電大計算方法試題參考小抄參考

1、計算方法試題參考一計算及推導（5*8）1已知，試確定近似的有效數(shù)字位數(shù)。2有效數(shù)，試確定的相對誤差限。3已知，試計算差商4給出擬合三點和的直線方程。5推導中矩形求積公式6試證明插值型求積公式的代數(shù)精確度至少是n次。7已知非線性方程在區(qū)間內(nèi)有一實根，試寫出該實根的牛頓迭代公式。8用三角分解法求解線性方程組二給出下列函數(shù)值表89420.479430.564640.644220.71736要用二次插值多項式計算的近似值，試選擇合適的插值節(jié)點進行計算，并說明所選用節(jié)點依據(jù)。（保留5位有效數(shù)字）（12分）三 已知方程在內(nèi)有一實根（1）給出求該實根的一個迭代公式，試之對

2、任意的初始近似迭代法都收斂，并證明其收斂性。（2）試用構造的迭代公式計算的近似值，要求。四 設有方程組（1） 當參數(shù)a滿足什么條件時，雅可比方法對任意的初始向量都收斂。（2） 寫出與雅可比方法對應的高斯賽德爾迭代公式。（12分）五用歐拉預估校正法求解初值問題取h=0.1，小數(shù)點后保留5位。（8分）六證明求解初值問題 的如下單步法是二階方法。（10分）七試證明復化梯形求積公式對任意多的積分節(jié)點數(shù)n+1,該公式都是數(shù)值穩(wěn)定的。（6分）2003-2004第一學期一填空（3*5）1近似數(shù)關于真值有-位有效數(shù)字。2的相對誤差為的相對誤差的-倍。3設可微，求根的牛頓迭代公式-。4插值型求積公式的代數(shù)精確度

3、至少是-次。5擬合三點和的常函數(shù)是-。二已知有如下的數(shù)據(jù)12324123試寫出滿足插值條件以及的插值多項式，并寫出誤差的表達形式。三（1）用復化辛浦森公式計算為了使所得的近似值有6位有效數(shù)字，問需要被積函數(shù)在多少個點上的函數(shù)值？ （2）取7個等距節(jié)點（包括端點）用復化辛浦森公式計算，小數(shù)點后至少保留4位。四曲線與在點（0.7，0.3）附近有一個交點，試用牛頓迭代公式計算的近似值，要求五 用雅可比方法解方程組是否對任意的初始向量都收斂，為什么？取，求出解向量的近似向量，要求滿足。六用校正一次的歐拉預估校正格式求解初值問題的解函數(shù)在處的近似值，要求寫出計算格式。（步長,小數(shù)點后保留5位有效數(shù)字）七

4、設有求解初值問題的如下格式如假設問常數(shù)為多少時使得該格式為二階格式？ 2005-2006第二學期一填空（3*5）1 設近似數(shù)都是四舍五入得到的，則相對誤差-。2 矛盾方程組的最小二乘解為-。3 近似數(shù)關于真值有幾位有效數(shù)字4 取，迭代過程是否穩(wěn)定？5 求積公式有幾次的代數(shù)精確度？二 取初值，用牛頓迭代法求的近似值，要求先論證收斂性。當時停止迭代。三用最小二乘法確定中的常數(shù)a和b，使該曲線擬合于下面的四個點（1，1.01）（2，7.04）（3，17.67）（4，31.74）（計算結果保留到小數(shù)點后4位）四用乘冪法求矩陣a的按模最大的特征值的第k次近似值及相應的特征向量，要求取初值且這里 a=五考

5、察用高斯賽德爾迭代法解方程組收斂性，并取，求近似解，使得（i=1，2，3）六已知單調連續(xù)函數(shù)的如下數(shù)據(jù)用插值法求方程在區(qū)間（0.00，1.80）內(nèi)根的近似值。（小數(shù)點后至少保留4位）七設有積分 取5個等距節(jié)點（包括端點），列出被積函數(shù)在這些節(jié)點上的函數(shù)值表（小數(shù)點后至少保留4位）用復化的simpson公式求該積分的近似值，并且由截斷誤差公式估計誤差大小。八給定初值問題寫出euler預估校正格式取步長為0.2，計算在1.4處的函數(shù)的近似值。九設矩陣a對稱正定，考慮迭代格式對任意的初始向量是否收斂到的解，為什么？ 計算方法2006-2007第二學期1 填空1）. 近似數(shù)關于真值有_為有效數(shù)字。2）

6、 適當選擇求積節(jié)點和系數(shù)，則求積公式的代數(shù)精確度最高可以達到_次.3） 設近似數(shù)，都是四舍五入得到的，則相對誤差 的相對誤差限_4) 近似值的相對誤差為的_ 倍。5） 擬合三點a(0,1), b(1,3)，c(2,2)的平行于軸的直線方程為_.2. 用迭代法求方程在（-1，0）內(nèi)的重根的近似值。要求1）說明所用的方法為什么收斂；2）誤差小于時迭代結束。3用最小二乘法確定中的和，使得該函數(shù)曲線擬合于下面四個點 (1.0,1.01), (1.5,2.45), (2.0,4.35), (2.5,6.71) (計算結果保留到小數(shù)點后4位)4 設函數(shù)有二階連續(xù)導數(shù)，在一些點上的值如下1.01.11.20

7、.010.110.24寫出中心差分表示的二階三點微分公式，并由此計算。5 已知五階連續(xù)可導函數(shù)的如下數(shù)據(jù)0101010試求滿足插值條件的四次多項式6 設有如下的常微分方程初值問題1） 寫出每步用歐拉法預估，用梯形法進行一次校正的計算格式。2） 取步長0.2用上述格式求解。7 設有積分1）取7個等距節(jié)點（包括端點），列出被積函數(shù)在這些點出的值（保留到小數(shù)點后4位）2）用復化simpson公式求該積分的近似值。8 用lu分解法求解線性代數(shù)方程組9 當常數(shù)c取合適的值時，兩條拋物線 與就在某點相切，試取出試點，用牛頓迭代法求切點橫坐標。誤差小于時迭代結束。參考答案； 1: (1)2, (2) 2n-

8、1 (3) 2.1457*10e-3 （4）1/5 (5) x=12 解：將方程變形為 即求在（-1，0）內(nèi)的根的近似值牛頓迭代格式為 收斂性證明； 局部收斂定理結果 。3 用最小二乘法 正則方程組為解得 a=1.0072; b=0.45634解 推導中心差分格式得到5 解 截斷誤差 6 7 0.68058 （0 1 0 1）9 解 兩條曲線求導 和切點橫坐標一定滿足=將等式變形為 牛頓迭代法 結果為 0.34781winger tuivasa-sheck, who scored two tries in the kiwis 20-18 semi-final win over england,

9、 has been passed fit after a lower-leg injury, while slater has been named at full-back but is still recovering from a knee injury aggravated against usa.both sides boast 100% records heading into the encounter but australia have not conceded a try since josh charnleys effort in their first pool mat

10、ch against england on the opening day.aussie winger jarryd hayne is the competitions top try scorer with nine, closely followed by tuivasa-sheck with eight.but it is recently named rugby league international federation player of the year sonny bill williams who has attracted the most interest in the

11、 tournament so far.the kiwi - with a tournament high 17 offloads - has the chance of becoming the first player to win the world cup in both rugby league and rugby union after triumphing with the all blacks in 2011.id give every award back in a heartbeat just to get across the line this weekend, said

12、 williams.the (lack of) air up there watch mcayman islands-based webb, the head of fifas anti-racism taskforce, is in london for the football associations 150th anniversary celebrations and will attend citys premier league match at chelsea on sunday.i am going to be at the match tomorrow and i have

13、asked to meet yaya toure, he told bbc sport.for me its about how he felt and i would like to speak to him first to find out what his experience was.uefa hasopened disciplinary proceedings against cskafor the racist behaviour of their fans duringcitys 2-1 win.michel platini, president of european foo

14、tballs governing body, has also ordered an immediate investigation into the referees actions.cska said they were surprised and disappointed by toures complaint. in a statement the russian side added: we found no racist insults from fans of cska. baumgartner the disappointing news: mission aborted.th

15、e supersonic descent could happen as early as sunda.the weather plays an important role in this mission. starting at the ground, conditions have to be very calm - winds less than 2 mph, with no precipitation or humidity and limited cloud cover. the balloon, with capsule attached, will move through t

16、he lower level of the atmosphere (the troposphere) where our day-to-day weather lives. it will climb higher than the tip of mount everest (5.5 miles/8.85 kilometers), drifting even higher than the cruising altitude of commercial airliners (5.6 miles/9.17 kilometers) and into the stratosphere. as he

17、crosses the boundary layer (called the tropopause),e can expect a lot of turbulence.the balloon will slowly drift to the edge of space at 120,000 feet ( then, i would assume, he will slowly step out onto something resembling an olympic diving platform.they blew it in 2008 when they got caught cold i

18、n the final and they will not make the same mistake against the kiwis in manchester.five years ago they cruised through to the final and so far history has repeated itself here - the last try they conceded was scored by englands josh charnley in the opening game of the tournament.that could be class

19、ed as a weakness, a team under-cooked - but i have been impressed by the kangaroos focus in their games since then.they have been concentrating on the sort of stuff that wins you tough, even contests - strong defence, especially on their own goal-line, completing sets and a good kick-chase. theyve been great at all the unglamorous stuff that often go