7楼: Originally posted by zhaoyanqiao at 2012-07-07 09:40:24
v=0:0.1:10;
y1=sin(v/2);
y2=3*v/14/pi;
plot(v,y1)
hold on
plot(v,y2)
grid on

可以看到，在v5.5左右时有一个交点，也就是我所需要的值...

2楼: Originally posted by 不是不想 at 2012-07-06 14:13:07
s=solve('sin（v/2)-3*v/14/pi=0 ')      试一下

3楼: Originally posted by luomia at 2012-07-06 15:02:39
>> s=solve('sin(v/2)=3*v/14/pi');
>> s

s =

0

5楼: Originally posted by zhaoyanqiao at 2012-07-06 21:20:36
对不起，程序运行下来是这样的：
>> s=solve('sin(v/2)=3*v/14/pi');
>> s

s =

14/3*sin(0)*pi

>>
仍然不是我想要的结果
我需要的是除了0之外的那个值。...

6楼: Originally posted by luomia at 2012-07-07 07:52:46
大哥，你这个问题只有0解。sin(v/2)=3*v/14/pi等价于v=14*sin(v/2)/(3*pi)。然后你可以到matlab里输入
>> v=0:50;
>> f=14*sin(v/2)/(3*pi);
>> plot(v,f)
>> grid on
>> axis ...

7楼: Originally posted by zhaoyanqiao at 2012-07-07 09:40:24
v=0:0.1:10;
y1=sin(v/2);
y2=3*v/14/pi;
plot(v,y1)
hold on
plot(v,y2)
grid on

可以看到，在v5.5左右时有一个交点，也就是我所需要的值...

8楼: Originally posted by luomia at 2012-07-07 11:09:53
你这个问题由于涉及到PI，需要指定计算精度。用常规迭代法很难收敛，以下程序希望能有所帮助。
>> for v=5:0.00000001:6;
f=14*pi*sin(v/2)/3;
if abs(f-v)<=0.000001
break;
end
end
>> vp ...