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FP1 2023 Jan Q4

A Level / Edexcel / FP1

IAL 2023 Jan Paper · Question 4

题目

Problem

 f(x)= 1−18x4 +27x7\begin{align*} &\,f(x)\\ =&\,1-\frac1{8x^4}\\ &\,\hspace{2pt}+\frac2{7\sqrt{x^7}} \end{align*} x>0x>0

The equation f(x)=0f(x) = 0 has a single root, α\alpha, that lies in the interval [0.15,0.25][0.15, 0.25]

(a) (i) Determine f′(x)f'(x)

(ii) Explain why 0.250.25 cannot be used as an initial approximation for α\alpha in the Newton-Raphson process.

(iii) Taking 0.150.15 as a first approximation to α\alpha apply the Newton-Raphson process once to f(x)f(x) to obtain a second approximation to α\alpha Give your answer to 33 decimal places.

(5)

(b) Use linear interpolation once on the interval [0.15,0.25][0.15, 0.25] to find another approximation to α\alpha Give your answer to 33 decimal places.

(3)
题目中文翻译  f(x)= 1−18x4 +27x7\begin{align*} &\,f(x)\\ =&\,1-\frac1{8x^4}\\ &\,\hspace{2pt}+\frac2{7\sqrt{x^7}} \end{align*} x>0x>0

方程 f(x)=0f(x) = 0 有唯一根 α\alpha，位于区间 [0.15,0.25][0.15, 0.25] 内。

(a) (i) 确定 f′(x)f'(x)

(ii) 解释为什么 0.250.25 不能作为 Newton-Raphson 法中 α\alpha 的初始近似值。

(iii) 取 0.150.15 作为 α\alpha 的第一个近似值，对 f(x)f(x) 应用一次 Newton-Raphson 法，求 α\alpha 的第二个近似值，答案保留 33 位小数。

(5)

(b) 在区间 [0.15,0.25][0.15, 0.25] 上使用一次线性插值法，求 α\alpha 的另一个近似值，答案保留 33 位小数。

(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)= 1−18x−4 +27x−7/2,\begin{align*} f(x)=&\,1-\frac18x^{-4}\\ &\,\hspace{2pt}+\frac27x^{-7/2}, \end{align*}

we obtain

 f′(x)= −18(−4)x−5 +27(−72)x−9/2= 12x5−1x9/2.\begin{align*} &\,f'(x)\\ =&\,-\frac18(-4)x^{-5}\\ &\,\hspace{2pt}+\frac27\Big(-\frac72\Big)x^{-9/2}\\ =&\,\boxed{\frac1{2x^5}-\frac1{x^{9/2}}}. \end{align*}

(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)，若它等于零，迭代式便没有定义。

答题过程

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At x=0.25=1/4x=0.25=1/4,

 f′(0.25)= 12(1/4)5 −1(1/4)9/2= 512−512= 0.\begin{align*} &\,f'(0.25)\\ =&\,\frac1{2(1/4)^5}\\ &\,\hspace{2pt}-\frac1{(1/4)^{9/2}}\\ =&\,512-512\\ =&\,0. \end{align*}

The Newton–Raphson formula

xn+1=xn−f(xn)f′(xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}

would therefore require division by zero. Hence 0.250.25 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 x0=0.15x_0=0.15，使用未舍入的函数值和导数值完成一次迭代，最后才取三位小数。

答题过程

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At x0=0.15x_0=0.15,

 f(0.15)= 1−18(0.15)4 +27(0.15)7/2= −27.332509…\begin{align*} &\,f(0.15)\\ =&\,1-\frac1{8(0.15)^4}\\ &\,\hspace{2pt}+\frac2{7(0.15)^{7/2}}\\ =&\,-27.332509\ldots \end{align*}

and

 f′(0.15)= 12(0.15)5 −1(0.15)9/2= 1484.137157….\begin{align*} &\,f'(0.15)\\ =&\,\frac1{2(0.15)^5}\\ &\,\hspace{2pt}-\frac1{(0.15)^{9/2}}\\ =&\,1484.137157\ldots. \end{align*}

Therefore,

 x1= 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*}

Hence, to 3 decimal places,

x1=0.168.\boxed{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 xx 轴的交点。函数值一正一负，因此可直接按两端函数值的比例定位近似根。

答题过程

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Using the value of f(0.15)f(0.15) from part (a),

 f(0.15)= −27.332509….\begin{align*} &\,f(0.15)\\ =&\,-27.332509\ldots. \end{align*}

At the other endpoint,

 f(0.25)= 1−18(0.25)4 +27(0.25)7/2= 5.571428….\begin{align*} &\,f(0.25)\\ =&\,1-\frac1{8(0.25)^4}\\ &\,\hspace{2pt}+\frac2{7(0.25)^{7/2}}\\ =&\,5.571428\ldots. \end{align*}

For the interpolation, let

N= 0−f(0.15)= 27.332509…,\begin{align*} N=&\,0-f(0.15)\\ =&\,27.332509\ldots, \end{align*}  D= f(0.25)−f(0.15)= 5.571428… +27.332509…= 32.903938…,\begin{align*} &\,D\\ =&\,f(0.25)-f(0.15)\\ =&\,5.571428\ldots\\ &\,\hspace{2pt}+27.332509\ldots\\ =&\,32.903938\ldots, \end{align*}

and

Δx= 0.25−0.15= 0.10.\begin{align*} \Delta x=&\,0.25-0.15\\ =&\,0.10. \end{align*}

Using linear interpolation once,

 α≈ 0.15+NDΔx= 0.15+27.332509…32.903938… ×0.10= 0.233067….\begin{align*} &\,\alpha\\ \approx&\,0.15+\frac ND\Delta x\\ =&\,0.15+ \frac{27.332509\ldots}{32.903938\ldots}\\ &\,\hspace{2pt}\times0.10\\ =&\,0.233067\ldots. \end{align*}

The unrounded endpoint values are used in this calculation. Therefore, to 3 decimal places,

α≈0.233.\boxed{\alpha\approx0.233}.