# 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?

• Random relevant comment: Taylor series are nice because you can write an arbitrary smooth function (which can be complicated) as a sum of polynomials (which are relatively simple). – mathematician Feb 10 '17 at 7:49
• Depends on you understanding of represents. Pointwise it is fairly obvious – Sebastian Bechtel Feb 10 '17 at 8:38
• @mathematician Do you count $e^{-1/x^2}$ as "arbitrary smooth"? – Antitheos Feb 10 '17 at 21:56
• @Antitheos I define smooth to be whatever makes me correct ;) So yes I should have said analytic. – mathematician Feb 10 '17 at 22:25

## 6 Answers

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.

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: $$\frac{(-1)^n}{n!}\sum _{k=0}^n{n \choose k}\left(-1\right)^{n-k}f(x)\left(\sin \left(x\right)+k\right)^n =f(x)$$

• But in order to evaulate the left hand side, you need to evaluate $f(x)$ which is not that such great deal of Express polynomial without polynomials because you stuck tuck it right back in there... – Laray Feb 10 '17 at 8:21
• @Laray That is actually how Mark's example works too. Strictly speaking, a polynomial multiplied by a non-polynomial is a non-polynomial, so this does satisfy the criteria – Brevan Ellefsen Feb 10 '17 at 8:22
• I was (as I read the Question) thinking about to evaluate the polynomial (and derivatives/Integrals/whatever you deem necessary) at a single point to get an approximation. Just like I do at a Taylor/Laurant-Series... – Laray Feb 10 '17 at 8:27
• @Laray True, this will not do that. However, that is not what the OP asked for. Some of these answers will do that to varying degrees anyway, but I don't have to make mine be a local approximation by any means. – Brevan Ellefsen Feb 10 '17 at 8:28
• Of course, this is because $\sum_{k=0}^n \binom{n}{k} (-1)^k k^r = 0$ for $r \in \{0,1,\cdots, n-1\}$, and $(-1)^n/n!$ for $r=n$, which was known to Euler. – Chappers Feb 10 '17 at 17:11

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

• your Fourier series on the RHS is not a polynomial even though it is equal to $x$ on the interval $(0, 2\pi)$ – hyportnex Feb 10 '17 at 1:26
• @hyportnex Yes...? – Simply Beautiful Art Feb 10 '17 at 1:28
• the question was how to represent a polynomial not a part of a polynomial – hyportnex Feb 10 '17 at 1:30
• but the RHS is periodic and the LHS is not over the whole axis, so the two sides can equal only on a restricted interval, the RHS is not equal to a polynomial over the whole axis for polynomials (except for the constant) are never periodic functions. – hyportnex Feb 10 '17 at 1:35
• @Dason One may also point out that most series expansions have a limited radius of convergence to the original function, so nothing is unreasonable here. – Simply Beautiful Art Feb 10 '17 at 14:33

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

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 $p(x).$

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.

• Already said, more generally. – Yves Daoust Feb 10 '17 at 10:15