# On the convergence of the function series $\sum_{n=0}^\infty(-1)^n\frac{f^{(n)}(x)}{n!}x^n$

Let $$f$$ be a smooth real function defined around origin. If we differentiate term by term the series

$$\hat{f}(x):=\sum_{n=0}^\infty(-1)^n\frac{f^{(n)}(x)}{n!}x^n$$, we get $$\frac{d}{dx}\hat{f}(x)=0$$. $$\begin{eqnarray}\frac{d}{dt}\hat{f}(t)&=& \sum_{n=0}^{\infty}(-1)^n\frac{f^{(n+1)}(t)}{n!}t^n+ \sum_{n=1}^{\infty}(-1)^n\frac{f^{(n)}(t)}{(n-1)!}t^{n-1}\nonumber\\&=& \sum_{n=0}^{\infty}(-1)^n\frac{f^{(n+1)}(t)}{n!}t^n-\sum_{n=0}^{\infty}(-1)^n\frac{f^{(n+1)}(t)}{n!}t^n \nonumber\\&=&0\nonumber \end{eqnarray}$$ Thus $$\hat{f}(t)$$ should be constant. But in fact we are not allowed to differentiate term by term from a series.

Next, suppose that $$f$$ is a smooth periodic function which by the Fourier analysis we know that it has Fourier expansion. That is we suppose $$f(t)=\sum_{m=-\infty}^\infty c_me^{im\omega t}.$$ Then it is well known that we can differentiate to get $$f^{(n)}(t)=\sum(im\omega)^n c_me^{im\omega t}.$$ Thus $$\begin{eqnarray}\hat{f}(t)&=&\sum_n\sum_m\frac{(-1)^n(im\omega)^{n}}{n!}c_me^{im\omega t}t^n\nonumber\\&=& \sum_m\sum_n(\frac{(-1)^n(im\omega)^{n}}{n!}t^n)c_me^{im\omega t}\nonumber\\&=& \sum_me^{-im\omega t}c_me^{im\omega t}\nonumber\\&=&\sum_{m=-\infty}^{\infty}c_m\nonumber\\&=&f(0).\nonumber \end{eqnarray}$$ Thus again it seems that the series should convergence to a constant. But in the above we have exchanged the order of two infinite sums which are not allowed.

The function $$f(t):=\sum_{m=0}^{\infty}\frac{1}{m!}e^{i2^mt}$$is smooth nowhere analytic, in the sense that convergence radius of the Taylor's series of $$f$$ at each point is zero and therefore $$\hat{f}(t)$$ diverges for all $$t\ne0$$.

The function $$f(t):=\sum_{m=1}^{\infty}\frac{1}{m!}e^{i2^{-m}t}$$is analytic at $$t=0$$ whose convergence radius is infinity. Thus $$\hat{f}(t)$$ converges for all $$t$$ to $$f(0)$$.

In fact one can show that

a) If $$f$$ is analytic at origin then the series $$\hat{f}$$ is convergent uniformly to the constant $$f(0)$$.

b)If $$f$$ is nowhere analytic in the sense that the radius of convergence of the Taylor's series is zero then of course the series is divergent. But if $$f$$ is nowhere analytic in the sense that the radius of convergence of the Taylor's series is positive but the Taylor's series does not converge to the function $$f$$ then the series $$\hat{f}$$ may converge.

c) About the function $$f(x):=e^{\frac{-1}{x^2}},f(0)=0$$ one can show that if the series is convergent then its sum is constant.

d) There are nowhere analytic functions such that the series is convergent in a dense subset to the constant $$f(0)$$ and there are nowhere analytic functions such that the series is divergent everywhere.

Now the main questions are.

1. Is there a smooth function $$f$$ which is not analytic at origin and the series $$\hat{f}$$ is convergent in an interval around origin and the sum is the constant $$f(0).$$?

2. Is there a smooth function $$f$$ which is not analytic at origin and the series $$\hat{f}$$ is convergent in an interval around origin and the sum is not constant.?

3. If we define a linear differential operator of infinite order $$f\mapsto \hat{f}-f(0)$$. Then in above we said that analytic functions at origin are contained in the space of eigenfunctions of the zero eigenvalue of this operator. Now the question arises that: are there nonzero eigenvalues for this operator?

4. For the function $$f(x):=e^{\frac{-1}{x^2}},f(0)=0$$, is the series $$\hat{f}$$ convergent? Please see the preprint arXiv:1105.2611v2 [math.GM] 5 Jun 2011 and the paper: Journal of Applied Analysis, Volume 25, Issue 2, Pages 131–139, DOI: https://doi.org/10.1515/jaa-2019-0014.

• If $f(x)=x$, then $\hat f(x)=-x\not=0=f(0)$. Is everything OK in the question?
– TaQ
Jul 16, 2013 at 9:30
• perhaps taking n = 0 in the lower limit would fix things? Jul 16, 2013 at 10:40
• Sorry you are right. In fact we have $\sum_{n=0}^{\infty}(-1)^n\frac{f^{(n)}(x)}{n!}x^n$ Jul 17, 2013 at 4:24
• This is series is the value of the (Taylor series of $f(t)$ in point $t_0=x$) at point $t=0$. So you may use the bounds for the remainder for Taylor series to estimate the differences between $f(0)$ and the partial sum of your series. Jun 21, 2020 at 8:52
• @Fedor Taylor series of $f$ may converge not to $f$ itself but to a function depending to $x$. Thus $\hat{f}(x)$ may converge to a sum depending to $x$. However, I do not know a function whose hat converges to a non-constant function! Jun 21, 2020 at 10:47

• The only thing which might be useful is that if $f$ satisfies $|f^{(n)}(x)|\le K^nM_n$ where $M_n$ is a sequence of positive numbers and $K$ is positive and if $L:=K\limsup_n\sqrt[n]{\frac{M_n}{n!}}<\infty$ then for $|x|<L^{-1}$ we have $\hat{f}(x)$ is convergent, see arXiv:1105.2611v2 [math.GM] 5 Jun 2011, Theorem 6. But the condition $L<\infty$ implies $M_n<A^nn!$ for some constant $A$ and then this implies that $f$ is analytic! Jul 17, 2013 at 7:43