题目要求











题目大意是让你用c系语言实现辛普森积分法对定积分的粗略估计，所谓辛普森积分法即为：

定义：辛普森法则（Simpson's rule）是一种数值积分方法，是牛顿-莱布尼茨公式的特殊形式，以二次曲线逼近的方式取代矩形或梯形积分公式，以求得定积分的数值近似解。其近似值如下：

 

                                          

 

那很明显可以看出，改进积分结果有两种方法，一是二分区间之后再次二分不断逼近，二是从积分间隔入手，不断缩小积分间隔

给出Matlab-C++代码

//Author：glm
#include<cstdio>
#include<cmath>
#include<iostream>
#include "mex.h"
#define ll long long int
#define rg register ll

inline double f(double x)
{
    if(x==0)return 1;
    return sin(x)/x;
}
inline double calculate(double a,double b)//int(f,a,b)=(b-a)/6*(f(a)+4*f((a+b)/2+f(b))
{
    double sum=0;
    for(rg i=0;i<(ll)((b-a)/0.025);i++)
    {
        double a1=a+0.0250*i,b1=a1+0.0250,mid=(a1+b1)/2.0;
        sum+=(mid-a1)/6*(f(a1)+4*f((a1+mid)/2)+f(mid))+(b1-mid)/6*(f(mid)+4*f((b1+mid)/2)+f(b1));
        
    }
    return sum;
   /* int parts=(int)((b-a)/0.01);
    double ans=0;
    for (int i=0;i<parts;i++)
        {
            double ai=a+i*0.01;
            double bi=ai+0.01;
            double mi=(ai+bi)/2;
            ans+=(mi-ai)/6*(f(ai)+4*f((ai+mi)/2)+f(mi))+(bi-mi)/6*(f(mi)+4*f((bi+mi)/2)+f(bi));
            //ans+=(bi-ai)/6*(f(ai)+4*f((ai+bi)/2)+f(bi));
        }
    return ans;*/
}
void mexFunction (int nlhs,mxArray *plhs[],int nrhs,const mxArray * prhs[])
{
    double *a;
    double b,c;
    plhs[0]=mxCreateDoubleMatrix(1,1,mxREAL);
    a=mxGetPr(plhs[0]);//
    b=*(mxGetPr(prhs[0]));
    c=*(mxGetPr(prhs[1]));
    *a=calculate(b,c);
}

报告的latex代码~

\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{mathtools}
\usepackage{graphics,epsfig}
\usepackage{xspace}
\usepackage{color}
\usepackage{amssymb}
\usepackage{subfigure}
\usepackage{multirow}
\usepackage{listings}
\usepackage{graphicx}
\usepackage{url}
%%uncomment following line if you are going to use ``Chinese''
%\usepackage{ctex} 

\definecolor{ballblue}{rgb}{0.13, 0.67, 0.8}
\definecolor{cornflowerblue}{rgb}{0.39,0.58,0.93}
\definecolor{babyblueeyes}{rgb}{0.63, 0.79, 0.95}

% preset-listing options
\lstset{
  backgroundcolor=\color{white},   
  basicstyle=\footnotesize,    
  language=matlab,
  breakatwhitespace=false,         
  breaklines=true,                 % sets automatic line breaking
  captionpos=b,                    % sets the caption-position to bottom
  commentstyle=\color{ballblue},    % comment style
  extendedchars=true,              
  frame=single,                    % adds a frame around the code
  keepspaces=true,                 
  keywordstyle=\color{blue},       % keyword style
  numbers=left,                    
  numbersep=5pt,                   
  numberstyle=\tiny\color{blue}, 
  rulecolor=\color{babyblueeyes},
  stepnumber=1,              
  stringstyle=\color{black},     % string literal style
  tabsize=4,                       % sets default tabsize to 4 spaces
  title=\lstname                   
}

\usepackage{geometry}
\geometry{
 a4paper,
 total={210mm,297mm},
 left=20mm,
 right=20mm,
 top=20mm,
 bottom=20mm,
 }

\marginparwidth = 10pt


\begin{document}
\title{Weekly Assignment:Simpson’s rule of calculation}
\author{ \\ Stdudent  Guo LiMin \\ Id: 22920182204174}
\maketitle


\section{Problem Description}

Given function  $f(x)=\cfrac{sinx}{x}$ , the integral range [a, b], please work out its numerical integral in range [a, b]. 
You are required to calculate the numerical integral by Simpson’s rule.
\\\\{\color{red}Requirements:}\\\\    1. Your function myQuad MUST be implemented in C and called by Matlab\\
2. The command interface in Matlab looks like: v = myQuad(a, b);\\
3. You are NOT allowed to use “recursive procedure” when you implement Simpson’s
rule\\
4. Given $F(x) =\int^{6}_{-2}\cfrac{sinx}{x}dx$, please plot out the curve of F(x) in the range [-2 6]\\
5. According to Simpson’s rule, the number of intervals will impact on the precision.
You are required to try different numbers of intervals and see how it impacts your
result. Please present an analysis about this in your report.

\section{Code of Solution}
The followings are codes implemented by C++,no fancy algorithm included,working out this problem only by using simple simulations\\
%
%if one wants to paste part of the code into the report
%one can put in following manner
\begin{lstlisting}[language=c++, linewidth=\linewidth,caption={main working function(myQuad.cpp)}]
#include<cstdio>
#include<cmath>
#include<iostream>
#include "mex.h"
#define ll long long int
#define rg register ll
inline double f(double x)
{
    return sin(x)/x;
}
inline double calculate(double a,double b)//int(f,a,b)=(b-a)/6*(f(a)+4*f((a+b)/2+f(b))
{
    double sum=0;
    for(rg i=0;i<(ll)((b-a)/0.001);i++)
    {
        double a1=a+0.001*i,b1=a1+0.001,mid=(a1+b1)/2.0;
        sum+=(mid-a1)/6*(f(a1)+4*f((a1+mid)/2)+f(mid))+(b1-mid)/6*(f(mid)+4*f((b1+mid)/2)+f(b1));
        
    }
    return sum;
   /* int parts=(int)((b-a)/0.01);
    double ans=0;
    for (int i=0;i<parts;i++)
        {
            double ai=a+i*0.01;
            double bi=ai+0.01;
            double mi=(ai+bi)/2;
            ans+=(mi-ai)/6*(f(ai)+4*f((ai+mi)/2)+f(mi))+(bi-mi)/6*(f(mi)+4*f((bi+mi)/2)+f(bi));
            //ans+=(bi-ai)/6*(f(ai)+4*f((ai+bi)/2)+f(bi));
        }
    return ans;*/
}
void mexFunction (int nlhs,mxArray *plhs[],int nrhs,const mxArray * prhs[])
{
    double *a;
    double b,c;
    plhs[0]=mxCreateDoubleMatrix(1,1,mxREAL);
    a=mxGetPr(plhs[0]);//
    b=*(mxGetPr(prhs[0]));
    c=*(mxGetPr(prhs[1]));
    *a=calculate(b,c);
}

\end{lstlisting}

\begin{lstlisting}[language=c++, linewidth=\linewidth,caption={Draw the contrast curve and main interface code(test.m)}]
mex myQuad.cpp
syms t
a=-2:0.27:6;
b=-2:0.27:6;
c=-2:0.27:6;
cnt=0;
for x=-2:0.27:6
y=int(sin(t)/t,-2,x);
cnt=cnt+1;
b(cnt)=y;
c(cnt)=myQuad(-2,x);
b(cnt),c(cnt);
end
plot(a,b,'r',a,c,'b')
    
\end{lstlisting}
\section{Experiment Theory and Results}
Given function f(x), and its range of x [a,b], it is possible to estimate its integral in range [a,b]
by quandratic interpolation based on “Simpson’s rule”. The “Simpson’s rule” is given as
follows.\\
\begin{equation}
\begin{split}$\int^{b}_{a}\cfrac{sinx}{x}dx &=\cfrac{(b-a)}{6}*[f(a)+4*f(\cfrac{a+b}{2})+f(b)]\\
            &=\cfrac{(m-6)}{6}*[f(a)+4*f(\cfrac{a+m}{2})+f(m)]+\cfrac{(b-m)}{6}*[f(m)+4*f(\cfrac{b+m}{2})+f(b)]\nonumber$   
\end{split}
\end{equation}
\\
What’s more,The more number of segments used in the range [a, b], the better is the approximation.
\\\\
To start with,we should learn how to compile c/c++ files with matlab and how to write fantastic and unified-formatted papers,
this aspect of learning has been updated to my personal blog  {\color{red}\url{newcoder-glm.blog.csdn.net}},so for more information,please go up to visit my blog.
\\\\
After mastering the mixtrue of C/C++& matlab programming technique(It's just an interface that calls an external compiler in essence),as easy as pie we can write the codes above and 
should curve figures(check out the attachment for more details) which shows graphs of the result and the exact result obtained by Simpson integral method when the interval of integral interval is divided differently(when the value is 0.05 0.25)

\begin{table}[h]
\caption{Circumstances when we apply different intervals}
\begin{center}
	\begin{tabular}{|c|c|c|c|}
	\hline
	Interval & 0.01 & 0.10 & 0.25\\ \hline
	Results & \textbf{2.551496047169967}& \textbf{2.551496048068467} & \textbf{2.551496047173387  } \\ \hline
	\end{tabular}
\end{center}
\end{table}


\subsection{further ideas}
We can see from the Simpson integral (the simple three-point formula) that there is some error, and it's easy to see that the smaller the interval, the more accurate the integral,
however,if we shorten the interval, of course the results' accuracy come up, but at the cost of higher time complexity.\\
So here we introduce "Adaptive Simpson’s rule" which is the integral can automatically control the size and length of the cutting interval, so that the accuracy can meet the requirements.
In particular, if the midpoint of [a,b] is c, the result will be directly returned if and only then {\color{red}$[S(a,c)+S(c,b)-S(a,b)]<15Eps$}  Otherwise, the recursive call will divide the interval again. The accuracy of Eps should be reduced by half when recursion is called.
\begin{figure}[h]
	\centering
	{\includegraphics[width=0.7\linewidth]{025.pdf}}
    \hspace{0.15in}
\end{figure}

\begin{figure}[h]
	\centering
	{\includegraphics[width=0.7\linewidth]{005.pdf}}
    \hspace{0.15in}
\end{figure}

\section{What I learned from the Project}
1.learning how to programing cpp files with powerful matlab,excuted with high efficiency and conciseness}
\\\\2.the ability of exploring new territory that never have been leart and mastering it(hand in report in time)}
\\\\3.learning how to write fantastic and unified-formatted papers with modernized Latex}

\section{Two attached pictures of point three}
\end{document}



报告的pdf~

 

 



 

 

 

 

数值积分
 复合梯形法
辛普森1/3法
CONTENTS
CONTENTS
1. 复合梯形法
1.1 代码
1.2 计算结果
1.3 图像结果
2. 辛普森1/3法
2.1 代码
2.2 计算结果
2.3 图像结果
3. 源码编辑器
1. 复合梯形法
1.1 代码
# -*- coding: utf-8 -*-
1.2 计算结果
运行代码即可得到结果，
该区间内的数值积分为
1.3 图像结果
2. 辛普森1/3法
2.1 代码
# -*- coding: utf-8 -*-
2.2 计算结果
该区间内的数值积分为
2.3 图像结果
3. 源码编辑器
Anaconda（强烈推荐） + Spyder 4（和MATLAB有着相似界面的编辑器）

复合辛普森积分

已知函数表达式与积分区间

精度esp正相关与1/num
%复合辛普森积分
%已知函数表达式与积分区间
clc;clear;
a=0;b=1;%积分范围
num=1000;%积分准确度
h=(b-a)/(2*num);
f=@(x)exp(-x);%积分表达式
I=0;%积分结果
I=f(a)-f(b);
for i=1:num
    I=I+(2*f(a+2*i*h)+4*f(a+(2*i-1)*h));
end
I=I*h/3

	

什么是龙贝格积分算法
 龙贝格(Romberg)积分算法也被称为逐次分半加速算法，通过把积分区间逐次分半的方法进行数值积分求解。由于其采用的是逐次分半计算，后一次计算是对前一次近似结果的修正，因此相对于辛普森和科特斯求积方法精度更高；并且其前一次分割计算的函数值在分半之后还可以被继续使用，因此提升了计算的效率，是一种精度较高的加速计算积分的方法。
龙贝格积分算法计算过程
龙贝格积分算法是一种将理查森外推加速法应用于复合梯形公式而形成的一种积分方法，递推公式可表示为
(1)取，求 (2)利用变步长梯形公式求，其中为区间的分半次数，即(3)按照次序依据递推公式逐个求出加速值  (4)比较递推矩阵相邻对角元素，两者之差绝对值小于预设精度要求时终止计算。  
龙贝格积分算法程序设计
龙贝格算法程序设计如下：
function [y,T]=romberg(f,a,b,e)k=0;h=b-a;T(1,1)=h/2*(f(a)+f(b));%第一步d=h;%初始化相邻两次误差while e<=d    k=k+1;    QH=0;    for j=0:2^(k-1)-1        QH=QH+f(a+(2*j+1)*(b-a)/2^k);    end    T(k+1,1)=T(k,1)/2+(b-a)/2^k*QH;%第二步    for j=1:k        T(k+1,j+1)=T(k+1,j)+(T(k+1,j)-T(k,j))/(4^j-1);%第三步    end   d=abs(T(k+1,k+1)-T(k,k));%第四步endy=T(k+1,k+1);
龙贝格积分算法算例分析
应用龙贝格算法计算积分
使其精度达到10-5，其精确值为0.25。
输入程序代码
f=inline('x^3')[y,T]=romberg(f,0,1,1e-5)
运行结果为
y =   0.250000000000000T =   0.500000000000000                  0                   0         0.312500000000000  0.250000000000000                   0         0.26562500000000   0.250000000000000   0.250000000000000
注：公式和代码左右滑动查看完整效果。