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$$f\left( x \right)=f\left( a \right)+\frac{{f}'\left( a \right)}{1!}\left( x-a \right)+\frac{{f}''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+\frac{{f}'''\left( a \right)}{3!}{{\left( x-a \right)}^{3}}+\cdots$$  $$x\in \left( a-r,a+r \right)$$

(2) $$\log 2=\frac{\ln 2}{\ln 10}$$  令 $$f\left( x \right)=\ln x$$
$${f}'\left( x \right)=\frac{1}{x},{f}''\left( x \right)=\frac{-1}{{{x}^{2}}},{f}'''\left( x \right)=\frac{2}{{{x}^{3}}},{{f}^{(4)}}\left( x \right)=\frac{-6}{{{x}^{4}}},\cdots \cdots$$

$$f\left( x \right)=0+\frac{\frac{1}{1}}{1!}\left( x-1 \right)+\frac{\frac{-1}{1}}{2!}{{\left( x-1 \right)}^{2}}+\frac{\frac{2}{{{1}^{3}}}}{3!}{{\left( x-1 \right)}^{3}}+\frac{\frac{-6}{{{1}^{4}}}}{4!}{{\left( x-1 \right)}^{4}}+\cdots \cdots$$
$$\Rightarrow f\left( x \right) = \left( {x - 1} \right) - \frac{1}{2}{\left( {x - 1} \right)^2} + \frac{1}{3}{\left( {x - 1} \right)^3} - \frac{1}{4}{\left( {x - 1} \right)^4} + \cdots \cdots$$

$$F\left( x \right)=f\left( x+1 \right)=\ln \left( x+1 \right)=x-\frac{1}{2}{{x}^{2}}+\frac{1}{3}{{x}^{3}}-\frac{1}{4}{{x}^{4}}+\cdots \cdots$$  $$-1<x<1$$
\begin{align} & G\left( x \right)=f\left( 1-x \right)=\ln \left( 1-x \right)=\left( -x \right)-\frac{1}{2}{{\left( -x \right)}^{2}}+\frac{1}{3}{{\left( -x \right)}^{3}}-\frac{1}{4}{{\left( -x \right)}^{4}}+\cdots \cdots \\ & \Rightarrow G\left( x \right)=f\left( 1-x \right)=\ln \left( 1-x \right)=-x-\frac{1}{2}{{x}^{2}}-\frac{1}{3}{{x}^{3}}-\frac{1}{4}{{x}^{4}}-\cdots \cdots \\ \end{align}  $$-1<x<1$$

\begin{align} & F\left( x \right)-G\left( x \right)=\ln \left( x+1 \right)-\ln \left( 1-x \right)=\ln \frac{1+x}{1-x} \\ & =2\left( x+\frac{1}{3}{{x}^{3}}+\frac{1}{5}{{x}^{5}}+\cdots \cdots \right) \\ \end{align}

(3)$$\frac{1+x}{1-x}=2\Rightarrow x=\frac{1}{3}$$    $$\ln 2\approx 2\left\{ \frac{1}{3}+\left( \frac{1}{3} \right){{\left( \frac{1}{3} \right)}^{3}} \right\}=\frac{168}{243}\approx 0.691$$
(4) $$\frac{1+x}{1-x}=10\Rightarrow x=\frac{9}{11}$$  $$\ln 10\approx 2\left\{ \frac{9}{11}+\left( \frac{1}{3} \right){{\left( \frac{9}{11} \right)}^{3}} \right\}=\frac{2664}{1331}\approx 2.0\sim$$
(5) $$\log 2=\frac{\ln 2}{\ln 10}\approx \frac{0.691}{2.0\sim }\approx 0.345$$  得證

[ 本帖最後由 shingjay176 於 2014-4-22 09:34 PM 編輯 ]

$$y = f(x) = \log x$$ 為嚴格遞增函數，

$$y = \log x = \frac{{\ln x}}{{\ln 10}}$$
$$y' = \frac{1}{{\ln 10}} \times \frac{1}{x} > 0$$            $$\ln 10 > 0$$   且真數 $$x$$恆正

[ 本帖最後由 shingjay176 於 2014-4-23 02:13 PM 編輯 ]

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