證明： \(0.3<{{\log }_{10}}2<0.4\)

解: (1) \(f\left( x \right)\)在\(x=a\) 處的泰勒展開式

重要注意的地方：是針對 \(x=a\) 處做高次切線的估計，
所以 如果是針對 \(x=a\) 處展開~則 帶入的 \(x\) 值也不可以跟 \(a\) 差太多，才可以得到正確的估計值

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

(2) \(\log 2=\frac{\ln 2}{\ln 10}\)  令 \(f\left( x \right)=\ln x\)
\({f}'\left( x \right)=\frac{1}{x},{f}''\left( x \right)=\frac{-1}{{{x}^{2}}},{f}'''\left( x \right)=\frac{2}{{{x}^{3}}},{{f}^{(4)}}\left( x \right)=\frac{-6}{{{x}^{4}}},\cdots \cdots \)
在\(a=1\) 附近的泰勒展開式
\(f\left( x \right)=0+\frac{\frac{1}{1}}{1!}\left( x-1 \right)+\frac{\frac{-1}{1}}{2!}{{\left( x-1 \right)}^{2}}+\frac{\frac{2}{{{1}^{3}}}}{3!}{{\left( x-1 \right)}^{3}}+\frac{\frac{-6}{{{1}^{4}}}}{4!}{{\left( x-1 \right)}^{4}}+\cdots \cdots \)
\(\Rightarrow f\left( x \right) = \left( {x - 1} \right) - \frac{1}{2}{\left( {x - 1} \right)^2} + \frac{1}{3}{\left( {x - 1} \right)^3} - \frac{1}{4}{\left( {x - 1} \right)^4} +  \cdots  \cdots\)  

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

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

(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