计算方法 [课程学习]
计算方法 [课程学习] 201809
有效数字的规格化形式：
计算有效数字：误差限，规格化，得到m,则有效数字是小数点后第m位到最左端第一非零数字之间的数字是有效数字，位数是有效数字位数。
一元非线性方程求根
求非线性方程
的近似解，图像如下（找出隔根区间）：
![计算方法 [课程学习]_第1张图片](http://img.e-com-net.com/image/info8/7115ce8ed5c042f0bcb203be88f7f631.png)
二分法计算步骤：
(1) 输入隔根区间的端点
预先给定的精度
;
(2)令
，并计算
(3)若
，则令
，否则令
，转向(4)；
(4)若
，则输出方程满足精度的根
，结束；否则转向(2);
double bitf(double x){
return (x*x*x - x*x - 1.0);
}
void bits(){
printf("二分法求方程x^3 - x^2 - 1 = 0的根\nx0 = 1.5\nx = x^3 - x^2 + x - 1\n");
double _begin_time = clock();
double l = 0, r = 3.0, mid;
int cas = 0;
while(!check(l,r)){
mid = (l+r)/2.0;
double f = bitf(mid), fl = bitf(l), fr = bitf(r);
if(check(f, 0)) {
printf("%d %.6lf\n", ++cas, mid);
break;
} else if(f*fl<0) r = mid;
else l = mid;
printf("%d %.6lf\n", ++cas, mid);
}
double _end_time = clock();
printf("迭代 %d 次 用时 time = %.10lf\n\n", cas, _end_time - _begin_time);
}迭代法 ：
1）方程改写：
，但是这样的形式有很多种不同形式的方程，并不是每个都能通过迭代法求得函数的近似解；
满足一下两个条件的
是我们要找的：
（1）对
，都有
（2）
2）这里选择函数
；
3）不断迭代，
，则可取
作为原方程的数值近似根；
![\varphi (x) \in [a,b]](http://img.e-com-net.com/image/info8/efc8b68388444c28bf1b300be29b8221.gif){kind=small}
//x = x - 1/8.0 * (x*x - 7);
void Iteration1(int t, double x, double esp = 0.0005) {
double It[10000];
It[0] = x;
for(int i = 1; i < t; i++) {
It[i] = It[i-1] - 1/8.0*(It[i-1]*It[i-1] - 7.0);
printf("第%d次迭代x*的值为%.10lf %.6lf\n", i, It[i], It[i] - It[i-1]);
if(fabs(It[i] - It[i-1])
不动点迭代的加速
松弛迭代法
松弛迭代公式如下：
注：这里的
是上文转换后的
double Relaxation_iterationf(double x){
return x*x*x - x*x + x - 1;
}
double Relaxation_iterationf1(double x){
return x*x*3.0 - 2.0*x + 1;
}
void Relaxation_iteration(){
printf("松弛迭代迭代法求方程x^3 - x^2 - 1 = 0的根\nx0 = 1.5\nx = x^3 - x^2 + x - 1\n");
double _begin_time = clock();
double prex = 1.5, nextx;
int cas = 0;
while(1){
double w = 1.0/(1.0 - Relaxation_iterationf1(prex));
nextx = (1.0 - w)*prex + w*Relaxation_iterationf(prex);
printf("%d %.6lf\n", ++cas, nextx);
if(fabs(prex - nextx) < esp) break;
prex = nextx;
}
double _end_time = clock();
printf("迭代 %d 次 用时 time = %.10lf\n\n", cas, _end_time - _begin_time);
}埃特金加速迭代法
这样构造出的加速迭代公式不含导数信息, 但它需要用两次迭代值
//函数 x^3 - x^2 - 1 + x = x
double Etkinf(double x){
return x*x*x - x*x + x - 1.0;
}
void Etkin(){
printf("埃特金加速迭代法求方程x^3 - x^2 - 1 = 0的根\nx0 = 1.5\nx = x^3 - x^2 + x - 1\n");
double _begin_time = clock();
double _x[100000], _x1[100000], prex = 1.5, nextx;
int i = 1;
while(1) {
_x[i] = Etkinf(prex);
_x1[i] = Etkinf(_x[i]);
nextx = _x1[i] - (_x1[i] - _x[i])*(_x1[i] - _x[i])/(_x1[i] - 2.0*_x[i] + prex);
printf("%d %.6lf\n", i++, nextx);
if(fabs(prex - nextx) < esp) break;
prex = nextx;
}
double _end_time = clock();
printf("迭代 %d 次 用时 time = %.10lf\n\n", i-1, _end_time - _begin_time);
} 牛顿迭代法
加速迭代公式：
注：这里的
指的是
的方程；
bool check(double x, double y){
return fabs(x-y) < esp;
}
//函数 f(x) = x^3 - x^2 - 1
double newtonf(double x) {
return (x*x*x - x*x - 1.0);
//return (x*x*x - 3*x - exp(x) + 2.0);
}
double newtonf1(double x) {
return (x*x*3.0 - 2.0*x);
//return (3*x*x - exp(x) - 3.0);
}
/*
x[k+1] = x[k] - f(x[k])/f1(x[k]);
*/
void newton(){
double prex = 1.5, nextx;
//double prex = 1.0, nextx;
printf("newton法求方程x^3 - x^2 - 1 = 0的根\nx0 = 1.5\nf(x) = x^3 - x^2 - 1\nf1(x) = 3*x^2 - 2*x\n");
double _begin_time = clock();
int cas = 0;
while(1){
nextx = prex - newtonf(prex)/newtonf1(prex);
printf("%d %.6lf\n", ++cas, nextx);
if(fabs(prex - nextx) <= esp) break;
prex = nextx;
}
double _end_time = clock();
printf("迭代 %d 次 用时 time = %.10lf\n\n", cas, _end_time - _begin_time);
}弦截法
弦截迭代加速：
注：这里的
//函数 f(x) = x^3 - x^2 - 1
double chordcutf(double x){
return (x*x*x - x*x - 1.0);
}
/*
x[k+1] = x[k] - (x[k] - x[k-1])/(f(x[k]) - d(x[k-1]))*f(x[k])
*/
void chordcut(){
printf("弦截法求方程x^3 - x^2 - 1 = 0的根\nx0 = 1.5\nf(x) = x^3 - x^2 - 1\n");
double _begin_time = clock();
double x0 = 1.5, x1 = chordcutf(x0), x2;
printf("1 %.6lf\n", x1);
int cas = 1;
while(1) {
x2 = x1 - (x1 - x0)/(chordcutf(x1) - chordcutf(x0))*chordcutf(x1);
printf("%d %.6lf\n", ++cas, x2);
if(check(x2, x1)) break;
x0 = x1; x1 = x2;
}
double _end_time = clock();
printf("迭代 %d 次 用时 time = %.10lf\n\n", cas, _end_time - _begin_time);
}求非线性方程
的近似解，图像如下（找出隔值区间）：
![计算方法 [课程学习]_第2张图片](http://img.e-com-net.com/image/info8/c156930cdaac47768d714d2a51cd6ab4.jpg)
这里选择函数
迭代法：
//x = x - 1/8.0*(x*x*x - x - 1)
void Iteration2(int t, double x, double esp = 0.0005) {
double It[10000];
It[0] = x;
for(int i = 1; i < t; i++) {
It[i] = It[i-1] - 1/8.0*(It[i-1]*It[i-1]*It[i-1] - It[i-1] - 1.0);
printf("第%d次迭代x*的值为%.10lf %.6lf\n", i, It[i], It[i] - It[i-1]);
if(fabs(It[i] - It[i-1])
所有的方法：
#include
#define ll long long
using namespace std;
double esp = 1e-6;
/*
计算方法作业20180921
*/
bool check(double x, double y){
return fabs(x-y) < esp;
}
//函数 f(x) = x^3 - x^2 - 1
double newtonf(double x) {
return (x*x*x - x*x - 1.0);
//return (x*x*x - 3*x - exp(x) + 2.0);
}
double newtonf1(double x) {
return (x*x*3.0 - 2.0*x);
//return (3*x*x - exp(x) - 3.0);
}
/*
x[k+1] = x[k] - f(x[k])/f1(x[k]);
*/
void newton(){
double prex = 1.5, nextx;
//double prex = 1.0, nextx;
printf("newton法求方程x^3 - x^2 - 1 = 0的根\nx0 = 1.5\nf(x) = x^3 - x^2 - 1\nf1(x) = 3*x^2 - 2*x\n");
double _begin_time = clock();
int cas = 0;
while(1){
nextx = prex - newtonf(prex)/newtonf1(prex);
printf("%d %.6lf\n", ++cas, nextx);
if(fabs(prex - nextx) <= esp) break;
prex = nextx;
}
double _end_time = clock();
printf("迭代 %d 次 用时 time = %.10lf\n\n", cas, _end_time - _begin_time);
}
//函数 f(x) = x^3 - x^2 - 1
double chordcutf(double x){
return (x*x*x - x*x - 1.0);
}
/*
x[k+1] = x[k] - (x[k] - x[k-1])/(f(x[k]) - d(x[k-1]))*f(x[k])
*/
void chordcut(){
printf("弦截法求方程x^3 - x^2 - 1 = 0的根\nx0 = 1.5\nf(x) = x^3 - x^2 - 1\n");
double _begin_time = clock();
double x0 = 1.5, x1 = chordcutf(x0), x2;
printf("1 %.6lf\n", x1);
int cas = 1;
while(1) {
x2 = x1 - (x1 - x0)/(chordcutf(x1) - chordcutf(x0))*chordcutf(x1);
printf("%d %.6lf\n", ++cas, x2);
if(check(x2, x1)) break;
x0 = x1; x1 = x2;
}
double _end_time = clock();
printf("迭代 %d 次 用时 time = %.10lf\n\n", cas, _end_time - _begin_time);
}
//函数 x^3 - x^2 - 1 + x = x
double Etkinf(double x){
return x*x*x - x*x + x - 1.0;
}
void Etkin(){
printf("埃特金加速迭代法求方程x^3 - x^2 - 1 = 0的根\nx0 = 1.5\nx = x^3 - x^2 + x - 1\n");
double _begin_time = clock();
double _x[100000], _x1[100000], prex = 1.5, nextx;
int i = 1;
while(1) {
_x[i] = Etkinf(prex);
_x1[i] = Etkinf(_x[i]);
nextx = _x1[i] - (_x1[i] - _x[i])*(_x1[i] - _x[i])/(_x1[i] - 2.0*_x[i] + prex);
printf("%d %.6lf\n", i++, nextx);
if(fabs(prex - nextx) < esp) break;
prex = nextx;
}
double _end_time = clock();
printf("迭代 %d 次 用时 time = %.10lf\n\n", i-1, _end_time - _begin_time);
}
double Relaxation_iterationf(double x){
return x*x*x - x*x + x - 1;
}
double Relaxation_iterationf1(double x){
return x*x*3.0 - 2.0*x + 1;
}
void Relaxation_iteration(){
printf("松弛迭代迭代法求方程x^3 - x^2 - 1 = 0的根\nx0 = 1.5\nx = x^3 - x^2 + x - 1\n");
double _begin_time = clock();
double prex = 1.5, nextx;
int cas = 0;
while(1){
double w = 1.0/(1.0 - Relaxation_iterationf1(prex));
nextx = (1.0 - w)*prex + w*Relaxation_iterationf(prex);
printf("%d %.6lf\n", ++cas, nextx);
if(fabs(prex - nextx) < esp) break;
prex = nextx;
}
double _end_time = clock();
printf("迭代 %d 次 用时 time = %.10lf\n\n", cas, _end_time - _begin_time);
}
double iterationf(double x){
return (x - (x*x*x - x*x - 1.0)/4.0);
}
void iteration(){
printf("普通迭代法求方程x^3 - x^2 - 1 = 0的根\nx0 = 1.5\nx = x^3 - x^2 + x - 1\n");
double _begin_time = clock();
double prex = 1.5, nextx;
int cas = 0;
while(1){
nextx = iterationf(prex);
printf("%d %.6lf\n", ++cas, nextx);
if(fabs(prex - nextx) < esp) break;
prex = nextx;
}
double _end_time = clock();
printf("迭代 %d 次 用时 time = %.10lf\n\n", cas, _end_time - _begin_time);
}
double bitf(double x){
return (x*x*x - x*x - 1.0);
}
void bits(){
printf("二分法求方程x^3 - x^2 - 1 = 0的根\nx0 = 1.5\nx = x^3 - x^2 + x - 1\n");
double _begin_time = clock();
double l = 0, r = 3.0, mid;
int cas = 0;
while(!check(l,r)){
mid = (l+r)/2.0;
double f = bitf(mid), fl = bitf(l), fr = bitf(r);
if(check(f, 0)) {
printf("%d %.6lf\n", ++cas, mid);
break;
} else if(f*fl<0) r = mid;
else l = mid;
printf("%d %.6lf\n", ++cas, mid);
}
double _end_time = clock();
printf("迭代 %d 次 用时 time = %.10lf\n\n", cas, _end_time - _begin_time);
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
double _begin_time = clock();
#endif
newton();
chordcut();
Etkin();
Relaxation_iteration();
iteration();
bits();
#ifndef ONLINE_JUDGE
double _end_time = clock();
printf("time = %6lf\n", _end_time - _begin_time);
#endif
return 0;
}