Simpson‘s rule Error analysis (辛普森积分方法的误差分析)
Introduction
Simpson’s rule is an integral approximate method. Instead of using the
original function f(x)f(x)f(x) to compute the integration, it uses a
polynomial function. If we have three points a,m,ba,m,ba,m,b and their values
f(a),f(m),f(b)f(a), f(m), f(b)f(a),f(m),f(b) where m=(a+b)/2m=(a+b)/2m=(a+b)/2, then the approximation is
f(x)≈P(x)=f(a)(x−m)(x−b)(a−m)(a−b)+f(m)(x−a)(x−b)(m−a)(m−b)+f(b)(x−a)(x−m)(b−a)(b−m)f(x)\approx P(x)=f(a) \frac{(x-m)(x-b)}{(a-m)(a-b)}+f(m) \frac{(x-a)(x-b)}{(m-a)(m-b)}+f(b) \frac{(x-a)(x-m)}{(b-a)(b-m)}f(x)≈P(x)=f(a)(a−m)(a−b)(x−m)(x−b)+f(m)(m−a)(m−b)(x−a)(x−b)+f(b)(b−a)(b−m)(x−a)(x−m)
and
∫abf(x)dx≈∫abP(x)dx\int_a^bf(x)dx\approx \int_{a}^{b}{P(x)dx}∫abf(x)dx≈∫abP(x)dx
[外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-KK6W0djP-1644459971010)(Simpsons.png)]{width=“0.4\linewidth”}
Next, we will analysis the error of this kind of simpson’s rule.
Error Analysis
E(f)=∫ab[f(x)−P(x)]dxE(f)=\int_{a}^{b}{[f(x)-P(x)]dx}E(f)=∫ab[f(x)−P(x)]dx
Note that the residual of lagrange interpolation is
R(x)=f(x)−P(x)=f(3)(ξ(x))3!(x−a)(x−m)(x−b).R(x)=f(x)-P(x)=\frac{f^{(3)}(\xi(x))}{3!}(x-a)(x-m)(x-b).R(x)=f(x)−P(x)=3!f(3)(ξ(x))(x−a)(x−m)(x−b).
Hence, we have
E(f)=∫ab(x−a)(x−m)(x−b)f(3)(ξ(x))3!dxE(f)=\int_{a}^{b}{(x-a)(x-m)(x-b) \frac{f^{(3)}(\xi(x))}{3!}dx}E(f)=∫ab(x−a)(x−m)(x−b)3!f(3)(ξ(x))dx
Let w(x)w(x)w(x) be
w(x)=∫ax(t−a)(t−m)(t−b)dt.w(x)=\int_{a}^{x}{(t-a)(t-m)(t-b)dt}.w(x)=∫ax(t−a)(t−m)(t−b)dt.
Then we have
-
w(a)=w(b)=0w(a)=w(b)=0w(a)=w(b)=0.
-
w(x)>0w(x)>0w(x)>0 for a<x<ba<x<ba<x<b.
Then we have
E(f)=∫abw′(x)f(3)(ξ(x))3!dx=[w(x)f(3)(ξ(x))3!]x=ax=b−∫abw(x)ddxf(3)(ξ(x))3!dx=−∫abw(x)ddxf(3)(ξ(x))3!dx.\left. \begin{aligned} E(f)&= \int_{a}^{b}{w^{'}(x) \frac{f^{(3)}(\xi(x))}{3!}dx}\\ &= \left[ w(x) \frac{f^{(3)}(\xi(x))}{3!} \right]_{x=a}^{x=b}-\int_{a}^{b}{w(x)\frac{d}{dx}\frac{f^{(3)}(\xi(x))}{3!}dx}\\ &= -\int_{a}^{b}{w(x)\frac{d}{dx}\frac{f^{(3)}(\xi(x))}{3!}dx}. \end{aligned} \right.E(f)=∫abw′(x)3!f(3)(ξ(x))dx=[w(x)3!f(3)(ξ(x))]x=ax=b−∫abw(x)dxd3!f(3)(ξ(x))dx=−∫abw(x)dxd3!f(3)(ξ(x))dx.
Using the divided difference, we have
f[a,m,b,x]=f(3)(ξ(x))3!.f[a,m,b,x]= \frac{f^{(3)}(\xi(x))}{3!}.f[a,m,b,x]=3!f(3)(ξ(x)).
Hence,
ddx(f(x)(ξ(x))3!)=limh→0f[a,m,b,x+h]−f[a,m,b,x]h=limh→0f[x,a,m,b,x+h]=f[x,a,m,b,x]=f[a,m,b,x,x]=f(4)(ξ(x))4!\left. \begin{aligned} \frac{d}{dx} \left( \frac{f^{(x)}(\xi(x))}{3!} \right) &= \lim_{h\rightarrow 0} \frac{f[a,m,b,x+h]-f[a,m,b,x]}{h}\\ &=\lim_{h\rightarrow 0} f[x,a,m,b,x+h]\\ &=f[x,a,m,b,x]\\ &=f[a,m,b,x,x]\\ &=\frac{f^{(4)}(\xi(x))}{4!} \end{aligned} \right.dxd(3!f(x)(ξ(x)))=h→0limhf[a,m,b,x+h]−f[a,m,b,x]=h→0limf[x,a,m,b,x+h]=f[x,a,m,b,x]=f[a,m,b,x,x]=4!f(4)(ξ(x))
Substituting the above formula into E(f)E(f)E(f), we have
E(f)=−∫abf(4)(ξ(x))4!∫ax(t−a)(t−m)(t−b)dtdx\left. \begin{aligned} E(f) &= - \int_{a}^{b}{\frac{f^{(4)}(\xi(x))}{4!} \int_{a}^{x}{(t-a)(t-m)(t-b)dt} dx} \end{aligned} \right.E(f)=−∫ab4!f(4)(ξ(x))∫ax(t−a)(t−m)(t−b)dtdx
Applying the weighted mean value theorem, we have
E(f)=−f(4)(η)4!∫ab∫ax(t−a)(t−m)(t−b)dtdx=−f(4)(η)4!1120(b−a)5=−f(4)(η)24[415h5].h=b−a2\left. \begin{aligned} E(f) &= - \frac{f^{(4)}(\eta)}{4!} \int_{a}^{b}{\int_{a}^{x}{(t-a)(t-m)(t-b)dt}dx}\\ &=-\frac{f^{(4)}(\eta)}{4!} \frac{1}{120}(b-a)^5\\ &=-\frac{f^{(4)}(\eta)}{24}\left[\frac{4}{15} h^{5}\right] . \quad h=\frac{b-a}{2} \end{aligned} \right.E(f)=−4!f(4)(η)∫ab∫ax(t−a)(t−m)(t−b)dtdx=−4!f(4)(η)1201(b−a)5=−24f(4)(η)[154h5].h=2b−a
where η∈[a,b]\eta\in [a,b]η∈[a,b].
Thus, we have
E[f]=−h590f(4)(η),η∈[a,b].E[f]=-\frac{h^5}{90}f^{(4)}(\eta), \eta\in [a,b].E[f]=−90h5f(4)(η),η∈[a,b].
有“AI”的1024 = 2048，欢迎大家加入2048 AI社区
更多推荐
8
0- 0
已为社区贡献4条内容
所有评论(0)