計(jì)算方法引論課后答案5777

精品文檔第一章誤差1.試舉例,說明什么是模型誤差,什么是方法誤差.A4r2計(jì)算其表面積,這個(gè)近似看為球體的過程產(chǎn)生解:例如,把地球近似看為一個(gè)標(biāo)準(zhǔn)球體,利用公式的誤差即為模型誤差.在計(jì)算過程中,要用到,我們利用無窮乘積公式計(jì)算的值:222...qq12其中q2,1q2q,n2,3,...n1n我們?nèi)∏?項(xiàng)的乘積作為的近似值,得3.141587725...這個(gè)去掉的無窮乘積公式中第9項(xiàng)后的部分產(chǎn)生的誤差就是方法誤差,也成為截?cái)嗾`差.2.按照四舍五入的原則,將下列各數(shù)舍成五位有效數(shù)字:816.95676.00001517.322501.23565193.182130.01523623解:816.966.000017.3231.235793.1820.0152363.下列各數(shù)是按照四舍五入原則得到的近似數(shù),它們各有幾位有效數(shù)字?81.8970.008136.320050.1800解:五位三位六位四位4.若1/4用0.25表示,問有多少位有效數(shù)字?解:兩位a5.若1.1062,b0.947,是經(jīng)過舍入后得到的近似值,問:ab,ab各有幾位有效數(shù)字?解:已知da110,db1103,422又ab0.2053210,隨意編輯精品文檔dabdadbdadb110411030.551031102,222所以ab有三位有效數(shù)字;因?yàn)閍b0.1047571410,dabbdaadb0.94711041.106211030.600451031102222所以ab有三位有效數(shù)字.1,1的計(jì)算值與真yy6.設(shè)y0.9863,y0.0062xx,是經(jīng)過舍入后作為,的近似值.求121212值的相對(duì)誤差限及yy與真值的相對(duì)誤差限.12xydx,xydx,dx=110-4,dx2110-4,解:已知22111221211041dxdxdrx20.98630.50104;dr11y1x1x1111041dxdxdrx20.00620.81102;dr22y2x2x22drxxdrxdrx0.501040.811020.82102.12127.正方形的邊長(zhǎng)約為100cm,應(yīng)該怎樣測(cè)量,才能使其面積的誤差不超過1cm2.解:設(shè)正方形面積為S,邊長(zhǎng)為a,則S=a.所以要使:dsda22ada1,則要求2da110.5102.所以邊長(zhǎng)的誤差不能超過0.5102cm.2a2008.用觀測(cè)恒星的方法求得某地維度為4502(讀到秒),試問:計(jì)算sin將有多大誤差?解:dsin1cosdcos4502.219.真空中自由落體運(yùn)動(dòng)距離s與時(shí)間的關(guān)系由公式s2gt2確定,g是重力加速度.現(xiàn)在假設(shè)g是準(zhǔn)確的,而對(duì)t的測(cè)量有0.1的誤差,證明t增加時(shí),距離的絕對(duì)誤差增加而相對(duì)誤差s隨意編輯精品文檔卻減小.1gtdt;dsgtdtgtdtdtds證明:因?yàn)?dsdgt2.ds與t成正比,與t成反比,212ssts2gt2所以當(dāng)dt固定的時(shí)候,t增加時(shí),距離的絕對(duì)誤差增加而相對(duì)誤差卻減小.x0,的相對(duì)誤差為,求lnx的絕對(duì)誤差.10.設(shè)x解:已知dxxdxx,所以lnx的絕對(duì)誤差dlnx.11.設(shè)的相對(duì)誤差為%,求xxn的相對(duì)誤差.解:dxnnxn1dxndxn%.xnxnx12.計(jì)算球的體積,為了使相對(duì)誤差限為1%,問度量半徑R時(shí)允許的相對(duì)誤差限如何?解:已知V4drRdRaRR3,設(shè),則要使得3drVdVVdlnVdlnR33dlnR3dlnR3drR3a1%,則a11%.3隨意編輯精品文檔第二章插值法與數(shù)值微分yx100,121,144三處的值是很容易求得的,試以這三個(gè)點(diǎn)建立yx的1.設(shè)x,在二次插值多項(xiàng)式,并用此多項(xiàng)式計(jì)算115的近似值,且給出誤差估計(jì).用其中的任意兩點(diǎn),構(gòu)造線性插值函數(shù),用得到的三個(gè)線性插值函數(shù),計(jì)算115的近似值,并分析其結(jié)果不同的原因.解:已知x100,x121,x144;y10,y11,y12,012012建立二次Lagrange插值函數(shù)可得:x121x144x100x144Lx11100121100144101211001211442x121x10012144121144100115L11510.7228.所以誤差2f,所以xxxxxx,x,x,x,012012Rx23!0.0006555R1150.0016312利用前兩個(gè)節(jié)點(diǎn)建立線性插值函數(shù)可得:x121x100Lx10121100111001211隨意編輯精品文檔115L11510.7143.所以所以所以1利用后兩個(gè)節(jié)點(diǎn)建立線性插值可得:x144x12112144121Lx111211441115L11510.7391.1利用前后兩個(gè)節(jié)點(diǎn)建立線性插值可得:x144x10012144100Lx101001442115L11510.6818.1與115的真實(shí)值比較,二次插值比線性插值效果好,利用前兩個(gè)節(jié)點(diǎn)的線性插值比其他兩個(gè)線性插值效果好.此說明,二次插值比線性插值效果好,內(nèi)插比外插效果好.2.利用(2.9)式證明2xxRxmaxfx,xxx01108xxx01證明:由(2.9)式Rxfxxxx,xx01012!當(dāng)xxx時(shí),01f12maxfxmaxxxxxxx10,401xxxxxx0101所以2xxRxmaxfx,xxx01108xxx010,1,...,3.若證明n為互異節(jié)點(diǎn),且有xjxx...xxj1xx...xxlx0j1nxx...xxj1xx...xxjj0jj1jjn隨意編輯精品文檔nxlxxk,k0,1,...,nkjjj0證明:由于1ij;0ij.lxjiijnk0,1,...,n,所以且xlx和xk都為k次多項(xiàng)式,而且在k+1個(gè)不同的節(jié)點(diǎn)處的函數(shù)值都相同jjkj0馬上有nxlxxk,k0,1,...,n.kjjj04.設(shè)給出sinx在,上的數(shù)值表,用二次插值進(jìn)行計(jì)算,若希望截?cái)嗾`差小于105,問函數(shù)表的步長(zhǎng)最大能取多少?解:記插值函數(shù)為p(x),則sinsinxpxxxxxxxi1i13!imaxsinxpcosxxxxxx所以3!i1ii1xcos1;令gxxxxxxx,設(shè)xxth,得i1ii1i1gxthhtt1t2,t0,23i13ttt1t2,t0,2的最大值為10.3849,所以有又30.38496maxsinxph3105xh0.0538.所以5.用拉格朗日插值和牛頓插值找經(jīng)過點(diǎn)3,1,0,2,3,2,6,10的三次插值公式.隨意編輯精品文檔解:Lagrange插值函數(shù):xxxxxxxxxxxx3Lxyy1123xxxxxx02xxxxxx13300102031012xxxxxxxxxxxx012yy3013xxxxxxxxxxxx2220212330313xx3x10x3x3x1016227x3xx3.x3xx102781/5牛頓插值:首先計(jì)算差商31021321.3330.388961040.88890.1420Nx1x30.3889x3x1.1420x3xx3.3也可以利用等距節(jié)點(diǎn)構(gòu)造,首先計(jì)算差分310233247610121623可得前插公式723Nxth13ttt1tt1t2;2630和后插公式162326Nxth1012ttt1tt1t2.33,使6.確定一次數(shù)不高于4的多項(xiàng)式x00,00,111,21.解:利用重節(jié)點(diǎn)計(jì)算差商0000011111110121011/21/4隨意編輯精品文檔則可構(gòu)造Hermite插值函數(shù)滿足題設(shè)條件:Hx00x01x0x01x0x0x141x0x0x1x1414x4x394x2.321個(gè)點(diǎn)x,滿足條件:H2n17.尋找過n,,...,的2n1次多項(xiàng)式xxx01nxfx,Hxfx,Hxfx,...,H2n1002n1112n1nnfx.Hx0fx,Hxfx,..,Hx2n102n1112n1nn解:和Lagrange插值函數(shù)的構(gòu)造類似,可將插值函數(shù)寫成hxyhxyn,in,ixnH2n1iii0其中,基函數(shù)滿足條件(1)hx,hxP2n1;,n,ini(2)hx,hx0;hx,hx0,ni,nin,ijijn,ijjijj則可由已知條件,可得2hx12lxxxlx;n,in,iiin,ihxxxlx2.n,iin,i所以可得xniH12lxxxlxyxxlxy222n1n,iiin,iiin,ii08.過0,1兩點(diǎn)構(gòu)造一個(gè)三次Hermite插值多項(xiàng)式,滿足條件:11f01,f0,f12,f122解:計(jì)算重節(jié)點(diǎn)的差商01011/21211/2121/2-1/21隨意編輯精品文檔馬上可得11Hx1x0x0x01x0x0x1223x332x212x19.過給定數(shù)組757677787980x2.7682.8332.9032.9793.062y3.153(1)作一分段線性插值函數(shù).(2)取第二類邊界條件,作三次樣條插值多項(xiàng)式.(3)用兩種插值函數(shù)分別計(jì)算x75.5,x78.3的函數(shù)值.解:(1)做分段線性插值函數(shù)可得:Ix2.768lx2.833lx2.903lx2.979lx3.062lx3.153lx5012345x75x75,7675,76;xx76lx0lx77xx76,77;其中,0x75,76.10x75,77.x76x76,77x77x77,78xx78,79;lx78xx77,78;lx79230x77,79.0x76,78.x78x78,7979,80;xx80lx80xx79,80;lx0x79,80.450x78,80.(2)把已知節(jié)點(diǎn)值帶入M關(guān)系式可得:隨意編輯精品文檔1M2M12M0.015201212M2M12M0.01812312M2M12M0.014234112M2MM0.0165234MM0,所以上面方程組變?yōu)榭汕蠼夥匠探M由邊界條件可得052M12M0.0151212M2M12M0.01812312M2M12M0.01423412M2M0.01634解得M0.0058,M0.0067,M0.0036,M0.0071.1234所以可得在每個(gè)區(qū)間上的三次樣條函數(shù)的表達(dá)式:xxyMMMMxxyxxj133sxj16xxj16jj66j1j1jjj(3)當(dāng)x75.5時(shí),I75.52.768l75.52.833l75.52.8005;501s75.50.00580.005875.5752.799675.5762.7687675.52.83336當(dāng)x78.3時(shí),I75.52.979l78.33.062l78.33.0039;534s78.30.00367978.330.007178.3783660.00360.007178.3783.0034.2.9797978.33.0626610.若給出sinx,cosx,tanx的函數(shù)表:cosxsinxtanxx隨意編輯精品文檔0.9999920.00379683263.411251.5671.5681.5691.5700.9999960.002796357.61106556.690981255.76559130.9999980.001796430.9999990.00079673用表上的數(shù)據(jù)和任一插值公式求:(1)用tanx表格直接計(jì)算tan1.5695.(2)用sin1.5695和cos1.5695來計(jì)算tan1.5695.并討論這兩個(gè)結(jié)果中誤差變化的原因.解:利用Lagrange插值直接用tan表計(jì)算得tan1.5695819.0342874999274;利用Lagrange插值計(jì)算sin得sin1.56950.99999917500000;利用Lagrange插值計(jì)算cos得cos1.56950.00129630000000;最后利用sin/cos計(jì)算tan得tan1.5695771.4257309264500.出現(xiàn)小除數(shù),誤差被放大.11.求三次樣條函數(shù)sx,已知0.250.500000.300.390.450.53xi0.54770.62450.67080.728yi和邊界條件隨意編輯精品文檔s0.251.0000,s0.530.6868解:把表中數(shù)據(jù)帶入M關(guān)系式可得59M2MM4.314314140