01.4实数系的构造P044命题1.4.9

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命题1.4.9

设 $\displaystyle x\geq 0$, 则

$$\begin{aligned} 0 < \alpha\leq 1&\Rightarrow (1+x)^\alpha\leq 1+\alpha x, \qquad(I)\\\\ \alpha\geq 1&\Rightarrow (1+x)^\alpha\geq 1+\alpha x,\qquad(II)\\\\ \alpha\geq 0&\Rightarrow (1+x)^\alpha\geq 1+\frac{\alpha x}{1+x}.\qquad(III) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$

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(1)、 对 $\displaystyle 0 < \alpha\leq 1$, 取

$$\begin{aligned} \mathbb{Z}\ni m > \frac{1}{1-\alpha}, n=[m\alpha]+1, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$

则

$$\begin{aligned} \alpha\leq \frac{n}{m}\leq \alpha+\frac{1}{m} < 1\Leftrightarrow \frac{n-1}{m}\leq \alpha\leq \frac{n}{m}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$

而

$$\begin{aligned} (1+x)^\alpha&\leq (1+x)^\frac{n}{m}=\left[1^{m-n} (1+x)^n\right]^\frac{1}{m} \stackrel{\tiny\mbox{均值}}{\leq} \frac{(m-n)+n(1+x)}{m}\\\\ &=1+\frac{n}{m}x=1+\alpha x+\frac{x}{m}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$

由 $\displaystyle m$ 的任意性即知 $\displaystyle (1+x)^\alpha\leq 1+\alpha x$.

(2)、

$$\begin{aligned} &(1+x)^\alpha\geq 1+\alpha x \Leftrightarrow 1+x\geq (1+\alpha x)^\frac{1}{\alpha} \Leftrightarrow (1+\alpha x)^\frac{1}{\alpha}\leq 1+\frac{1}{\alpha}\cdot \alpha x\Leftarrow (I). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$

(3)、

$$\begin{aligned} (1+x)^\alpha\geq 1+\frac{\alpha x}{1+x} \Leftrightarrow& (1+x)^\alpha\geq \frac{1+(1+\alpha)x}{1+x}\\\\ \Leftrightarrow& (1+x)^{1+\alpha}\geq 1+(\alpha+1)x\Leftarrow (II). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$

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