MATLAB中的數值計算與符號計算

1、微積分實驗(2) MATLAB中的數值計算與符號計算,函數復合 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,設f(sin(x/2) = cos(x) + 1, 求f(x)與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做不到這一點,但這個積分在Mathematica中的計算結果為,先畫出圖形,syms x y; y=2*x3-6*x2-18*x+7; ezplot(y,-5,5,再求出駐點(有2個,分別為-1及3) solve(diff(y,x),x) ans = -1 3 求函數的2階導數 y=diff(y,x,2) y = 12*x-12,計算

7、2階導數在駐點處的函數值 x=-1; eval(y) ans = -24 x=3; eval(y) ans = 24 由高等數學定理知,在-1處取得極大值, 在3處取得極小值,這個積分無法計算出理論解,只能計算數值解 syms x; int(sin(sin(x),1,10) ans = int(sin(sin(x),x = 1 . 10,第一種方法,用梯形法計算,其命令為trapz(x,y), 其中x, y是具有相同長度的向量,表示用梯形法分割時,數據點上的值. 此積分的實際值為0.199(精確到20位) x=1:0.1:10; y=sin(sin(x); trapz(x,y) ans = 1

8、.1981,第二種方法,用Simpson(辛普森法)計算,其命令是quad(f,a,b,tol), 其中a,b為積分的上下限,tol為迭代誤差,其默認值為10(-3) quad(sin(sin(x),1,10,10(-7) ans = 1.1990 第三種方法,用Newton-cotes法(牛頓-科特茨法),其命令是quad8(f,a,b,tol),其用法與Simpson法一樣 quad8(sin(sin(x),1,10,10(-3) ans = 1.1990 Newton-cotes法的計算精度比其它2種要高,例如,在相同的計算量下,比較 quad(sin(sin(x),1,10,10(-3

9、) ans = 1.1976,命令: dblquad(f, xmin, xmax, ymin, ymax) 先畫出區(qū)域的圖形 syms x y; ezplot(y2-x,0,5,-3,3); hold on; ezplot(x-2,0,5,-3,3); hold off,再求2條曲線的2個交點 syms x y; x,y=solve(y2-x=0,x-2=y) x = 1 4 y = -1 2 因此,求出的2個交點是(1,-1)及(4,2),這是一個左右型區(qū)域, 應該先對x積分,再對y積分比較簡單,因此有 syms x y; dblquad(x*y,y2,y+2,-1,2) 結果,MATLAB

10、會提示出現錯誤,是語法錯誤嗎? 將上面改成,syms x y; dblquad(x*y,0,1,-1,2) ans = 0.7500 這說明,不是語法錯誤. 實際上,這個命令只能計算矩形區(qū)域的積分,如果積分區(qū)域不上矩形區(qū)域,那么,你就得定義一個二元函數,讓這個函數在積分區(qū)域外為0,比如,對此積分,定義,而且,它也只能計算二重積分的數值解,關于微分方程的求解,命令格式: dsolve(eqn1, eqn2, .); 如果不額外說明, 默認的變量是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,關于數項級數求和問題,命令格式: symsum(f, a, b), 計算表達式f從