your requirement of being analytic nowhere. – Jayanth R Varma Mar 19 '25 at 04:59