? 提问：数值分析是做什么用的？
数值分析输入复杂问题或运算
.,,,,,),(,)(
,,ln,,
xf
dx
ddxxf
bxAxax
b
a
x
计算机近似解第一章 误差 /* Error */
§ 1 误差的背景介绍 /* Introduction */
1,来源与分类 /* Source & Classification */
从实际问题中抽象出数学模型
—— 模型误差 /* Modeling Error */
通过测量得到模型中参数的值
—— 观测误差 /* Measurement Error */
求近似解 —— 方法误差 (截断误差 /* Truncation Error */ )
机器字长有限 ——舍入误差 /* Roundoff Error */
§ 1 Introduction,Source & Classification
The following problem can be solved either the easy way or the hard way.
Two trains 200 miles apart are moving toward each other; each one is going
at a speed of 50 miles per hour,A fly starting on the front of one of them
flies back and forth between them at a rate of 75 miles per hour,It does this
until the trains collide and crush the fly to death,What is the total distance
the fly has flown?
The fly actually hits each train an infinite number of times before it gets
crushed,and one could solve the problem the hard way with pencil and
paper by summing an infinite series of distances,The easy way is as follows,
Since the trains are 200 miles apart and each train is going 50 miles an hour,
it takes 2 hours for the trains to collide,Therefore the fly was flying for two
hours,Since the fly was flying at a rate of 75 miles per hour,the fly must
have flown 150 miles,That's all there is to it.
When this problem was posed to John von Neumann,
he immediately replied,"150 miles."
"It is very strange," said the poser,"but nearly everyone
tries to sum the infinite series."
"What do you mean,strange?" asked Von Neumann,"That's how I did it!"
§ 1 Introduction,Source & Classification
dxe x10 2 近似计算:例大家一起猜？
dxe 2x10
11 / e
解法之一,将 作 Taylor展开后再积分2xe?
9
1
!4
1
7
1
!3
1
5
1
!2
1
3
1
1
)
!4!3!2
1(
1
0
864
21
0
dx
xxx
xdxe
2x
S4 R4 /* Remainder */
,10 4 Sdxe 2x取则
11
1
!5
1
9
1
!4
1
4R
称为 截断误差 /* Truncation Error */
0 0 5091!414,R这里
7 4 300 2 40103 3 30142110 13114,...S
0 0 1020 0 0 50,,| 舍入误差 /* Roundoff Error */ |
0 0 600 0 100 0 5010 2,..dxe -x 的总体误差计算
= 0.747… …
由截去部分
/* excluded terms */
引起由留下部分
/* included terms */
引起
§ 1 Introduction,Spread & Accumulation
2,传播与积累 /* Spread & Accumulation */
例,蝴蝶效应 ——纽约的一只蝴蝶翅膀一拍，风和日丽的北京就刮起台风来了？！
NY BJ
以上是一个 病态问题 /* ill-posed problem*/
关于本身是病态的问题，我们还是留给数学家去头痛吧！
§ 1 Introduction,Spread & Accumulation
.,,,,,2101 10,,,n,dxexeI xnn
例：计算
11 nn InI
公式一：
注意此公式 精确 成立
6 3 2 1 2 0 5 60111 100,edxeeI x
记为
*0I
8000 1050,IIE则初始误差
1
1
1
111 11
0
01
0 nI)e ( ndxexeIdxexe n
n
n
n
39141423151
9 5 9 4 2 494141
2 2 7 6 4 8 07131
6 3 2 8 9 6 0 00121
0 3 0 5 9 2 0 00111
0 8 8 1 2 8 0 00101
............
3 6 7 8 7 9 4 4011
1415
*
13
*
14
*
12
*
13
*
11
*
12
*
10
*
11
*
9
*
10
*
0
*
1
.II
.II
.II
.II
.II
.II
.II
!
! !
What
happened
!
§ 1 Introduction,Spread & Accumulation
考察第 n步的误差 nE
|)1()1(||||| * 11* nnnnn nInIIIE ||! 01 En||En n
我们有责任改变。
造成这种情况的是 不稳定的算法 /* unstable algorithm */
迅速积累，误差呈递增走势。可见初始的小扰动 8
0 1050||,E
)1(11 11 nnnn InIInI 公式二：
注意此公式与公式一在理论上 等价 。
方法：先估计一个 IN,再反推要求的 In ( n << N )。
1
1
)1(
1
NINe N?
NN INNeI
1
1
)1(
1
2
1*可取
0* NNN IIEN,时当
§ 1 Introduction,Spread & Accumulation
6 3 2 1 2 0 5 60)1(
1
1
3 6 7 8 7 9 4 40)1(
2
1
0 8 3 8 7 7 1 1 50)1(
11
1
0 7 7 3 5 1 7 3 20)1(
12
1
0 7 1 7 7 9 2 1 40)1(
13
1
0 6 6 8 7 0 2 2 00)1(
14
1
0 6 3 8 1 6 9 1 80)1(
15
1
0 4 2 7 4 6 2 3 30
16
1
16
1
2
1
*
1
*
0
*
2
*
1
*
11
*
10
*
12
*
11
*
13
*
12
*
14
*
13
*
15
*
14
*
15
.II
.II
.II
.II
.II
.II
.II
.
e
I
取
We just got lucky?
§ 1 Introduction,Spread & Accumulation
考察反推一步的误差：
||1)1(1)1(1|| *1 NNNN ENININE
以此类推，对 n < N 有：
||)1(...)1( 1|| Nn EnNNE
误差逐步递减,这样的算法称为 稳定的算法 /* stable algorithm */
在我们今后的讨论中,误差 将不可回避，
算法的 稳定性 会是一个非常重要的话题。
§ 2 误差与有效数字 /* Error and Significant Digits */
绝对误差 /* absolute error */
xxe ** 其中 x为精确值,x*为 x的近似值。
Hey isn’t it simple?
Oh yeah? Then tell
me the absolute error
of
7 4 3.010 2 dxe x
Oops!
|| *e *ε
10 0 0 607 4 302,.dxe x** εxx,例如：工程上常记为
，称为 绝对误差限 /* accuracy */，的上限记为注,e* 理论上讲是唯一确定的,可能取正,也可能取负。
e 0 不唯一,当然 e? 越小越具有参考价值。
I can tell that this
part’s diameter is
20cm?1cm.
I can tell that distance
between two planets is
1 million light year ± 1
light year.
Of course mine is more
accurate ! The accuracy
relates to not only the
absolute erro,but also to
the size of the exact value.
§ 2 Error and Significant Digits
相对误差 /* relative error */
x
ee
r
**?
Now I wouldn’t call it simple,
Say … what is the relative error
of 20cm± 1cm?
Don’t tell me it’s 5% because…But what kind of information does that 5% give us anyway?
|| *
*
*
x
εε
r?
x 的 相对误差上限 /* relative accuracy */ 定义为
A mathematician,a physicist,and an engineer were traveling
through Scotland when they saw a black sheep through the
window of the train,
"Aha," says the engineer,"I see that Scottish sheep are black."
"Hmm," says the physicist,"You mean that some Scottish sheep are black."
"No," says the mathematician,"All we know is that there is at least one
sheep in Scotland,and that at least one side of that one sheep is black!"
注：从 的定义可见,实际上被 偷换 成了,而后才考察其上限。那么这样的偷换是否 合法？
严格的说法是,与 是否反映了 同一数量级 的误差？
关于此问题的详细讨论可见教材第 3页。
*re *re
*
*
x
e
x
e*
*
*
x
e
§ 2 Error and Significant Digits
有效数字 /* significant digits */
用科学计数法，记 （其中 ）。若
（即 的截取按四舍五入规则），则称为有 n 位有效数字，精确到 。
mnaa.ax 100 21* 01?a
nm.xx 1050|| * na
*x nm?10
1415.3*;8979321415926535.3例：
问,有几位有效数字？请证明你的结论。*?
*
10501050*a n d
103 1 4 1 50
413
1
,.π|| π
,.π*?
证明：
有 位有效数字，精确到小数点后第 位。4 3
注,0.2300有 4位有效数字，而 00023只有 2位有效。 12300如果写成 0.123?105，则表示只有 3位有效数字。
数字末尾的 0不可随意省去！
§ 2 Error and Significant Digits
有效数字与相对误差的关系
有效数字?相对误差限
1
1
121
10
2
1
02
10
100
1050
*
n
n
m
n
nm
r
a
.aaa.a
.
x*
ε*
ε
已知 x* 有 n 位 有效数字,则其 相对误差限 为
相对误差限?有效数字
nmm
n
m
n
r
.a
a
a.a
a
xεxx
105010)1(
)1(2
10
100
)1(2
10
|*|*|*|
1
1
1
1
21
1
1
1
1
10)1(2 1* nr aε已知 x* 的 相对误差限 可写为则可见 x* 至少 有 n 位 有效数字 。
§ 2 Error and Significant Digits
例：为使 的相对误差小于 0.001%,至少应取几位有效数字？*π
解：假设?* 取到 n 位有效数字，则其相对误差上限为
1
1
102 1* nr aε
要保证其相对误差小于 0.001%，只要保证其上限满足
%001.0102 1* 1
1
nr aε
已知 a1 = 3，则从以上不等式可解得 n > 6? log6，即
n? 6，应取?* = 3.14159。
§ 3 函数的误差估计 /*Error Estimation for Functions*/
问题,对于 y = f (x)，若用 x* 取代 x，将对 y 产生什么影响？
分析,e*(y) = f (x*)? f (x) e*(x) = x*? xMean Value Theorem
= f ’(? )(x*? x)
x* 与 x 非常接近时，可认为 f ’(? )? f ’(x*),则有：
|e*(y)|? | f ’(x*)|·|e*(x)|
即,x*产生的误差经过 f 作用后被放大 /缩小了 | f ’(x*)|
倍。故称 | f ’(x*)|为 放大因子 /* amplification factor */ 或绝对条件数 /* absolute condition number */.
§ 3 Error Estimation for Functions
*)(
)(*|)(|
xf
yey*e
r? *
)(*|)(|
x
xex*e
r?
)(*
*)(
*)(*
*
*
*)(
*
*
)(*)(
xe
xf
xfx
x
xx
xf
x
xx
xfxf
r?
相对误差条件数
/* relative condition number*/
f 的条件数在某一点是 小 \大,则称 f 在该点是 好条件的
/* well-conditioned */ \坏条件的 /* ill-conditioned */。
注：关于多元函数 的讨论，请参阅教材第 5,6页。
)...,( 21 nx,x,xfy?
§ 3 Error Estimation for Functions
例,计算 y = ln x。若 x? 20，则取 x 的几位有效数字可保证
y 的相对误差 < 0.1%?
*ln
|)(*||)(*|
*)(
*)(*|)(|
x
xexe
xy
xyxy*e r
rr
解：设截取 n 位有效数字后得 x*? x，则估计 x 和 y 的相对误差上限满足近似关系
)(**ln)(* yxx rr ee
%1.0*ln102 1 1
1
xa n
不知道怎么办啊？
x 可能是 20.#，也可能是 19.#，取最坏情况，
即 a1 = 1。
n? 4
例：计算,取 4 位有效，即,则相对误差
9820ln )8920l n (,
%..
.
101002
9
820ln
9
820ln8920ln
5
§ 4 几点注意事项 /* Remarks */
1,避免相近二数相减 (详细分析请参阅教材 p.6 - p.7)
例,a1 = 0.12345,a2 = 0.12346，各有 5位有效数字。
而 a2? a1 = 0.00001，只剩下 1位有效数字。
几种经验性避免方法：;xεx εxεx ;1lnlnln xεxεx
当 | x | << 1 时：;2s i n2c o s1 2 xx
,..612111 2xxxe x
更多技巧请见教材第 8页习题 6。
§ 4 Remarks
2,避免小分母,分母小会造成浮点溢出 /* over flow */
3,避免大数 吃 小数例：用单精度计算 的根。 010)110(
992 xx
精确解为
110 291 x,x
算法 1,利用求根公式
a
acbbx
2
42
在计算机内,109存为 0.1?1010,1存为 0.1?101。 做加法时，
两加数的指数先向大指数对齐，再将浮点部分相加。即 1
的指数部分须变为 1010，则,1 = 0.0000000001? 1010，取单精度时就成为：
109+1=0.10000000?1010+0.00000000?1010=0.10000000?1010
大数 吃 小数
02 4,102 4
2
2
9
2
1?
a
acbbx
a
acbbx
§ 4 Remarks
算法 2,先解出再利用
9
2
1 102
4)(
a
acbbs i g nbx
11010 9
9
1
221 xa
cx
a
cxx
注,求和时 从小到大 相加，可使和的误差减小。
例：按从小到大、以及从大到小的顺序分别计算
1 + 2 + 3 + … + 40 + 10 9
4,先化简再计算，减少步骤，避免误差积累。
一般来说，计算机处理下列运算的速度为e x p,,
5,选用稳定的算法。 HW,p.8-9 #1,#7Self-study Ch.2-1
Excuses for not
doing homework
I accidentally divided by zero
and my paper burst into flames,
Lab 01,Numerical Summation of a Series
Produce a table of the values of the series
(1)
for the 3001 values of x,x = 0.0,0.1,0.2,…,300.00,All entries of the
table must have an absolute error less than 1.0e-10,This problem is
based on a problem from Hamming (1962),when mainframes were very
slowby today's microcomputer standards.
Input
There is no input.
Output
The output is to be formatted as two columns with the values of x and
(x) printed as in the C fprintf:
fprintf(outfile,"%6.2f?%16.12f\n",x,psix); /* here?represents a space */
1 )( 1)( k xkkx?
As an example,the sample output below shows 4 acceptable lines out of
3001,which might appear in the output file,The values of x should
start at 0.00 and increase by 0.1 until the line with x = 300.00 is output,
Sample Output (?represents a space)
0.001.644934066848
0.101.534607244904
...
1.001.000000000000
...
2.000.750000000000
...
300.000.020942212934
To improve the convergence
of the summation process
note that
1
11
)1(
1
kkkk
(2)
which implies?(1) =1.0,One can then
produce a series for?(x) –?(1) which
converges faster than the original series.
This series not only converges much
faster,it also reduces roundoff loss.
This process of finding a faster
converging series may be repeated
again on the second series to produce a
third sequence,which converges even
more rapidly than the second,
The following equation is helpful in
determining how may items are
required in summing the series above,
1 a n d 1f o r 11 1
rkdxxk n r
nk r
数值分析输入复杂问题或运算
.,,,,,),(,)(
,,ln,,
xf
dx
ddxxf
bxAxax
b
a
x
计算机近似解第一章 误差 /* Error */
§ 1 误差的背景介绍 /* Introduction */
1,来源与分类 /* Source & Classification */
从实际问题中抽象出数学模型
—— 模型误差 /* Modeling Error */
通过测量得到模型中参数的值
—— 观测误差 /* Measurement Error */
求近似解 —— 方法误差 (截断误差 /* Truncation Error */ )
机器字长有限 ——舍入误差 /* Roundoff Error */
§ 1 Introduction,Source & Classification
The following problem can be solved either the easy way or the hard way.
Two trains 200 miles apart are moving toward each other; each one is going
at a speed of 50 miles per hour,A fly starting on the front of one of them
flies back and forth between them at a rate of 75 miles per hour,It does this
until the trains collide and crush the fly to death,What is the total distance
the fly has flown?
The fly actually hits each train an infinite number of times before it gets
crushed,and one could solve the problem the hard way with pencil and
paper by summing an infinite series of distances,The easy way is as follows,
Since the trains are 200 miles apart and each train is going 50 miles an hour,
it takes 2 hours for the trains to collide,Therefore the fly was flying for two
hours,Since the fly was flying at a rate of 75 miles per hour,the fly must
have flown 150 miles,That's all there is to it.
When this problem was posed to John von Neumann,
he immediately replied,"150 miles."
"It is very strange," said the poser,"but nearly everyone
tries to sum the infinite series."
"What do you mean,strange?" asked Von Neumann,"That's how I did it!"
§ 1 Introduction,Source & Classification
dxe x10 2 近似计算:例大家一起猜？
dxe 2x10
11 / e
解法之一,将 作 Taylor展开后再积分2xe?
9
1
!4
1
7
1
!3
1
5
1
!2
1
3
1
1
)
!4!3!2
1(
1
0
864
21
0
dx
xxx
xdxe
2x
S4 R4 /* Remainder */
,10 4 Sdxe 2x取则
11
1
!5
1
9
1
!4
1
4R
称为 截断误差 /* Truncation Error */
0 0 5091!414,R这里
7 4 300 2 40103 3 30142110 13114,...S
0 0 1020 0 0 50,,| 舍入误差 /* Roundoff Error */ |
0 0 600 0 100 0 5010 2,..dxe -x 的总体误差计算
= 0.747… …
由截去部分
/* excluded terms */
引起由留下部分
/* included terms */
引起
§ 1 Introduction,Spread & Accumulation
2,传播与积累 /* Spread & Accumulation */
例,蝴蝶效应 ——纽约的一只蝴蝶翅膀一拍，风和日丽的北京就刮起台风来了？！
NY BJ
以上是一个 病态问题 /* ill-posed problem*/
关于本身是病态的问题，我们还是留给数学家去头痛吧！
§ 1 Introduction,Spread & Accumulation
.,,,,,2101 10,,,n,dxexeI xnn
例：计算
11 nn InI
公式一：
注意此公式 精确 成立
6 3 2 1 2 0 5 60111 100,edxeeI x
记为
*0I
8000 1050,IIE则初始误差
1
1
1
111 11
0
01
0 nI)e ( ndxexeIdxexe n
n
n
n
39141423151
9 5 9 4 2 494141
2 2 7 6 4 8 07131
6 3 2 8 9 6 0 00121
0 3 0 5 9 2 0 00111
0 8 8 1 2 8 0 00101
............
3 6 7 8 7 9 4 4011
1415
*
13
*
14
*
12
*
13
*
11
*
12
*
10
*
11
*
9
*
10
*
0
*
1
.II
.II
.II
.II
.II
.II
.II
!
! !
What
happened
!
§ 1 Introduction,Spread & Accumulation
考察第 n步的误差 nE
|)1()1(||||| * 11* nnnnn nInIIIE ||! 01 En||En n
我们有责任改变。
造成这种情况的是 不稳定的算法 /* unstable algorithm */
迅速积累，误差呈递增走势。可见初始的小扰动 8
0 1050||,E
)1(11 11 nnnn InIInI 公式二：
注意此公式与公式一在理论上 等价 。
方法：先估计一个 IN,再反推要求的 In ( n << N )。
1
1
)1(
1
NINe N?
NN INNeI
1
1
)1(
1
2
1*可取
0* NNN IIEN,时当
§ 1 Introduction,Spread & Accumulation
6 3 2 1 2 0 5 60)1(
1
1
3 6 7 8 7 9 4 40)1(
2
1
0 8 3 8 7 7 1 1 50)1(
11
1
0 7 7 3 5 1 7 3 20)1(
12
1
0 7 1 7 7 9 2 1 40)1(
13
1
0 6 6 8 7 0 2 2 00)1(
14
1
0 6 3 8 1 6 9 1 80)1(
15
1
0 4 2 7 4 6 2 3 30
16
1
16
1
2
1
*
1
*
0
*
2
*
1
*
11
*
10
*
12
*
11
*
13
*
12
*
14
*
13
*
15
*
14
*
15
.II
.II
.II
.II
.II
.II
.II
.
e
I
取
We just got lucky?
§ 1 Introduction,Spread & Accumulation
考察反推一步的误差：
||1)1(1)1(1|| *1 NNNN ENININE
以此类推，对 n < N 有：
||)1(...)1( 1|| Nn EnNNE
误差逐步递减,这样的算法称为 稳定的算法 /* stable algorithm */
在我们今后的讨论中,误差 将不可回避，
算法的 稳定性 会是一个非常重要的话题。
§ 2 误差与有效数字 /* Error and Significant Digits */
绝对误差 /* absolute error */
xxe ** 其中 x为精确值,x*为 x的近似值。
Hey isn’t it simple?
Oh yeah? Then tell
me the absolute error
of
7 4 3.010 2 dxe x
Oops!
|| *e *ε
10 0 0 607 4 302,.dxe x** εxx,例如：工程上常记为
，称为 绝对误差限 /* accuracy */，的上限记为注,e* 理论上讲是唯一确定的,可能取正,也可能取负。
e 0 不唯一,当然 e? 越小越具有参考价值。
I can tell that this
part’s diameter is
20cm?1cm.
I can tell that distance
between two planets is
1 million light year ± 1
light year.
Of course mine is more
accurate ! The accuracy
relates to not only the
absolute erro,but also to
the size of the exact value.
§ 2 Error and Significant Digits
相对误差 /* relative error */
x
ee
r
**?
Now I wouldn’t call it simple,
Say … what is the relative error
of 20cm± 1cm?
Don’t tell me it’s 5% because…But what kind of information does that 5% give us anyway?
|| *
*
*
x
εε
r?
x 的 相对误差上限 /* relative accuracy */ 定义为
A mathematician,a physicist,and an engineer were traveling
through Scotland when they saw a black sheep through the
window of the train,
"Aha," says the engineer,"I see that Scottish sheep are black."
"Hmm," says the physicist,"You mean that some Scottish sheep are black."
"No," says the mathematician,"All we know is that there is at least one
sheep in Scotland,and that at least one side of that one sheep is black!"
注：从 的定义可见,实际上被 偷换 成了,而后才考察其上限。那么这样的偷换是否 合法？
严格的说法是,与 是否反映了 同一数量级 的误差？
关于此问题的详细讨论可见教材第 3页。
*re *re
*
*
x
e
x
e*
*
*
x
e
§ 2 Error and Significant Digits
有效数字 /* significant digits */
用科学计数法，记 （其中 ）。若
（即 的截取按四舍五入规则），则称为有 n 位有效数字，精确到 。
mnaa.ax 100 21* 01?a
nm.xx 1050|| * na
*x nm?10
1415.3*;8979321415926535.3例：
问,有几位有效数字？请证明你的结论。*?
*
10501050*a n d
103 1 4 1 50
413
1
,.π|| π
,.π*?
证明：
有 位有效数字，精确到小数点后第 位。4 3
注,0.2300有 4位有效数字，而 00023只有 2位有效。 12300如果写成 0.123?105，则表示只有 3位有效数字。
数字末尾的 0不可随意省去！
§ 2 Error and Significant Digits
有效数字与相对误差的关系
有效数字?相对误差限
1
1
121
10
2
1
02
10
100
1050
*
n
n
m
n
nm
r
a
.aaa.a
.
x*
ε*
ε
已知 x* 有 n 位 有效数字,则其 相对误差限 为
相对误差限?有效数字
nmm
n
m
n
r
.a
a
a.a
a
xεxx
105010)1(
)1(2
10
100
)1(2
10
|*|*|*|
1
1
1
1
21
1
1
1
1
10)1(2 1* nr aε已知 x* 的 相对误差限 可写为则可见 x* 至少 有 n 位 有效数字 。
§ 2 Error and Significant Digits
例：为使 的相对误差小于 0.001%,至少应取几位有效数字？*π
解：假设?* 取到 n 位有效数字，则其相对误差上限为
1
1
102 1* nr aε
要保证其相对误差小于 0.001%，只要保证其上限满足
%001.0102 1* 1
1
nr aε
已知 a1 = 3，则从以上不等式可解得 n > 6? log6，即
n? 6，应取?* = 3.14159。
§ 3 函数的误差估计 /*Error Estimation for Functions*/
问题,对于 y = f (x)，若用 x* 取代 x，将对 y 产生什么影响？
分析,e*(y) = f (x*)? f (x) e*(x) = x*? xMean Value Theorem
= f ’(? )(x*? x)
x* 与 x 非常接近时，可认为 f ’(? )? f ’(x*),则有：
|e*(y)|? | f ’(x*)|·|e*(x)|
即,x*产生的误差经过 f 作用后被放大 /缩小了 | f ’(x*)|
倍。故称 | f ’(x*)|为 放大因子 /* amplification factor */ 或绝对条件数 /* absolute condition number */.
§ 3 Error Estimation for Functions
*)(
)(*|)(|
xf
yey*e
r? *
)(*|)(|
x
xex*e
r?
)(*
*)(
*)(*
*
*
*)(
*
*
)(*)(
xe
xf
xfx
x
xx
xf
x
xx
xfxf
r?
相对误差条件数
/* relative condition number*/
f 的条件数在某一点是 小 \大,则称 f 在该点是 好条件的
/* well-conditioned */ \坏条件的 /* ill-conditioned */。
注：关于多元函数 的讨论，请参阅教材第 5,6页。
)...,( 21 nx,x,xfy?
§ 3 Error Estimation for Functions
例,计算 y = ln x。若 x? 20，则取 x 的几位有效数字可保证
y 的相对误差 < 0.1%?
*ln
|)(*||)(*|
*)(
*)(*|)(|
x
xexe
xy
xyxy*e r
rr
解：设截取 n 位有效数字后得 x*? x，则估计 x 和 y 的相对误差上限满足近似关系
)(**ln)(* yxx rr ee
%1.0*ln102 1 1
1
xa n
不知道怎么办啊？
x 可能是 20.#，也可能是 19.#，取最坏情况，
即 a1 = 1。
n? 4
例：计算,取 4 位有效，即,则相对误差
9820ln )8920l n (,
%..
.
101002
9
820ln
9
820ln8920ln
5
§ 4 几点注意事项 /* Remarks */
1,避免相近二数相减 (详细分析请参阅教材 p.6 - p.7)
例,a1 = 0.12345,a2 = 0.12346，各有 5位有效数字。
而 a2? a1 = 0.00001，只剩下 1位有效数字。
几种经验性避免方法：;xεx εxεx ;1lnlnln xεxεx
当 | x | << 1 时：;2s i n2c o s1 2 xx
,..612111 2xxxe x
更多技巧请见教材第 8页习题 6。
§ 4 Remarks
2,避免小分母,分母小会造成浮点溢出 /* over flow */
3,避免大数 吃 小数例：用单精度计算 的根。 010)110(
992 xx
精确解为
110 291 x,x
算法 1,利用求根公式
a
acbbx
2
42
在计算机内,109存为 0.1?1010,1存为 0.1?101。 做加法时，
两加数的指数先向大指数对齐，再将浮点部分相加。即 1
的指数部分须变为 1010，则,1 = 0.0000000001? 1010，取单精度时就成为：
109+1=0.10000000?1010+0.00000000?1010=0.10000000?1010
大数 吃 小数
02 4,102 4
2
2
9
2
1?
a
acbbx
a
acbbx
§ 4 Remarks
算法 2,先解出再利用
9
2
1 102
4)(
a
acbbs i g nbx
11010 9
9
1
221 xa
cx
a
cxx
注,求和时 从小到大 相加，可使和的误差减小。
例：按从小到大、以及从大到小的顺序分别计算
1 + 2 + 3 + … + 40 + 10 9
4,先化简再计算，减少步骤，避免误差积累。
一般来说，计算机处理下列运算的速度为e x p,,
5,选用稳定的算法。 HW,p.8-9 #1,#7Self-study Ch.2-1
Excuses for not
doing homework
I accidentally divided by zero
and my paper burst into flames,
Lab 01,Numerical Summation of a Series
Produce a table of the values of the series
(1)
for the 3001 values of x,x = 0.0,0.1,0.2,…,300.00,All entries of the
table must have an absolute error less than 1.0e-10,This problem is
based on a problem from Hamming (1962),when mainframes were very
slowby today's microcomputer standards.
Input
There is no input.
Output
The output is to be formatted as two columns with the values of x and
(x) printed as in the C fprintf:
fprintf(outfile,"%6.2f?%16.12f\n",x,psix); /* here?represents a space */
1 )( 1)( k xkkx?
As an example,the sample output below shows 4 acceptable lines out of
3001,which might appear in the output file,The values of x should
start at 0.00 and increase by 0.1 until the line with x = 300.00 is output,
Sample Output (?represents a space)
0.001.644934066848
0.101.534607244904
...
1.001.000000000000
...
2.000.750000000000
...
300.000.020942212934
To improve the convergence
of the summation process
note that
1
11
)1(
1
kkkk
(2)
which implies?(1) =1.0,One can then
produce a series for?(x) –?(1) which
converges faster than the original series.
This series not only converges much
faster,it also reduces roundoff loss.
This process of finding a faster
converging series may be repeated
again on the second series to produce a
third sequence,which converges even
more rapidly than the second,
The following equation is helpful in
determining how may items are
required in summing the series above,
1 a n d 1f o r 11 1
rkdxxk n r
nk r
