OI不得不知的那些数学定理

Binomial theorem

One can define\[{r \choose k}=\frac{r\,(r-1) \cdots (r-k+1)}{k!} =\frac{(r)_k}{k!}\]
Then, if \(x\) and \(y\) are real numbers with \(|x| > |y|\)( This is to guarantee convergence. Depending on \(r\), the series may also converge sometimes when \(|x| = |y|\).), and \(r\) is any complex number, one has
\[(x+y)^r =\sum_{k=0}^\infty {r \choose k} x^{r-k} y^k \]

Valid for \(|x| < 1\):\[(1+x)^{-1} = \frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - x^5 + \cdots\]

Lagrange polynomial

\[ L(x) := \sum_{j=0}^{k} y_j \prod_{\begin{smallmatrix}0\le m\le k\\ m\neq j\end{smallmatrix}} \frac{x-x_m}{x_j-x_m}\]

Lucas' theorem

For non-negative integers m and n and a prime p, the following congruence relation holds:\[\binom{m}{n}\equiv\prod_{i=0}^k\binom{m_i}{n_i}\pmod p,\]
where\[m=m_kp^k+m_{k-1}p^{k-1}+\cdots +m_1p+m_0,\]
and\[n=n_kp^k+n_{k-1}p^{k-1}+\cdots +n_1p+n_0\]
are the base \(p\) expansions of m and n respectively. This uses the convention that \(\tbinom{m}{n} = 0\) if \(m < n\).

A binomial coefficient \(\tbinom{m}{n}\) is divisible by a prime \(p\) if and only if at least one of the base \(p\) digits of \(n\) is greater than the corresponding digit of \(m\).

The p-th power mapping