MATLAB中的数值计算与符号计算

1、微积分实验(2) MATLAB中的数值计算与符号计算,函数复合 命令格式: compose(f,g,x), 即求f(g(x); 例子: syms f, x; f=sin(x);compose(f,f,x) ans = sin(sin(x) syms f, x; f=sin(x); g=1+x2; compose(f,g,x) ans = sin(1+x2,设f(sin(x/2) = cos(x) + 1, 求f(x)与f(cos(x/2). syms x, f; old=cos(x); new=1-2*sin(x/2)2; subs(f(sin(x/2)=cos(x)+1,old,new) a

2、ns = f(sin(x/2)=(1-2*sin(x/2)2)+1 syms u; subs(ans,sin(x/2),u) ans = f(u) = 2-2*u2,Syms x,f,g; f=x/sqrt(1+x2);g=x; g=simple(compose(f,f,x) f =x/(1+x2)(1/2) g =x/(2*x2+1)(1/2) g=simple(compose(g,f,x) g =x/(3*x2+1)(1/2) g=simple(compose(g,f,x) g =x/(4*x2+1)(1/2) g=simple(compose(g,f,x) g =x/(5*x2+1)(1

3、/2,syms x,f; f=sin(x)2/(1+cot(x)+cos(x)2/(1+tan(x); diff(f,x) Ans= 2*sin(x)/(1+cot(x)*cos(x)-sin(x)2/(1+cot(x)2*(-1-cot(x)2)-2*cos(x)/(1+tan(x)*sin(x)-cos(x)2/(1+tan(x)2*(1+tan(x)2) simple(ans) ans =-2*cos(x)2+1 diff(f,x,2) ans=2*cos(x)2/(1+cot(x)-4*sin(x)/(1+cot(x)2*cos(x)*(-1-cot(x)2)-2*sin(x)2/(1

4、+cot(x)+2*sin(x)2/(1+cot(x)3*(-1-cot(x)2)2+2*sin(x)2/(1+cot(x)2*cot(x)*(-1-cot(x)2)+2*sin(x)2/(1+tan(x)+4*cos(x)/(1+tan(x)2*sin(x)*(1+tan(x)2)-2*cos(x)2/(1+tan(x)+2*cos(x)2/(1+tan(x)3*(1+tan(x)2)2-2*cos(x)2/(1+tan(x)2*tan(x)*(1+tan(x)2) simple(ans) ans =4*sin(x)*cos(x,syms x y z f g; f=exp(z)-z+x*y-

5、3; g=jacobian(f,x, y, z) g= y, x, exp(z)-1 g(1) ans = y g(2) ans = x g(3) ans = exp(z)-1 g(1)/g(3) ans = y/(exp(z)-1,syms x y t a; x=a*cos(t)3; y=a*sin(t)3; y=simple(diff(y,t)/diff(x,t) y = -tan(t) simple(diff(y,t)/diff(x,t) ans = 1/3/cos(t)4/a/sin(t,syms x; int(1-sin(x)+cos(x)/(1+sin(x)-cos(x) ans

6、= 2*log(tan(1/2*x)-2*log(tan(1/2*x)+1)-2*atan(tan(1/2*x,simple(ans) ans=2*log(tan(1/2*x)-2*log(tan(1/2*x)+1)-2*atan(tan(1/2*x,MATLAB做不到这一点,但这个积分在Mathematica中的计算结果为,先画出图形,syms x y; y=2*x3-6*x2-18*x+7; ezplot(y,-5,5,再求出驻点(有2个,分别为-1及3) solve(diff(y,x),x) ans = -1 3 求函数的2阶导数 y=diff(y,x,2) y = 12*x-12,计算

7、2阶导数在驻点处的函数值 x=-1; eval(y) ans = -24 x=3; eval(y) ans = 24 由高等数学定理知,在-1处取得极大值, 在3处取得极小值,这个积分无法计算出理论解,只能计算数值解 syms x; int(sin(sin(x),1,10) ans = int(sin(sin(x),x = 1 . 10,第一种方法,用梯形法计算,其命令为trapz(x,y), 其中x, y是具有相同长度的向量,表示用梯形法分割时,数据点上的值. 此积分的实际值为0.199(精确到20位) x=1:0.1:10; y=sin(sin(x); trapz(x,y) ans = 1

8、.1981,第二种方法,用Simpson(辛普森法)计算,其命令是quad(f,a,b,tol), 其中a,b为积分的上下限,tol为迭代误差,其默认值为10(-3) quad(sin(sin(x),1,10,10(-7) ans = 1.1990 第三种方法,用Newton-cotes法(牛顿-科特茨法),其命令是quad8(f,a,b,tol),其用法与Simpson法一样 quad8(sin(sin(x),1,10,10(-3) ans = 1.1990 Newton-cotes法的计算精度比其它2种要高,例如,在相同的计算量下,比较 quad(sin(sin(x),1,10,10(-3

9、) ans = 1.1976,命令: dblquad(f, xmin, xmax, ymin, ymax) 先画出区域的图形 syms x y; ezplot(y2-x,0,5,-3,3); hold on; ezplot(x-2,0,5,-3,3); hold off,再求2条曲线的2个交点 syms x y; x,y=solve(y2-x=0,x-2=y) x = 1 4 y = -1 2 因此,求出的2个交点是(1,-1)及(4,2),这是一个左右型区域, 应该先对x积分,再对y积分比较简单,因此有 syms x y; dblquad(x*y,y2,y+2,-1,2) 结果,MATLAB

10、会提示出现错误,是语法错误吗? 将上面改成,syms x y; dblquad(x*y,0,1,-1,2) ans = 0.7500 这说明,不是语法错误. 实际上,这个命令只能计算矩形区域的积分,如果积分区域不上矩形区域,那么,你就得定义一个二元函数,让这个函数在积分区域外为0,比如,对此积分,定义,而且,它也只能计算二重积分的数值解,关于微分方程的求解,命令格式: dsolve(eqn1, eqn2, .); 如果不额外说明, 默认的变量是t dsolve(Dy=y2*(1-y) Warning: Explicit solution could not be found; implicit solution returned. In C:MATLABR12toolboxsymbolicdsolve.m at line 292 ans = t+1/y-log(y)+log(-1+y)+C1=0 dsolve(D2y = -a2*y, y(0) = 1, Dy(pi/a) = 0) ans =cos(a*t) x,y=dsolve(Dx = y, Dy = -x, x(0)=0, y(0)=1) x =sin(t) y =cos(t,关于数项级数求和问题,命令格式: symsum(f, a, b), 计算表达式f从