数值分析第四章数值积分和数值微分习题集答案解析
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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:
101210121
12120
(1)()()(0)();
(2)()()(0)();
(3)()[(1)2()3()]/3;
(4)()[(0)()]/2[(0)()];
h
h
h
h h
f x dx A f h A f A f h f x dx A f h A f A f h f x dx f f x f x f x dx h f f h ah f f h -----≈-++≈-++≈-++''≈++-⎰⎰
⎰⎰
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)()h
h
f x dx A f h A f A f h --≈-++⎰
令()1f x =,则
1012h A A A -=++
令()f x x =,则
110A h A h -=-+
令2
()f x x =,则
3
221123
h h A h A -=+ 从而解得
011431313A h A h A h -⎧=⎪⎪
⎪
=⎨⎪
⎪=⎪⎩
令3
()f x x =,则
3()0h
h
h
h
f x dx x dx --==⎰
⎰
101()(0)()0A f h A f A f h --++=
令4
()f x x =,则
455
1012()5
2
()(0)()3
h
h
h
h
f x dx x dx h A f h A f A f h h ---==
-++=⎰
⎰
故此时,
101()()(0)()h
h
f x dx A f h A f A f h --≠-++⎰
故
101()()(0)()h h
f x dx A f h A f A f h --≈-++⎰
具有3次代数精度。 (2)若
21012()()(0)()h
h
f x dx A f h A f A f h --≈-++⎰
令()1f x =,则
1014h A A A -=++
令()f x x =,则
110A h A h -=-+
令2
()f x x =,则
3
2211163
h h A h A -=+ 从而解得
1143
8383A h A h A h -⎧=-⎪⎪
⎪
=⎨⎪
⎪=⎪⎩
令3
()f x x =,则
22322()0h
h
h
h
f x dx x dx --==⎰
⎰
101()(0)()0A f h A f A f h --++=
令4
()f x x =,则
2245
2264()5
h
h
h h f x dx x dx h --==
⎰⎰
510116
()(0)()3
A f h A f A f h h --++=
故此时,
21012()()(0)()h
h
f x dx A f h A f A f h --≠-++⎰
因此,
21012()()(0)()h h
f x dx A f h A f A f h --≈-++⎰
具有3次代数精度。 (3)若
1
121
()[(1)2()3()]/3f x dx f f x f x -≈-++⎰
令()1f x =,则
1
121
()2[(1)2()3()]/3f x dx f f x f x -==-++⎰
令()f x x =,则
120123x x =-++
令2
()f x x =,则
22
122123x x =++
从而解得
12
0.28990.5266x x =-⎧⎨
=⎩或120.6899
0.1266x x =⎧⎨=⎩ 令3
()f x x =,则
1
1
31
1
()0f x dx x dx --==⎰
⎰
12[(1)2()3()]/30f f x f x -++≠
故
1
121
()[(1)2()3()]/3f x dx f f x f x -=-++⎰
不成立。
因此,原求积公式具有2次代数精度。 (4)若
20
()[(0)()]/2[(0)()]h
f x dx h f f h ah f f h ''≈++-⎰