复化积分
复化梯形公式
$[x_i,x_{i+1}]$ 复化梯形公式
等距节点 $h=\frac{b-a}n$
\[x_i=a+ih, \quad i=0,1,\cdots,n
\]
$[x_i, x_{i+1}]$
\[\int_{x_i}^{x_{i+1}}f(x)dx=\frac{h}2(f(x_i)+f(x_{i+1}))-\frac{h^3}{12}f''(\xi_i)
\]
\[\begin{aligned}
I(f)&=\int_a^bf(x)dx=\sum_{i=0}^{n-1}\int_{x_i}^{x_{i+1}}f(x)dx \\
&=\sum_{i=0}^{n-1}\frac{h}2(f(x_i)+f(x_{i+1}))-\sum_{i=0}^{n-1}\frac{h^3}{12}f''(\xi_i) \\
&=h\left(\frac12f(a)+\sum_{i=1}^{n-1}f(x_i)+\frac12f(b)\right)-\sum_{i=0}^{n-1}\frac{h^3}{12}f''(\xi_i)
\end{aligned}
\]
等距节点 复化梯形公式
\[T_n(f)=h\left(\frac12f(a)+\sum_{i=1}^{n-1}f(x_i)+\frac12f(b)\right)
\]
$f\in C^2[a,b]$
\[R=-\sum_{i=0}^{n-1}\frac{h^3}{12}f''(\xi_i)=-n\frac{h^3}{12}f''(\xi)
=-h^2\frac{(b-a)}{12}f''(\xi)
\]
注 收敛 $\displaystyle\lim_{h\to0}R=0$
例 1 . $f(x)=\frac4{1+x^2}$
$x_i$ $f(x_i)$
解 .
\[\begin{aligned}
I=&\frac{0.1}2(f(0)+f(0.1))+\frac{0.1}2(f(0.1)+f(0.2)) \\
&+\frac{0.2}2(f(0.2)+f(0.4))+\frac{0.2}2(f(0.4)+f(0.6)) \\
&+\frac{0.2}2(f(0.6)+f(0.8))+\frac{0.2}2(f(0.8)+f(1))
\end{aligned}
\]
3.1386580254636933
复化Simpson积分
复化Simpson积分
$h=\frac{b-a}n$ $n=2m$ $[x_{2i},x_{2i+2}]$
\[\begin{aligned}
\int_{x_{2i}}^{x_{2i+2}}f(x)dx=&\frac{2h}6(f(x_{2i})+4f(x_{2i+1})+f(x_{2i+2})) \\
&-\frac{(2h)^5}{2880}f^{(4)}(\xi_i)
\end{aligned}
\]
\[\begin{aligned}
\int_a^bf(x)dx=&\sum_{i=0}^m\int_{x_{2i}}^{x_{2i+2}}f(x)dx \\
=&\sum_{i=0}^m\frac{2h}6(f(x_{2i})+4f(x_{2i+1})+f(x_{2i+2}))\\
&-\sum_{i=0}^m\frac{(2h)^5}{2880}f^{(4)}(\xi_i)
\end{aligned}
\]
等距节点 复化Simpson公式
\[S_m(f)=\frac{h}3\left(f(a)+4\sum_{i=0}^{m-1}f(x_{2i+1})+2\sum_{i=1}^{m-1}f(x_{2i})+f(b)\right)
\]
$f\in C^4[a,b]$
\[\begin{aligned}
R&=-\sum_{i=0}^m\frac{(2h)^5}{2880}f^{(4)}(\xi_i)=-m\frac{(2h)^5}{2880}f^{(4)}(\xi) \\
&=-{\color{red}{h^4}}\frac{b-a}{180}f^{(4)}(\xi)
\end{aligned}
\]
定义 1 .
\[\lim_{h\to0}\frac{R}{h^p}=C, C\neq 0
\]
p阶收敛
例 2 . $f(x)=\frac4{1+x^2}$
$x_i$ $f(x_i)$
解 .
\[\begin{aligned}
I=&\frac{0.2}6(f(0)+4f(0.1)+f(0.2)) \\
&+\frac{0.4}6(f(0.2)+4f(0.4)+f(0.6)) \\
&+\frac{0.4}6(f(0.6)+4f(0.8)+f(1))
\end{aligned}
\]
3.1414384532577757
例 3 . $\pi=\int_0^1\frac{4}{1+x^2}dx$
解 . $[0,1]$ $x_k=\frac{k}8$
复化梯形公式
\[T_8(f)=\frac18\left(\frac12f(0)+\sum_{k=1}^7f(x_k)+\frac12f(1)\right)=3.1{\color{red}{38988494}}
\]
复化Simpson公式
\[\begin{aligned}
S_4(f)&=\frac18\frac13\left(f(0)+4\sum_{odd}f(x_k)+2\sum_{even}f(x_k)+f(1)\right) \\
&=3.141592{\color{red}{502}}
\end{aligned}
\]
程序作业-复化积分
\[I(f)=\int_1^5\sin(x)dx
\]
$x_i=1+i\times\frac{4}{2^N}$ $N$ $1,2,\cdots,12$
复化梯形公式
N=1, 误差 0.0008560379230928561
N=2，误差 0.00021390241995433712 阶: 2.0020020
...
复化Simpson公式
N=1, 误差 -2.2921544006182515e-06
N=2，误差 -1.4274775839151488e-07 阶: 4.0572407
...
$E=O(h^p)$ $h$
$\frac{h}2$
\[E_2=C_2\left(\frac{h}2\right)^p
\]
$C_1\approx C_2$ $\frac{E_1}{E_2} \approx 2^p$
\[p\approx \frac{\ln(|\frac{E_1}{E_2}|)}{\ln2}
\]
问题 . $\displaystyle\int_a^b f(x)dx$ $\epsilon$ $h$
事后误差估计
\[I(f)-T_n(f)=-h^2\frac{(b-a)}{12}f''(\xi)
\]
\[I(f)-T_{2n}(f)=-(\frac{h}2)^2\frac{(b-a)}{12}f''(\xi_2)
\]
假定$f''(\xi)\approx f''(\xi_2)$
\[\frac{I(f)-T_{n}(f)}{I(f)-T_{2n}(f)}\approx 2^2
\]
\[{I(f)-T_{n}(f)}\approx 2^2({I(f)-T_{2n}(f)})
\]
\[-T_{n}(f)\approx (2^2-1)I(f)-2^2T_{2n}(f)
\]
\[\begin{aligned}
T_{2n}(f)-T_{n}(f)\approx& (2^2-1)I(f)-(2^2-1)T_{2n}(f) \\
=&(2^2-1)[I(f)-T_{2n}(f)]
\end{aligned}
\]
\[I(f)-T_{2n}(f)\approx\frac1{2^2-1}(T_{2n}(f)-T_{n}(f))
\]
\[I(f)-S_{n}(f)\approx 2^4(I(f)-S_{2n}(f))
\]
\[\begin{aligned}
S_{2n}(f)-S_{n}(f)\approx& (2^4-1)I(f)-(2^4-1)S_{2n}(f) \\
=&(2^4-1)[I(f)-S_{2n}(f)]
\end{aligned}
\]
复化梯形公式 事后误差估计
\[I(f)-T_{2n}(f)\approx\frac1{2^2-1}(T_{2n}(f)-T_{n}(f))
\]
复化Simpson公式 事后误差估计
\[I(f)-S_{2n}(f)\approx\frac1{2^4-1}(S_{2n}(f)-S_{n}(f))
\]
Romberg积分
Richardson外推
\[T_{2n}(f)+\frac13(T_{2n}(f)-T_n(f))=S_n(f)
\]
$O(h^2)$ $O(h^4)$
\[S_{2n}(f)+\frac1{15}(S_{2n}(f)-S_n(f))=C_n(f)
\]
$O(h^4)$ $O(h^6)$
定理 1 . (Euler-MacLaurin定理 ) $I^{(m)}$ $2m$ $I(f)=I^{(m)}(h)+O(h^{2m})$
\[I^{(m+1)}(\frac{h}2)=I^{m}(\frac{h}2)+\frac{I^{(m)}(\frac{h}2)-I^{(m)}(h)}{2^{2m}-1}
\]
$2m+2$ $I(f)=I^{(m+1)}(h)+O(h^{2m+2})$
注
$R^{(k)}$ $k=1,2,3,\cdots$ $R^{(k)}_j$ $R^{(1)}_0$ $R^{(0)}_1$ $R^{(0)}_2$
定义 2 . Romberg积分公式
\[\displaystyle R^{(k)}_j=R^{(k-1)}_j+\frac{R^{(k-1)}_j-R^{(k-1)}_{j-1}}{2^{2k-2}-1}
\]
例 4 . $I=\int_0^1\frac{4}{1+x^2}dx$
解 .
$T_1=\frac12(f(0)+f(1))=3$ $T_2=\frac12T_1+\frac12f(\frac12)=3.1$ $S_1=\frac{4T_2-T_1}{2^2-1}=3.133333$ $T_4=\frac12T_2+\frac14(f(\frac14)+f(\frac34))=3.131176$ $S_2=\frac{4T_4-T_2}{2^2-1}=3.14156863$ $C_1=\frac{16S_2-S_1}{2^4-1}=3.14211765$ $T_8=\frac12T_4+\frac18(f(\frac18)+f(\frac38)+f(\frac58)+f(\frac78))=3.138988$ $S_4=\frac{4T_8-T_4}{2^2-1}=3.1415925$ $C_2=\frac{16S_4-S_2}{2^4-1}=3.141594$ $R_1=\frac{64C_2-C_2}{2^6-1}=3.141586$ $T_{16}=3.140942$ $S_8=3.14159265$ $C_4=3.14159266$ $R_2=3.14159264$ $3.14159267$ $T_{32}=3.141430$ $S_{16}=3.14159265$ $C_8=3.141593$ $R_4=3.141593$ $3.14159265$ $3.14159265$
$n=1$ $n=2$ $n=4$ $n=8$ $n={16}$ $n={32}$
积分的自适应计算
问题 .
基本思路：
def recursive_asr(f,a,b,eps,simpson_quad):
"Recursive implementation of adaptive Simpson's rule."
c = (a+b) / 2.0
left = simpsons_rule(f,a,c)
right = simpsons_rule(f,c,b)
if abs(left + right - simpson_quad) <= 15*eps:
return left + right + (left + right - simpson_quad)/15.0
return recursive_asr(f,a,c,eps/2.0,left) + recursive_asr(f,c,b,eps/2.0,right)
例 5 .
\[\int_0^4 13x(1-x)e^{-\frac32x}dx
\]
