MATLAB作業(yè)2參考答案(2018).doc

MATLAB作業(yè)二參考答案1、試求下面線性代數(shù)方程的解析解與數(shù)值解，并檢驗解的正確性。【求解】求出A, A;B 兩個矩陣的秩，可見二者相同，所以方程不是矛盾方程，應該有無窮多解。 A=2,-9,3,-2,-1; 10,-1,10,5,0; 8,-2,-4,-6,3; -5,-6,-6,-8,-4;B=-1,-4,0; -3,-8,-4; 0,3,3; 9,-5,3;rank(A), rank(A B)ans =4 4用下面的語句可以求出方程的解析解，并可以驗證該解沒有誤差。 x0=null(sym(A);x_analytical=sym(A)B; syms a;x=a*x0 x0 x0+x_analyticalx = a+967/1535, a-943/1535, a-159/1535 -1535/1524*a, -1535/1524*a, -1535/1524*a -3659/1524*a-1807/1535,-3659/1524*a-257/1535,-3659/1524*a-141/1535 1321/508*a+759/1535, 1321/508*a-56/1535, 1321/508*a-628/1535 -170/127*a-694/307, -170/127*a+719/307, -170/127*a+103/307 A*x-Bans = 0, 0, 0 0, 0, 0 0, 0, 0 0, 0, 0用數(shù)值解方法也可以求出方程的解，但會存在誤差，且與任意常數(shù)a 的值有關。 x0=null(A); x_numerical=AB; syms a;x=a*x0 x0 x0+x_numerical; vpa(x,10)ans = .2474402553*a+.1396556436, .2474402553*a-.6840666849, .2474402553*a-.1418420333-.2492262414*a+.4938507789,-.2492262414*a+.7023776988e-1,-.2492262414*a+.3853511888e-1 -.5940839201*a, -.5940839201*a, -.5940839201*a .6434420813*a-.7805411315, .6434420813*a-.2178190763,.6434420813*a-.5086089095-.3312192394*a-1.604263460, -.3312192394*a+2.435364854, -.3312192394*a+.3867176824 A*x-Bans = 1/18014398509481984*a, 1/18014398509481984*a, 1/18014398509481984*a -5/4503599627370496*a, -5/4503599627370496*a, -5/4503599627370496*a -25/18014398509481984*a, -25/18014398509481984*a, -25/18014398509481984*a 13/18014398509481984*a, 13/18014398509481984*a, 13/18014398509481984*a2、求解下面的聯(lián)立方程，并檢驗得出的高精度數(shù)值解（準解析解）的精度。【求解】給出的方程可以由下面的語句直接求解，經(jīng)檢驗可見，精度是相當高的。 x1,x2=solve(x12-x2-1=0,(x1-2)2+(x2-0.5)2-1=0,x1,x2)x1 =-1.3068444845633173592407431426632-1.2136904451605911320167045558746*sqrt(-1)-1.3068444845633173592407431426632+1.2136904451605911320167045558746*sqrt(-1)1.06734608580668971340859731280701.5463428833199450050728889725194x2 =-.76520198984055124633885565606586+3.1722093284506318231178646481143*sqrt(-1)-.76520198984055124633885565606586-3.1722093284506318231178646481143*sqrt(-1).139227666886861440483624988051411.3911763127942410521940863240803 norm(double(x1.2-x2-1 (x1-2).2+(x2-0.5).2-1)ans =6.058152713871457e-0313、用Jacobi、Gauss-Seidel迭代法求解方程組 ，給定初值為?！厩蠼狻浚壕帉慗acobi、Gauss-Seidel函數(shù)計算，function y=jacobi(a,b,x0)D=diag(diag(a); U=-triu(a,1); L=-tril(a,-1);B=D(L+U); f=Db;y=B*x0+f;n=1;while norm(y-x0)=1.0e-6 x0=y; y=B*x0+f; n=n+1;endn a=10,-1,-2;-1,10,-2;-1,-1,5;b=72;83;42; jacobi(a,b,0;0;0)n = 17ans = 11.0000 12.0000 13.0000function y=seidel(a,b,x0)D=diag(diag(a);U=-triu(a,1);L=-tril(a,-1);G=(D-L)U ;f=(D-L)b;y=G*x0+f; n=1;while norm(y-x0)=1.0e-6 x0=y; y=G*x0+f; n=n+1;endn seidel(a,b,0;0;0)n = 10ans = 11.0000 12.0000 13.00004555、假設已知一組數(shù)據(jù)，試用插值方法繪制出區(qū)間內的光滑函數(shù)曲線，比較各種插值算法的優(yōu)劣。-2-1.7-1.4-1.1-0.8-0.5-0.20.10.40.711.3.10289.11741.13158.14483.15656.16622.17332.1775.17853.17635.17109.163021.61.92.22.52.83.13.43.744.34.64.9.15255.1402.12655.11219.09768.08353.07015.05786.04687.03729.02914.02236【求解】用下面的語句可以立即得出給定樣本點數(shù)據(jù)的三次插值與樣條插值，得出的結果如，可見，用兩種插值方法對此例得出的結果幾乎一致，效果均很理想。 x=-2,-1.7,-1.4,-1.1,-0.8,-0.5,-0.2,0.1,0.4,0.7,1,1.3,.1.6,1.9,2.2,2.5,2.8,3.1,3.4,3.7,4,4.3,4.6,4.9;y=0.10289,0.11741,0.13158,0.14483,0.15656,0.16622,0.17332,.0.1775,0.17853,0.17635,0.17109,0.16302,0.15255,0.1402,.0.12655,0.11219,0.09768,0.08353,0.07019,0.05786,0.04687,.0.03729,0.02914,0.02236;x0=-2:0.02:4.9;y1=interp1(x,y,x0,cubic);y2=interp1(x,y,x0,spline);plot(x0,y1,:,x0,y2,x,y,o)5、 假設已知實測數(shù)據(jù)由下表給出，試對在(0.1,0.1)(1.1,1.1)區(qū)域內的點進行插值，并用三維曲面的方式繪制出插值結果。00.10.20.30.40.50.60.70.80.911.10.1.83041.82727.82406.82098.81824.8161.81481.81463.81579.81853.823040.2.83172.83249.83584.84201.85125.86376.87975.89935.92263.94959.98010.3.83587.84345.85631.87466.89867.9284.963771.00451.05021.11.15290.4.84286.86013.88537.91865.959851.00861.06421.12531.19041.2571.32220.5.85268.88251.92286.973461.03361.10191.17641.2541.33081.40171.46050.6.86532.91049.968471.03831.1181.20461.29371.37931.45391.50861.53350.7.88078.943961.02171.11181.21021.3111.40631.48591.53771.54841.50520.8.89904.982761.0821.19221.30611.41381.50211.55551.55731.49151.3460.9.920061.02661.14821.27681.40051.50341.56611.56781.48891.31561.04541.943811.07521.21911.36241.48661.56841.58211.50321.3151.0155.624771.1.970231.12791.29291.44481.55641.59641.53411.34731.0321.61268.14763【求解】直接采用插值方法可以解決該問題，得出的插值曲面。 x,y=meshgrid(0.1:0.1:1.1);z=0.83041,0.82727,0.82406,0.82098,0.81824,0.8161,0.81481,0.81463,0.81579,0.81853,0.82304;0.83172,0.83249,0.83584,0.84201,0.85125,0.86376,0.87975,0.89935,0.92263,0.94959,0.9801;0.83587,0.84345,0.85631,0.87466,0.89867,0.9284,0.96377,1.0045,1.0502,1.1,1.1529;0.84286,0.86013,0.88537,0.91865,0.95985,1.0086,1.0642,1.1253,1.1904,1.257,1.3222;0.85268,0.88251,0.92286,0.97346,1.0336,1.1019,1.1764,1.254,1.3308,1.4017,1.4605;0.86532,0.91049,0.96847,1.0383,1.118,1.2046,1.2937,1.3793,1.4539,1.5086,1.5335;0.88078,0.94396,1.0217,1.1118,1.2102,1.311,1.4063,1.4859,1.5377,1.5484,1.5052;0.89904,0.98276,1.082,1.1922,1.3061,1.4138,1.5021,1.5555,1.5573,1.4915,1.346;0.92006,1.0266,1.1482,1.2768,1.4005,1.5034,1.5661,1.5678,1.4889,1.3156,1.0454;0.94381,1.0752,1.2191,1.3624,1.4866,1.5684,1.5821,1.5032,1.315,1.0155,0.62477;0.97023,1.1279,1.2929,1.4448,1.5564,1.5964,1.5341,1.3473,1.0321,0.61268,0.14763;x1,y1=meshgrid(0.1:0.02:1.1);z1=interp2(x,y,z,x1,y1,spline);surf(x1,y1,z1)axis(0.1,1.1,0.1,1.1,min(z1(:),max(z1(:)其實，若光需要插值曲面而不追求插值數(shù)值的話，完全可以直接采用MATLAB 下的shading interp 命令來實現(xiàn)?？梢?，這樣的插值方法更好，得出的插值曲面更光滑。 surf(x,y,z); shading interp6、習題4和5給出的數(shù)據(jù)分別為一元數(shù)據(jù)和二元數(shù)據(jù)，試用分段三次樣條函數(shù)和B樣條函數(shù)對其進行擬合?！厩蠼狻肯瓤紤]習題4，相應的三次樣條插值和B-樣條插值原函數(shù)與導數(shù)函數(shù)分別為： x=-2,-1.7,-1.4,-1.1,-0.8,-0.5,-0.2,0.1,0.4,0.7,1,1.3,.1.6,1.9,2.2,2.5,2.8,3.1,3.4,3.7,4,4.3,4.6,4.9;y=0.10289,0.11741,0.13158,0.14483,0.15656,0.16622,0.17332,.0.1775,0.17853,0.17635,0.17109,0.16302,0.15255,0.1402,.0.12655,0.11219,0.09768,0.08353,0.07019,0.05786,0.04687,.0.03729,0.02914,0.02236;S=csapi(x,y); S1=spapi(6,x,y);fnplt(S); hold on; fnplt(S1) Sd=fnder(S); Sd1=fnder(S1);fnplt(Sd), hold on; fnplt(Sd1)再考慮習題5中的數(shù)據(jù)，原始數(shù)據(jù)不能直接用于樣條處理，因為meshgrid() 函數(shù)產(chǎn)生的數(shù)據(jù)格式與要求的ndgrid() 函數(shù)不一致，所以需要對數(shù)據(jù)進行處理，其中需要的x 和y 均應該是向量，而z 是原來z 矩陣的轉置，所以用下面的語句可以建立起三次樣條和B-樣條的插值模型，函數(shù)的表面圖所示，可見二者得出的結果很接近。 x,y=meshgrid(0:0.1:1.1);z=0.83041,0.82727,0.82406,0.82098,0.81824,0.8161,0.81481,0.81463,0.81579,0.81853,0.82304;0.83172,0.83249,0.83584,0.84201,0.85125,0.86376,0.87975,0.89935,0.92263,0.94959,0.9801;0.83587,0.84345,0.85631,0.87466,0.89867,0.9284,0.96377,1.0045,1.0502,1.1,1.1529;0.84286,0.86013,0.88537,0.91865,0.95985,1.0086,1.0642,1.1253,1.1904,1.257,1.3222;0.85268,0.88251,0.92286,0.97346,1.0336,1.1019,1.1764,1.254,1.3308,1.4017,1.4605;0.86532,0.91049,0.96847,1.0383,1.118,1.2046,1.2937,1.3793,1.4539,1.5086,1.5335;0.88078,0.94396,1.0217,1.1118,1.2102,1.311,1.4063,1.4859,1.5377,1.5484,1.5052;0.89904,0.98276,1.082,1.1922,1.3061,1.4138,1.5021,1.5555,1.5573,1.4915,1.346;0.92006,1.0266,1.1482,1.2768,1.4005,1.5034,1.5661,1.5678,1.4889,1.3156,1.0454;0.94381,1.0752,1.2191,1.3624,1.4866,1.5684,1.5821,1.5032,1.315,1.0155,0.62477;0.97023,1.1279,1.2929,1.4448,1.5564,1.5964,1.5341,1.3473,1.0321,0.61268,0.14763; x0=0.0:0.1:1; y0=x0; z=z;S=csapi(x0,y0,z); fnplt(S)figure; S1=spapi(5,5,x0,y0,z); fnplt(S1) S1x=fnder(S1,0,1); fnplt(S1x)figure; S1y=fnder(S1,0,1); fnplt(S1y)7、 重新考慮習題4中給出的數(shù)據(jù)，試考慮用多項式插值的方法對其數(shù)據(jù)進行逼近，并選擇一個能較好擬合原數(shù)據(jù)的多項式階次。 【求解】可以選擇不同的多項式階次，例如選擇3,5,7,9,11，則可以對其進行多項式擬合，并繪制出曲線。 x=-2,-1.7,-1.4,-1.1,-0.8,-0.5,-0.2,0.1,0.4,0.7,1,1.3,.1.6,1.9,2.2,2.5,2.8,3.1,3.4,3.7,4,4.3,4.6,4.9;y=0.10289,0.11741,0.13158,0.14483,0.15656,0.16622,0.17332,.0.1775,0.17853,0.17635,0.17109,0.16302,0.15255,0.1402,.0.12655,0.11219,0.09768,0.08353,0.07019,0.05786,0.04687,.0.03729,0.02914,0.02236;x0=-2:0.02:4.9;p3=polyfit(x,y,3); y3=polyval(p3,x0);p5=polyfit(x,y,5); y5=polyval(p5,x0);p7=polyfit(x,y,7); y7=polyval(p7,x0);p9=polyfit(x,y,9); y9=polyval(p9,x0);p11=polyfit(x,y,11); y11=polyval(p11,x0);plot(x0,y3; y5; y7; y9; y11)從擬合的結果可以發(fā)現(xiàn)，選擇5 次多項式就能較好地擬合原始數(shù)據(jù)。8、 假設習題4中給出的數(shù)據(jù)滿足原型，試用最小二乘法求出的值，并用得出的函數(shù)將函數(shù)曲線繪制出來，觀察擬合效果。 【求解】令則可以將原型函數(shù)寫成這時可以寫出原型函數(shù)為 f=inline(exp(-(x-a(1).2/2/a(2)2)/(sqrt(2*pi)*a(2),a,x);由原型函數(shù)則可以用下面的語句擬合出待定參數(shù)。這樣，擬合曲線得出的擬合效果是滿意的。 x=-2,-1.7,-1.4,-1.1,-0.8,-0.5,-0.2,0.1,0.4,0.7,1,1.3,.1.6,1.9,2.2,2.5,2.8,3.1,3.4,3.7,4,4.3,4.6,4.9;y=0.10289,0.11741,0.13158,0.14483,0.15656,0.16622,0.17332,.0.1775,0.17853,0.17635,0.17109,0.16302,0.15255,0.1402,.0.12655,0.11219,0.09768,0.08353,0.07019,0.05786,0.04687,.0.03729,0.02914,0.02236;a=lsqcurvefit(f,1,1,x,y)a =0.34605753554886 2.23400202798747 x0=-2:0.02:5; y0=f(a,x0);plot(x0,y0,x,y,o)9、 假設習題5中數(shù)據(jù)的原型函數(shù)為，試用最小二乘方法識別出的數(shù)值。 【求解】用下面的語句可以用最小二乘的得出 x,y=meshgrid(0.1:0.1:1.1);z=0.83041,0.82727,0.82406,0.82098,0.81824,0.8161,0.81481,0.81463,0.81579,0.81853,0.82304;0.83172,0.83249,0.83584,0.84201,0.85125,0.86376,0.87975,0.89935,0.92263,0.94959,0.9801;0.83587,0.84345,0.85631,0.87466,0.89867,0.9284,0.96377,1.0045,1.0502,1.1,1.1529;0.84286,0.86013,0.88537,0.91865,0.95985,1.0086,1.0642,1.1253,1.1904,1.257,1.3222;0.85268,0.88251,0.92286,0.97346,1.0336,1.1019,1.1764,1.254,1.3308,1.4017,1.4605;0.86532,0.91049,0.96847,1.0383,1.118,1.2046,1.2937,1.3793,1.4539,1.5086,1.5335;0.88078,0.94396,1.0217,1.1118,1.2102,1.