MIT_单变量微积分_12

逃离我推掉我的手 2022-03-12 01:58 95阅读 0赞

### 1. 牛顿迭代法 ###

`Ex1:`  
 x 2 = 5 , f ( x ) = x 2 − 5 , 求 f ( x ) = 0 时 的 x x^2=5,f(x)=x^2-5,求f(x)=0时的x x2=5,f(x)=x2−5,求f(x)=0时的x.  
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y − y 0 = m ( x − x 0 ) y-y\_0=m(x-x\_0) y−y0=m(x−x0)  
 y = 0 , ⇒ y=0,\\Rightarrow y=0,⇒

0 − y 0 = m ( x 1 + x 0 ) x 1 = − y 0 m + x 0 = x 0 − y 0 m = x 0 − f ( x 0 ) f ′ ( x 0 ) 0-y\_0=m(x\_1+x\_0)\\\\ x\_1=\\frac\{-y\_0\}\{m\}+x\_0\\\\ =x\_0-\\frac\{y\_0\}\{m\}\\\\ =x\_0-\\frac\{f(x\_0)\}\{f&\#x27;(x\_0)\} 0−y0=m(x1\+x0)x1=m−y0\+x0=x0−my0=x0−f′(x0)f(x0)

`牛顿法：`  
 x n + 1 = x n − f ( x 0 ) f ′ ( x 0 ) x\_\{n+1\}=x\_n-\\frac\{f(x\_0)\}\{f&\#x27;(x\_0)\} xn\+1=xn−f′(x0)f(x0)  
 x 0 = 2 , f ( x ) = x 2 − 5 x\_0=2,f(x)=x^2-5 x0=2,f(x)=x2−5  
 f ′ ( x ) = 2 x f&\#x27;(x)=2x f′(x)=2x  
 x 1 = x 0 − x 0 2 − 5 2 x 0 = x 0 − 1 2 x 0 + 5 2 x 0 = 1 2 x 0 + 5 2 x 0 x\_1=x\_0-\\frac\{x\_0^2-5\}\{2x\_0\}\\\\=x\_0-\\frac\{1\}\{2\}x\_0+\\frac\{5\}\{2x\_0\}\\\\ =\\frac\{1\}\{2\}x\_0+\\frac\{5\}\{2x\_0\} x1=x0−2x0x02−5=x0−21x0\+2x05=21x0\+2x05  
依次进行 x 2 = 161 72 x\_2=\\frac\{161\}\{72\} x2=72161  
 x 3 = 1 2 ∗ 161 72 + 5 2 ∗ 72 161 = ？ ？ ？ ？ ？ ？ x\_3=\\frac\{1\}\{2\}\*\\frac\{161\}\{72\}+\\frac\{5\}\{2\}\*\\frac\{72\}\{161\}=？？？？？？ x3=21∗72161\+25∗16172=？？？？？？.

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