【证毕QED】一个积分六个平均数

[证毕QED]一个积分六个平均数

参考文献: Chen, Hongwei. “Means Generated by an Integral.” Mathematics Magazine, vol. 78, no. 5, 2005, pp. 397–99.

引入

$$\begin{array} { | c | c | c | } \hline \text{算术平均数} & \text{几何平均数} & \text{调和平均数} \\ \hline A(a,b)=\dfrac{a+b}{2} & G(a,b)=\sqrt{ab} & H(a,b)=\dfrac{1}{\frac{1}{a}+\frac{1}{b}} \\ \hline \text{对数平均数} & \text{中心平均数} & \text{希罗平均数} \\ \hline L(a,b)=\dfrac{b-a}{\ln{b}-\ln{a}} & T(a,b)=\dfrac{2(a^2+ab+b^2)}{3(a+b)} & N(a,b)=\dfrac{a+\sqrt{ab}+b}{3} \\ \hline \end{array}$$

↓

$$f \left ( t \right )=\frac{\int_{a}^{b}x^{t+1}\mathrm{d}x}{\int_{a}^{b}x^{t}\mathrm{d}x}$$

对数平均数

$$\begin{aligned} L(a, b) & =\frac{b-a}{\ln b-\ln a} =\frac{\left.x\right|_{a} ^{b}}{\left.(\ln x)\right|_{a} ^{b}} =\frac{\int_{a}^{b} \mathrm{~d} x}{\int_{a}^{b} \frac{1}{x} \mathrm{~d} x} =\frac{\int_{a}^{b} x^{0} \mathrm{~d} x}{\int_{a}^{b} x^{-1} \mathrm{~d} x} \\ & =f(-1) \end{aligned}$$

算术平均数

$$\begin{aligned} A(a,b)& =\frac{b+a}{2} =\frac{(b+a)(b-a)}{2(b-a)} =\frac{b^{2}-a^{2}}{2(b-a)} =\frac{\left.\frac{1}{2} x^{2}\right|_{a} ^{b}}{\left.x\right|_{a} ^{b}} =\frac{\int_{a}^{b} x \mathrm{~d} x}{\int_{a}^{b} \mathrm{~d} x} =\frac{\int_{a}^{b} x^{1} \mathrm{~d} x}{\int_{a}^{b} x^{0} \mathrm{~d} x}\\ & =f(0) \end{aligned}$$

调和平均数

$$\begin{aligned} H(a,b)& =\frac{2}{\frac{1}{b}+\frac{1}{a}} =\frac{2\left(\frac{1}{b}-\frac{1}{a}\right)}{\left(\frac{1}{b}+\frac{1}{a}\right)\left(\frac{1}{b}-\frac{1}{a}\right)} =\frac{\frac{1}{b}-\frac{1}{a}}{\frac{1}{2}\left(\frac{1}{b^{2}}-\frac{1}{a^{2}}\right)} =\frac{\left. -x^{-1}\right|_{a} ^{b}}{\left.-\frac{1}{2}x^{-2}\right|_{a} ^{b}} =\frac{\int_{a}^{b} x^{-2} \mathrm{~d} x}{\int_{a}^{b} x^{-3} \mathrm{~d} x} \\ & = f(-3) \end{aligned}$$

几何平均数

$$\begin{aligned} G(a,b)& =\sqrt{a b} =\frac{\sqrt{b}-\sqrt{a}}{\frac{\sqrt{b}-\sqrt{a}}{\sqrt{a b}}} =\frac{\sqrt{b}-\sqrt{a}}{\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}} =\frac{\left.2 x^{\frac{1}{2}}\right|_{a} ^{b}}{-\left.2 x^{-\frac{1}{2}}\right|_{a} ^{b}} =\frac{\int_{a}^{b} x^{-\frac{1}{2}} \mathrm{~d} x}{\int_{a}^{b} x^{-\frac{3}{2}} \mathrm{~d} x} \\ & =f\left(-\frac{3}{2}\right) \end{aligned}$$

中心平均数

$$\begin{aligned} T(a, b) & = \frac{2\left(a^{2}+a b+b^{2}\right)}{3(a+b)} =\frac{2\left(b^{3}-a^{3}\right)}{3\left(b^{2}-a^{2}\right)} =\frac{\left.\frac{1}{3} x^{3}\right|_{a} ^{b}}{\left.\frac{1}{2} x^{2}\right|_{a} ^{b}} =\frac{\int_{a}^{b} x^{2} \mathrm{~d} x}{\int_{a}^{b} x \mathrm{~d} x}\\ & =f(1) \end{aligned}$$

希罗平均数

$$\begin{aligned} N(a, b)& =\frac{a+\sqrt{a b}+b}{3} =\frac{\frac{2}{3}\left(b^{\frac{3}{2}}-a^{\frac{3}{2}}\right)}{2\left(b^{\frac{1}{2}}-a^{\frac{1}{2}}\right)} =\frac{\left.\frac{2}{3} x^{\frac{3}{2}}\right|_{a} ^{b}}{\left.2 x^{\frac{1}{2}}\right|_{a} ^{b}} =\frac{\int_{a}^{b} x^{\frac{1}{2}} \mathrm{~d} x}{\int_{a}^{b} x^{-\frac{1}{2}} \mathrm{~d} x}\\ & = f\left(-\frac{1}{2}\right) \end{aligned}$$

大小关系证明

$$\begin{array} { | c | c | c | } \hline \text{调和平均数} & \text{几何平均数} \\ \hline f(-3)=H(a,b)=\dfrac{1}{\frac{1}{a}+\frac{1}{b}}& f\left(-\dfrac{3}{2}\right)=G(a,b)=\sqrt{ab} \\ \hline \text{对数平均数} & \text{希罗平均数} \\ \hline f(-1)=L(a,b)=\dfrac{b-a}{\ln{b}-\ln{a}}& f\left(-\dfrac{1}{2}\right)=N(a,b)=\dfrac{a+\sqrt{ab}+b}{3} \\ \hline \text{算术平均数} & \text{中心平均数} \\ \hline f(0)=A(a,b)=\dfrac{a+b}{2}& f(1)=T(a,b)=\dfrac{2(a^2+ab+b^2)}{3(a+b)} \\ \hline \end{array}$$

在 $b>a>0$ 时, $f(t)$ 在 $\mathbb{R}$ 上单调递增, 且 $\lim \limits_{t \to -\infty} f(t)=a$, $\lim \limits_{t \to \infty} f(t)=b$.

$$H(a,b)\leq G(a,b)\leq L(a,b)\leq N(a,b)\leq A(a,b)\leq T(a,b)$$

平方平均数

t=2