數(shù)學(xué)實驗課后習(xí)題解答

1、數(shù)學(xué)實驗課后習(xí)題解答配套教材：王向東 戎海武 文翰 編著數(shù)學(xué)實驗王汝軍編寫實驗一曲線繪圖【練習(xí)與思考】畫出下列常見曲線的圖形。以直角坐標(biāo)方程表示的曲線：1. 立方曲線clear;x=-2:0.1:2;y=x.3;plot(x,y) 2. 立方拋物線clear;y=-2:0.1:2;x=y.3;plot(x,y)grid on 3. 高斯曲線clear;x=-3:0.1:3;y=exp(-x.2);plot(x,y);grid on%axis equal 以參數(shù)方程表示的曲線4. 奈爾拋物線clear;t=-3:0.05:3;x=t.3;y=t.2;plot(x,y)axis equalgrid

2、 on 5. 半立方拋物線clear;t=-3:0.05:3;x=t.2;y=t.3;plot(x,y)%axis equalgrid on 6. 迪卡爾曲線clear;a=3;t=-6:0.1:6;x=3*a*t./(1+t.2);y=3*a*t.2./(1+t.2);plot(x,y) 7. 蔓葉線clear;a=3;t=-6:0.1:6;x=3*a*t.2./(1+t.2);y=3*a*t.3./(1+t.2);plot(x,y) 8. 擺線clear;clc;a=1;b=1;t=0:pi/50:6*pi;x=a*(t-sin(t);y=b*(1-cos(t);plot(x,y);axi

3、s equalgrid on 9. 內(nèi)擺線（星形線）clear;a=1;t=0:pi/50:2*pi;x=a*cos(t).3;y=a*sin(t).3;plot(x,y) 10. 圓的漸伸線（漸開線）clear;a=1;t=0:pi/50:6*pi;x=a*(cos(t)+t.*sin(t);y=a*(sin(t)+t.*cos(t);plot(x,y)grid on 11. 空間螺線cleara=3;b=2;c=1;t=0:pi/50:6*pi;x=a*cos(t);y=b*sin(t);z=c*t;plot3(x,y,z)grid on 以極坐標(biāo)方程表示的曲線：12. 阿基米德線clea

4、r;a=1;phy=0:pi/50:6*pi;rho=a*phy;polar(phy,rho,r-*) 13. 對數(shù)螺線clear;a=0.1;phy=0:pi/50:6*pi;rho=exp(a*phy);polar(phy,rho) 14. 雙紐線clear;a=1;phy=-pi/4:pi/50:pi/4;rho=a*sqrt(cos(2*phy);polar(phy,rho)hold onpolar(phy,-rho) 15. 雙紐線clear;a=1;phy=0:pi/50:pi/2;rho=a*sqrt(sin(2*phy);polar(phy,rho)hold onpolar(p

5、hy,-rho) 16. 四葉玫瑰線clear;closea=1;phy=0:pi/50:2*pi;rho=a*sin(2*phy);polar(phy,rho) 17. 三葉玫瑰線clear;closea=1;phy=0:pi/50:2*pi;rho=a*sin(3*phy);polar(phy,rho) 18. 三葉玫瑰線clear;closea=1;phy=0:pi/50:2*pi;rho=a*cos(3*phy);polar(phy,rho) 實驗二極限與導(dǎo)數(shù)【練習(xí)與思考】1 求下列各極限（1） （2） （3）clear;syms ny1=limit(1-1/n)n,n,inf)y2=

6、limit(n3+3n)(1/n),n,inf)y3=limit(sqrt(n+2)-2*sqrt(n+1)+sqrt(n),n,inf) y1 =1/exp(1)y2 =3y3 =0 （4） （5） （6）clear;syms x ;y4=limit(2/(x2-1)-1/(x-1),x,1)y5=limit(x*cot(2*x),x,0)y6=limit(sqrt(x2+3*x)-x,x,inf) y4 =-1/2y5 =1/2y6 =3/2 （7） （8） （9）clear;syms x my7=limit(cos(m/x),x,inf)y8=limit(1/x-1/(exp(x)-1)

7、,x,1)y9=limit(1+x)(1/3)-1)/x,x,0) y7 =1y8 =(exp(1) - 2)/(exp(1) - 1)y9 =1/3 2 考慮函數(shù)作出圖形，并說出大致單調(diào)區(qū)間；使用diff求，并求確切的單調(diào)區(qū)間。clear;close;syms x;f=3*x2*sin(x3);ezplot(f,-2,2)grid on 大致的單調(diào)增區(qū)間：-2,-1.7,-1.3,1.2,1.7,2;大致的單點減區(qū)間：-1.7,-1.3,1.2,1.7; f1=diff(f,x,1)ezplot(f1,-2,2)line(-5,5,0,0)grid onaxis(-2.1,2.1,-60,1

8、20)f1 =6*x*sin(x3) + 9*x4*cos(x3) 用fzero函數(shù)找的零點，即原函數(shù)的駐點x1=fzero(6*x*sin(x3) + 9*x4*cos(x3),-2,-1.7)x2=fzero(6*x*sin(x3) + 9*x4*cos(x3),-1.7,-1.5)x3=fzero(6*x*sin(x3) + 9*x4*cos(x3),-1.5,-1.1)x4=fzero(6*x*sin(x3) + 9*x4*cos(x3),0)x5=fzero(6*x*sin(x3) + 9*x4*cos(x3),1,1.5)x6=fzero(6*x*sin(x3) + 9*x4*co

9、s(x3),1.5,1.7)x7=fzero(6*x*sin(x3) + 9*x4*cos(x3),1.7,2) x1 = -1.9948x2 = -1.6926x3 = -1.2401x4 = 0x5 = 1.2401x6 = 1.6926x7 = 1.9948 確切的單調(diào)增區(qū)間：-1.9948,-1.6926,-1.2401,1.2401,1.6926,1.9948確切的單調(diào)減區(qū)間：-2,-1.9948,-1.6926,-1.2401,1.2401,1.6926,1.9948,23 對于下列函數(shù)完成下列工作，并寫出總結(jié)報告，評論極值與導(dǎo)數(shù)的關(guān)系，(i) 作出圖形，觀測所有的局部極大、局部極

10、小和全局最大、全局最小值點的粗略位置；(ii) 求所有零點（即的駐點）；(iii) 求出駐點處的二階導(dǎo)數(shù)值;(iv) 用fmin求各極值點的確切位置;(v) 局部極值點與有何關(guān)系?(1) (2) (3) clear;close;syms x;f=x2*sin(x2-x-2)ezplot(f,-2,2)grid on f =x2*sin(x2 - x - 2) 局部極大值點為：-1.6,局部極小值點為為：-0.75，-1.6全局最大值點為為：-1.6，全局最小值點為：-3f1=diff(f,x,1)ezplot(f1,-2,2)line(-5,5,0,0)grid onaxis(-2.1,2.1

11、,-6,20) f1 =2*x*sin(x2 - x - 2) + x2*cos(x2 - x - 2)*(2*x - 1) 用fzero函數(shù)找的零點，即原函數(shù)的駐點x1=fzero(2*x*sin(x2-x-2)+x2*cos(x2-x-2)*(2*x-1),-2,-1.2)x2=fzero(2*x*sin(x2-x-2)+x2*cos(x2-x-2)*(2*x-1),-1.2,-0.5)x3=fzero(2*x*sin(x2-x-2)+x2*cos(x2-x-2)*(2*x-1),-0.5,1.2)x4=fzero(2*x*sin(x2-x-2)+x2*cos(x2-x-2)*(2*x-1

12、),1.2,2)x1 = -1.5326x2 = -0.7315x3 = -3.2754e-027x4 = 1.5951 ff=(x) x.2.*sin(x.2-x-2)ff(-2),ff(x1),ff(x2),ff(x3),ff(x4),ff(2) ff = (x)x.2.*sin(x.2-x-2)ans = -3.0272ans = 2.2364ans = -0.3582ans = -9.7549e-054ans = -2.2080ans = 0 實驗三級數(shù)【練習(xí)與思考】1. 用taylor命令觀測函數(shù)的maclaurin展開式的前幾項, 然后在同一坐標(biāo)系里作出函數(shù)和它的taylor展開式

13、的前幾項構(gòu)成的多項式函數(shù)的圖形，觀測這些多項式函數(shù)的圖形向的圖形的逼近的情況(1) clear;syms xy=asin(x);y1=taylor(y,0,1)y2=taylor(y,0,5)y3=taylor(y,0,10)y4=taylor(y,0,15)x=-1:0.1:1;y=subs(y,x);y1=subs(y1,x);y2=subs(y2,x);y3=subs(y3,x);y4=subs(y4,x);plot(x,y,x,y1,:,x,y2,-.,x,y3,-,x,y4,:,linewidth,3) y1 =0y2 =x3/6 + xy3 =(35*x9)/1152 + (5*x

14、7)/112 + (3*x5)/40 + x3/6 + xy4 =(231*x13)/13312 + (63*x11)/2816 + (35*x9)/1152 + (5*x7)/112 + (3*x5)/40 + x3/6 + x (2) clear;syms xy=atan(x);y1=taylor(y,0,3)y2=taylor(y,0,5),y3=taylor(y,0,10),y4=taylor(y,0,15)x=-1:0.1:1;y=subs(y,x);y1=subs(y1,x);y2=subs(y2,x);y3=subs(y3,x);y4=subs(y4,x);plot(x,y,x,

15、y1,:,x,y2,-.,x,y3,-,x,y4,:,linewidth,3) y1 =xy2 =x - x3/3y3 =x9/9 - x7/7 + x5/5 - x3/3 + xy4 =x13/13 - x11/11 + x9/9 - x7/7 + x5/5 - x3/3 + x (3) clear;syms xy=exp(x2);y1=taylor(y,0,3)y2=taylor(y,0,5)y3=taylor(y,0,10)y4=taylor(y,0,15)x=-1:0.1:1;y=subs(y,x);y1=subs(y1,x);y2=subs(y2,x);y3=subs(y3,x);y

16、4=subs(y4,x);plot(x,y,x,y1,:,x,y2,-.,x,y3,-,x,y4,:,linewidth,3) y1 =x2 + 1y2 =x4/2 + x2 + 1y3 =x8/24 + x6/6 + x4/2 + x2 + 1y4 =x14/5040 + x12/720 + x10/120 + x8/24 + x6/6 + x4/2 + x2 + 1 (4) clear;syms xy=sin(x)2;y1=taylor(y,0,1)y2=taylor(y,0,5)y3=taylor(y,0,10)y4=taylor(y,0,15)x=-pi:0.1:pi;y=subs(y

17、,x);y1=subs(y1,x);y2=subs(y2,x);y3=subs(y3,x);y4=subs(y4,x);plot(x,y,x,y1,:,x,y2,-.,x,y3,-,x,y4,:,linewidth,3) y1 =0y2 =x2 - x4/3y3 =- x8/315 + (2*x6)/45 - x4/3 + x2y4 =(4*x14)/42567525 - (2*x12)/467775 + (2*x10)/14175 - x8/315 + (2*x6)/45 - x4/3 + x2 (5) clear;syms xy=exp(x)/(1-x);y1=taylor(y,0,3)y

18、2=taylor(y,0,5)y3=taylor(y,0,10)y4=taylor(y,0,15)x=-1:0.1:0;y=subs(y,x);y1=subs(y1,x);y2=subs(y2,x);y3=subs(y3,x);y4=subs(y4,x);plot(x,y,x,y1,:,x,y2,-.,x,y3,-,x,y4,:,linewidth,3) y1 =(5*x2)/2 + 2*x + 1y2 =(65*x4)/24 + (8*x3)/3 + (5*x2)/2 + 2*x + 1y3 =(98641*x9)/36288 + (109601*x8)/40320 + (685*x7)/2

19、52 + (1957*x6)/720 + (163*x5)/60 + (65*x4)/24 + (8*x3)/3 + (5*x2)/2 + 2*x + 1y4 =(47395032961*x14)+ (8463398743*x13)/3113510400 + (260412269*x12)/95800320 + (13563139*x11)/4989600 + (9864101*x10)/3628800 + (98641*x9)/36288 + (109601*x8)/40320 + (685*x7)/252 + (1957*x6)/720 + (163*x5)/60

20、 + (65*x4)/24 + (8*x3)/3 + (5*x2)/2 + 2*x + 1 (6) clear;syms xy=log(x+sqrt(1+x2);y1=taylor(y,0,3)y2=taylor(y,0,5)y3=taylor(y,0,10)y4=taylor(y,0,15)x=-1:0.1:1;y=subs(y,x);y1=subs(y1,x);y2=subs(y2,x);y3=subs(y3,x);y4=subs(y4,x);plot(x,y,x,y1,:,x,y2,-.,x,y3,-,x,y4,:,linewidth,3) y1 =xy2 =x - x3/6y3 =(3

21、5*x9)/1152 - (5*x7)/112 + (3*x5)/40 - x3/6 + xy4 =(231*x13)/13312 - (63*x11)/2816 + (35*x9)/1152 - (5*x7)/112 + (3*x5)/40 - x3/6 + x 2. 求公式中的數(shù)的值.k=4 5 6 7 8;syms nsymsum(1./n.(2*k),1,inf) ans = pi8/9450, pi10/93555, (691*pi12)/638512875, (2*pi14)/18243225, (3617*pi16)/325641566250 3. 利用公式來計算的近似值。精確到

22、小數(shù)點后100位,這時應(yīng)計算到這個無窮級數(shù)的前多少項?請說明你的理由.解：matlab代碼為clear;clc;closeepsl=1.0e-100;ep=1;fn=1;a=1;n=1;while epepsla=a+fn;n=n+1;fn=fn/n;ep=fn;endfnvpa(a,100)n fn = 8.3482e-101ans =2.71828182845904553488480814849026501178741455078125n = 70 精確到小數(shù)點后100位,這時應(yīng)計算到這個無窮級數(shù)的前71項,理由是誤差小于10的負(fù)100次方，需要最后一項小于10的負(fù)100次方，由上述循環(huán)知

23、n=70時最后一項小于10的負(fù)100次方，故應(yīng)計算到這個無窮級數(shù)的前71項.4. 用練習(xí)3中所用觀測法判斷下列級數(shù)的斂散性(1) clear;clc;epsl=0.000001;n=50000;p=1000;syms nun=1/(n2+n3);s1=symsum(un,1,n);s2=symsum(un,1,n+p);sa=vpa(s2-s1);sa=setstr(sa);sa=str2num(sa);fprintf(級數(shù))disp(un)if saepsl disp(收斂)else disp(發(fā)散)end 級數(shù)1/(n3 + n2)收斂 clear;closesyms ns=;for k=

24、1:100s(k)=symsum(1/(n3 + n2),1,k);endplot(s,.) (2) clear;clc;epsl=0.000001;n=50000;p=1000;syms nun=1/(n*2n);s1=symsum(un,1,n);s2=symsum(un,1,n+p);sa=vpa(s2-s1);sa=setstr(sa);sa=str2num(sa);fprintf(級數(shù))disp(un)if saepsl disp(收斂)else disp(發(fā)散)end 級數(shù)1/(2n*n)收斂 clear;closesyms ns=;for k=1:100s(k)=symsum(1

25、/(2n*n),1,k);endplot(s,.) (3) clear;clc;epsl=0.00000000000001;n=50000;p=100;syms nun=1/sin(n);s1=symsum(un,1,n);s2=symsum(un,1,n+p);sa=vpa(s2-s1);sa=setstr(sa);sa=str2num(sa);fprintf(級數(shù))disp(un)if abs(sa)epsl disp(收斂)else disp(發(fā)散)end 級數(shù)1/sin(n)發(fā)散 clear;closesyms ns=;for k=1:100s(k)=symsum(1/sin(n),1

26、,k);endplot(s,.) 發(fā)散 (4) clear;clc;epsl=0.0000001;n=50000;p=1000;syms nun=log(n)/(n3);s1=symsum(un,1,n);s2=symsum(un,1,n+p);sa=vpa(s2-s1);sa=setstr(sa);sa=str2num(sa);fprintf(級數(shù))disp(un)if saepsl disp(收斂)else disp(發(fā)散)end 級數(shù)log(n)/n3收斂 clear;closesyms ns=;for k=1:100s(k)=symsum(log(n)/n3,1,k);endplot(

27、s,.) (5) clear;closesyms ns=;he=0;for k=1:100he=he+factorial(k)/kk;s(k)=he;endplot(s,.) (6) clear;clc;epsl=0.0000001;n=50000;p=1000;syms nun=1/log(n)n;s1=symsum(un,3,n);s2=symsum(un,3,n+p);sa=vpa(s2-s1);sa=setstr(sa);sa=str2num(sa);fprintf(級數(shù))disp(un)if saepsl disp(收斂)else disp(發(fā)散)end 級數(shù)1/log(n)n收斂

28、clear;closesyms ns=;for k=3:100s(k)=symsum(1/log(n)n,3,k);endplot(s,.) (7) clear;clc;epsl=0.0000001;n=50000;p=100;syms nun=1/(log(n)*n);s1=symsum(un,3,n);s2=symsum(un,3,n+p);sa=vpa(s2-s1);sa=setstr(sa);sa=str2num(sa);fprintf(級數(shù))disp(un)if (sa)epsl disp(收斂)else disp(發(fā)散)end 級數(shù)1/(n*log(n)發(fā)散 clear;close

29、syms ns=;for k=3:300s(k)=symsum(1/(n*log(n),2,k);endplot(s,.) (8) clear;clc;epsl=0.0000001;n=50000;p=100;syms nun=(-1)n*n/(n2+1);s1=symsum(un,3,n);s2=symsum(un,3,n+p);sa=vpa(s2-s1);sa=setstr(sa);sa=str2num(sa);fprintf(級數(shù))disp(un)if (sa)epsl disp(收斂)else disp(發(fā)散)end 級數(shù)(-1)n*n)/(n2 + 1)收斂 clear;closes

30、yms ns=;for k=3:300s(k)=symsum(-1)n*n/(n2+1),2,k);endplot(s,.) 實驗四積分【練習(xí)與思考】1(不定積分)用int計算下列不定積分，并用diff驗證，解：matlab代碼為：syms xy1=x*sin(x2);y2=1/(1+cos(x);y3=1/(exp(x)+1);y4=asin(x);y5=sec(x)3;f1=int(y1)f2=int(y2)f3=int(y3)f4=int(y4)f5=int(y5) dy=simplify(diff(f1;f2;f3;f4;f5) dy = x*sin(x2) tan(x/2)2/2 +

31、 1/2 1/(exp(x) + 1) asin(x) (cot(pi/4 + x/2)*(tan(pi/4 + x/2)2/2 + 1/2)/2 + 1/(2*cos(x) + tan(x)2/cos(x) f1 =-cos(x2)/2f2 =tan(x/2)f3 =x - log(exp(x) + 1)f4 =x*asin(x) + (1 - x2)(1/2)f5 =log(tan(pi/4 + x/2)/2 + tan(x)/(2*cos(x) 2(定積分)用trapz,quad,int計算下列定積分,解：matlab代碼為clear;x=(0+eps):0.05:1;y1=sin(x)

32、./x;f1=trapz(x,y1) f1 =0.9460 fun1=(x)sin(x)./x;f12=quad(fun1,0+eps,1) f12 = 0.9461 f13=vpa(int(sin(x)/x,0,1),5) f13 =0.94608 3(橢圓的周長) 用定積分的方法計算橢圓的周長解：橢圓的參數(shù)方程為由參數(shù)曲線的弧長公式得matlab代碼為s=vpa(int(sqrt(5*sin(t)2+4),t,0,2*pi),5) s =15.865 4（二重積分）計算數(shù)值積分解：fxy=(x,y)1+x+y;ylow=(x)1-sqrt(1-x.2);yup=(x)1+sqrt(1-x.

33、2);s=quad2d(fxy,-1,1,ylow,yup) s =6.2832 或符號積分法：syms x yxi=int(1+x+y,y,1-sqrt(1-x2),1+sqrt(1-x2);s=int(xi,x,-1,1) s =2*pi 5（假奇異積分）用trapz,quad8計算積分，會出現(xiàn)什么問題？分析原因，并求出正確的解。解：matlab代碼為clearx=-1:0.05:1;y=x.(1/3).*cos(x);s1=trapz(x,y)fun5=(x)x.(1/3).*cos(x);s2=quad(fun5,-1,1)int(x(1/3)*cos(x),x,-1,1) s1 =

34、0.9036 + 0.5217is2 = 0.9114 + 0.5262iwarning: explicit integral could not be found. ans =int(x(1/3)*cos(x), x = -1.1) ，原函數(shù)不存在，不能用int函數(shù)運算。用梯形法和辛普森法計算數(shù)值積分時，由于對負(fù)數(shù)的開三次方運算結(jié)果為復(fù)數(shù)，所以導(dǎo)致結(jié)果錯誤且為復(fù)數(shù)；顯然被積函數(shù)為奇函數(shù)，在對稱區(qū)間上的積分等于0，此時可以這樣處理：（1）重新定義被積函數(shù)%fun5.mfunction y=fun5(x)m,n=size(x);for k=1:mfor l=1:ny(k,l)=nthroot(x

35、(k,l),3)*cos(x(k,l);endendend用辛普森法：s=quad(fun5,-1,1) s = 0 用梯形法clear;x=-1:0.01:1;y=fun5(x);s=trapz(x,y) s = -1.3878e-017 6（假收斂現(xiàn)象）考慮積分，（1）用解析法求;clear;syms x k;ik=int(abs(sin(x),0,k*pi) warning: explicit integral could not be found. ik =int(abs(sin(x), x = 0.pi*k) （2）分別用trapz,quad和quad8求和,發(fā)現(xiàn)什么問題？clear

36、;for k=4:2:8;x=0:pi/1000:k*pi;y=abs(sin(x);trapz(x,y)end ans = 8.0000ans = 12.0000ans = 16.0000 for k=4:2:8fun6=(x)abs(sin(x);quad(fun6,0,k*pi)end ans = 8.0000ans = 12.0000ans = 16.0000 7(simpson積分法)編制一個定步長simpson法數(shù)值積分程序.計算公式為其中為偶數(shù),解：matlab代碼為%fun7.mfunction y=fun7(f_name,a,b,n)%f_name為被積函數(shù)%a,b為積分區(qū)間

37、%n為偶數(shù)，用來確定步長h=(b-a)/nif mod(n,2)=0 disp(n必須為偶數(shù)) return;endif nargin4n=100;endif nargin in fmincon at 445local minimum possible. constraints satisfied.fmincon stopped because the predicted change in the objective functionis less than the default value of the function tolerance and constraints were sa

38、tisfied to within the default value of the constraint tolerance.no active inequalities.x = 161.9676 182.0320fval = -715.4403 heigh和height兩個函數(shù)分別定義如下：(應(yīng)寫在m文件中)%heigh.mfunction f=heigh(beta,xdata)xx1=xdata(:,1);xx2=xdata(:,2);f=beta(1)+beta(2)*xx1+beta(3)*xx2+beta(4)*xx1.2+beta(5)*xx2.2;end%height.mfun

39、ction y=height(x)y=-(538.4375+1.4901*x(1)+0.6189*x(2)-0.0046*x(1).2-0.0017*x(2).2);end實驗六多元函數(shù)的極值【練習(xí)與思考】1.求的極值，并對圖形進行觀測。解：maltab代碼為syms x y;z=x4+y4-4*x*y+1;dzx=diff(z,x);dzy=diff(z,y);s=solve(dzx,dzy,x,y);x=s.x.y=s.y. x = 0, 1, -1, (-1)(3/4), -(-1)(3/4), -i, i, -(-1)(3/4)*i, (-1)(3/4)*iy = 0, 1, -1, (-1)(1/4), -(-1)(1/4), i, -i, (-1)(1/4)*i, -(-1)(1/4)*i 經(jīng)計算可知，函數(shù)的駐點為(0,0)、(1,1)、(-1,-1)ezmeshc(z,-2,2,-2,2) 從圖形上觀測可知，(1,1)、(-1,-1)為極值點，(0,0)不是極值點。clearsyms x y;z=x4+y4-4*x*y+1;dzx=diff(z