Quiz
约 616 个字 16 张图片 预计阅读时间 3 分钟
Quiz 1
（a）Exact：\(\frac{4}{5}\times\frac{1}{3}=\frac{4}{15}\)
Three-Digit Chopping：\(\frac{4}{5}=0.800,\frac{1}{3}\approx 0.333,\frac{4}{5}\times\frac{1}{3}=0.8\times 0.333=0.2664\approx 0.266\)
Three-Digit Rounding：\(\frac{4}{5}=0.800,\frac{1}{3}\approx 0.333,\frac{4}{5}\times\frac{1}{3}=0.8\times 0.333=0.2664\approx 0.266\)
Relative Error：\(e=\frac{|p-p^*|}{|p|}=\frac{|\frac{4}{15}-0.266|}{\frac{4}{15}}=0.0025\)
（b）Exact：\((\frac{1}{3}+\frac{3}{11})-\frac{3}{20}=\frac{301}{660}\)
Three-Digit Chopping：\(\frac{1}{3}\approx 0.333,\frac{3}{11}\approx 0.272,\frac{3}{20}=0.150,0.333+0.272-0.150=0.455\)
Three-Digit Rounding：\(\frac{1}{3}\approx 0.333,\frac{3}{11}\approx 0.273,\frac{3}{20}=0.150,0.333+0.272-0.150=0.456\)
Chopping Error：\(e=\frac{|p-p^*|}{|p|}=\frac{|\frac{301}{660}-0.455|}{\frac{301}{660}}=0.002325\)
Rounding Error：\(e=\frac{|p-p^*|}{|p|}=\frac{|\frac{301}{660}-0.456|}{\frac{301}{660}}=0.000133\)
Quiz 2
构造 \(g(x)=x=(x+1)^{\frac{1}{3}},g'(x)=\frac{1}{3(x+1)^{\frac{2}{3}}}\)，在 \([1,2]\) 中 \(|g'(x)|\leq|g'(1)|\approx 0.21<1\)，所以存在不动点且可以迭代计算
\[
\begin{aligned}
p_1=g(p_0)=2^{\frac{1}{3}}\approx 1.2599&,p_2=g(p_1)=(1+1.2599)^{\frac{1}{3}}\approx 1.3123\\
\Rightarrow e\leq\frac{0.21}{1-0.21}|1.3123&-1.2599|=0.0139>10^{-2}\\
p_3=g(p_2)&=(1+1.3123)^{\frac{1}{3}}\approx 1.3224\\
\Rightarrow e=\frac{0.21^2}{1-0.21}|1.3123&-1.2599|=2.925\times 10^{-3}<10^{-2}
\end{aligned}
\]
综上计算答案为 \(1.3224\)
Quiz 3
（a）
\[
\begin{aligned}
\overline{A}=\begin{bmatrix}
1 & 1 & 0 & 1 & 2\\
2 & 1 & -1 & 1 & 1\\
4 & -1 & -2 & 2 & 0\\
3 & -1 & -1 & 2 & -3
\end{bmatrix}
\Rightarrow\begin{bmatrix}
1 & 1 & 0 & 1 & 2\\
0 & 1 & 1 & 1 & 3\\
0 & 0 & 3 & 3 & 7\\
0 & 0 & 0 & 0 & -4
\end{bmatrix}
\end{aligned}
\]
可以看到方程无解，不需要行交换
（b）
\[
\begin{aligned}
\overline{A}=\begin{bmatrix}
1 & 1 & 0 & 1 & 2\\
2 & 1 & -1 & 1 & 1\\
-1 & 2 & 3 & -4 & 4\\
3 & -1 & -1 & 2 & -3
\end{bmatrix}
\Rightarrow\begin{bmatrix}
1 & 1 & 0 & 1 & 2\\
0 & 1 & 1 & 1 & 3\\
0 & 0 & 0 & -6 & -3\\
0 & 0 & 3 & 3 & 3\\
\end{bmatrix}
\Rightarrow\begin{bmatrix}
1 & 1 & 0 & 1 & 2\\
0 & 1 & 1 & 1 & 3\\
0 & 0 & 1 & 1 & 1\\
0 & 0 & 0 & -6 & -3
\end{bmatrix}
\end{aligned}
\]
方程有唯一解 \(x_1=-\frac{1}{2},x_2=2,x_3=\frac{1}{2},x_4=\frac{1}{2}\)，需要进行行交换
Quiz 4
能观察到第一次消元第二行前三个元素均为 0，所以需要进行行交换
\[
\begin{aligned}
P&=\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0
\end{bmatrix}
\Rightarrow PA=\begin{bmatrix}
1 & -2 & 3 & 0\\
2 & 1 & 3 & -1\\
1 & -2 & 2 & -2\\
1 & -2 & 3 & 1
\end{bmatrix}\\
L_1&=\begin{bmatrix}
1 & 0 & 0 & 0\\
-2 & 1 & 0 & 0\\
-1 & 0 & 1 & 0\\
-1 & 0 & 0 & 1
\end{bmatrix}
\Rightarrow L_1PA=\begin{bmatrix}
1 & -2 & 3 & 0\\
0 & 5 & -3 & -1\\
0 & 0 & -1 & -2\\
0 & 0 & 0 & 1
\end{bmatrix}=U\\
\Rightarrow A&=P^{-1}L_1^{-1}U,P'=P^{-1}=\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0
\end{bmatrix}\\
L&=L_1^{-1}=\begin{bmatrix}
1 & 0 & 0 & 0\\
2 & 1 & 0 & 0\\
1 & 0 & 1 & 0\\
1 & 0 & 0 & 1
\end{bmatrix},U=\begin{bmatrix}
1 & -2 & 3 & 0\\
0 & 5 & -3 & -1\\
0 & 0 & -1 & -2\\
0 & 0 & 0 & 1
\end{bmatrix}
\end{aligned}
\]
Quiz 5
\[
\begin{aligned}
x_1^{(1)}&=x_1^{(0)}+\frac{1.1}{3}(1-3x_1^{(0)}+x_2^{(0)}-x_3^{(0)})=\frac{1.1}{3}\approx 0.367\\
x_2^{(1)}&=x_2^{(0)}+\frac{1.1}{6}(0-3x_1^{(1)}-6x_2^{(0)}-2x_3^{(0)})=-\frac{1.21}{6}\approx -0.202\\
x_3^{(1)}&=x_3^{(0)}+\frac{1.1}{7}(0-3x_1^{(1)}-3x_2^{(1)}-7x_3^{(0)})\approx 0.551\\
\therefore x^{(1)}&=\begin{bmatrix}
0.367\\
-0.202\\
0.551
\end{bmatrix}
\end{aligned}
\]
Quiz 6
（a）
\[
\begin{aligned}
\det(A)&=\frac{1}{2}\times\frac{1}{4}-\frac{1}{3}\times\frac{1}{3}=\frac{1}{72}\\
A^{-1}&=72\times\begin{bmatrix}
\frac{1}{4} & -\frac{1}{3}\\
-\frac{1}{3} & \frac{1}{2}
\end{bmatrix}=\begin{bmatrix}
18 & -24\\
-24 & 36
\end{bmatrix}\\
K(A)&=\|A\|\|A^{-1}\|=\frac{5}{6}\times 60=50
\end{aligned}
\]
（b）
\[
\begin{aligned}
\det(A)&=1\times 2-1.00001\times 2=-0.00002\\
A^{-1}&=-50000\times\begin{bmatrix}
2 & -2\\
-1.00001 & 1
\end{bmatrix}=\begin{bmatrix}
-100000 & 100000\\
50000.5 & -50000
\end{bmatrix}\\
K(A)&=\|A\|\|A^{-1}\|=3.00001\times 200000=600002
\end{aligned}
\]
Quiz 7
| \(x_0=0.0\) |
1.0000 |
|
|
|
|
| \(x_1=0.2\) |
1.22140 |
1.1070 |
|
|
|
| \(x_2=0.4\) |
1.49182 |
1.3521 |
0.61275 |
|
|
| \(x_3=0.6\) |
1.82212 |
1.6515 |
0.7485 |
0.22625 |
|
| \(x_4=0.8\) |
2.22554 |
2.0171 |
0.914 |
0.27583 |
0.061975 |
\[
\begin{aligned}
\therefore f(0.05)&=f(0)+1.1070*(0.05-0)+0.61275*(0.05-0)*(0.05-0.2)\\
&+0.22625*(0.05-0)*(0.05-0.2)*(0.05-0.4)\\
&+0.061975*(0.05-0)*(0.05-0.2)*(0.05-0.4)*(0.05-0.6)\approx 1.0513
\end{aligned}
\]
Quiz 8
| \(F(l)\) |
\(l\) |
\(x= l-E\) |
| 2 |
7.0 |
1.7 |
| 4 |
9.4 |
4.1 |
| 6 |
12.3 |
7.0 |
对误差 \(S=\sum\limits_{n=1}^3[F-kx]^2\) 求偏导得 \(\frac{\partial S}{\partial k}=\sum\limits_{i=1}^3-2x(F-kx)=0\Rightarrow k=\frac{\sum\limits_{i=1}^nFx}{\sum\limits_{i=1}^nx^2}\approx 0.89956\)
Quiz 9
\(\int_{-1}^1 1dx=2,\int_{-1}^1 x^2dx=\frac{2}{3},\int_{-1}^1(x^2-2x+3)dx=\frac{20}{3},\int_{-1}^1(x^2-2x+3)xdx=-\frac{4}{3}\)
\(\therefore f(x)\approx \frac{10}{3}-2x=3.33333-2x\)
Quiz 10
\[
\begin{aligned}
LHS&=\int_{-1}^1\frac{[\cos(n\arccos x)]^2}{\sqrt{1-x^2}}dx\\
&=\int_{-1}^1[\cos(n\arccos x)]^2d(\arccos x)\\
&=\int_0^{\pi}[\cos ny]^2dy=\int_0^{\pi}\frac{\cos 2ny+1}{2}dy=\frac{\pi}{2}
\end{aligned}
\]
或者换元 \(\theta=\arccos(x)\) 也可
Quiz 11
根据题目条件我们得到方程组：
\[
\begin{cases}
2f(0)=12\\
1\times (f(-0.5)+f(0.5))=5\\
\frac{1}{3}[f(-1)+4f(0)+f(1)]=6\\
f(-1)=f(1)\\
f(-0.5)=f(0.5)-1
\end{cases}
\]
解得 \(f(-1)=-3,f(-0.5)=2,f(0)=6,f(0.5)=3,f(1)=-3\)
Quiz 12
给定节点 \(x_1,x_2,\cdots,x_n\)，构造 \(w(x)=\prod_{i=1}^n(x-x_i),P(x)=w^2(x)=(x-x_1)^2(x-x_2)^2\cdots(x-x_n)^2\)
计算 \(P(x)\) 在区间 \([a,b]\) 上的精确积分值 \(I=\int_a^bP(x)dx=\int_a^b[w(x)]^2dx>0\)
应用到公式当中，\(Q(P)=\sum\limits_{i=1}^nc_iP(x_i)=0\neq I\)
所以存在一个 2n 次多项式使得求积公式不精确
Quiz 13
对于中点法：
\[
\begin{aligned}
w_{i+1}&=w_i+h[-w_i-\frac{h}{2}(-w_i+t_i+1)+t_i+\frac{h}{2}+1]\\
&=w_i(1-h+\frac{h^2}{2})+t_ih(1-\frac{h}{2})+h
\end{aligned}
\]
对于改进欧拉法：
\[
\begin{aligned}
w_{i+1}&=w_i+\frac{h}{2}\cdot[f(t_i,w_i)+f(t_i+h,w_i+h\cdot f(t_i,w_i))]\\
&=w_i+\frac{h}{2}\cdot[-w_i+t_i+1+-w_i-h(-w_i+t_i+1)+t_i+h+1]\\
&=w_i(1-h+\frac{h^2}{2})+t_ih(1-\frac{h}{2})+h
\end{aligned}
\]
所以两者迭代公式相同，对任意 \(h\) 估计也相同
Quiz 14
使用牛顿向后差分公式，在 \((t_i,f_i),(t_{i-1},f_{i-1})\) 上对 \(f\) 插值：
\[
P_1(t_i+sh) = f_i + s \nabla f_i = f_i + s(f_i - f_{i-1})
\]
得到 \(w_{i+1} = w_i + h \int_0^1 [f_i + s(f_i - f_{i-1})]ds = w_i + \frac{h}{2}(3f_i - f_{i-1})\)
Quiz 15
对于测试方程 \(f(t,y)=\lambda y\)，代入梯形方法公式：
\[
\begin{aligned}
w_{i+1}&=w_i+\frac{h}{2}(\lambda w_i+\lambda w_{i+1})\\
\Rightarrow w_{i+1}-\frac{h\lambda}{2}w_{i+1}&=w_i+\frac{h\lambda}{2}w_i\\
\Rightarrow w_{i+1}&=\frac{1+\frac{h\lambda}{2}}{1-\frac{h\lambda}{2}}w_i
\end{aligned}
\]
令 \(z=h\lambda\)，则解得比例因子为 \(R(z)=\frac{1+\frac{z}{2}}{1-\frac{z}{2}}\)，由稳定性要求 \(|R(z)|\leq 1\)：
\[
\begin{aligned}
\bigg|\frac{1+\frac{z}{2}}{1-\frac{z}{2}}\bigg|&\leq 1\\
\bigg|1+\frac{z}{2}\bigg|^2&\leq\bigg|1-\frac{z}{2}\bigg|^2\\
(1+\frac{z}{2})(1+\frac{\overline{z}}{2})&\leq(1-\frac{z}{2})(1-\frac{\overline{z}}{2})\\
\Rightarrow\frac{z+\overline{z}}{2}&\leq -\frac{z+\overline{z}}{2}\\
\Rightarrow Re(z)&\leq 0
\end{aligned}
\]
因此稳定性要求 \(Re(\lambda)\leq 0\)