matlab解方程组方程组1:(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)-e*sin(f))^2-(m/2-n*sin(c/2+d)+e*cos(f-b))^2-(h+n*cos(c/2+d)-e*sin(f-b))^2=0;方程组2:(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)-e*sin(f))^2-(m/2+e*cos(a+f)-n*sin(c/2-d))^2-(h
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![matlab解方程组方程组1:(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)-e*sin(f))^2-(m/2-n*sin(c/2+d)+e*cos(f-b))^2-(h+n*cos(c/2+d)-e*sin(f-b))^2=0;方程组2:(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)-e*sin(f))^2-(m/2+e*cos(a+f)-n*sin(c/2-d))^2-(h](/uploads/image/z/993322-10-2.jpg?t=matlab%E8%A7%A3%E6%96%B9%E7%A8%8B%E7%BB%84%E6%96%B9%E7%A8%8B%E7%BB%841%EF%BC%9A%28m%2F2-n%2Asin%28c%2F2%29%2Be%2Acos%28f%29%29%5E2%2B%28h%2Bn%2Acos%28c%2F2%29-e%2Asin%28f%29%29%5E2-%28m%2F2-n%2Asin%28c%2F2%2Bd%29%2Be%2Acos%28f-b%29%29%5E2-%28h%2Bn%2Acos%28c%2F2%2Bd%29-e%2Asin%28f-b%29%29%5E2%3D0%3B%E6%96%B9%E7%A8%8B%E7%BB%842%EF%BC%9A%28m%2F2-n%2Asin%28c%2F2%29%2Be%2Acos%28f%29%29%5E2%2B%28h%2Bn%2Acos%28c%2F2%29-e%2Asin%28f%29%29%5E2-%28m%2F2%2Be%2Acos%28a%2Bf%29-n%2Asin%28c%2F2-d%29%29%5E2-%28h)
matlab解方程组方程组1:(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)-e*sin(f))^2-(m/2-n*sin(c/2+d)+e*cos(f-b))^2-(h+n*cos(c/2+d)-e*sin(f-b))^2=0;方程组2:(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)-e*sin(f))^2-(m/2+e*cos(a+f)-n*sin(c/2-d))^2-(h
matlab解方程组
方程组1:(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)-e*sin(f))^2-(m/2-n*sin(c/2+d)+e*cos(f-b))^2-(h+n*cos(c/2+d)-e*sin(f-b))^2=0;
方程组2:(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)-e*sin(f))^2-(m/2+e*cos(a+f)-n*sin(c/2-d))^2-(h+n*cos(c/2-d)-e*sin(a+f))^2=0;
我想建立个函数 function f=myfun(b,c,e,f,h,m,n)
能把 两个方程组中的 ‘a’,‘d’作为未知数,其他字母通过参数传递,解方程组.
要求:输入 myfun(1,2,3,4,5,6,7)能得出 a和b的结果(不要复数解).
注:这2次带三角函数的方程组.a或者b 应该有两组(或者两组以上)实数解.
matlab解方程组方程组1:(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)-e*sin(f))^2-(m/2-n*sin(c/2+d)+e*cos(f-b))^2-(h+n*cos(c/2+d)-e*sin(f-b))^2=0;方程组2:(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)-e*sin(f))^2-(m/2+e*cos(a+f)-n*sin(c/2-d))^2-(h
function F=mymagic(x,b,c,e,f,h,m,n)
F=[(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)-e*sin(f))^2-(m/2-n*sin(c/2+x(2))+e*cos(f-b))^2-(h+n*cos(c/2+x(2))-e*sin(f-b))^2
(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)-e*sin(f))^2-(m/2+e*cos(x(1)+f)-n*sin(c/2-x(2)))^2-(h+n*cos(c/2-x(2))-e*sin(x(1)+f))^2];
fsolve(@(x) mymagic(x,1,2,3,4,5,6,7),[0;0])
上面是函数,下面是调用的语句,其中最后的[0;0]是迭代的初值,这里使用了fsolve进行数值求解,求解的方法就是牛顿迭代法!
祝你学习愉快!