作业
1. 用如下数值表构造不超过3次的插值多项式
2. P55 11题. 给出概率积分y =
2
⎰
x
e
-x
2
dx
的数据表
用2次插值计算,试问:
(1) 当x = 0.472时,积分值等于多少? (2) 当x 为何值时,积分值等于0.5? 解:(1) 取x 0 = 0.47, x 1 = 0.48, x 2 = 0.49
y (0. 472) ≈y (x -
1)(x -2
) +(x -0)(x -
2
) 0)(x -
1
)
(x 1
+y (x -2
-x 1
)(x 0
-x 2
) y (x 1
-
x 0
)(x 1
-x 2
) (x 2
-x
)(x 2-x
1
)
=
L
2
(0. 472)
=0.4937452⨯(0. 472-0. 48)(0. 472-0. 49) (0. 47-0. 48)(0. 472-0. 49) +0.5027498
⨯
(0. 472-0. 47)(0. 472-0. 49) (0. 48-0. 47)(0. 48-0. 49)
+0.5116683
⨯
(0. 472-0. 47)(0. 472-0. 48) (0. 49-0. 47)(0. 49-0. 48)
=0.355496544+0.180989928-0.040933464=0.495553008
1
(2)
x (0. 5) ≈=
2
2
00
x
(y -
(y -
y y
11
)(y -)(
y
-
2
)
2
y
y
)
+
x
(y -
1
(y -
1
y y
00
(y -y y )
+x
)(y -y ) (y -y )(y -
1
2
2
)(y -)(
y
-
1
)
1
y
2
y
)
L
2
(0. 5)
(0. 5-0.5027498)(0. 5-0.5116683)
(0.4937452(0.5027498(0.5116683
-0.5027498)(0.4937452-0.4937452)(0.5027498-0.4937452)(0.5116683
-0.5116683) -0.5116683) -0.5027498)
=0.47⨯
+0.48⨯ +0.49⨯
(0. 5-0.4937452)(0. 5-0.5116683) (0. 5-0.4937452)(0. 5-0.5027498)
=0.093439176+0.436220438-0.052723675=0.476935939
3. 证明方程e x +10x -2=0在区间[0,1]内有一个根,如果使用二分法求该区间内的根,且误差不超过10-6,试问需要二分区间[0,1]多少次?
4. 设x t =451.01为准确值,x a =451.023为x t 的近似值,试求出x a 有效数字的位数及相对误差 作业答案
1. 解:N 2(x ) = f (0)+f [0,1](x -0)+ f [0,1,2](x -0) (x -1) 1+1×(x -0) +3×(x -0) (x -1) =3x 2-2x +1 为求得P 3(x ) ,根据插值条件知,P 3(x ) 应具有下面的形式 P 3(x ) =N 2(x ) +k (x -0) (x -1) (x -2) ,这样的P 3(x ) 自然满足:
P 3(x i )= f (x i )
由P 3’(1 )=3
P 3’(1 )= N2’(1 )+k (1-0) (1-2) =N2’(1 ) -k = 4-k=3
2
∴ k =1
∴ P 3(x ) =N 2(x ) + (x -0) (x -1) (x -2)=x 3+1 3. 证明 令f (x ) =e x +10x -2,
∵ f (0)=-1 0
∴ f (x )= ex +10x -2 =0在[0,1]有根。又
f '(x )= ex +10 >0(x ∈[0,1]),故f (x ) =0在区间[0,1]内有唯一实根。 给定误差限10-6,有
k ≥
ln(b -a ) -ln ε
6ln 10ln 2
-1=
ln 2
-1
只要取k =19次.
4. 解: |x a -x t |=|451.023-451.01|=0.013
x a =451.023=0.451023×103 ,3-l =-1,∴l =4 ∴x a 有4位有效数字
e x a -x t
0. 013r =
x =t
451. 01
=0. 0029%
取ε1-(n -1)
r =
2α*10
1
e *x
*
≤εr =
ε
x
称为相对误差限。
3
作业
1. 用如下数值表构造不超过3次的插值多项式
2. P55 11题. 给出概率积分y =
2
⎰
x
e
-x
2
dx
的数据表
用2次插值计算,试问:
(1) 当x = 0.472时,积分值等于多少? (2) 当x 为何值时,积分值等于0.5? 解:(1) 取x 0 = 0.47, x 1 = 0.48, x 2 = 0.49
y (0. 472) ≈y (x -
1)(x -2
) +(x -0)(x -
2
) 0)(x -
1
)
(x 1
+y (x -2
-x 1
)(x 0
-x 2
) y (x 1
-
x 0
)(x 1
-x 2
) (x 2
-x
)(x 2-x
1
)
=
L
2
(0. 472)
=0.4937452⨯(0. 472-0. 48)(0. 472-0. 49) (0. 47-0. 48)(0. 472-0. 49) +0.5027498
⨯
(0. 472-0. 47)(0. 472-0. 49) (0. 48-0. 47)(0. 48-0. 49)
+0.5116683
⨯
(0. 472-0. 47)(0. 472-0. 48) (0. 49-0. 47)(0. 49-0. 48)
=0.355496544+0.180989928-0.040933464=0.495553008
1
(2)
x (0. 5) ≈=
2
2
00
x
(y -
(y -
y y
11
)(y -)(
y
-
2
)
2
y
y
)
+
x
(y -
1
(y -
1
y y
00
(y -y y )
+x
)(y -y ) (y -y )(y -
1
2
2
)(y -)(
y
-
1
)
1
y
2
y
)
L
2
(0. 5)
(0. 5-0.5027498)(0. 5-0.5116683)
(0.4937452(0.5027498(0.5116683
-0.5027498)(0.4937452-0.4937452)(0.5027498-0.4937452)(0.5116683
-0.5116683) -0.5116683) -0.5027498)
=0.47⨯
+0.48⨯ +0.49⨯
(0. 5-0.4937452)(0. 5-0.5116683) (0. 5-0.4937452)(0. 5-0.5027498)
=0.093439176+0.436220438-0.052723675=0.476935939
3. 证明方程e x +10x -2=0在区间[0,1]内有一个根,如果使用二分法求该区间内的根,且误差不超过10-6,试问需要二分区间[0,1]多少次?
4. 设x t =451.01为准确值,x a =451.023为x t 的近似值,试求出x a 有效数字的位数及相对误差 作业答案
1. 解:N 2(x ) = f (0)+f [0,1](x -0)+ f [0,1,2](x -0) (x -1) 1+1×(x -0) +3×(x -0) (x -1) =3x 2-2x +1 为求得P 3(x ) ,根据插值条件知,P 3(x ) 应具有下面的形式 P 3(x ) =N 2(x ) +k (x -0) (x -1) (x -2) ,这样的P 3(x ) 自然满足:
P 3(x i )= f (x i )
由P 3’(1 )=3
P 3’(1 )= N2’(1 )+k (1-0) (1-2) =N2’(1 ) -k = 4-k=3
2
∴ k =1
∴ P 3(x ) =N 2(x ) + (x -0) (x -1) (x -2)=x 3+1 3. 证明 令f (x ) =e x +10x -2,
∵ f (0)=-1 0
∴ f (x )= ex +10x -2 =0在[0,1]有根。又
f '(x )= ex +10 >0(x ∈[0,1]),故f (x ) =0在区间[0,1]内有唯一实根。 给定误差限10-6,有
k ≥
ln(b -a ) -ln ε
6ln 10ln 2
-1=
ln 2
-1
只要取k =19次.
4. 解: |x a -x t |=|451.023-451.01|=0.013
x a =451.023=0.451023×103 ,3-l =-1,∴l =4 ∴x a 有4位有效数字
e x a -x t
0. 013r =
x =t
451. 01
=0. 0029%
取ε1-(n -1)
r =
2α*10
1
e *x
*
≤εr =
ε
x
称为相对误差限。
3