Taylor series and tempered distributions Suppose we have a function $\psi$ in $\mathbb{R}$ other than a polynomial which is equal to its Taylor Series for every point in $\mathbb{R}$.  Why is the following statement valid?
When we interpret both the $n$th sum of the Taylor series $T_n$ as a tempered distribution and $\psi$ as a tempered distribution, $\psi$ is NEVER the pointwise limit of $T_n$.
The only guess I have is it is because each $T_n$ is not a tempered distribution?  I am new to tempered distributions so please explain in detail not assuming much prior knowledge.
Edit:  I also know that a distribution is tempered iff it is a finite sum of derivatives of continuous functions growing at infinity slower than some polynomial. So maybe that can help?  
THANKS! 
 A: The trouble is much more obvious when one looks at the sequence of Fourier transforms $\widehat{T}_n$ of your sequence $T_n$. Let me be precise. Suppose that $\{a_k\}$ is the sequence of Taylor coefficients for $\psi$, where $\psi$ is a real analytic function that is not a polynomial, so that $\psi(x) = \sum_{k\geq 0}a_kx^k$ and $a_k\not = 0$ for infinitely many $k$. Following your notation, write $T_n$ for the $n$th partial sum of this power series. The Fourier transform of $T_n$ is going to be
$$
\widehat{T}_n(x) = \sum_{k = 0}^n a_ki^k\partial^k\delta_0,
$$
since the Fourier transform of $x^k$ is $i^k\partial^k \delta_0$ (depending on the normalization you use for the Fourier transform this formula may be off by a constant factor, but this is irrelevant); here $\partial$ just denotes differentiation and $\delta_0$ is the standard Dirac delta function at the origin. Now, if you assume that $T_n \to \psi$ in the sense of distributions, then you must also assume that $\widehat{T}_n \to \hat{\psi}$ in the sense of distributions; in particular, you must assume that
$$
\lim_{n\to \infty} \widehat{T}_n
$$
exists as a tempered distribution. And this is where things fall apart. If we choose a function $\varphi$ in the Schwartz space, then $\partial^{(k)}\delta_0(\varphi) = (-1)^k\varphi^{(k)}(0)$ and
$$
\widehat{T}_n(\varphi) = \sum_{k = 0}^n a_ki^k\partial^k\delta_0(\varphi) = \sum_{k = 0}^n a_k (-i)^k\varphi^{(k)}(0).
$$
So if $T_n(\varphi)$ is to converge to a finite limit for every $\varphi$, then it must be the case that
$$
\sum_{k = 0}^\infty a_k (-i)^k\varphi^{(k)}(0) \tag{1}
$$
is a convergent series for every Schwarz function $\varphi$. One way to see that this can never be the case is to invoke Borel's theorem, which ensures that we can find a Schwartz function $\varphi$ such that, for instance,
$$
\varphi^{(k)}(0) = \left\{\begin{array}{ll}
\frac{1}{(-i)^ka_k} & \text{if $a_k\not = 0$,}\\
0 & \text{if $a_k = 0$.}
\end{array}
\right.
$$
Then since infinitely many of the $a_k$ are nonzero by assumption, inserting this $\varphi$ into $(1)$ gives us a divergent series.

It may be worth mentioning, however, that whenever $\varphi$ is compactly supported, we do in fact get the convergence $\lim_{n\to \infty} T_n(\varphi) = \psi(\varphi)$; actually, if you take $\psi = e^{-itx}$, you'll be able to show with an argument similar to the one above that the Fourier transform $\hat{\varphi}$ must be real analytic (i.e., that it has an extension to an entire function on the complex plane).
