求定积分有什么好的算法吗？

lordbyorn

It turn out that the equation I show above can satisfy all condition. It can converge with all z from minus infinite to infinite.

So done.

Of course, while |z| is very large, n should be very large too. :p

minus273

一次把区间分成10个试试
20,30,50,100...
再求面积
PS: 求导数甚么方法较好?

lordbyorn

PS: 求导数甚么方法较好?

f'=df/dx

lordbyorn

Though the result is very small, you have to deal with large number. You should use some skill.
Integral can be another way.

I can write a funtion for you if you want.

pupilzeng

最初由 lordbyorn 发表
Though the result is very small, you have to deal with large number. You should use some skill.
Integral can be another way.

I can write a funtion for you if you want.

谢谢，如果不占用你很多时间的话，那就请你写给我吧。
:thank :thank

lordbyorn

What is the max |z| do you want? If you just want to calculate some value, you don't have to create table.

lordbyorn

How about this:

1. 
   
2. #include<stdio.h>
   
3. #include<math.h>
   
4. #define TERM 30
   
5. main(int argc,char ** argv){
   
6. double x,y;
   
7. int c;
   
8. double tmp;
   
9. double poly[TERM];
   
10. sscanf(argv[1],"%lf",&x);
    
11. poly[0]=x;
    
12. y=0;
    
13. for(c=1;c<TERM;c++){
    
14.         tmp=x*x/(c*2);
    
15.         poly[c]=poly[c-1]*tmp;
    
16.         }
    
17. for(c=1;c<TERM;c++){
    
18.         poly[c]=poly[c]/(2*c+1);
    
19.         }
    
20. for(c=1;c<TERM;c+=2){
    
21.         poly[c]=-poly[c];
    
22.         }
    
23. printf("x=%lf\n",x);
    
24. for(c=0;c<TERM;c++){
    
25.         y+=poly[c];
    
26.         }
    
27. y=0.5+0.39894228*y;
    
28. printf("y=%lf\n",y);
    
29. return 0;
    
30. }
    
31.                                                                                                                
    
32. ~
    

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minus273

f:=dy/dx不够精确，I like some cool methods

lordbyorn

With very little improvement

1. 
   
2. #include<stdio.h>
   
3. #include<math.h>
   
4. 
   
5. #define TERM 100
   
6. #define CONST 0.39894228
   
7. 
   
8. main(int argc,char ** argv){
   
9. double x,y;
   
10. int c;
    
11. double tmp;
    
12. double poly[TERM];
    
13. sscanf(argv[1],"%lf",&x);
    
14. poly[0]=x;
    
15. y=0;
    
16. for(c=1;c<TERM;c++){
    
17.         tmp=x*x/(c*2);
    
18.         poly[c]=poly[c-1]*tmp;
    
19.         }
    
20. for(c=1;c<TERM;c++){
    
21.         poly[c]=poly[c]/(2*c+1);
    
22.         }
    
23. for(c=0,tmp=1;c<TERM;c++){
    
24.         y+=poly[c]*tmp;
    
25.         tmp*=-1;
    
26.         }
    
27. y=0.5+CONST*y;
    
28. printf("y=%lf\n",y);
    
29. return 0;
    
30. }
    

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lordbyorn

Good job. fast, accurate.

1. 
   
2. #include<stdio.h>
   
3. #include<math.h>
   
4. #define TERM 100
   
5. 
   
6. double poly[TERM];
   
7. 
   
8. double zhentai(double x);
   
9. 
   
10. main(int argc,char ** argv){
    
11. double z,step;
    
12. for(z=0,step=0.0001;z<=3.0;z+=step){
    
13. printf("z=%lf,y=%lf\n",z,zhentai(z));
    
14. }
    
15. 
    
16. return 0;
    
17. }
    
18. 
    
19. double zhentai(double x){
    
20. double y;
    
21. int c;
    
22. double tmp;
    
23. poly[0]=x;
    
24. y=0;
    
25. for(c=1;c<TERM;c++){
    
26.         tmp=x*x/(c*2);
    
27.         poly[c]=poly[c-1]*tmp;
    
28.         }
    
29. for(c=1;c<TERM;c++){
    
30.         poly[c]=poly[c]/(2*c+1);
    
31.         }
    
32. for(c=0,tmp=1;c<TERM;c++){
    
33.         y+=poly[c]*tmp;
    
34.         tmp*=-1;
    
35.         }
    
36. y=0.5+0.39894228*y;
    
37. return y;
    
38. }
    
39. 
    

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