How do I best formally verify or disprove the following statement:
If for an arbitrarily often differentiable function $f(x)$, the Taylor Polynomials are constant
$$T_n(x)=const. \forall n\in \mathbb{N}_0$$
then
$$f(x)=const. \text{on } \mathbb{R}$$
How can an arbitrarily often differentiable function be constant? I am stuck on this question.
No, it's false. Counter-example:
Consider the function $f$ defined by $$\begin{cases}f(x)=\mathrm e^{-\tfrac 1{x^2}}&\text{if }x\ne0,\\f(0)=0.\end{cases}$$
One checks that $f$ is $\mathcal C^\infty$ on $\mathbf R$ and that $f^{(n)}(0)=0$ for all $n$, so is Taylor series (and its Taylor polynomials) is $0$. Yet $f(x)\ne 0$ if $x\ne 0$.