Can a polynomial be expressed as an infinite sum of non-polynomial functions? The Taylor series represents a non-polynomial function as an infinite series of polynomials, so is it possible to express a polynomial function as an infinite series of non-polynomial functions?
 A: Sure it's possible!  Take a simple fourier series.  One of my favorite:
$$x=\pi-2\left(\frac{\sin(x)}1+\frac{\sin(2x)}2+\frac{\sin(3x)}3+\dots\right)$$
For $x\in(0,2\pi)$.
$$y=-2\left(\frac{\sin(y)}1-\frac{\sin(2y)}2+\frac{\sin(3y)}3-\dots\right)$$
For $y\in(-\pi,\pi)$.
A: $\frac{x^2}{(1-x)^k}$ is not a polynomial for any $k>0$.
Yet $$
\sum_{k=1}^\infty \frac{x^2}{(1-x)^k} = -x
$$
is a polynomial.
If you want a series which converges on the entire real axis, try 
$$
\sum_{k=1}^\infty x^2
\left(e^{x^2}-1\right)e^{-kx^2} = x^2
$$
A: Here's an easy example. The hard part is actually just making sure none of the infinite number of functions in the sum is a polynomial, assuming you consider the zero function to be a polynomial.
Let $p(x)$ be an arbitrary polynomial. Then let
\begin{align}
f_0(x) &= \sin x \\
f_1(x) &= \begin{cases}  p(x)   & x < 0 \\
                        -\sin x & x \geq 0 \end{cases} \\
f_2(x) &= \begin{cases} -\sin x & x < 0 \\
                         p(x)   & x \geq 0 \end{cases} \\
f_n(x) &= \begin{cases} \dfrac{\sin x}{2^{n+1}} & \text{$n>2$ and $n$ odd} \\
                        \dfrac{\sin x}{2^n} & \text{$n>2$ and $n$ even}
          \end{cases} \\
\end{align}
None of these functions is a polynomial, since each function has an infinite number of zeros. But $p(x) = f_0(x) + f_1(x) + f_2(x).$
The functions $f_n(x)$ for $n > 2$ are defined in order to satisfy the requirement to express $p(x)$ as an infinite number of non-polynomial functions (which I think is the most difficult part of this problem). 
Each successive pair of functions has sum zero, so
$f_3(x) + \cdots + f_{2k}(x) = 0$ for any integer $k$.
For $n>2,$ therefore, 
the partial sum $f_0(x) + \cdots + f_n(x)$ is $p(x)$ if $n$ is even
and is $p(x) + \frac{\sin x}{2^{n+1}}$ if $n$ is odd.
The functions $\frac{\sin x}{2^{n+1}}$ converge to zero as $n\to\infty,$
so the sum converges to $p(x).$
A: Consider the sum of the sequence of characteristic functions of all intervals [n,n+1) for all integers n. None of these functions is a polynomial, yet their sum is one.
A: Without constraints on the functions, the answer is trivial. Take any family of non-polynomial functions $\phi_n(x),n>0$ such that their sum converges to some non-polynomial $\sigma(x)$. 
Then define
$$\phi_0(x):=P(x)-\sigma(x)$$ and you have it:
$$\sum_{n=0}^\infty\phi_n(x)=P(x).$$

The set of non-polynomial functions is much richer than that of polynomials, so there is no symmetry between Taylor and "reverse Taylor".

Another very simple example is the family of functions that equal the desired polynomial in range $[n,n+1)$ and zero elsewhere.
A: Here is an interesting example I found when I messed around with this same topic a while back. This sum gives you any polynomial you want:
$$\frac{(-1)^n}{n!}\sum _{k=0}^n{n \choose k}\left(-1\right)^{n-k}f(x)\left(\sin \left(x\right)+k\right)^n =f(x)$$
