取整函数

设$x$为任意实数，$[x]$表示不超过$x$的最大整数。函数$y=[x]$称为取整函数
$[1.2]=1$

从数轴上向左看。

$[-0.32]=-1$---
$\begin{array}{rl}& [x+3]=[x]+3\\ & [1.2+3]=4=[1.2]+3\end{array}$

图像

$$y=[x]$$

取整函数.excalidraw

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单调性：单调不减
奇偶性：非奇非偶
有界性：无界
周期性：无周期性

$$y=x-[x](x的小数部分)$$

Floor funcion小数部分.excalidraw

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单调性：单调增区间$[n,n+1],n\in Z$
奇偶性：非奇非偶
有界性：有界（下届为0，上届为1）
周期性：T=1

结论一

$\left[\begin{array}{c}x+n\end{array}\right]=\left[\begin{array}{c}x\end{array}\right]+n,n\text{为整数}.$
对于任意一个实数x，它可以分成整数部分和小数部分。
$x=[x]+\{x\}$
则$[x]=[[x]+\{x\}]$
所以证明过程为：
$[x+n]=[[x]+\{x\}+n]=[x]+n$

结论二

$x-1<[x]\le x<[x]+1$
证明

$$\begin{array}{r}x-[x]=\{x\}<1\\ 则x-1<[x]\\ x<[x]+1\end{array}$$

数列极限引入

设n，N都是正整数，如果当 n > N时，必有$n>\frac{1}{ϵ}(ϵ>0)$，则 N 可取？

n 必须是 N+1 才能满足等式。

Drawing 2026-01-02 18.02.39.excalidraw

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N 可以在 1/ \epslion 左边

如果 epslion > 1

N就可能为0，不满足是正整数的条件。

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设n，N都是正整数，如果当 n > N时，必有$n>\square$，则 N 可取$\square +1$

设n，N都是正整数，如果当 n > N时，必有$\frac{1}{n^2}<ϵ(ϵ>0)$，则 N 可取

$$\begin{array}{r}\frac{1}{n^2}<ϵ\\ 1<ϵ\cdot n^2\\ n^2>\frac{1}{ϵ}\\ n>\frac{1}{\sqrt{ϵ}}\\ \therefore N可取[\frac{1}{\sqrt{ϵ}}]+1\end{array}$$