Homework 09

约 608 个字 3 张图片 预计阅读时间 3 分钟

8.1.5

（a）我们有 \(b_0=\sum\limits_{i=1}^m 1=10,b_1=\sum\limits_{i=1}^mx_i=54.1,b_2=\sum\limits_{i=1}^mx_i^2=303.39,c_0=\sum\limits_{i=1}^my_i=1958.39,c_1=\sum\limits_{i=1}^my_ix_i=11366.843\)

\[ \begin{bmatrix} 10 & 54.1\\ 54.1 & 303.39 \end{bmatrix}\begin{bmatrix} a_0\\ a_1 \end{bmatrix}=\begin{bmatrix} 1958.39\\ 11366.843 \end{bmatrix} \]

解得 \(a_0=-194.138,a_1=72.0845\)，即 \(P(x)=72.0845x-194.138,\text{error}=329\)

（b）\(b_3=1759.831,b_4=10523.1207,c_2=68006.6811\)

\[ \begin{bmatrix} 10 & 54.1 & 303.39\\ 54.1 & 303.39 & 1759.831\\ 303.39 & 1759.831 & 10523.1207 \end{bmatrix}\begin{bmatrix} a_0\\ a_1\\ a_2 \end{bmatrix}=\begin{bmatrix} 1958.39\\ 11366.843\\ 68006.6811 \end{bmatrix} \]

解得 \(a_0=1.23556,a_1=-1.14352,a_2=6.61821\)，即 \(P(x)=6.61821x^2-1.14352x+1.23556,\text{error}=1.44\times 10^{-3}\)

（c）\(b_5=64607.97751,b_6=405616.743519,c_3=417730.09823\)

\[ \begin{bmatrix} 10 & 54.1 & 303.39 & 1759.831\\ 54.1 & 303.39 & 1759.831 & 10523.1207\\ 303.39 & 1759.831 & 10523.1207 & 64607.97751\\ 1759.831 & 10523.1207 & 64607.97751 & 405616.743519 \end{bmatrix}\begin{bmatrix} a_0\\ a_1\\ a_2\\ a_3 \end{bmatrix}=\begin{bmatrix} 1958.39\\ 11366.843\\ 68006.6811\\ 417730.09823 \end{bmatrix} \]

解得 \(a_0=3.42904,a_1=-2.37919,a_2=6.84557,a_3=-0.0136742\)，即 \(P(x)=-0.0136742x^3+6.84557x^2-2.37919x+3.42904,\text{error}=5.27\times 10^{-4}\)

（d）\(\ln y=\ln b+ax\)，\(c_0=52.0336,c_1=285.4898\)

\[ \begin{bmatrix} 10 & 54.1\\ 54.1 & 303.39 \end{bmatrix}\begin{bmatrix} a_0\\ a_1 \end{bmatrix}=\begin{bmatrix} 52.0336\\ 285.4898 \end{bmatrix} \]

解得 \(a_0=3.1888,a_1=0.372382\)，即 \(P(x)=24.2588e^{0.372382x},\text{error}=418\)

（e）\(\ln y=\ln b+a\ln x\)，\(b_0=10, b_1=16.6995,b_2=28.2537,c_0=52.0336,c_1=87.6334\)

\[ \begin{bmatrix} 10 & 16.6995\\ 16.6995 & 28.2537 \end{bmatrix}\begin{bmatrix} a_0\\ a_1 \end{bmatrix}=\begin{bmatrix} 52.0336\\ 87.6334 \end{bmatrix} \]

解得 \(a_0=1.8308,a_1=2.01954\)，即 \(P(x)=6.23903x^{2.01954},\text{error}=0.00703\)

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8.2.3

（a）\(\int_{-1}^1 1dx=2,\int_{-1}^1 x^2dx=\frac{2}{3},\int_{-1}^1(x^2-2x+3)dx=\frac{20}{3},\int_{-1}^1(x^2-2x+3)xdx=-\frac{4}{3}\)

\(\therefore f(x)\approx \frac{10}{3}-2x=3.33333-2x\)

（b）\(\int_{-1}^1x^3dx=0,\int_{-1}^1x^4dx=\frac{2}{5}\)

\(\therefore f(x)\approx 0.6x\)

（c）\(\int_{-1}^1\frac{1}{x+2}dx=\ln 3,\int_{-1}^1\frac{x}{x+2}dx=2-2\ln 3\)

\(\therefore f(x)\approx \frac{\ln 3}{2}+(3-3\ln 3)x=0.54931-0.29584x\)

（d）\(\int_{-1}^1e^xdx=e-\frac{1}{e},\int_{-1}^1xe^xdx=\frac{2}{e}\)

\(\therefore f(x)\approx 1.17520+1.10364x\)

（e）\(\int_{-1}^1(\frac{1}{2}\cos x+\frac{1}{3}\sin 2x)dx=\sin 1,\int_{-1}^1(\frac{1}{2}x\cos x+\frac{1}{3}x\sin 2x)dx=-\frac{1}{6}(2\cos 2-\sin 2)\)

\(\therefore f(x)\approx 0.42074+0.42540x\)

（f）\(\int_{-1}^1\ln(x+2)dx=3\ln 3-2,\int_{-1}^1x\ln(x+2)dx=\frac{1}{2}(4-3\ln 3)\)

\(\therefore f(x)=0.64792+0.52812x\)

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8.2.11

\[ \begin{aligned} B_1&=\frac{\int_0^{\infty}xe^{-x}dx}{\int_0^{\infty}e^{-x}dx}=1\quad L_1(x)=(x-B_1)L_0(x)=x-1\\ B_2&=\frac{\int_0^{\infty}(x-1)^2xe^{-x}dx}{\int_0^{\infty}(x-1)^2e^{-x}dx}=3\quad C_2=\frac{\int_0^{\infty}x(x-1)e^{-x}dx}{\int_0^{\infty}e^{-x}dx}=1\\ L_2(x)&=(x-B_2)L_1(x)-C_2L_0(x)=x^2-4x+2\\ B_3&=\frac{\int_0^{\infty}x(x^2-4x+2)^2e^{-x}dx}{\int_0^{\infty}(x^2-4x+2)^2e^{-x}dx}=5\quad C_3=\frac{\int_0^{\infty}x(x^2-4x+2)(x-1)e^{-x}dx}{\int_0^{\infty}(x-1)^2e^{-x}dx}=4\\ L_3(x)&=(x-B_3)L_2(x)-C_3L_1(x)=(x-5)(x^2-4x+2)-4(x-1)=x^3-9x^2+18x-6\\ \end{aligned} \]

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