插值
拉格朗日 :
L n ( x i ) = y i , i = 0 , 1... n L_n(x_i) = y_i, i=0,1...n Ln(xi)=yi,i=0,1...n ， 其中 L n L_n Ln为 ≤ n \leq n ≤n次多项式
解答
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L_n(x) = l_0(x)\cdot y_0 + l_1(x)\cdot y_1 + ... + l_n(x)\cdot y_n \\ l_i(x) = \frac{w(x)}{w'(x_i)\cdot(x-x_i)} \\
Ln(x)=l0(x)⋅y0+l1(x)⋅y1+...+ln(x)⋅ynli(x)=w′(xi)⋅(x−xi)w(x)
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w(x) = (x-x_0)\cdot(x-x_1)\cdot...(x-x_n) \\ w'(x_i) = (x_i-x_0)\cdot(x_i-x_{i-1})\cdot...(x_i-x_{i+1})\cdot(x_i-x_n) \\
w(x)=(x−x0)⋅(x−x1)⋅...(x−xn)w′(xi)=(xi−x0)⋅(xi−xi−1)⋅...(xi−xi+1)⋅(xi−xn)
方法误差
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R_n(x) \leq |\frac{f^{(n+1)}(\eta)}{(n+1)!}w(x)| \\ (pf): R_n(x) = f(x) - L_n(x) = K(x)\cdot w(x)\\ let\space \phi(t) = f(t) - L_n(t) - R_n(t) \\ \phi(x)=\phi(x_0)=...=\phi(x_n) = 0 \\ \phi有n+2个零点 \Rightarrow \phi^{(n+1)}有1个零点 \Rightarrow \phi^{(n+1)}(\eta)=f^{(n+1)}(\eta)-K(x)\cdot(n+1)! \\ K(x) = \frac{f^{(n+1)}(\eta)}{(n+1)!}, 又\space R(x) = K(x)\cdot w(x)
Rn(x)≤∣(n+1)!f(n+1)(η)w(x)∣(pf):Rn(x)=f(x)−Ln(x)=K(x)⋅w(x)let ϕ(t)=f(t)−Ln(t)−Rn(t)ϕ(x)=ϕ(x0)=...=ϕ(xn)=0ϕ有n+2个零点⇒ϕ(n+1)有1个零点⇒ϕ(n+1)(η)=f(n+1)(η)−K(x)⋅(n+1)!K(x)=(n+1)!f(n+1)(η),又 R(x)=K(x)⋅w(x)
例题: 函数
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R_1(x) \leq max_{x_k\leq x\leq x_{k+1}}|\frac{f''(x)}{2!}w(x)| \leq |\frac{1}{2}\cdot\frac{h^2}{4}| \leq \frac{h^2}{8}
R1(x)≤maxxk≤x≤xk+1∣2!f′′(x)w(x)∣≤∣21⋅4h2∣≤8h2
牛顿插值
k阶均差 :
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定义: f[x_0,...,x_k] = \frac{f[x_0,...,x_{k-2},x_k]-f[x_0,...,x_{k-1}]}{x_k-x_{k-1}} \\ 性质1: f[x_0,x_1...,x_k] = f[x_1,x_0...,x_k] = f[x_k,...,x_1,x_0] \\ 性质2: f[x_0,...,x_k] = \frac{f[x_0,...,x_{k-1}]-f[x_1,...,x_k]}{x_k-x_0} \\ 性质3: f[x_0,...,x_k] = \frac{f^{(n)}(\eta)}{n!}
定义:f[x0,...,xk]=xk−xk−1f[x0,...,xk−2,xk]−f[x0,...,xk−1]性质1:f[x0,x1...,xk]=f[x1,x0...,xk]=f[xk,...,x1,x0]性质2:f[x0,...,xk]=xk−x0f[x0,...,xk−1]−f[x1,...,xk]性质3:f[x0,...,xk]=n!f(n)(η)
证明(性质三): (第二个等号成立是因为唯一性)
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R_n(x) = f(x)-P_n(x) = \frac{f^{(n)}(\eta)}{n!}w(x) = f[x_0,...,x_k]\cdot w(x)
Rn(x)=f(x)−Pn(x)=n!f(n)(η)w(x)=f[x0,...,xk]⋅w(x)
插值公式
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f(x) = f(x_0) + f[x_0, x_1](x-x_0) + f[x_0, x_1, x_2](x-x_0)(x-x_1) + f[x_0,...,x_n](x-x_0)\cdot\cdot\cdot(x-x_{n-1})
f(x)=f(x0)+f[x0,x1](x−x0)+f[x0,x1,x2](x−x0)(x−x1)+f[x0,...,xn](x−x0)⋅⋅⋅(x−xn−1)
等距牛顿插值
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f[x_k,x_{k+1},...,x_{k+m}] = \frac{1}{m!} \frac{1}{h^m}\Delta^m f_k = \frac{f^{(n)}(\eta)}{n!} \\ \Delta^n f_k = \Delta^{n-1}f_{k+1} - \Delta^{n-1}f_k
f[xk,xk+1,...,xk+m]=m!1hm1Δmfk=n!f(n)(η)Δnfk=Δn−1fk+1−Δn−1fk
f ( x 0 + t h ) = f 0 + t Δ f 0 + t ( t − 1 ) 2 ! Δ 2 f 0 + . . . + t ( t − 1 ) . . . ( t − n + 1 ) n ! Δ n f 0 f(x_0+th) = f_0 + t\Delta f_0 + \frac{t(t-1)}{2!}\Delta^2f_0 +...+ \frac{t(t-1)...(t-n+1)}{n!}\Delta^nf_0 f(x0+th)=f0+tΔf0+2!t(t−1)Δ2f0+...+n!t(t−1)...(t−n+1)Δnf0
埃米尔特插值
第一种: $f(x_0) = y_0, f(x_1) = y_1, f(x_2) = y_2, f’(x_1)=m $
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P(x) = f(x_0) + f[x_0,x_1](x-x_0) + f[x_0,x_1,x_2](x-x_0)(x-x_1) + A(x-x_0)(x-x_1)(x-x_2) \\ A = \frac{f'(x_1) - f[x_0,x_1] - f[x_0,x_1,x_2](x_1-x_0)}{(x_1-x_0)(x_1-x_2)}
P(x)=f(x0)+f[x0,x1](x−x0)+f[x0,x1,x2](x−x0)(x−x1)+A(x−x0)(x−x1)(x−x2)A=(x1−x0)(x1−x2)f′(x1)−f[x0,x1]−f[x0,x1,x2](x1−x0)
其中A的取值可以用
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P′(x1)=f′(x1)求得。类似拉格朗日方法，设
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R=k(x)w(x),ϕ(x)=f−P−R，套罗尔定理，求出
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K套回
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R(x) = \frac{1}{4!}f^{(4)}(\eta)(x-x_0)(x-x_1)^2(x-x_2)
R(x)=4!1f(4)(η)(x−x0)(x−x1)2(x−x2)
第二种(两点三次): f ( x 0 ) = y 0 , f ( x 1 ) = y 1 , f ′ ( x 0 ) = m 0 , f ′ ( x 1 ) = m 1 f(x_0) = y_0, f(x_1) = y_1, f'(x_0) = m_0, f'(x_1)=m_1 f(x0)=y0,f(x1)=y1,f′(x0)=m0,f′(x1)=m1
给出假设
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H(x) = \alpha_0(x) y_0 + \alpha_1(x) y_1 + \beta_0(x) m_0 + \beta_1(x) m_1 \\ \alpha_i(x_j) = \begin{cases} 1, i=j \\ 0,i\neq j\end{cases}, \space \alpha_i'(x_j) = 0 \\ \beta_i(x_j) = 0 , \space \beta_i'(x_j) = \begin{cases} 1, i=j \\ 0,i\neq j\end{cases} \\
H(x)=α0(x)y0+α1(x)y1+β0(x)m0+β1(x)m1αi(xj)={1,i=j0,i=j, αi′(xj)=0βi(xj)=0, βi′(xj)={1,i=j0,i=j
经过一通推导，得公式
误差同拉格朗日有
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R_3(x) = \frac{1}{4!}f^{(4)}(\eta)(x-x_0)^2(x-x_1)^2
R3(x)=4!1f(4)(η)(x−x0)2(x−x1)2
分段插值
分段线性
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P_h(x) = \frac{x-x_{k+1}}{x_k-x_{k+1}} y_k + \frac{x-x_{k}}{x_{k+1}-x_{k}} y_{k+1} \\ R(x) \leq |\frac{M}{8}h^2| \\ lim_{h\rightarrow 0} P_h(x) = f(x)
Ph(x)=xk−xk+1x−xk+1yk+xk+1−xkx−xkyk+1R(x)≤∣8Mh2∣limh→0Ph(x)=f(x)
分段埃尔米特(两点三次埃尔米特公式)
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R(x) \leq \frac{M}{384}h^4
R(x)≤384Mh4
三次样条:
条件 (4n-2)、 边界条件 (2)
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S_i(x_i) = y_i,\space S_{i}(x_{i+1})=y_{i+1},\space S_i'(x_i)=S_{i+1}'(x),\space S_i''(x_i)=S_{i+1}''(x) \\ S_0'(a) = S_{n-1}'(b) = 0
Si(xi)=yi, Si(xi+1)=yi+1, Si′(xi)=Si+1′(x), Si′′(xi)=Si+1′′(x)S0′(a)=Sn−1′(b)=0
公式 (其中
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mj=S′(xj)是未知量，需要靠二阶导相等规则来求取)
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S(x)=j=0∑nαjyj+βjmj
误差
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R(x) \leq |\frac{5M}{384}h^4|
R(x)≤∣3845Mh4∣
逼近
范数
无穷范数 : ∣ f ( x ) − P ( x ) ∣ ∞ = m a x a ≤ x ≤ b ∣ f ( x ) − P ( x ) ∣ |f(x)-P(x)|_\infty = max_{a\leq x\leq b} |f(x)-P(x)| ∣f(x)−P(x)∣∞=maxa≤x≤b∣f(x)−P(x)∣
二范数 : ∣ f ( x ) − P ( x ) ∣ 2 = ∫ a b ( f ( x ) − P ( x ) ) 2 |f(x) - P(x)|_2 = \sqrt{\int_a^b(f(x)-P(x))^2} ∣f(x)−P(x)∣2=∫ab(f(x)−P(x))2
定理
维尔斯特拉
设 f ( x ) ∈ C [ a , b ] , ∀ ϵ > 0 , ∃ P n ( x ) f(x) \in C[a,b], \forall\epsilon>0, \exist P_n(x) f(x)∈C[a,b],∀ϵ>0,∃Pn(x) 使得 ∣ f ( x ) − P ( x ) ∣ < ϵ |f(x)-P(x)|<\epsilon ∣f(x)−P(x)∣<ϵ在区间上一致成立 ( ϵ \epsilon ϵ越小、 n n n越大)
最佳一致逼近
∀ f ( x ) ∈ C [ a , b ] , \forall f(x)\in C[a,b], ∀f(x)∈C[a,b], 存在唯一最佳一致逼近多项式 P ∗ ( x ) , s . t Δ ( f , P ∗ ) = min P max x ∣ f ( x ) − P ( x ) ∣ P^*(x),\space s.t\space\Delta(f,P^*) = \min_P\max_x |f(x) - P(x)| P∗(x), s.t Δ(f,P∗)=minPmaxx∣f(x)−P(x)∣
P n ∗ ( x ) P^*_n(x) Pn∗(x)是 f ( x ) ∈ C [ a , b ] f(x)\in C[a,b] f(x)∈C[a,b]的最佳一致逼近多项式 ⇒ \Rightarrow ⇒ P n ∗ ( x ) P^*_n(x) Pn∗(x)是 f ( x ) f(x) f(x)的插值多项式
切比雪夫
P n ∗ ( x ) P_n^*(x) Pn∗(x)是 f ( x ) ∈ C [ a , b ] f(x)\in C[a,b] f(x)∈C[a,b]的最佳一致逼近多项式 ⇔ \Leftrightarrow ⇔ P n ∗ ( x ) P_n^*(x) Pn∗(x)与 f ( x ) f(x) f(x)在 [ a , b ] [a,b] [a,b]中至少有 n + 2 n+2 n+2个正负偏差点
最佳一致逼近求解
例题: f ( x ) = ∣ x ∣ , x ∈ [ − 1 , 2 ] f(x) = |x|, x\in[-1,2] f(x)=∣x∣,x∈[−1,2], 求最佳一致逼近多项式 P ( x ) = a x + b P(x) = ax+ b P(x)=ax+b,
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由 [ P ( x 1 ) − f ( x 1 ) ] ′ = 0 [P(x_1)-f(x_1)]'=0 [P(x1)−f(x1)]′=0求得 x 1 x_1 x1、根据边界条件求得 x 0 , x 2 x_0, x_2 x0,x2
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{ f ( x 0 ) − P ( x 0 ) = f ( x 2 ) − P ( x 2 ) f ( x 0 ) − P ( x 0 ) = − [ f ( x 1 ) − P ( x 1 ) ] \begin{cases} f(x_0)-P(x_0) = f(x_2)-P(x_2) \\ f(x_0)-P(x_0) = -[f(x_1) - P(x_1)] \end{cases} {f(x0)−P(x0)=f(x2)−P(x2)f(x0)−P(x0)=−[f(x1)−P(x1)]，求得 a 0 , a 1 . . . a_0, a_1... a0,a1...
李米兹迭代逼近: (最佳一致逼近计算量大，求取近似最佳逼近)
- 令 P ( x ) = a 0 + a 1 x + a 2 x 2 . . . P(x) = a_0 + a_1 x + a_2 x^2... P(x)=a0+a1x+a2x2...，随机初始化 x 0 , . . . , x n + 1 x_0, ..., x_{n+1} x0,...,xn+1
- 根据 Δ x k = − Δ x k + 1 \Delta_{x_k} = -\Delta_{x_{k+1}} Δxk=−Δxk+1 算出 a 0 , . . a n a_0,..a_n a0,..an
- 根据边界条件以及$[P(x_1)-f(x_1)]’=0 $ 求出 x 1 , x 2 . . . x_1, x_2... x1,x2...
- 迭代到收敛
切比雪夫多项式
定义:
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T_n(x) = \cos(n\cdot \arccos(x)), x\in[-1,1]
Tn(x)=cos(n⋅arccos(x)),x∈[−1,1]
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T_0(x) = 1 \\ T_1(x) = x \\ T_2(x) = 2x^2 - 1 \\ T_3(x) = 4x^3 - 3x \\ T_4(x) = 8x^4 - 8x^2 + 1 \\ T_5(x) = 16x^5 - 20x^3 + 5x \\
T0(x)=1T1(x)=xT2(x)=2x2−1T3(x)=4x3−3xT4(x)=8x4−8x2+1T5(x)=16x5−20x3+5x
性质 :
- T n + 1 ( x ) = 2 x T n ( x ) − T n − 1 ( x ) T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x) Tn+1(x)=2xTn(x)−Tn−1(x)
- T n ( x ) T_n(x) Tn(x)首相为 2 n − 1 2^{n-1} 2n−1、 n n n为奇数时为奇函数、 n n n为偶数时为偶函数
- ∣ T n ( x ) ∣ ≤ 1 |T_n(x)|\leq 1 ∣Tn(x)∣≤1，在 x k = cos ( k π n ) x_k = \cos(\frac{k\pi}{n}) xk=cos(nkπ)处轮流取 ± 1 \pm 1 ±1
- T n ( x ) T_n(x) Tn(x)在 [ − 1 , 1 ] [-1,1] [−1,1]上有 n n n个零点, n + 1 n+1 n+1个偏差点
- T n ( x ) T_n(x) Tn(x)在 [ − 1 , 1 ] [-1,1] [−1,1]上带权 1 1 − x 2 \frac{1}{\sqrt{1-x^2}} 1−x21正交
- 函数正交: f , g f,g f,g在 [ a , b ] [a,b] [a,b]区间上带权 w w w正交 ⇒ \Rightarrow ⇒ $\int f(x)\cdot g(x)\cdot w(x) dx = 0 $
例题 :
f n ( x ) − P n − 1 ( x ) = k T n ( x ) f_n(x) - P_{n-1}(x) = kT_n(x) fn(x)−Pn−1(x)=kTn(x)、 f f f比 P P P 高一阶、可以列出 n + 1 n+1 n+1个方程，求解 k , a 0 , . . . , a n − 1 k,a_0,...,a_{n-1} k,a0,...,an−1
最小零偏差 w n ( x ) = 1 2 n − 1 T ( x ) w_n(x)= \frac{1}{2^{n-1}}T(x) wn(x)=2n−11T(x)与零偏差最小且 Δ ( w n , 0 ) = 1 2 n − 1 \Delta(w_n, 0) = \frac{1}{2^{n-1}} Δ(wn,0)=2n−11
拉格朗日插值: 取插值节点
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R(x) \leq |\frac{M}{n!}max_xw_n(x)| = |\frac{M}{n!}\frac{1}{2^{n-1}}|
R(x)≤∣n!Mmaxxwn(x)∣=∣n!M2n−11∣
区间变换:
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t = \cos(\frac{(2k+1)\pi}{2n}), t\in[-1,1] \\ x = \frac{a+b}{2} + \frac{a-b}{2}t, x\in[a,b] \\ R(x) \leq |\frac{M}{n!}\frac{1}{2^{n-1}}(\frac{b-a}{2})^n| = |\frac{M}{n!}\frac{(b-a)^n}{2^{2n-1}}|
t=cos(2n(2k+1)π),t∈[−1,1]x=2a+b+2a−bt,x∈[a,b]R(x)≤∣n!M2n−11(2b−a)n∣=∣n!M22n−1(b−a)n∣
勒让德多项式
定义:
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P_n(x) = \frac{n!}{(2n)!}\frac{d^n}{dx^n}(x^2-1)^n
Pn(x)=(2n)!n!dxndn(x2−1)n
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P_0(x) = 1,\space P_1(x) = x \\ P_2(x) = (3x^2-1)/2 \\ P_3(x) = (5x^3-3x)/2 \\ P_4(x) = (35x^4-30x^2+3)/8 \\ P_5(x) = (63x^5-70x^3+15x)/8 \\
P0(x)=1, P1(x)=xP2(x)=(3x2−1)/2P3(x)=(5x3−3x)/2P4(x)=(35x4−30x2+3)/8P5(x)=(63x5−70x3+15x)/8
性质
- 正交性: ∫ − 1 1 1 ⋅ P i ( x ) ⋅ P j ( x ) d x = 0 , i ≠ j \int_{-1}^1 1\cdot P_i(x)\cdot P_j(x) dx = 0, i\neq j ∫−111⋅Pi(x)⋅Pj(x)dx=0,i=j
- 迭代: ( n + 1 ) P n + 1 = ( 2 n + 1 ) x P n ( x ) − n P n − 1 ( x ) (n+1)P_{n+1} = (2n+1)xP_n(x) - nP_{n-1}(x) (n+1)Pn+1=(2n+1)xPn(x)−nPn−1(x)
- 奇偶性: 当n是奇数 P n ( x ) P_n(x) Pn(x)是奇函数，当n是偶数 P n ( x ) P_n(x) Pn(x)是偶函数
- n个零点
- 在所有首次项为1的多项式中， P n ( x ) P_n(x) Pn(x)与零的平方误差最小
最佳平方逼近
内积
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(f,g) = \int_a^b\rho(x)f(x)g(x) \\ (f,g) = (g,f) \\ c(f,g) = (cf, g) \\ (f_1+f_2,g) = (f_1,g) + (f_2, g) \\ (f,f) \geq 0 \\ (f,f) = 0 \Leftrightarrow f = 0
(f,g)=∫abρ(x)f(x)g(x)(f,g)=(g,f)c(f,g)=(cf,g)(f1+f2,g)=(f1,g)+(f2,g)(f,f)≥0(f,f)=0⇔f=0
求解
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f(x)\in[a,b], P(x) = a_1x+a_0, \rho(x)=1 \\ \begin{cases} (\phi_0,\phi_0)a_0 + (\phi_0,\phi_1)a_1 = (\phi_0, f) \\ (\phi_1,\phi_0)a_0 + (\phi_1,\phi_1)a_1 = (\phi_1, f) \\ \end{cases} \\ (\phi_0,\phi_1) = \int_a^b xdx \\ (\phi_1,f) = \int_a^b xf(x)dx \\ R(x) = (f,f) - \sum_j a_j(\phi_j, f)
f(x)∈[a,b],P(x)=a1x+a0,ρ(x)=1{(ϕ0,ϕ0)a0+(ϕ0,ϕ1)a1=(ϕ0,f)(ϕ1,ϕ0)a0+(ϕ1,ϕ1)a1=(ϕ1,f)(ϕ0,ϕ1)=∫abxdx(ϕ1,f)=∫abxf(x)dxR(x)=(f,f)−j∑aj(ϕj,f)
正交多项式:
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ϕ选择正交多项式
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(\phi_i, \phi_j) = 0, i\neq j \\ a_i = \frac{(f,\phi_i)}{(\phi_i, \phi_i)} \\ R(x) = (f,f) - \sum_j \frac{(f,\phi_j)^2}{(\phi_j,\phi_j)}
(ϕi,ϕj)=0,i=jai=(ϕi,ϕi)(f,ϕi)R(x)=(f,f)−j∑(ϕj,ϕj)(f,ϕj)2
切比雪夫
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f(x)\in[-1, 1], \phi=span\{T_0,T_1,...,T_n\} \rho(x) = \frac{1}{\sqrt{1-x^2}} \\ S^*(x)=\sum_j a_jT_j = \frac{1}{\pi}(f,T_0) + \sum_{j=1}^n \frac{2}{\pi}(f,T_j) \\ a_0 = \frac{1}{\pi}(f,T_0),\space a_k = \frac{2}{\pi}(f,T_k) \\ R(x) = f -S^* = \frac{2}{\pi}(f,T_{n+1})T_{n+1} \\
f(x)∈[−1,1],ϕ=span{T0,T1,...,Tn}ρ(x)=1−x21S∗(x)=j∑ajTj=π1(f,T0)+j=1∑nπ2(f,Tj)a0=π1(f,T0), ak=π2(f,Tk)R(x)=f−S∗=π2(f,Tn+1)Tn+1
勒让德
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f(x)\in[-1, 1], \phi=span\{P_0,P_1,...,P_n\}, \rho(x) = 1 \\ S^*(x)=\sum_j a_jP_j = \sum_{j=0}^n \frac{2k+1}{2}(f,P_j) \\ R(x) = (f,f) - \sum_j \frac{2j+1}{2}(f,P_j)^2 \\
f(x)∈[−1,1],ϕ=span{P0,P1,...,Pn},ρ(x)=1S∗(x)=j∑ajPj=j=0∑n22k+1(f,Pj)R(x)=(f,f)−j∑22j+1(f,Pj)2
数值积分
基本概念
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近似方法
∫ a b f ( x ) d x = ∫ a b L n ( x ) d x = ∫ a b ∑ k l k ( x ) f ( x k ) d x = ∑ k A k f ( x k ) A k = ∫ a b l k ( x ) d x R ( x ) = ∫ a b f ( n ) ( η ) n ! w ( x ) \int_a^b f(x)dx = \int_a^b L_n(x)dx = \int_a^b \sum_k l_k(x)f(x_k)dx = \sum_k A_kf(x_k) \\ A_k = \int_a^b l_k(x) dx \\ R(x) = \int_a^b \frac{f^{(n)}(\eta)}{n!}w(x) ∫abf(x)dx=∫abLn(x)dx=∫abk∑lk(x)f(xk)dx=k∑Akf(xk)Ak=∫ablk(x)dxR(x)=∫abn!f(n)(η)w(x) -
代数精度
∫ a b f ( x ) d x = ∑ k n A k f ( x k ) \int_a^b f(x)dx = \sum_k^n A_kf(x_k) ∫abf(x)dx=∑knAkf(xk)对所有 ≤ n \leq n ≤n次多项式都成立 ⇒ \Rightarrow ⇒ 公式具有n次代数精度
公式具有n次代数精度 ⇒ \Rightarrow ⇒ A k = ∫ a b l k ( x ) d x A_k = \int_a^b l_k(x) dx Ak=∫ablk(x)dx
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稳定性
当f存在误差时，近似积分得到的值误差足够小
牛顿-科特斯
公式 (其中n+1是插值节点个数、k是idx)
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I_n = (b-a)\sum_{k=0}^n C_k^{(n)}f(x_k) \\ C_k^{(n)} = \frac{(-1)^{n-k}}{nk!(n-k)!}\int_0^n\Pi_{j\neq k}(t-j)dt \\ C_0^{(1)} = C_1^{(1)} = \frac{1}{2} \\ C_0^{(2)} = \frac{1}{6}, C_1^{(2)} = \frac{2}{3}, C_2^{(2)} = \frac{1}{6} \\ C_0^{(3)} = \frac{1}{8}, C_1^{(3)} = \frac{3}{8}, C_2^{(3)} = \frac{3}{8}, C_2^{(3)} = \frac{1}{8} \\
In=(b−a)k=0∑nCk(n)f(xk)Ck(n)=nk!(n−k)!(−1)n−k∫0nΠj=k(t−j)dtC0(1)=C1(1)=21C0(2)=61,C1(2)=32,C2(2)=61C0(3)=81,C1(3)=83,C2(3)=83,C2(3)=81
代数精度: 当
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方法误差:
矩形: ( b − a ) 3 24 f ′ ′ ( η ) \frac{(b-a)^3}{24}f''(\eta) 24(b−a)3f′′(η)
梯形: − ( b − a ) 3 12 f ′ ′ ( η ) -\frac{(b-a)^3}{12}f''(\eta) −12(b−a)3f′′(η)
辛普森: − ( b − a ) 5 2880 f ( 4 ) ( η ) -\frac{(b-a)^5}{2880}f^{(4)}(\eta) −2880(b−a)5f(4)(η)
复化积分
复化梯形公式( n + 1 n+1 n+1个点): I = h 2 [ f ( a ) + 2 ∑ k = 1 n − 1 f ( x k ) + f ( b ) ] I = \frac{h}{2}[f(a) + 2\sum_{k=1}^{n-1} f(x_k) + f(b)] I=2h[f(a)+2∑k=1n−1f(xk)+f(b)]
复化梯形误差: R ( f ) = − n h 3 12 f ′ ′ ( η ) = − ( b − a ) h 2 12 f ′ ′ ( η ) R(f) = \frac{-nh^3}{12}f''(\eta) = \frac{-(b-a)h^2}{12}f''(\eta) R(f)=12−nh3f′′(η)=12−(b−a)h2f′′(η)
复化辛普森公式( 2 n + 1 2n+1 2n+1个点): I = h 6 [ f ( a ) + 2 ∑ k = 1 n − 1 f ( x k ) + 4 ∑ k = 0 n − 1 f ( x k + h 2 ) + f ( b ) ] I = \frac{h}{6}[f(a) + 2\sum_{k=1}^{n-1} f(x_k) + 4\sum_{k=0}^{n-1}f(x_k+\frac{h}{2}) + f(b)] I=6h[f(a)+2∑k=1n−1f(xk)+4∑k=0n−1f(xk+2h)+f(b)]
复化辛普森误差: R ( f ) = − n h 5 2880 f ′ ′ ( η ) = − ( b − a ) h 4 2880 f ′ ′ ( η ) R(f) = \frac{-nh^5}{2880}f''(\eta) = \frac{-(b-a)h^4}{2880}f''(\eta) R(f)=2880−nh5f′′(η)=2880−(b−a)h4f′′(η)
龙贝格&外推加速
递推梯形公式
$T_{2n} = \frac{1}{2} T_n + \frac{h}{2}\sum_{k=0}^{n-1}f(x_{k+\frac{h}{2}}) $
龙贝格 (从梯形公式
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\frac{I-T_{2n}}{I-T_n} = \frac{-\frac{(b-a)}{12}(\frac{h}{2})^2f''(\eta)}{-\frac{(b-a)}{12}h^2f''(\eta)} = \frac{1}{4} \\ I = \frac{1}{3}(4T_{2n}-T_n) \\
I−TnI−T2n=−12(b−a)h2f′′(η)−12(b−a)(2h)2f′′(η)=41I=31(4T2n−Tn)
**理查孙外推 **(每加速一次收敛速度提高两阶)
T m ( k ) = 4 m 4 m − 1 T m − 1 ( k + 1 ) + 1 4 m − 1 T m − 1 ( k ) T_m^{(k)} = \frac{4^m}{4^m-1} T_{m-1}^{(k+1)} + \frac{1}{4^m-1} T_{m-1}^{(k)} Tm(k)=4m−14mTm−1(k+1)+4m−11Tm−1(k)
高斯求积
- 理论
高斯求积:
性质: 公式 ∫ a b f ( x ) d x = ∑ k = 0 n A k f ( x k ) \int_a^b f(x)dx = \sum_{k=0}^nA_k f(x_k) ∫abf(x)dx=∑k=0nAkf(xk)有 A k , x k A_k,x_k Ak,xk等 2 n + 2 2n+2 2n+2个未知数，可以达到 2 n + 1 2n+1 2n+1阶精度
方法: 分别令 f ( x ) = 1 , x , . . . , x 2 n + 1 f(x)=1,x,...,x^{2n+1} f(x)=1,x,...,x2n+1，可得 2 n + 1 2n+1 2n+1个方程，求解得到 A k , x k A_k,x_k Ak,xk取值
高斯点
定义: 能使公式具有 2 n + 1 2n+1 2n+1次精度的 x 0 , . . , x k x_0,..,x_k x0,..,xk
性质: w ( x ) = ( x − x 0 ) ⋅ ⋅ ⋅ ( x − x n ) w(x)=(x-x_0)\cdot\cdot\cdot(x-x_n) w(x)=(x−x0)⋅⋅⋅(x−xn)与 p n ( x ) = 1 , x , . . . , x n p_n(x) = 1,x,...,x^n pn(x)=1,x,...,xn带权 ρ ( x ) \rho(x) ρ(x)正交 ⇒ \Rightarrow ⇒ x 0 , . . , x n x_0,..,x_n x0,..,xn是高斯点
A k ( x ) ≥ 0 A_k(x) \geq 0 Ak(x)≥0
- 高斯-勒让德:
方法: 用勒让德多项式的根作为 x 0 , . . . , x n x_0,...,x_n x0,...,xn
例题: ∫ − 1 1 f ( x ) d x \int_{-1}^1 f(x)dx ∫−11f(x)dx的两点高斯-勒让德，由 P 2 ( x ) = 1 2 ( 3 x 2 − 1 ) P_2(x) = \frac{1}{2} (3x^2-1) P2(x)=21(3x2−1)可以得到$x_0=\sqrt{\frac{1}{3}}, x_1=-\sqrt{\frac{1}{3}} $
区间变换: ∫ a b f ( x ) d x \int_a^b f(x)dx ∫abf(x)dx -> b − a 2 ∫ − 1 1 f ( b − a 2 t + b + a 2 ) d t \frac{b-a}{2}\int_{-1}^1 f(\frac{b-a}{2}t + \frac{b+a}{2}) dt 2b−a∫−11f(2b−at+2b+a)dt
误差: R [ f ] = f ( 2 n + 2 ) ( η ) ( 2 n + 2 ) ! ∫ a b w 2 ( x ) d x = 2 2 n + 3 [ ( n + 1 ) ! ] 4 ( 2 n + 3 ) [ ( 2 n + 2 ) ! ] 3 f ( 2 n + 2 ) ( η ) R[f] = \frac{f^{(2n+2)}(\eta)}{(2n+2)!}\int_a^b w^2(x)dx = \frac{2^{2n+3}[(n+1)!]^4}{(2n+3)[(2n+2)!]^3}f^{(2n+2)}(\eta) R[f]=(2n+2)!f(2n+2)(η)∫abw2(x)dx=(2n+3)[(2n+2)!]322n+3[(n+1)!]4f(2n+2)(η)
- 高斯-切比雪夫
方法: 用切比雪夫多项式的根作为 x 0 , . . . , x n x_0,...,x_n x0,...,xn
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