How to write $f(x)=x^n+(1-x)^n$ in terms of $r=x(1-x)$

Question

The functions $f_n(x) = x^n + (1-x)^n$ with $n\in\mathbb N$ obviously have $f_n(x)=f_n(1-x)$.

Thus we can write $f_n(x)=\sum_k a_{n,k} r^k$ with $r=x\cdot(1-x)$.

Is there a nice way to determine the $a_{n,k}$?

Answer 1

Yes, there is a nice way, it is called Girard-Newton-Waring formula.

More generally, the symmetric polynomial $x_1^n+x_2^n$ can be written as a polynomial in the elementary symmetric polynomials $x_1+x_2$ and $x_1x_2$: $$x_1^n+x_2^n=\sum_{k=0}^{\lfloor n/2\rfloor} a_{n,k}(x_1+x_2)^{n-2k}(x_1x_2)^k.$$ where $$a_{n,k}=\frac{(-1)^k n}{n-k}\binom{n-k}{k}\qquad \text{for $k=0,\dots, \lfloor n/2\rfloor$}.$$ In your case $x_1=x$ and $x_2=1-x$ and therefore $x_1+x_2=1$ and $x_1x_2=x(1-x)$.

See also A132460 as a reference.

Answer 2

$$(x+1-x)^n=1^n=x^n+(1-x)^n+p[x(1-x), n]$$

where $P[x(1-x), n]$ is a polynomial in terms of $r=x(1-x)$, so we can write:

$f_n(x)=1-P(r)$