① ②这里用到了极限与不等关系 ③如果a≠b,那么便不会有$\lim _{n\rightarrow \infty }\left| I_n \right| =0$ ④如果还存在一点c在 内,那么同样也不会有$\lim _{n\rightarrow \infty }\left| I_n \right| =0$ 希望深入了解闭区间套定理(Nested intervals theorem),请看讲解2:http://www.cnblogs.com/iMath/p/6260953.html…

①确界与极限,看完这篇你才能明白 http://www.cnblogs.com/iMath/p/6265001.html ②这个批注由这个问题而来 表示$c$可能在$\bigcap_{n=1}^{\infty} (a_{n},b_{n})$或$\bigcap_{n=1}^{\infty} (a_{n},b_{n}]$或$\bigcap_{n=1}^{\infty} [a_{n},b_{n})$或$\bigcap_{n=1}^{\infty} [a_{n},b_{n}]$内,$\bigcap_{n…

① ②这里用到了极限与不等关系 ③如果a≠b,那么便不会有$\lim _{n\rightarrow \infty }\left| I_n \right| =0$ ④如果还存在一点c在内,那么同样也不会有$\lim _{n\rightarrow \infty }\left| I_n \right| =0$…

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1. 问题 Karatsuba 大整数的快速乘积算法的运行时间(时间复杂度的递推关系式)为 T(n)=O(n)+4⋅T(n/2),求其最终的时间复杂度. 2. 主定理的内容 3. 分析 所以根据主定理的判别方法,可知对于 T(n)=O(n)+4⋅T(n/2),a=4,b=2,则 f(n)=O(n)<nlogab=2,符合第一个判别式,因此,T(n)=O(n2)…

兰道定理的内容: 一个竞赛图强连通的充要条件是:把它的所有顶点按照入度d从小到大排序,对于任意\(k\in [0,n-1]\)都不满足\(\sum_{i=0}^k d_i=\binom{k+1}{2}\). 兰道定理的证明: 引理: 一个竞赛图强连通的充要条件是对于任意\(S \subsetneq 点集V\),都存在一个点\(u \notin S\),满足u到S有边. 证明: 1.必要性:比较显然 2.充分性:假设我们现在已经得到了\(V\)中的一个强连通子集\(S\),想办法不断扩展\(S\)…

1. 几种形式 ∮∂SPdx+Qdy+Rdz=∬S∣∣∣∣∣∣cosα∂∂xPcosβ∂∂yQcosγ∂∂zR∣∣∣∣∣∣dS ∮∂Ωw=∬Ωdw 左边是内积: 右边是外积: 物理上的应用: ∮∂SE⃗ ⋅dℓ⃗ =∬S(∇×E⃗ )⋅dA⃗  场函数 E⃗  沿边界曲线(Γ=∂S),等于其旋度(\nabla\times \vec E\right)在曲面 S 的二重积分:…

Dividing an unit interval \([0,1]\) into two equal subintervals by the midpoint \(\dfrac {0+1} {2}=\dfrac {1} {2}\), denote the left subinterval by $I_{1}=\left[ 0,\dfrac {1} {2^{1}}\right] $, next, divide \(I_{1}\) into two equal parts by its midpoi…

From the perspective of analytical geometry, an interval is composed of infinitely many points, while after the length of an interval was defined, it is intuitively to believe its length is the sum of the length of all points within it, then it becom…

Given an infinite sequence (a1, a2, a3, ...), a series is informally the form of adding all those terms together: a1 + a2 + a3 + ···. To emphasize that there are an infinite number of terms, a series is often called an infinite series. 值得注意的是等式右边并不是左…