6

Because $x^n$ and $(x-1)^n$ are relatively prime in $\mathbb{Z}[x]$, so $p(x)x^n+q(x)(1-x)^n=1$ for some $p(x), q(x)\in \mathbb{Z}[x]$. What are the explicit formulas of $p(x), q(x)$?

HeHe
  • 213
  • @Macavity I try Euclidean algorithm， but the formula seems to be very complicated. – HeHe Mar 15 '18 at 04:57
  • @Macavity Also, because $\mathbb{Z}$ is not a field, I am not even sure whether Euclidean algorithm makes sense. – HeHe Mar 15 '18 at 04:59
  • Perhaps you need to convince yourself Euclidean algorithm works for integers first, then look at $\mathbb Z[x]$. – Macavity Mar 15 '18 at 05:34
  • I don't know of explicit formulas, but you can get recursive formulas. The general principle at hand is that if $a$ and $b$ are two elements of a commutative ring $R$ satisfying $aR + bR = R$, then $a^nR + b^mR = R$ for any nonnegative integers $n$ and $m$. There is an easy proof of this in two steps: first, prove it for the case $n = 1$ by induction on $m$; then, prove it for general $n$ by induction on $n$. Both inductions lead to recursive formulas for the required cofactors. Whether you can solve the recursions explicitly is a different story. – darij grinberg Mar 15 '18 at 06:05
  • I think the adjugate of the Sylvester matrix of $x^n$ and $\left(1-x\right)^n$ also has something interesting to say, but I'm lacking the time to find out what. – darij grinberg Mar 15 '18 at 06:09
  • A nice question. :-) – user90369 Mar 15 '18 at 11:02

2 Answers2

8

Hint:

Start with the equality $$x+(1-x)=1$$ raise to a large enough power ( $2n-1$ will do it ), and use Newton binomial formula. Separate the LHS into a multiple of $x^n$ and a multiple of $(1-x)^n$.

orangeskid
  • 53,909
6

You have a closed form in the more general case when you have two distinct exponents $n$ and $m$ :

$$ (1-x)^n\Bigg(\sum_{j=0}^{m-1} \binom{n+j-1}{j}x^j\Bigg) -x^m\Bigg(n\binom{n+m-1}{n}\sum_{k=0}^{n-1} \frac{(-1)^{k+1}\binom{n-1}{k}}{m+k}x^k\Bigg)=1. $$

You can also rewrite this more compactly as

$$ (1-x)^n\Bigg(\sum_{j=0}^{m-1} \binom{n+j-1}{j}x^j\Bigg) +x^m\Bigg(\sum_{k=0}^{n-1} \binom{m+k-1}{k}(1-x)^k\Bigg)=1. $$

Ewan Delanoy
  • 61,600