实验三 数值积分（android）

实验二博客地址：http://blog.csdn.net/double2hao/article/details/51217356

实验一博客地址：http://blog.csdn.net/double2hao/article/details/51152843

一、实验内容

分别写出变步长梯形法和romberge法计算定积分的算法，编写程序上机调试出结果，要求所编程序适用于任何类型的定积分，即能解决这一类问题，而不是某一个问题。

试验中以下列数据验证程序的正确性。

求  (sinx)/x的积分，积分区间为[0,1]

效果：（源码在文章底部）


  


主要工作：

1、添加了ThreeFragment的xml界面

2、掌握理解变步长梯形法和龙贝格法

主要逻辑代码：

ThreeFragment:

package com.example.double2.numericcalculationtest;

import android.app.Fragment;
import android.os.Bundle;
import android.view.LayoutInflater;
import android.view.View;
import android.view.ViewGroup;
import android.widget.ArrayAdapter;
import android.widget.Button;
import android.widget.Spinner;
import android.widget.TextView;

/**
* 项目名称：NumericCalculationTest
* 创建人：Double2号
* 创建时间：2016/4/13 21:41
* 修改备注：
*/
public class ThreeFragment extends Fragment {

private View views;
private Spinner mSpinner;
private final String[] spinnerChose = {"变步长梯形法", "龙贝格法"};
private Button btnSure;
private TextView tvResult;
private String stringResult;

@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
views = inflater.inflate(R.layout.fra_three, null);
initView();
return views;
}

private void initView() {
mSpinner = (Spinner) views.findViewById(R.id.sp_three_chose);
btnSure = (Button) views.findViewById(R.id.btn_three_sure);
tvResult = (TextView) views.findViewById(R.id.tv_three_result);

mSpinner.setAdapter(new ArrayAdapter<String>(getActivity(),
android.R.layout.simple_list_item_1, spinnerChose));
btnSure.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
caculating();
}
});
}

private void caculating() {
if (mSpinner.getSelectedItemPosition() == 0) {
TrapezoidCaculation();
} else {
RombergCaculation();
}
}

//变步长梯形法求值
private void TrapezoidCaculation() {
Double a = 0.0, b = 1.0, h;
int m = 1, n = 0;
Double[] T = new Double[100];//固定只能计算100次

T
= (b - a) * (f(a) + f(b)) / 2;
do {
h = (b - a) / m;
double s = 0.0;
for (int k = 0; k < m; k++) {
s += f(a + (k + 0.5) * h);
}
T[n + 1] = T
/ 2 + h / 2 * s;
m = 2 * m;
n++;
} while (Math.abs(T
- T[n - 1]) >= 0.0000001);

stringResult = "n= " + n +
"\nS= " + T

+ "\n误差为 " + Math.abs(T
- T[n - 1]);
tvResult.setText(stringResult);
}

//龙贝格法求值
private void RombergCaculation() {
Double a = 0.0, b = 1.0, h;
Double[][] T = new Double[100][100];//固定只能计算100次

int i = 0;
T[i][i] = (b - a) * (f(a) + f(b)) / 2;
do {
i++;
double s = 0;
for (int j = 0; j <= Math.pow(2, i - 1) - 1; j++) {
s += f(a + (2 * j + 1) * (b - a) / Math.pow(2, i));
}
s = s * (b - a) / Math.pow(2, i);
s += T[i - 1][0] / 2;
T[i][0] = s;
for (int m = 1; m <= i; m++) {
T[i][m] = (Math.pow(4, m) * T[i][m - 1] - T[i - 1][m - 1]) / (Math.pow(4, m) - 1);
}
} while (Math.abs(T[i][i] - T[i - 1][i - 1]) >= 0.0000001);
stringResult = "n= " + i +
"\nS= " + T[i][i]
+ "\n误差为 " + Math.abs(T[i][i] - T[i - 1][i - 1]);
tvResult.setText(stringResult);

}

double f(double x) {
double y;
//如果x等于0，就直接返回1
if (x == 0)
y = 1;
else y = Math.sin(x) / x;
return (y);
}
}

源码地址：http://download.csdn.net/detail/double2hao/9513415