数值分析答案VIP免费

第四章数值积分与数值微分1.确定下列求积公式中的特定参数，使其代数精度尽量高，并指明所构造出的求积公式所具有的代数精度：10121012112120(1)()()(0)();(2)()()(0)();(3)()[(1)2()3()]/3;(4)()[(0)()]/2[(0)()];hhhhhfxdxAfhAfAfhfxdxAfhAfAfhfxdxffxfxfxdxhffhahffh解：求解求积公式的代数精度时，应根据代数精度的定义，即求积公式对于次数不超过m的多项式均能准确地成立，但对于m+1次多项式就不准确成立，进行验证性求解。（1）若101(1)()()(0)()hhfxdxAfhAfAfh令()1fx，则1012hAAA令()fxx，则110AhAh令2()fxx，则3221123hhAhA从而解得011431313AhAhAh令3()fxx，则3()0hhhhfxdxxdx101()(0)()0AfhAfAfh故101()()(0)()hhfxdxAfhAfAfh成立。令4()fxx，则4551012()52()(0)()3hhhhfxdxxdxhAfhAfAfhh故此时，101()()(0)()hhfxdxAfhAfAfh故101()()(0)()hhfxdxAfhAfAfh具有3次代数精度。（2）若21012()()(0)()hhfxdxAfhAfAfh令()1fx，则1014hAAA令()fxx，则110AhAh令2()fxx，则32211163hhAhA从而解得011438383AhAhAh令3()fxx，则22322()0hhhhfxdxxdx101()(0)()0AfhAfAfh故21012()()(0)()hhfxdxAfhAfAfh成立。令4()fxx，则22452264()5hhhhfxdxxdxh510116()(0)()3AfhAfAfhh故此时，21012()()(0)()hhfxdxAfhAfAfh因此，21012()()(0)()hhfxdxAfhAfAfh具有3次代数精度。（3）若1121()[(1)2()3()]/3fxdxffxfx令()1fx，则1121()2[(1)2()3()]/3fxdxffxfx令()fxx，则120123xx令2()fxx，则22122123xx从而解得120.28990.5266xx或120.68990.1266xx令3()fxx，则11311()0fxdxxdx12[(1)2()3()]/30ffxfx故1121()[(1)2()3()]/3fxdxffxfx不成立。因此，原求积公式具有2次代数精度。（4）若20()[(0)()]/2[(0)()]hfxdxhffhahffh令()1fx，则0(),hfxdxh2[(0)()]/2[(0)()]hffhahffhh令()fxx，则200221()21[(0)()]/2[(0)()]2hhfxdxxdxhhffhahffhh令2()fxx，则23002321()31[(0)()]/2[(0)()]22hhfxdxxdxhhffhahffhhah故有33211232112hhaha令3()fxx，则340024441()41111[(0)()]/2[(0)()]12244hhfxdxxdxhhffhhffhhhh令4()fxx，则450025551()51111[(0)()]/2[(0)()]12236hhfxdxxdxhhffhhffhhhh故此时，201()[(0)()]/2[(0)()],12hfxdxhffhhffh因此，201()[(0)()]/2[(0)()]12hfxdxhffhhffh具有3次代数精度。2.分别用梯形公式和辛普森公式计算下列积分：120121091260(1),8;4(1)(2),10;(3),4;(4)4sin,6;xxdxnxedxnxxdxndn解：21(1)8,0,1,,()84xnabhfxx复化梯形公式为781[()2()()]0.111402kkhTfafxfb复化辛普森公式为7781012[()4()2()()]0.111576kkkkhSfafxfxfb121(1)(2)10,0,1,,()10xenabhfxx复化梯形公式为9101[()2()()]1.391482kkhTfafxfb复化辛普森公式为99101012[()4()2()()]1.454716kkkkhSfafxfxfb(3)4,1,9,2,(),nabhfxx复化梯形公式为341[()2()()]17.227742kkhTfafxfb复化辛普森公式为33410122[()4()2()()]17.322226(4)6,0,,,()4sin636kkkkhSfafxfxfbnabhfx复化梯形公式为561[()2()()]1.035622kkhTfafxfb复化辛普森公式为5561012[()4()2()()]1.035776kkkkhSfafxfxfb3。直接验证柯特斯教材公式（2。4）具有5交代数精度。证明：柯特斯公式为01234()[7()32()12()32()7()]90babafxdxfxfxfxfxfx令()1fx，则01234()90[7()32()12()32()7()]90babafxdxbafxfxfxfxfxba令()fxx，则2222012341()()21[7()32()12()32()7()]()902bbaafxdxxdxbabafxfxfxfxfxba令2()fxx，则23333012341()()31[7()32()12()32()7()]()903bbaafxdxxdxbabafxfxfxfxfxba令3()fxx，则34444012341()()41[7()32()12()32()7()]()904bbaafxdxxdxbabafxfxfxfxfxba令4()fxx，则45555012341()()51[7()32()12()32()7()]()905bbaafxdxxdxbabafxfxfxfxfxba令5()fxx，则56666012341()()61[7()32()12()32()7()]()906bbaafxdxxdxbabafxfxfxfxfxba令6()fxx，则012340()[7()32()12()32()7()]90hbafxdxfxfxfxfxfx因此，该柯特斯公式具有5次代数精度。4。用辛普森公式求积分10xedx并估计误差。解：辛普森公式为[()4()()]62baabSfaffb此时，0,1,(),xabfxe从而有1121(14)0.632336See误差为4(4)04()()()1802110.00035,(0,1)1802babaRffe5。推导下列三种矩形求积公式：223()()(...