數(shù)學(xué)實(shí)驗-線性代數(shù)方程組的數(shù)值解

實(shí)驗5線性代數(shù)方程組的數(shù)值解法數(shù)學(xué)實(shí)驗報告實(shí)驗5線性代數(shù)方程組的數(shù)值解法數(shù)學(xué)實(shí)驗報告5線性代數(shù)方程組的數(shù)值解法1黃浩2011011743一、實(shí)驗?zāi)康腗ATLAB定性作初步分析；通過實(shí)例學(xué)習(xí)用線性代數(shù)方程組解決簡化的實(shí)際問題。二、實(shí)驗內(nèi)容《數(shù)學(xué)實(shí)驗》第二版（問題敘述：通過求解線性方程組??1??=??1,??2??=??2，理解條件數(shù)的意義和方程組性態(tài)對解的影響，其中??1是n階范德蒙矩陣，即1

??20

… ????10?? = 1

??2

… ????1

??=1+

0,1,…,?? 11 …

1 … … ??[1

??2??1

… ????1]??1]??2nb1，b2??1??2的行和。編程構(gòu)造??1（??2可直接用命令產(chǎn)生）和你能預(yù)先知道方程組??1??=和??2??=??2（用預(yù)先知道的解可驗證程序。n=5,7,9，…，計算??1和??2的條件數(shù)。為觀察他們是否病態(tài)，做以和??2的元素??1(n分別加擾動ε后求解；??1和??2不變，b1，b2b1(n),b2(n)分別加擾動εAbε10^-10,10^-8,10^-6。經(jīng)擾動得到的解記做‖????‖??‖模型轉(zhuǎn)換及實(shí)驗過程：小題.b1,b2??1,??2的行和，可知方程組??1??=和??2??=??2n1n=5matlab（.11,2x1,x2excel算而得）為：xx1e1x2e211.0000000000000800.0000000000000801.0000000000000200.00000000000002010.999999999999463-0.0000000000005370.999999999999558-0.00000000000044211.0000000000009700.0000000000009701.0000000000020100.00000000000201010.999999999999288-0.0000000000007120.999999999996863-0.00000000000313711.0000000000001800.0000000000001801.0000000000015600.000000000001560由上表可見，當(dāng)n=5時，所得的解都接近真值，誤差在10^-12的量級左右。（2）小題n=5,7,9,11,13,151和2c1c2（程序見四結(jié)果如下：n579111315c13.574E+058.739E+072.274E+106.518E+122.059E+158.291E+17c24.766E+054.754E+084.932E+115.227E+146.263E+173.675E+17??1和??2b)（.3，結(jié)果如下：(為使結(jié)果清晰簡潔，在此僅列出n=5,9,13的情況，n=7,11,15略去)ε=10^-10時：nxx1xx250.9999999284994020.9999999370003031.0000002510851691.0000012599939300.9999996704145830.9999943300273341.0000001916677021.0000088199574730.9999999583331440.999995590021263nxx1xx290.9999975548050940.9999829278035681.0000149244303261.0012291970545280.9999603223369260.9784890657946091.0000600154317051.1577467696177110.9999435079279020.4084498576816771.0000338876411232.2304238808731590.9999873483708230.4354941236522751.0000026878626961.8788737437097660.9999997511934060.780281609027228nxx1xx2131.0007784033869290.9999998767583270.9937743987274311.0000261185116641.0226872410459610.9987794790026500.9501840256039321.0233044427317511.0734106035353810.7678410435503330.9235081059886532.3669384594613431.0577906781685654.0983444092691000.96809962810653313.4983690949393491.01276967899037019.4039025417116410.99638464083603822.9640269119150101.00068723718260013.9700648285337990.9999212432336596.8528369055604181.0000041151939490.000188901053803=10^-8nxx1xx250.9999928500302660.9999937027771001.0000251082268421.0001259444579840.9999670418066980.9994332499390881.0000191665851081.0008816112058550.9999958333510870.999559194397072nxx1xx290.9997569218861830.9999249646886321.0014841958519581.0054025423430740.9960527298593050.9054555097425781.0059726416246391.6933262585415040.9943760878476931.5999734614530311.0033746909787466.4079447885959670.9987396985502615.3092689110300421.0002678327406924.8628176967655060.9999752006605240.034295576497007nxx1xx2130.9980046593334370.9999998767282631.0164113342717761.0000261243251180.9384398540328910.9987792205553051.1392634494355081.0233092230919600.7883771891288210.7677944673667701.2275767009039882.3672082173739580.8224007796560294.0993377962873331.10134460129783313.5007796714360190.95802934170017619.4078052094758501.01230286523370722.9681988664444180.99757692077678013.9728915924515431.0002879131499591.0002879131499590.9999843910790946.8539364954676760.000001889299579ε=10^-6時：nxx1xx250.9992852977929170.9993966095201831.0025097875871361.0120678095963350.9967055393606440.9456948568165061.0019158683868581.084474667174326n0.999583506872445xx10.957762666412834xx290.9757501420061600.9999223275065541.1480658167117691.0055924194987990.6062136691657740.9021326589030921.5958409910913531.7176938349075140.4389487716335661.6913518835480071.3366650248743926.5980119249310320.8742701611529635.5310139213940691.0267194461452354.998579957366803n0.997525977218788xx10.000355009332673xx2130.7071952366081760.9999998767279653.3904929638964631.0000261243832617.8992555052967080.99877921797036220.9772809601275481.02330927090318629.1195663172520300.76779400153262433.1329169063691252.36721091537023223.8745993777145214.09934773169287015.07902114649785113.5008037809264344.78290586997535419.4078442421848052.68118057389295422.9682405924371480.67162490808792313.9729198644592591.0386948092493446.8539474930660780.9979195655092280.000000018893025由上表可見：n=96n=1322ε。n<101.6n>1033（因此，我認(rèn)為第一版上的參考答案不夠全面）Ab大，隨ε的增加，解的偏差也有增加的趨勢。方程組的解表格在此略去，程序也類似，其具體影響及分析見下一小問。（3）小題A、b以通過計算誤差‖?????‖，來與條件數(shù)估計的誤差比較，來進(jìn)一步得出結(jié)論?！??‖2-norm（x）bA(n,n)以微小的擾動（程序見四.4（E1、E2分別為兩方程解的誤差‖?????‖）：‖??‖nε=10^10ε=10^8ε=10^6條件數(shù)E1E2E1E2E1E2Cond(A1)Cond(A2)52.07E075.12E062.07E055.12E042.07E034.90E023.57E+054.77E+0573.19E062.10E033.19E041.89E013.19E021.74E+008.74E+074.75E+0893.30E057.28E013.30E033.20E+003.30E013.31E+002.27E+104.93E+11112.19E045.94E+002.56E026.02E+002.55E+006.02E+006.52E+125.23E+14132.51E024.71E+001.25E014.71E+001.59E+014.71E+002.06E+156.26E+17152.31E+007.84E+001.79E+007.51E+006.19E+015.96E+008.29E+173.68E+17由上表可得出結(jié)論：第一，隨著階數(shù)的增加，誤差值有增加的趨勢第二，ε的變化對誤差大小也有影響，但不是絕對的，對于n=5,7,9,11,13，εn=15，找不到εE第三，實(shí)際算得的誤差遠(yuǎn)遠(yuǎn)小于條件數(shù)指出的誤差上限第四，當(dāng)n<10時，微小擾動對范德蒙矩陣的影響較小，但當(dāng)n>10時，微小n=15爾伯特陣類似。再保持??1和??2不變，b1，b2b1(n),b2(n)分別加擾動ε（程序見四.5）結(jié)果如下：E1E2E1E2E1E2Cond(A1)Cond(A2)52.07E-075.12E-062.07E-055.12E-042.07E-035.12E-023.57E+054.77E+0573.19E-062.10E-033.19E-042.10E-013.19E-022.10E+018.74E+074.75E+0893.33E-059.33E-013.30E-039.33E+013.30E-019.33E+032.27E+104.93E+11111.62E-044.31E+022.57E-024.31E+042.56E+004.31E+066.52E+125.23E+14133.55E-022.49E+041.31E-012.49E+061.59E+012.49E+082.06E+156.26E+17151.06E+013.73E+049.61E-013.73E+067.01E+013.73E+088.29E+173.68E+17nε=10^10εnε=10^10ε=10^8ε=10^6條件數(shù)第一，隨著階數(shù)的增加，誤差值有增加的趨勢第二，ε越n=5,7,9,11,13εεE第三，實(shí)際算得的誤差遠(yuǎn)遠(yuǎn)小于條件數(shù)指出的誤差上限n<10n>10bA是，最終范德蒙陣的病態(tài)程度也沒有趕上希爾伯特陣bA造成的結(jié)果偏差更大《數(shù)學(xué)實(shí)驗》第二版（問題敘述：已知方程組Ax=b，其中A∈R^(20×20)，定義為：試通過迭代法求解此方程組，認(rèn)識迭代法收斂的含義以及迭代初值和方程組系數(shù)矩陣性質(zhì)對收斂速度的影響。實(shí)驗要求：選取不同的初始向量x(0)b用雅可比迭代法和高斯-賽德爾迭代法計算，觀測得到的迭代向量序列是否均收斂？若收斂，記錄迭代次數(shù)，分析計算結(jié)果并得出你的結(jié)論；bx(0)A非主對角線元素不變，每次用雅可比迭代法計算，要求迭代誤差滿足‖x(k+1)x(k)‖ <比較收斂速度，分析現(xiàn)象并得出你的結(jié)論?！弈Ｐ娃D(zhuǎn)換與實(shí)驗過程：小題matlab設(shè)計雅克比迭代法和高斯-賽德爾迭代法的函數(shù)，設(shè)為JacDD1GSDD1（.6、四.7Abx0mtol，xJ（xG）k。首先，保持常數(shù)項向量b=(1,1,1,…,1)T不變，分別設(shè)定初始向量x(0)(0,0,0…,0)T(5,5,5,…,5)T(100,100,100,…,100)T（程序見四.8，所得結(jié)果如下：x(0)x(0)解向量(0,0,0,…,0)T(5,5,5,…,5)T(100,100,100,…,100)TxJxGxJxGxJxG0.4816350110.4816335080.4816371820.4816409430.4816366890.4816398830.5734131580.5734116110.5734168540.5734210570.5734160170.5734194990.6327989970.6327976730.632804280.6328083480.6328030890.6328062750.6520986790.6520977760.6521053730.6521088050.6521038710.6521062630.6609418470.6609414470.6609497970.6609523360.6609480230.6609494090.6643052030.6643053610.664314220.6643156670.664312220.6643125230.6657142460.6657150040.6657241290.6657243440.6657219490.6657211990.6662729030.6662742810.666283440.6662823630.6662811270.6662794310.6664962310.6664982130.6665072060.6665048770.6665048050.6665023220.6665735250.6665760480.6665847190.666581280.6665822740.666579190.6665735250.6665764810.6665847190.6665803960.6665822740.6665787840.6664962310.6664994830.6665072060.6665022740.6665048050.6665010990.6662729030.6662762940.666283440.6662781920.6662811270.6662773790.6657142460.6657176190.6657241290.665718850.6657219490.6657183160.6643052030.6643084110.664314220.664309170.664312220.6643088380.6609418470.6609447570.6609497970.66094520.6609480230.6609450050.6520986790.6521011780.6521053730.6521014210.6521038710.6521013130.6327989970.6328009950.632804280.6328011180.6328030890.6328010640.5734131580.5734145680.5734168540.5734146220.5734160170.5734145980.4816350110.4816358440.4816371820.4816358630.4816366890.481635855171119122415迭代次數(shù)由上表可見，隨著初值x(0)迭代次數(shù)但是二者的結(jié)果相近，都在誤差上限之內(nèi)。為了證明兩種求解方法的收斂性，我們以x(0)=(5,5,5,…,5)??時的高斯-賽德（程序見四似，在此略去：k=0k=1k=2k=3k=4k=5k=651.5833333330.8097993830.5867626890.5159863770.4929626410.48538750751.8472222220.9783629120.7052563650.6168153850.5877768120.57818087752.0231481481.084426440.7813237890.6819209230.6490964610.63821656752.0744598771.1206981240.8071642940.7035546140.6692025920.65779093752.0976723251.1372138040.8191053420.7135320420.6784443550.66677151852.1058170441.1436399750.8237874540.7173972820.6819885220.67019847552.1091088681.1463317770.8257794470.7190360460.683482730.67163852852.1103362321.1474109130.8265897910.7196988670.6840822740.67221365952.1108151111.1478512890.8269263590.7199736250.6843293130.67244801252.1109972041.1480285630.8270643090.7200860580.684429750.67252176652.111067461.1481001480.8271211030.7201323670.68445080.67243591352.1110943441.1481289050.8271443740.7201513680.6842499120.67203420152.1111046791.148140450.827153910.719918050.6832395860.67110569252.1111086421.1481450730.8271578080.7178314910.6805376480.66952080652.1111101641.1481469220.8242658810.7102614320.675688690.66701021152.1111107471.148147660.805458660.6951998230.6686267110.66268609152.1111109721.1134257320.7566388370.6735497460.6567218090.65311633152.1111110580.9629628860.6907658520.6445060140.6352482410.63333137351.6944444240.7185570680.600404170.5787075670.5745170960.57365217950.7916666590.5333397520.4909645160.4834934290.4820235360.481719644k=7k=8k=9k=10k=11k=120.4828815270.4820500490.4817736970.4816817550.4816511440.4816409430.5749989270.5739417540.573590110.5734730570.5734340650.5734210570.6346027310.6334008590.6330008350.632867610.6328231840.6328083480.6539954920.6527322070.6523115110.6521712730.6521244310.6521088050.6628857660.6615916860.6611604250.6610163940.6609682580.6609523360.6662714780.6649629750.6645261480.6643799480.6643313890.6643156670.6676920040.6663752910.6659342820.6657871530.6657392950.6657243440.6682561060.6669308290.666486750.666341410.666295940.6662823630.6684736390.6671370010.6666960450.6665576910.6665165820.6665048770.6685091180.6671722840.6667505750.666625920.6665908350.666581280.6683888310.6671001890.6667211670.6666159640.666587780.6665803960.6680849140.6669276810.6666119130.6665289910.6665076820.6665022740.6675508740.6666014440.6663582090.666297130.6662819480.6662781920.6666577650.6659471590.6657736310.6657315260.6657213240.665718850.6649462430.6644591990.6643443560.6643171770.6643107140.664309170.6613428570.6610368410.6609663530.6609499580.660946110.66094520.6523289070.6521531220.6521132330.6521040570.6521019220.6521014210.6329187230.6328276480.6328071450.6328024570.6328013720.6328011180.5734671370.5734264380.5734173010.5734152160.5734147340.5734146220.4816544160.4816400440.4816368120.4816360740.4816359030.481635863迭代次數(shù)增加k越來越小，是逐漸收斂的。然后，保持初始向量x(0)=(2,2,2,…,2)T不變，分別設(shè)定常數(shù)項向量b(0.01,0.01,0.01,…,0.01)T(1,1,1…,1)T(5,5,5…,5)T（xJ，b(0.01,0.01,0.01,…,0.01)T(1,1,1,…,1)Tb(0.01,0.01,0.01,…,0.01)T(1,1,1,…,1)T(5,5,5,…,5)T解向量kxJxGxJxGxJxG0.0048163670.0048164390.48163670.4816405622.4081724482.4081652010.005734160.0057342440.5734160320.5734205922.8670613492.8670550730.006328030.0063281150.6328031040.6328078783.1639886493.1639849690.0065210380.0065211130.6521038780.6521084383.2604853863.2604853060.006609480.0066095390.6609480160.6609521713.304699743.3047035850.0066431220.0066431640.6643121940.6643157723.321515273.3215231710.0066572190.0066572440.6657219010.665724723.3285594793.3285715020.0066628110.0066628190.6662810590.6662829453.3313520093.3313681320.0066650470.0066650420.6665047220.6665055613.3324681483.3324881660.0066658220.0066658060.6665821830.6665819633.3328543633.3328778060.0066658220.0066657970.6665821830.6665810013.3328543633.3328804820.0066650470.0066650170.6665047220.6665027633.3324681483.3324959760.0066628110.0066627780.6662810590.6662785563.3313520093.3313804540.0066572190.0066571860.6657219010.6657191013.3285594793.3285874180.0066431220.006643090.6643121940.6643093323.321515273.3215416260.006609480.0066094510.6609480160.6609452983.304699743.3047235330.0065210380.0065210140.6521038780.6521014763.2604853863.2605057490.006328030.0063280110.6328031040.6328011463.1639886493.1640049050.005734160.0057341460.5734160320.5734146342.8670613492.8670728070.0048163670.0048163590.48163670.4816358682.4081724482.40817921251518111510由上表可見，隨著b少，而且雅克比迭代法比高斯-賽德爾迭代法所需的迭代次數(shù)要多，但是二者的結(jié)果相近，都在誤差上限之內(nèi)。x=（0,0,0,…，0）T1042545綜上所述，我們可以得到以下幾點(diǎn)結(jié)論：x(0)迭代次數(shù)增多。這是因為迭代收斂時，迭代值逐次逼近真值，因而初值與真值越接近，就可以在較少迭代次數(shù)的情況下滿足誤差上限。在迭代收斂的情況下，當(dāng)b0迭代次數(shù)減少。迭代次數(shù)要多。因矩陣的譜半徑不超過（任一種）范數(shù)，可求得：ρ(Bj)≤‖????‖2=0.4893062,ρ (Bg)≤‖????‖2=0.3292286顯然，二者都是收斂的，而且后者的收斂速度快。小題我們固定常數(shù)項向量b=初始向量x(0)=(1,1,1,…,1)T，將A的主對角線元素??????解向量3,9,15,21,27,331,3,5,7,9,11（程序見四.10??????解向量39152127330.4816367890.1219637270.0703231230.0494345270.0381191020.0310207030.5734161790.1289623430.0727104420.0506266140.0388319270.0314944760.6328033050.1327656850.073965230.0512443130.039198370.0317368180.6521041190.1331791860.0740476840.0512734260.0392118290.0317441330.6609482860.1333090390.0740714350.0512814990.0392154770.0317460970.664312480.1333280570.0740736230.0512820480.0392156720.0317461930.6657221950.1333327940.0740740980.0512821620.0392157110.0317462150.6662813540.133333610.0740741530.0512821740.0392157140.0317462180.6665050150.1333337970.0740741640.0512821760.0392157140.0317462190.6665824740.1333338330.0740741660.0512821770.0392157140.031746220.6665824740.1333338330.0740741660.0512821770.0392157140.031746220.6665050150.1333337970.0740741640.0512821760.0392157140.0317462190.6662813540.133333610.0740741530.0512821740.0392157140.0317462180.6657221950.1333327940.0740740980.0512821620.0392157110.0317462150.664312480.1333280570.0740736230.0512820480.0392156720.0317461930.6609482860.1333090390.0740714350.0512814990.0392154770.0317460970.6521041190.1331791860.0740476840.0512734260.0392118290.0317441330.6328033050.1327656850.073965230.0512443130.039198370.0317368180.5734161790.1289623430.0727104420.0506266140.0388319270.0314944760.4816367890.1219637270.0703231230.0494345270.0381191020.031020703迭代迭代次數(shù)1687665由上表可得出結(jié)論幅度加快，但是減小的幅度也變得平緩?！稊?shù)學(xué)實(shí)驗》第二版（問題敘述：入產(chǎn)出關(guān)系、外部需求、初始投入等如下表所示：投入產(chǎn)出投入產(chǎn)出農(nóng)業(yè)制造業(yè)服務(wù)業(yè)外部需求總產(chǎn)出農(nóng)造業(yè)301045115200服務(wù)業(yè)2060070150初始投入3511075總投入100200150如果今年對農(nóng)業(yè)、制造業(yè)和服務(wù)業(yè)的外部需求分別為50，150，100元，問這三個部門的總產(chǎn)出分別應(yīng)為多少？如果三個部門的外部需求分別增加1增加多少？模型轉(zhuǎn)換及實(shí)驗過程：小題.設(shè)一定時期內(nèi)第i個部門的總產(chǎn)出為????，其中對第j個部門的投入為設(shè)消耗系數(shù)矩陣A=aij ，其中??????=??????÷則由題中表格，可作出三個部門間投入系數(shù)aij的表格：投入產(chǎn)出農(nóng)業(yè)制造業(yè)服務(wù)業(yè)農(nóng)業(yè)0.15 0.10.2制造業(yè)0.3 0.050.3服務(wù)業(yè)0.2 0.30設(shè)產(chǎn)出向量x= x1,x2,x3T，需求向量d= d1,d2,d3T，根據(jù)每個部門的總產(chǎn)出等于總投入，可以得到下列關(guān)系：x=Ax+d即：(I?A)x=dmatlab（程序見四.11果為：

d=(139.280,267.606,208.138)T50，150，100139.280、267.606、208.138小題.因為(IA)x=d，即x=(IA)?1d，因此：xx=(I?A)?1(d+J),J分別為(1,0,0)??(0,1,0)??(0,0,1)??即：總產(chǎn)出的增加額dx=xx?x=(I?A)?1J需求增加產(chǎn)出增加農(nóng)業(yè)制造業(yè)服務(wù)業(yè)matlab(I)x=J需求增加產(chǎn)出增加農(nóng)業(yè)制造業(yè)服務(wù)業(yè)農(nóng)業(yè)1.350.560.44制造業(yè)0.251.270.43服務(wù)業(yè)0.340.491.22得出結(jié)論：11.35、0.56、0.4410.25、1.27、0.4310.34、0.49、1.22三、實(shí)驗總結(jié)本次數(shù)學(xué)實(shí)驗是求解線性代數(shù)方程組的數(shù)值解。因素，在方程不合適時，收斂步數(shù)可能會達(dá)到上千步甚至更多。習(xí)中，通過更實(shí)際化的數(shù)模練習(xí)，相信會更加熟練地掌握這一技能。四、程序清單1.第一題（1）n=5時的高斯消元解n=5;k=0:n-1;x=1+0.1*k;fori=0:(n-1)A1(i+1,:)=x.^i;endA1=A1';A2=hilb(n);x1=A1\b1x2=A2\b22.第一題（2）輸出n=5,7,9,11,13,15的條件數(shù)forj=1:6n=3+2*j;A1=ones(n);k=0:n-1;x=1+0.1*k;fori=0:(n-1)A1(i+1,:)=x.^i;C

endA1=A1';A2=hilb(n);c1=cond(A1);c2=cond(A2);C(j,:)=[nc1c2];x（n=5,9,13的情況，n=7,11,15bAAb10^-10）clear;forj=1:3n=1+4*j;A1=ones(n);k=0:n-1;x=1+0.1*k;fori=0:(n-1)A1(i+1,:)=x.^i;

endA1=A1';A11=A1;A11(n,n)=A11(n,n)+10^-10;A2=hilb(n);A22=A2;A22(n,n)=A22(n,n)+10^-10;b1=A1*ones(n,1);b2=A2*ones(n,1);x11=A11\b1;x22=A22\b2;[x11x22]第一題（3）bAclear;forj=1:6n=3+2*j;A1=ones(n);k=0:n-1;x=1+0.1*k;fori=0:(n-1)A1(i+1,:)=x.^i;endA1=A1';