I'd greatly appreciate it if someone could provide examples of the following:

1) A infinitely differentiable function whose Taylor series does not converge to the function.

2) An infinitely differentiable function whose Taylor series diverges.

My differential equations text says that "most" smooth functions are of one of these types. What can be done if we weaken the infinitely differentiable condition?

(Please do not use $f(x) = e^{-1/x^2}$ as an example!)

  • $\begingroup$ See mathoverflow.net/questions/72/… . There, the example is $e^{-1/x}$ for $x>0$, $0$ for $x\leq0$. $\endgroup$ – user122283 Mar 1 '14 at 3:51
  • $\begingroup$ I'm looking for an example different from $e^{-1/x^2}$. That is more or less the same. $\endgroup$ – Ayesha Mar 1 '14 at 3:52
  • 1
    $\begingroup$ If your function is not infinitely differentiable, you only get Taylor polynomials approximating the function at each point. Naturally, they do not agree with the function on any neighborhood (or else the function would be infinitely differentiable after all). There is Peano's example, here, see also the other answers at that link. For some references to the great variety we can have, see here and links therein. $\endgroup$ – Andrés E. Caicedo Mar 1 '14 at 4:05
  • $\begingroup$ See my post for a conjectured monotonic infinitely differentiable function whose Taylor series at each point has a zero radius of convergence. math.stackexchange.com/questions/675555/… $\endgroup$ – Sheldon L Mar 5 '14 at 22:07

Infinitely differentiable functions whose Taylor series diverges except at $0$:

  1. $\displaystyle f(x)=\int_0^\infty e^{-t}\cos(t^2 x)\;dt$.

Source: A primer of real functions by R. Boas Jr.

  1. $\displaystyle f(x)=\sum_{n=1}^\infty f_n(x)$, where $f_n(x)=\phi_{n,n-1}(x)$, where $$\phi_{n1}(x)=\int_0^x\phi_{n0}(t)\;dt, \quad \phi_{n2}(x)=\int_0^x\phi_{n1}(t)\;dt,\quad \cdots \quad \phi_{n,n-1}(x)=\int_0^x\phi_{n,n-2}(t)\;dt,$$ where $$\phi_{n0}(x)=\left\{\begin{aligned} ((n-1)!)^2,&\quad \text{if}\quad 0\leq |x|\leq \frac{1}{2^{n}(n!)^2}\\ 0,&\quad \text{if}\quad |x|\geq \frac{1}{2^{n-1}(n!)^2}. \end{aligned}\right.$$

Source: Counterexamples in Analysis by B. Gelbaum and J. Olmsted.


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