题目
Problem
f ( x ) = 1 − 1 8 x 4 + 2 7 x 7 x > 0 f(x) = 1 - \frac{1}{8x^4} + \frac{2}{7\sqrt{x^7}} \quad x > 0 f ( x ) = 1 − 8 x 4 1 + 7 x 7 2 x > 0
The equation f ( x ) = 0 f(x) = 0 f ( x ) = 0 α \alpha α [ 0.15 , 0.25 ] [0.15, 0.25] [ 0.15 , 0.25 ]
(a) (i) Determine f ′ ( x ) f'(x) f ′ ( x )
(ii) Explain why 0.25 0.25 0.25 α \alpha α
(iii) Taking 0.15 0.15 0.15 α \alpha α f ( x ) f(x) f ( x ) α \alpha α 3 3 3
(5)
(b) Use linear interpolation once on the interval [ 0.15 , 0.25 ] [0.15, 0.25] [ 0.15 , 0.25 ] α \alpha α 3 3 3
(3)
题目中文翻译
f ( x ) = 1 − 1 8 x 4 + 2 7 x 7 x > 0 f(x) = 1 - \frac{1}{8x^4} + \frac{2}{7\sqrt{x^7}} \quad x > 0 f ( x ) = 1 − 8 x 4 1 + 7 x 7 2 x > 0
方程 f ( x ) = 0 f(x) = 0 f ( x ) = 0 α \alpha α [ 0.15 , 0.25 ] [0.15, 0.25] [ 0.15 , 0.25 ]
(a) (i) 确定 f ′ ( x ) f'(x) f ′ ( x )
(ii) 解释为什么 0.25 0.25 0.25 α \alpha α
(iii) 取 0.15 0.15 0.15 α \alpha α f ( x ) f(x) f ( x ) α \alpha α 3 3 3
(b) 在区间 [ 0.15 , 0.25 ] [0.15, 0.25] [ 0.15 , 0.25 ] α \alpha α 3 3 3
解答
(a)(i)
解法一
思路
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先把分式和根式改写成负指数幂，再逐项求导。这样指数变化和系数约分会更清楚。
答题过程
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Writing
f ( x ) = 1 − 1 8 x − 4 + 2 7 x − 7 / 2 , f(x)=1-\frac18x^{-4}+\frac27x^{-7/2}, f ( x ) = 1 − 8 1 x − 4 + 7 2 x − 7/2 , we obtain
f ′ ( x ) = − 1 8 ( − 4 ) x − 5 + 2 7 ( − 7 2 ) x − 9 / 2 = 1 2 x 5 − 1 x 9 / 2 . \begin{align*}
f'(x)
=&\,-\frac18(-4)x^{-5}
+\frac27\left(-\frac72\right)x^{-9/2}\\
=&\,\boxed{\frac1{2x^5}-\frac1{x^{9/2}}}.
\end{align*} f ′ ( x ) = = − 8 1 ( − 4 ) x − 5 + 7 2 ( − 2 7 ) x − 9/2 2 x 5 1 − x 9/2 1 .
(a)(ii)
解法一
思路
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Newton–Raphson 公式需要除以初始点处的导数。计算 f ′ ( 0.25 ) f'(0.25) f ′ ( 0.25 )
答题过程
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At x = 0.25 = 1 / 4 x=0.25=1/4 x = 0.25 = 1/4
f ′ ( 0.25 ) = 1 2 ( 1 / 4 ) 5 − 1 ( 1 / 4 ) 9 / 2 = 512 − 512 = 0. \begin{align*}
f'(0.25)
=&\,\frac1{2(1/4)^5}
-\frac1{(1/4)^{9/2}}\\
=&\,512-512\\
=&\,0.
\end{align*} f ′ ( 0.25 ) = = = 2 ( 1/4 ) 5 1 − ( 1/4 ) 9/2 1 512 − 512 0. The Newton–Raphson formula
x n + 1 = x n − f ( x n ) f ′ ( x n ) x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} x n + 1 = x n − f ′ ( x n ) f ( x n ) would therefore require division by zero. Hence
0.25 cannot be used as the initial approximation . \boxed{0.25\text{ cannot be used as the initial approximation}}. 0.25 cannot be used as the initial approximation .
(a)(iii)
解法一
思路
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在 Newton–Raphson 迭代式中代入 x 0 = 0.15 x_0=0.15 x 0 = 0.15
答题过程
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At x 0 = 0.15 x_0=0.15 x 0 = 0.15
f ( 0.15 ) = − 27.332509 … f(0.15)=-27.332509\ldots f ( 0.15 ) = − 27.332509 … and
f ′ ( 0.15 ) = 1484.137157 … . f'(0.15)=1484.137157\ldots. f ′ ( 0.15 ) = 1484.137157 … . Therefore,
x 1 = 0.15 − f ( 0.15 ) f ′ ( 0.15 ) = 0.15 − − 27.332509 … 1484.137157 … = 0.168416 … . \begin{align*}
x_1
=&\,0.15-\frac{f(0.15)}{f'(0.15)}\\
=&\,0.15-
\frac{-27.332509\ldots}{1484.137157\ldots}\\
=&\,0.168416\ldots.
\end{align*} x 1 = = = 0.15 − f ′ ( 0.15 ) f ( 0.15 ) 0.15 − 1484.137157 … − 27.332509 … 0.168416 … . Hence, to 3 decimal places,
x 1 = 0.168 . \boxed{x_1=0.168}. x 1 = 0.168 .
(b)
解法一
思路
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在端点对应的两点之间作割线，用一次线性插值求割线与 x x x
答题过程
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The endpoint values are
f ( 0.15 ) = − 27.332509 … f(0.15)=-27.332509\ldots f ( 0.15 ) = − 27.332509 … and
f ( 0.25 ) = 5.571428 … . f(0.25)=5.571428\ldots. f ( 0.25 ) = 5.571428 … . Using linear interpolation once,
α ≈ 0.15 + 0 − f ( 0.15 ) f ( 0.25 ) − f ( 0.15 ) × ( 0.25 − 0.15 ) = 0.15 + 27.332509 … 5.571428 … + 27.332509 … × ( 0.10 ) = 0.233067 … . \begin{align*}
\alpha
\approx&\,0.15+
\frac{0-f(0.15)}{f(0.25)-f(0.15)}\\
&\,\hspace{2pt}\times(0.25-0.15)\\
=&\,0.15+
\frac{27.332509\ldots}
{5.571428\ldots+27.332509\ldots}\\
&\,\hspace{2pt}\times(0.10)\\
=&\,0.233067\ldots.
\end{align*} α ≈ = = 0.15 + f ( 0.25 ) − f ( 0.15 ) 0 − f ( 0.15 ) × ( 0.25 − 0.15 ) 0.15 + 5.571428 … + 27.332509 … 27.332509 … × ( 0.10 ) 0.233067 … . Therefore, to 3 decimal places,
α ≈ 0.233 . \boxed{\alpha\approx0.233}. α ≈ 0.233 . 