produce the linear and quadratic taylor polynomials for the following cases graph functions and these taylor polynomials 1. f(x)=√ x , a=1 2. f(x)=e^cos(x) , a=0
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http://zh.wikipedia.org/wiki/%E6%B3%B0%E5%8B%92%E7...
泰勒級數的公式請參考這邊
n就是你要展開的次數
用f(^n)表示f的微分次數
f(^0) = f
f(^1) = f'
f(^2) = f''
f(x)的1次泰勒展開 = f(^0)(a)/0!.(x - a)^0 + f(^1)(a)/1!.(x - a)^1
f(x)的2次泰勒展開 = f(^0)(a)/0!.(x - a)^0 + f(^1)(a)/1!.(x - a)^1 + f(^2)(a)/2!.(x - a)^2
a) f(x)=√ x , a=1
f(^0)(x) = f(x)=√ x
=> f(^0)(a) = 1
f(^1)(x) = f'(x) = 1/2.1/√x
=> f(^1)(a) = 1/2
f(^2)(x) = f''(x) = -1/2.1/2.1/(√x)^3
=> f(^2)(a) = -1/4
√ x的1次泰勒展開
= f(^0)(a)/0!.(x - a)^0 + f(^1)(a)/1!.(x - a)^1
= 1/1.1 + (1/2)/ 1.(x-1)
= 1 + 1/2.(x - 1)
√ x的2次泰勒展開
= f(^0)(a)/0!.(x - a)^0 + f(^1)(a)/1!.(x - a)^1 + f(^2)(a)/2!.(x - a)^2
= 1/1.1 + (1/2)/ 1.(x-1) + (-1/4)/2.(x-1)^2
= 1 + 1/2.(x - 1) - 1/8.(x-1)^2
--
b) f(x)=e^cos(x), a = 0
cos(0) = 1, sin(0) = 0
f(^0)(x) = f(x)=e^cos(x)
=> f(^0)(a) = e^cos(0) = e
f(^1)(x) = f'(x) =e^cos(x).sin(x)
=> f(^1)(a) = e^cos(0).sin(0) = 0
f(^2)(x) = f''(x) =e^cos(x).[sin(x)]^2 + e^cos(x).cos(x)
=> f(^2)(a) = e^cos(0).[sin(0)]^2 + e^cos(0).cos(0) = e
= e/1 + 0/1.(x-0)
= e
= e/1 + 0/1.(x-0) + e/2.(x-0)^2
= e + ex^2/2
圖自己畫看看,利用微分的方法來畫
2013-09-07 02:37:07 補充:
https://www.dropbox.com/s/gsu0vq75arqj7tc/20130906...
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http://zh.wikipedia.org/wiki/%E6%B3%B0%E5%8B%92%E7...
泰勒級數的公式請參考這邊
n就是你要展開的次數
用f(^n)表示f的微分次數
f(^0) = f
f(^1) = f'
f(^2) = f''
f(x)的1次泰勒展開 = f(^0)(a)/0!.(x - a)^0 + f(^1)(a)/1!.(x - a)^1
f(x)的2次泰勒展開 = f(^0)(a)/0!.(x - a)^0 + f(^1)(a)/1!.(x - a)^1 + f(^2)(a)/2!.(x - a)^2
a) f(x)=√ x , a=1
f(^0)(x) = f(x)=√ x
=> f(^0)(a) = 1
f(^1)(x) = f'(x) = 1/2.1/√x
=> f(^1)(a) = 1/2
f(^2)(x) = f''(x) = -1/2.1/2.1/(√x)^3
=> f(^2)(a) = -1/4
√ x的1次泰勒展開
= f(^0)(a)/0!.(x - a)^0 + f(^1)(a)/1!.(x - a)^1
= 1/1.1 + (1/2)/ 1.(x-1)
= 1 + 1/2.(x - 1)
√ x的2次泰勒展開
= f(^0)(a)/0!.(x - a)^0 + f(^1)(a)/1!.(x - a)^1 + f(^2)(a)/2!.(x - a)^2
= 1/1.1 + (1/2)/ 1.(x-1) + (-1/4)/2.(x-1)^2
= 1 + 1/2.(x - 1) - 1/8.(x-1)^2
--
b) f(x)=e^cos(x), a = 0
cos(0) = 1, sin(0) = 0
f(^0)(x) = f(x)=e^cos(x)
=> f(^0)(a) = e^cos(0) = e
f(^1)(x) = f'(x) =e^cos(x).sin(x)
=> f(^1)(a) = e^cos(0).sin(0) = 0
f(^2)(x) = f''(x) =e^cos(x).[sin(x)]^2 + e^cos(x).cos(x)
=> f(^2)(a) = e^cos(0).[sin(0)]^2 + e^cos(0).cos(0) = e
√ x的1次泰勒展開
= f(^0)(a)/0!.(x - a)^0 + f(^1)(a)/1!.(x - a)^1
= e/1 + 0/1.(x-0)
= e
√ x的2次泰勒展開
= f(^0)(a)/0!.(x - a)^0 + f(^1)(a)/1!.(x - a)^1 + f(^2)(a)/2!.(x - a)^2
= e/1 + 0/1.(x-0) + e/2.(x-0)^2
= e + ex^2/2
圖自己畫看看,利用微分的方法來畫
2013-09-07 02:37:07 補充:
https://www.dropbox.com/s/gsu0vq75arqj7tc/20130906...