Can a polynomial be expressed as an infinite sum of non-polynomial functions?

Question

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?

Answer 1

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.

Answer 2

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 $y\in(-\pi,\pi)$.

Answer 3

$\frac{x^2}{(1-x)^k}$ 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 $$ \sum_{k=1}^\infty x^2 \left(e^{x^2}-1\right)e^{-kx^2} = x^2 $$

Answer 4

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 $p(x).$

Answer 5

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.