MATLAB實(shí)驗(yàn)練習(xí)題(計(jì)算機(jī))南郵MATLAB數(shù)學(xué)實(shí)驗(yàn)大作業(yè)答案

1、MATLA練習(xí)題要求：抄題、寫由操作命令、運(yùn)行結(jié)果，并根據(jù)要求，貼上運(yùn)行圖。1、求ex 3x2 =0的所有根。(先畫圖后求解)(要求貼圖)>> solve('exp(x)-3*xA2',0) ans =-2*lambertw(-1/6*3A(1/2)-2*lambertw(-1,-1/6*3A(1/2)-2*lambertw(1/6*3A(1/2)2、求下列方程的根。5_1) x 5x 1 =0a=solve('xA5+5*x+1',0);a=vpa(a,6)1.10447+1.05983*i-1.00450+1.06095*i-.199936-1.

2、00450-1.06095*i1.10447-1.05983*i>> fzero('x*sin(x)-1/2', 3) ans =2.9726>> fzero('x*sin(x)-1/2',-3) ans =-2.9726>> fzero('x*sin(x)-1/2',0) ans =-0.74083) sin xcosx x2 =0 所有根>> fzero('sin(x)*cos(x)-xA2',0) ans =>> fzero('sin(x)*cos(x)-x

3、A2',0.6) ans =0.70223、求解下列各題:1)limx _0x -sin x3x>> sym x;>> limit(x-sin(x)/xA3) ans =1/6x(10)2) y =e cosx,求 y>> sym x;>> diff(exp(x)*cos(x),10) ans =(-32)*exp(x)*sin(x)1/223） e ex dx（精確到17位有效數(shù)字）> > sym x;>> vpa(int(exp(xA2),x,0,1/2),17)ans =0.54498710418362222

4、4)dxx225 4x2> > sym x;> > int(xA4/(25+xA2),x)ans =125*atan(x/5) - 25*x + *人3/35)求由參數(shù)方程Ix ="后F所確定的函數(shù)的一階導(dǎo)數(shù)dy與二階導(dǎo)數(shù) 嗎I y =arctantdxdx> > sym t;> > x=log(sqrt(1+tA2);y=atan；> > diff(y,t)/diff(x,t)ans =1/t6)設(shè)函數(shù)y=f(x)由方程xy +3，=3所確定，求y'(x)。> > syms x y;f=x*y+exp(

5、y)-exp(1);> > -diff(f,x)/diff(f,y)ans =-y/(x + exp(y), x7)e- sin 2xdxo> > syms x;> > y=exp(-x)*sin(2*x);> > int(y,0,inf)ans =2/58） 將。1十x在x=0展開（最（Wj次帚為 8）>> syms xf=sqrt(1+x);taylor(f,0,9)ans =-(429*xA8)/32768 + (33*xA7)/2048 - (21*xA6)/1024 + (7*xA5)/256 - (5*xA4)/128 +

6、 xA3/16 -xA2/8 + x/2 + 1isin(3)9) y =e x 求 y (2)> > syms x y;> > y=exp(sin(1/x);> > dy=subs(diff(y,3),x,2)dy =-0.5826 X2 .,、.，I、,.，10)求變上限函數(shù) Ja+tdt對(duì)變量x的導(dǎo)致。'x> > syms a t;> > diff(int(sqrt(a+t),t,x,xA2)Warning: Explicit integral could not be found.ans =2*x*(xA2 + a)A

7、(1/2) - (a + xF(1/2)x-3vz-1,、4、求點(diǎn)(1,1,4)到直線L: =-= 的距離-102>> M0=1,1,4;M1=3,0,1;M0M1=M1-M0;v=-1,0,2;d=norm(cross(M0M1,v)/norm(v)d =1.095411x J5、已知f(x) =-=e 2犬，分別在下列條件下畫出f(x)的圖形：(要求貼圖).2 二二(1)仃=1時(shí)，N=0,1 1 ,在同一坐標(biāo)系里作圖> > syms x;>> fplot('(1/sqrt(2*pi)*exp(-(x)A2)/2)',-3,3,'r

8、')> > hold on>> fplot('(1/sqrt(2*pi)*exp(-(x-1)A2)/2)',-3,3,'y')> > hold on>> fplot('(1/sqrt(2*pi)*exp(-(x+1)A2)/2)',-3,3,'g')> > hold offD4D35 03D25 02 015D.1門 JD133匡 0(2)卜=0時(shí),仃=1,2,4,在同一坐標(biāo)系里作圖。>> syms x;fplot('(1/sqrt(2*pi

9、)*exp(-(x)A2)/2)',-3,3,T) hold onfplot('(1/(sqrt(2*pi)*2)*exp(-(x)A2)/(2*2A2)',-3,3,y) hold onfplot('(1/(sqrt(2*pi)*4)*exp(-(x)A2)/(2*4A2)',-3,3,'g') hold off(要求貼圖)0 _t _200 _u _2x = u 由耐,y = u cosCt), z = t/46、畫下列函數(shù)的圖形：x =u sint(1) «y = u costt z =一 工4>> ezmes

10、h('u*sin(t)','u*cos(t)','t/4',0,20Q2)(2) z =sin(xy) 0 £x £3,0 My 三3>> x=0:0.1:3;y=x;X Y=meshgrid(x,y);Z=sin(X*Y);>> mesh(X,Y,Z)x =sint (3 cosu)(3) «y =cost (3+cosu)I|z =sin u0 _t _2二0_u _2二ezmesh('sin(t)*(3+cos(u)','cos(t)*(3+cos(u)'

11、,'sin(u)',0,2*pi,0,2*pi)x =p+c-Dsfu), y = ci&sfl) (3-hcos(u),: = wi"力口.5-7、已知A =4-3-201-24-3 ,在 MATLAB 命令窗口中建立 A、B矩陣并1對(duì)其進(jìn)行以下操作：(1)計(jì)算矩陣A的行列式的值det(A)>> A=4,-2,2;-3,0,5;1,5,3;>> det(A) ans =12 TB, AB ,A B,A2, AT-158(2)分別計(jì)算下列各式：2A - B, A* B,A.*>> A=4,-22-3Q5;1,5,3;B=1,

12、3,4;-2Q-3;2,-1,1;>> 2*A-Bans =7-70-40130115>> A*Bans =1210247-14-7-30-8>> A.*Bans =4-6860-152-53>> A*inv(B)ans =-0.0000-0.00002.0000-2.7143-8.0000-8.14292.42863.00002.2857>> inv(A)*Bans =0.48730.41141.00000.3671-0.43040.0000-0.10760.24680.0000>> A*Aans =2424-7319-

13、81336>> Aans =4-31-205253>>8、 在MATLAB中分別利用矩陣的初等變換及函數(shù)rank、函數(shù)inv求下列矩陣的秩:門(1) A= 3-6 3 2一5 4 0 ,求 rank(A)=?-11 2 4，>> A=1,-6,3,2;3,-5,4,0;-1,-11,2,4;>> rank(A)ans =(2) B =I1 20 10 02 00 2>> B=3,5,0,1;1,2,0,0;1,0,2,0;1,2,0,2>> inv(B)ans =2.0000-1.0000-1.00000-4.00002.

14、50002.0000-0.5000-0.00000.00000.50000-1.00000.50000.50000.50009、在MATLAB中判斷下列向量組是否線性相關(guān)，并找出向量組«1 =(1132)T,% =(11 13)T,ot3 =(5 289)T,o(4=(1317)T 中的一個(gè)最大線性無關(guān)組。>> a1=1 1 3 2'a2=-1 1 -1 3'a3=5 -2 8 9'a4=-1 3 1 7'A= a1, a2 ,a3 ,a4 ;R jb=rref(A)a1 =1132 a2 =-11-13 a3 =5-289 a4 =-13

15、171.0000001.09091.7879-0.060601.00000001.00000jb 二1>> A(:,jb)ans =113210、在-11-13MATLAB5-289中判斷下列方程組解的情況，若有多個(gè)解，寫出通解。X x1 f 4x3 -2x4 = 0 (1)- x1 X2 X3 +2人=0| 3x1 , X2 , 7 X3 - 2 X4 = 0x1 -3x2 -12x3 6x4 =0>> A=1,-1,4,2;1,-1,-1,2;3,1,7,-2;1,-3,-12,6;>> rank(A)ans =3>> rref(A)ans

16、=1000010-200100000>> A=1,-1,4,2;1,-1,-1,2;3,1,7,-2;1,-3,-12,6;>> format ratn=4;RA=rank(A)RA =3>> if(RA=n)fprintf('%方程只有零解)elseb=null(A,'r') endb =0201>> syms k X=k*bX =02*k0 k2x1 3x2 x3 =4x1 -2x2 4x3 = -5(2)3x1 8x2 -2x3 =134x1 -x2 9x3 = -6> > A=2 3 1;1 -2 4;

17、3 8 -2;4 -1 9;b=4 -5 13 -6'B=A b;> > n=3;> > RA=rank(A)RA =2>> RB=rank(B)RB =rref(B) ans =102-101-1200000000>> format ratif RA=RB&RA=n % 判斷有唯一解X=Abelseif RA=RB&RA<n % 判斷有無窮解X=Ab %求特解C=null(A,'r') %求AX=0的基礎(chǔ)解系 else X='equition no solve' % 判斷無解 end

18、Warning: Rank deficient, rank = 2, tol =8.9702e-015.X =03/2-1/2C =-211-2 1 1 、-._ 111、求矩陣A= 02 0的逆矩陣A及特征值和特征向量。T 1 3A=-2 1 1;0 2 0;-4 1 3;>> a1=inv(A)a1 =1/2-3/21/201/20-21/21>> P,R=eig(A)P =-985/1393-528/2177379/125700379/419-985/1393-2112/2177379/1257-100020002A的三個(gè)特征值是：r1=-1 , r2=2, r3

19、=2。三個(gè)特征值分別對(duì)應(yīng)的特征向量是P1=1 0 1；p2=1 0 4;p3=1 3 12 22-2 ,12、化方陣A= 25M為對(duì)角陣。2 Y 5>> A=2 2 -2;2 5 -4;-2 -4 5;P,D=eig(A)P =- 0.29810.89440.3333- 0.5963-0.44720.6667- 0.74540-0.6667D =1.00000001.000000010.0000>> B=inv(P)*A*PB =1.0000-0.00000.00000.00001.00000.0000-0.0000程序說明：010.0000所求得的特征值矩陣 D即為矩

20、陣A對(duì)角化后的角矩陣， D和A相似。22213、求一個(gè)正交變換，將二次型= 5x1 +5x2 +3x3 -2x1x2 +6x1x3 -6x2x3 化為標(biāo)準(zhǔn)型。>> A=5 -1 3;-1 5 -3;3 -3 3;>> syms y1 y2 y3 y=y1;y2;y3;P,D=eig(A)881/2158985/1393-780/1351-881/2158985/1393780/1351-780/1351-881/1079>>x=P*y(6A(1/2)*y1)/6 + (2A(1/2)*y2)/2 - (30(1/2)*丫3)/3(2A(1/2)*y2)/2

21、- (6A(1/2)*y1)/6 + (3A(1/2)*y3)/3-(3A(1/2)*y3)/3 - (2人(1/2)*3人(1/2)*丫1)/3>> f=y1 y2 y3*D*y-y1A2/2251799813685248 + 4*y2A2 + 9*y3A27xn 1=(xn -)/214、6位有效設(shè)«xn，數(shù)列 xn是否收斂？若收斂，其值為多少？精確到X =3數(shù)字。f=inline('(x+7/x)/2');>> x0=3;>> fo門=1:20x0=f(x0);fprintf('%g,%gn',i,x0);en

22、d1,2.666672,2.645833,2.645754,2.645755,2.645756,2.645757,2.645758,2.645759,2.6457510,2.6457511,2.6457512,2.6457513,2.6457514,2.6457515,2.6457516,2.6457517,2.6457518,2.6457519,2.6457520,2.64575該數(shù)列收斂于三，它的值是11115、設(shè)xn =1+下+二+.+ = , xn是否收斂？右收斂， 其值為多少？精確到 17位有效2 P3pnp數(shù)字。(注：學(xué)號(hào)為單號(hào)的取p = 7 ,學(xué)號(hào)為雙號(hào)的取p = 8 )>

23、>f=inline('1/(xA8)');x0=0;for i=1:20x0=(x0+f(i);fprintf('%g , %.16fn',i,x0);end1 , 1.00000000000000002 , 1.00390625000000003 , 1.00405866579027594,1.00407392457933845 ,1.00407648457933846,1.00407707995351927 , 1.00407725342004488 , 1.00407731302468969 , 1.004077336255262610 , 1.00

24、4077346255262611 , 1.004077350920336512 , 1.004077353246016813 , 1.004077354471911514,1.004077355149515015 , 1.004077355539699316,1.004077355772530017 , 1.004077355915883518 , 1.004077356006628119 , 1.004077356065508520 , 1.0040773561045711>>16、求二重極限limx 1y 0ln(x ey).x2 y2>> clear> &g

25、t; syms x y;> > f=(log(x+exp(y)/sqrt(xA2+yA2);> > 僅=limit(f,'x',1);>>僅丫=1而讓(僅'丫',0)fxy =Iog(2)17、已知 ex-xyz=0,求名。 Fx>> clearsyms x y z;>> F=exp(x)-x*y*z;>> Fx= diff(F, 'x')Fx =exp(x) - y*z>> Fz= diff(F, 'z')Fz =-x*y>> G=-

26、Fx/FzG =(exp(x) - y*z)/(x*y)22218、已知函數(shù) f(x, y,z)=x +2y +3z +xy+3x 3y 6z ,求梯度。一：>> clearsyms x y z;> > f=xA2+2*yA2+3*zA2+x*y+3*x-3*y-6*z;> > dxyz=jacobian(f)dxyz =2*x + y + 3, x + 4*y - 3, 6*z - 6二：> > clear> > syms x y z;> > f=xA2+2*yA2+3*zA2+x*y+3*x-3*y-6*z;>

27、> gr=jacobian(f)gr =2*x + y + 3, x + 4*y - 3, 6*z - 6119、計(jì)算積分I =口一(2xy)dxdy,其中D由直線y=x與y = x2圍成。 d 2>> A=int(int (2-x-y),'y',xA2,x),'x',0,1)/2A =11/120220、計(jì)算曲線積分fzyds,其中曲線 C:x = cost,y =sint,z = t tw0,2n。c x yclearsyms x y z tx=cos(t);y=sin(t);z=t;dx=diff(x,t);dy=diff(y,t);dz

28、=diff(z,t);ds=sqrt(dxA2+dyA2+dzA2);f=zA2/(xA2+yA2);I=int(f*ds,t,0,2*pi)I =(8*2人(1*piA3)/321、計(jì)算曲面積分f f(x + y + z)dS ,其中 S: z = Ja2 -x2 - y2。S> > clear> > syms x y z a;> > z=sqrt(aA2-xA2-yA2);> > f=x+y+z;>> I=int(int(f,'y',0,sqrt(aA2-xA2),'x',0,a)I=1/2*aA3

29、+1/4*aA3*pi+1/3*aA2*(aA2)A(1/2)+1/3*(-1/2-1/4*pi)*aA32x6.3322、求解二階微分方程：y -10y 9y -e2x, y(0) =-,y(0) = y o> > clear>> syms x y;>> d_equa='D2y-10*Dy+9*y=exp(2*x)'d_equa =D2y-10*Dy+9*y=exp(2*x)>> Condit= 'y(0)=6/7,Dy(0)=33/7'Condit =y(0)=6/7,Dy(0)=33/7 >> y

30、1=dsolve( d_equa , Condit , 'x')y1 =exp(9*x)/2 - exp(2*x)/7 + exp(x)/2 一二 1 ，一23、求數(shù)項(xiàng)級(jí)數(shù)I =£的和。nj(n 1)> > clear> > syms n;> > f=1/(n*(n+1);> > I=symsum(f,n,1,inf)I =11 一24、將函數(shù)f (x)=展開為(x -3)的帚級(jí)數(shù)。 x> > clear> > syms x;> > f=1/x;>> taylor(f,10

31、,x,3) ans =(x - 3)A2/27 - x/9 - (x - 3)A3/81 + (x - 3)A4/243 - (x - 3)A5/729 + (x - 3)A6/2187 - (x - 3)A7/6561 +(x - 3)人8/19683 - (x - 349/59049 + 2/325、能否找到一個(gè)分式線性函數(shù)axb ,使它產(chǎn)生的迭代序列收斂到給定的數(shù)？用這種辦法cx d近似計(jì)算2o>> f=inline('(2+xA2)/(2*x)');x1=2;for i=1:20x1=f(x1);fprintf('%g,%gn',i,x1);

32、end;1,1.5 2,1.416673,1.414224,1.414215,1.414216,1.414217,1.414218,1.414219,1.4142110,1.4142111,1.4142112,1.4142113,1.4142114,1.4142115,1.4142116,1.4142117,1.4142118,1.4142119,1.4142120,1.414212x,0 <x < 126、函數(shù)f(x)=«2的迭代是否會(huì)廣生混沌?12(1 -x), x _1 2>> x1=0:0.05:0.5;y1=2*x1;x2=0.5:0.05:1;y2

33、=2*(1-x2);figureplot(x1,y1,x2,y2)gtext('2*x,)gtext('2*(1-x)')27、函數(shù)f (x) =ax(1 x) (0EXE1)稱為Logistic映射，試從 蜘蛛網(wǎng)”圖觀察它取初值為x0 =0.5產(chǎn)生的迭代序列的收斂性，將觀察記錄填人下表，作出圖形。若出現(xiàn)循環(huán)，請(qǐng)指出它的周期。(要求貼圖)0.90.B Q7 HE 0.5 Q4 0.3 Q2 0.1IJ0 Cl1 D.2 D3D.5D.6 Q7 £1B0.91f=inline('3.3*x*(1-x)');x=linspace(1,202,202

34、);y=linspace(1,202,202);x(1)=0.5;y(1)=0;x(2)=x(1);y(2)=x(1);3工值for i=1:100x(1+2*i)=x(2*i);x(2+2*i)=f(x(1+2*i);y(1+2*i)=x(2+2*i);y(2+2*i)=y(1+2*i);endplot(x,y,'r');hold on;syms x y;y=x;ezplot(x,0,1);ezplot(f(x),0,1);axis(0,1,0,3.3/4);T=0.35hold offhold onf=inline('3.5*x*(1-x)');x=lins

35、pace(1,202,202);y=linspace(1,202,202);x(1)=0.5;y(1)=0;x(2)=x(1);y(2)=x(1);fo門=1:100x(1+2*i)=x(2*i);x(2+2*i尸f(x(1+2*i);y(1+2*i)=x(2+2*i);y(2+2*i)=y(1+2*i);endplot(x,y,'r');hold on;syms x y;y=x;4T=0.4ezplot(x,0,1);ezplot(f(x),0,1);axis(0,1,0,3.5/4);hold offhold onf=inline('3.56*x*(1-x)'

36、;);x=linspace(1,202,202);y=linspace(1,202,202);x(1)=0.5;y(1)=0;x(2)=x(1);y(2)=x(1);fo門=1:100x(1+2*i)=x(2*i);x(2+2*i)=f(x(1+2*i);y(1+2*i)=x(2+2*i);y(2+2*i)=y(1+2*i);endplot(x,y,'r');hold on;syms x y;y=x;ezplot(x,0,1);DI (1201 Q5口日 DSezplot(f(x),0,1);axis(0,1,0,3.56/4);hold offhold onf=inline(

37、'3.568*x*(1-x)');x=linspace(1,202,202);y=linspace(1,202,202);x(1)=0.5;y(1)=0;x(2)=x(1);y(2)=x(1);fo門=1:100x(1+2*i)=x(2*i);x(2+2*i)=f(x(1+2*i);y(1+2*i)=x(2+2*i);y(2+2*i)=y(1+2*i);endplot(x,y,'r');hold on;syms x y;y=x;ezplot(x,0,1);ezplot(f(x),0,1);axis(0,1,0,3.568/4);hold onf=inline(&

38、#39;3.6*x*(1-x)');x=linspace(1,202,202);y=linspace(1,202,202);x(1)=0.5;y(1)=0;x(2)=x(1);y(2)=x(1);fo門=1:100x(1+2*i)=x(2*i);x(2+2*i)=f(x(1+2*i);y(1+2*i)=x(2+2*i);y(2+2*i)=y(1+2*i);endplot(x,y,'r');hold on;syms x y;y=x;ezplot(x,0,1);ezplot(f(x),0,1);axis(0,1,0,3.6/4);hold offhold onf=inlin

39、e('3.84*x*(1-x)');x=linspace(1,202,202);y=linspace(1,202,202);x(1)=0.5;y(1)=0;x(2)=x(1);y(2)=x(1);fo門=1:100x(1+2*i)=x(2*i);x(2+2*i)=f(x(1+2*i);y(1+2*i)=x(2+2*i);y(2+2*i)=y(1+2*i);endplot(x,y,'r');hold on;syms x y;y=x;ezplot(x,0,1);ezplot(f(x),0,1);axis(0,1,0,3.84/4); hold off表 Logist

40、ic迭代的收斂性ot3.33.53.563.5683.63.84序列收斂情況不收斂不收不收斂不收斂不收斂不收斂I 斂 I 28、由函數(shù)f (x, y) =y 一sgnxj|bx _c|與g(x,y) =a _x構(gòu)成的二維迭代 Martin迭代。現(xiàn)觀 察其當(dāng)a =4.5 25 55 120, b =2, c = 400時(shí)取初值為(0,0)所得到的二維迭代散點(diǎn)圖有什 么變化。(要求貼圖)function Martin (a,b,c N)f=(x,y)(y-sign(x)*sqrt(abs(a*x-c);g=(x)(a-x);m=0;0;for n=1:Nm(:,n+1)=f(m(1,n),m(2,

41、n),g(m(1,n); endplot(m(1,:),m(2,:),'kx');axis equalMartin(4.52555120,2,-300,500)書上62頁(Xi(0),x20)t =(1,2)T ,求出平面映射(x1(n),x2n)T的通項(xiàng)，并畫出這些點(diǎn)的散點(diǎn)圖。A=4,2;1,3;t=;for i=1:20x=2*rand(2,1)-1;t(length(t)+1,1:2)=x;for j=1:40x=A*x;t(length(t)+1,1:2)=x;endendplot(t(:,1),t(:,2),'*')grid('on')

42、30、對(duì) A=4 1及隨機(jī)給出的.(0)(0).T，叱(Xi)，x2),觀察數(shù)列 jn).該數(shù)列有極限嗎？31、若該地區(qū)的天氣分為三種狀態(tài)：晴、陰、雨。對(duì)應(yīng)的轉(zhuǎn)移矩陣為:明天晴2=陰雨今天晴陰雨一3/41/21/41/81/41/21/81/41/4-且p(0) =(0.5,0.25.0.25)T ,試根據(jù)這些數(shù)據(jù)來求出若干天之后的天氣狀態(tài)，并找出其特點(diǎn)(取4位有效數(shù)字)。>> A1=3/4,1/2,1/4;1/8,1/4,1/2;1/8,1/4,1/4;p=0.5;0.25;0.25;fo門=1:20p(:,i+1)=A1*p(:,i);endPP =Columns 1 thro

43、ugh 70.50000.56250.59380.60350.60690.60810.60850.25000.25000.22660.22070.21850.21780.21750.25000.18750.17970.17580.17460.17410.1740Columns 8 through 140.60860.60870.60870.60870.60870.60870.60870.21740.21740.21740.21740.21740.21740.21740.17390.17390.17390.17390.17390.17390.1739Columns 15 through 210.60870.60870.60870.60870.60870.60870.60870.21740.21740.21740.21740.21740.21740.21740.17390.17390.17390.17390.17390.17390.173932、對(duì)于上例中的 A,求出矩陣 A的特征值與特征向量，并將特征向量與上例中的結(jié)論作對(duì)比。>> A=3/4 1/2 1/4;1/8 1/4 1/2;1/8 1/4 1/4;>> P,R=eig(A)P =