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通过微分求近似值

时间:2022-10-04 15:02:31浏览次数:55  
标签:11 approx frac 通过 sqrt 120 近似值 微分 Delta

Definition

计算 \(\Delta y\) 和函数在 \(x=x_{0}\) 处附近一点 \(x=x_{1}\) 函数值的近似值

设函数 \(y=f(x)\) 在点\(x_{0}\)处可微分,且\(f'(x_{0}) \ne 0,\) 由微分之定义, 当 \(\left| \Delta x \right|\) 极小时,有:

\(\Delta y =f(x_{0}+\Delta x)-f(x_{0}) \approx f'(x_{0}) \Delta x\)

可表示为:

  • \(f(x_{0}+\Delta x)\approx f(x_{0}) +f'(x_{0}) \cdot \Delta x\)

    • ps: \(f(x_{0}+\Delta x)=\Delta y\)

  • \(f(x_{1})\approx f(x_{0})+f'(x_{0})(x_{1}-x_{0})\)

    • ps: \(x_{1}=x+\Delta x\)



Instance



\[应用例: 求 \sqrt[]{120} 之近似值 \]

\[\\ \\ \]

\[\sqrt[]{120}=\sqrt[]{121-1}=(121-1)^{\frac{1}{2}} =[(11)^{2}-1]^{\frac{1}{2}} \\ 提11而出: \quad \sqrt[]{120}=11(1-\frac{1}{11^{2}})^{\frac{1}{2}} \]

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\[设f(x) \approx \sqrt[]{1+x}, \quad x_{0}=0, x_{1}=-\frac{1}{11^{2}} \]

\[题意即为: \quad 求 f(x) 在 x_{0} 处附近一点 x_{1} 之近似值 \]

\[\\ \\ \]

\[由公式得: \quad f(-\frac{1}{11^{2}}) \approx f(0)+f'(0)\cdot (-\frac{1}{11^{2}}-0) \]

\[\\ \\ \]

\[设1+x=k,\quad f'(x)=(\sqrt[]{1+x})'=(k)'\cdot (\sqrt[]{k})' \]

\[\\ \\ \]

\[(k)'=(1)'+(x)'=0+1=1 \]

\[\\ \\ \]

\[由公式, (\sqrt[]{k})' =\frac{1}{2\sqrt[]{k}} =\frac{1}{2\sqrt[]{1+x}} \]

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\[代入 x_{0}=0, \quad \therefore f'(0)=\frac{1}{2} \]

\[\\ \\ \]

\[\therefore f(-\frac{1}{11^{2}}) \approx 1+\frac{1}{2} \cdot -\frac{1}{11^{2}}=\frac{241}{242} \]

\[\\ \\ \]

\[\because \sqrt[]{120} \approx 11\cdot f(x_{1}) \]

\[\\ \\ \]

\[\therefore \sqrt[]{120} \approx 11\cdot \frac{241}{242} \]

\[\\ \\ \]

\[\sqrt[]{120} \approx \frac{241}{22} \]

标签:11,approx,frac,通过,sqrt,120,近似值,微分,Delta
From: https://www.cnblogs.com/Preparing/p/16753700.html

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