Finite Precision Computation ( 续 )

武汉大学数学与统计学院向华

num=0EPS=1while (1+EPS)gt1EPS=EPS2num=num+1end

x=zeros(110)for n=110x(n)=sin(npi10)end

EPS=1 for num=11000EPS=EPS2if (1+EPS)lt=1EPS=EPS2BreakEndend

控制语句控制语句

for variable=expression Statementsend

while expression Statementsend

If expression Statementsend

switch x case 1 statement1 case statement2 otherwiseend

关系运算符关系运算符==

~=

amp

|

~

XOR

Matlab 编程

注释

续行 hellip

例

x=zeros(110)for n=110 x(n)=sin(npi10)end

num=0EPS=1while (1+EPS)gt1 EPS=EPS2 num=num+1end

EPS=1 for num=11000 EPS=EPS2 if (1+EPS)lt=1 EPS=EPS2 break endend

g=2for k=1100 g=1+1gendg

111

1

11

11

11

11

11

11

11

)(3

8)()()(

)()()()(28)(

)()(10)(

3213

31212

121

tytytytydt

d

tytytytytydt

d

tytytydt

d

function yprime = lorenzeq(ty)yprime = [ 10( y(2)-y(1) ) 28y(1) -y(2) -y(1)y(3) y(1)y(2) -83 y(3) ]

ODE solvert = [0 50]y0= [010][ty] = ode45(lorenzeqty0)plot( y(1)y(3) )xlabel(y_1)ylabel(y_2)title( Lorenz equationsFontSize16)

2^(1-53)ans = 2220446049250313e-016xmin=2^(-1021-1)xmin = 2225073858507201e-308xmin=2^(-1021-1-52)xmin = 4940656458412465e-324

eps=2220446049250313e-016realmax=1797693134862316e+308realmin=2225073858507201e-308 11realmax=Inf

00=NaN infinf=NaN

realmineps=4940656458412465e-324

realmineps01=0

MATLAB

double precision (64-bit word)

16 significant decimal digits

F(253-10211024true)

In[4]= N1 25321024 20Out[4]= 17976931348623157081 10308

)( maxmin denormeepbF

1min

min ebxpebx min

min

max)1(maxep bbx

a=1b=1while a+b~=a b=b2end

a=10e+308b=11e+308c=-1001e+308a+(b+c)a+b+c

numerical cancellationx=1e-15((1+x)-1)x

num=0EPS=1while (1+EPS)gt1EPS=EPS2num=num+1end

x=zeros(110)for n=110x(n)=sin(npi10)end

EPS=1 for num=11000EPS=EPS2if (1+EPS)lt=1EPS=EPS2BreakEndend

控制语句控制语句

for variable=expression Statementsend

while expression Statementsend

If expression Statementsend

switch x case 1 statement1 case statement2 otherwiseend

关系运算符关系运算符==

~=

amp

|

~

XOR

Matlab 编程

注释

续行 hellip

例

x=zeros(110)for n=110 x(n)=sin(npi10)end

num=0EPS=1while (1+EPS)gt1 EPS=EPS2 num=num+1end

EPS=1 for num=11000 EPS=EPS2 if (1+EPS)lt=1 EPS=EPS2 break endend

g=2for k=1100 g=1+1gendg

111

1

11

11

11

11

11

11

11

)(3

8)()()(

)()()()(28)(

)()(10)(

3213

31212

121

tytytytydt

d

tytytytytydt

d

tytytydt

d

function yprime = lorenzeq(ty)yprime = [ 10( y(2)-y(1) ) 28y(1) -y(2) -y(1)y(3) y(1)y(2) -83 y(3) ]

ODE solvert = [0 50]y0= [010][ty] = ode45(lorenzeqty0)plot( y(1)y(3) )xlabel(y_1)ylabel(y_2)title( Lorenz equationsFontSize16)

2^(1-53)ans = 2220446049250313e-016xmin=2^(-1021-1)xmin = 2225073858507201e-308xmin=2^(-1021-1-52)xmin = 4940656458412465e-324

eps=2220446049250313e-016realmax=1797693134862316e+308realmin=2225073858507201e-308 11realmax=Inf

00=NaN infinf=NaN

realmineps=4940656458412465e-324

realmineps01=0

MATLAB

double precision (64-bit word)

16 significant decimal digits

F(253-10211024true)

In[4]= N1 25321024 20Out[4]= 17976931348623157081 10308

)( maxmin denormeepbF

1min

min ebxpebx min

min

max)1(maxep bbx

a=1b=1while a+b~=a b=b2end

a=10e+308b=11e+308c=-1001e+308a+(b+c)a+b+c

numerical cancellationx=1e-15((1+x)-1)x

控制语句控制语句

for variable=expression Statementsend

while expression Statementsend

If expression Statementsend

switch x case 1 statement1 case statement2 otherwiseend

关系运算符关系运算符==

~=

amp

|

~

XOR

Matlab 编程

注释

续行 hellip

例

x=zeros(110)for n=110 x(n)=sin(npi10)end

num=0EPS=1while (1+EPS)gt1 EPS=EPS2 num=num+1end

EPS=1 for num=11000 EPS=EPS2 if (1+EPS)lt=1 EPS=EPS2 break endend

g=2for k=1100 g=1+1gendg

111

1

11

11

11

11

11

11

11

)(3

8)()()(

)()()()(28)(

)()(10)(

3213

31212

121

tytytytydt

d

tytytytytydt

d

tytytydt

d

function yprime = lorenzeq(ty)yprime = [ 10( y(2)-y(1) ) 28y(1) -y(2) -y(1)y(3) y(1)y(2) -83 y(3) ]

ODE solvert = [0 50]y0= [010][ty] = ode45(lorenzeqty0)plot( y(1)y(3) )xlabel(y_1)ylabel(y_2)title( Lorenz equationsFontSize16)

2^(1-53)ans = 2220446049250313e-016xmin=2^(-1021-1)xmin = 2225073858507201e-308xmin=2^(-1021-1-52)xmin = 4940656458412465e-324

eps=2220446049250313e-016realmax=1797693134862316e+308realmin=2225073858507201e-308 11realmax=Inf

00=NaN infinf=NaN

realmineps=4940656458412465e-324

realmineps01=0

MATLAB

double precision (64-bit word)

16 significant decimal digits

F(253-10211024true)

In[4]= N1 25321024 20Out[4]= 17976931348623157081 10308

)( maxmin denormeepbF

1min

min ebxpebx min

min

max)1(maxep bbx

a=1b=1while a+b~=a b=b2end

a=10e+308b=11e+308c=-1001e+308a+(b+c)a+b+c

numerical cancellationx=1e-15((1+x)-1)x

g=2for k=1100 g=1+1gendg

111

1

11

11

11

11

11

11

11

)(3

8)()()(

)()()()(28)(

)()(10)(

3213

31212

121

tytytytydt

d

tytytytytydt

d

tytytydt

d

function yprime = lorenzeq(ty)yprime = [ 10( y(2)-y(1) ) 28y(1) -y(2) -y(1)y(3) y(1)y(2) -83 y(3) ]

ODE solvert = [0 50]y0= [010][ty] = ode45(lorenzeqty0)plot( y(1)y(3) )xlabel(y_1)ylabel(y_2)title( Lorenz equationsFontSize16)

2^(1-53)ans = 2220446049250313e-016xmin=2^(-1021-1)xmin = 2225073858507201e-308xmin=2^(-1021-1-52)xmin = 4940656458412465e-324

eps=2220446049250313e-016realmax=1797693134862316e+308realmin=2225073858507201e-308 11realmax=Inf

00=NaN infinf=NaN

realmineps=4940656458412465e-324

realmineps01=0

MATLAB

double precision (64-bit word)

16 significant decimal digits

F(253-10211024true)

In[4]= N1 25321024 20Out[4]= 17976931348623157081 10308

)( maxmin denormeepbF

1min

min ebxpebx min

min

max)1(maxep bbx

a=1b=1while a+b~=a b=b2end

a=10e+308b=11e+308c=-1001e+308a+(b+c)a+b+c

numerical cancellationx=1e-15((1+x)-1)x

2^(1-53)ans = 2220446049250313e-016xmin=2^(-1021-1)xmin = 2225073858507201e-308xmin=2^(-1021-1-52)xmin = 4940656458412465e-324

eps=2220446049250313e-016realmax=1797693134862316e+308realmin=2225073858507201e-308 11realmax=Inf

00=NaN infinf=NaN

realmineps=4940656458412465e-324

realmineps01=0

MATLAB

double precision (64-bit word)

16 significant decimal digits

F(253-10211024true)

In[4]= N1 25321024 20Out[4]= 17976931348623157081 10308

)( maxmin denormeepbF

1min

min ebxpebx min

min

max)1(maxep bbx

a=1b=1while a+b~=a b=b2end

a=10e+308b=11e+308c=-1001e+308a+(b+c)a+b+c

numerical cancellationx=1e-15((1+x)-1)x

a=1b=1while a+b~=a b=b2end

a=10e+308b=11e+308c=-1001e+308a+(b+c)a+b+c

numerical cancellationx=1e-15((1+x)-1)x

Chaque calcul simple est ensuite fait au mieux le processeur renvoie lemeilleur resultat possible (norme IEEE-754) etant donnes ses imperatifs3142 = 9859 6 rarr 986Chaque resultat de calcul est donc arrondi

M^eme si un calcul est presque juste une succession de calculs est parfois fausse π2 rarr 3142 = 9859 6 rarr 986Mais π 2 = 9869 604 hellip donc le nombre ottant le plus proche est 987

19 192 194 196 198 2 202 204 206 208 21-15

-1

-05

0

05

1

15x 10

-10 Polynomial evaluated in factor form

19 192 194 196 198 2 202 204 206 208 21-15

-1

-05

0

05

1

15x 10

-10 Polynomial evaluated in series form

7899 14418)2()( xxxxxp

舍入误差的影响舍入误差的影响

82982839)( 23 xxxxP

82982839)( 23 xxxxP

Orientation de 3 points (G Melquiond)Etant donnes 3 points du plan p q et r On veut savoir si pqr sont alignes dans le sens horaire ou dans le sens anti-horaire

float det = ( qx - px ) ( r y - py )1048576 - ( qy - py ) ( r x - px ) i f ( det gt 0) return POSITIVE i f ( det lt 0) return NEGATIVE return ZERO

p

r

q

pyrypyqy

pxrxpxqxsignerqporient

)(2

Premiere guerre du Golfe - 100h plus tard28 GI morts et 98 blesses

iraquiens US

antiminusmissile Patriotmissile Scud

Premiere guerre du Golfe - explicationLanti-missile Patriot est prevu pour fonctionner pendant quelques heuresMais ca marchait tellement bien quils lont laisse brancheLhorloge interne ajoute 01s a chaque tic Mais 01 nest pas exact en binaire rarr une petite erreur a chaque ticrarr au bout de 100h une erreur susante pour rater le missile

Mon banquier ma propose cet investissement vous me donnez e=271828 currenlannee suivante je prends 1curren de frais et je multiplie par 1lannee suivante je prends 1curren de frais et je multiplie par 2lannee suivante je prends 1curren de frais et je multiplie par 3 apres n ans je prends 1curren de frais et je multiplie par nPour recuperer mon argent il y a 1curren de fraisDans 50 ans pour ma retraite combien dargent aurai-je

6500127190887506580397803840893860464266619976132998982702001269694679091261802058160555785871702272129904252549070848000000000

Matlab 或 C(format double)

x = 27182818284590452354for annee = 1 50 x = (x - 1) anneeend x = -4396803930182069e+048

00200

x

Fori 1 i 51 i x x 1ix SimplifyxNx 1 10

MATLAB1718281828459045e+0001436563656918090e+0001309690970754271e+0001238763883017082e+0001193819415085411e+0001162916490512465e+0001140415433587258e+0001123323468698061e+0001109911218282548e+0001099112182825479e+0001090234011080270e+0001082808132963237e+0001076505728522079e+0001071080199309108e+0001066202989636622e+0001059247834185953e+0001007213181161205e+0001298372609016951e-001-1653309204286779e+001-3506618408573559e+002-7384898658004473e+003

1718281828459045

1436563656918090

1309690970754271

1238763883017086

1193819415085428

1162916490512569

1140415433587986

1123323468703890

1109911218335008

1099112183350075

1090234016850830

1082808202209954

1076506628729406

1071092802211678

1066392033175171

1062272530802742

1058633023646621

1055394425639172

1052494087144259

1049881742885176

1047516600588701

Mathematica

includeltmathhgtmain()int n=1double x=10 h=10 deriv=cos(x) diffquoerror f(x)=sin(x) printf(deriv=136enderiv)printf(h diffquo abs(deriv-diffquo) n)

while(nlt=20)h=h10diffquo=(sin(x+h)-sin(x))herror=fabs(deriv-diffquo)printf(51e 136e 136e n h diffquo error)n++

h

xfhxfxf

)()()(

Approximating a derivative by a difference quotient

Error (absolute value of derivative minus difference quotient) as a function of h (Log-Log scale)

The truncation error of the finite difference approximation is bounded by Mh2 Assuming the error in function values is bounded by ε the rounding error is bounded by 2 εh Total computational error is

h

Mh 22

8102 Mh

Where M is a bound on |frdquo(t)| for t near x There is a tradeoff between truncation error and rounding error in choosing the step size h It is minimized when

x=ones(351)x(1)=112x(2)=6111for i=335 x(i)=111-(1130-3000x(i-2))x(i-1)endx(35)

Consider the recurrence

In exact arithmetic the xk form a monotonically increasing sequence that converges to 6 Implement the recurrence and compare the computed x34 with the true value 5998 (to four correct signicant figures)

1161211

)30001130(111

10

11

xx

xxx kkk

1161211

)30001130(111

10

11

xx

xxx kkk

30001130111)( 23 xxxxp

)100()55(3025251130111)( 223 xxxxxxq

The mathematical convergence depends on the initial conditions x0 x1convergence to 100 for almost all (x0 x1)convergence to 6 x0(11- x1)=30convergence to 5 x0=x1=5 (exactly in machine representation)

fplot(x^3-111x^2+1130x-3000[-400400])

fplot(x^3-111x^2+1130x-3000[47]) fplot(x^3-111x^2+1130x-3000[-10110])

syms x1 x2 x3x1=sym(11)sym(2)x2=sym(6111)for i=335 x3=111-(1130-3000x1)x2 x1=x2 x2=x3endx3x3=1721981182794095961389986301287093876567205105910375321asymp599797252168491

Array[x35]x[1]=112x[2]=6111Do[x[i+2]=111-(1130-3000x[i])x[i+1]i133](For[i=1ilt=33i++x[i+2]=111-(1130-3000x[i])x[i+1]])(While[ilt=33x[i+2]=111-(1130-3000x[i])x[i+1]i++] )x[35]

037532167205105912870938765

8998630179409596131721981182

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

19 192 194 196 198 2 202 204 206 208 21-15

-1

-05

0

05

1

15x 10

-10 Polynomial evaluated in factor form

19 192 194 196 198 2 202 204 206 208 21-15

-1

-05

0

05

1

15x 10

-10 Polynomial evaluated in series form

7899 14418)2()( xxxxxp

舍入误差的影响舍入误差的影响

82982839)( 23 xxxxP

82982839)( 23 xxxxP

Orientation de 3 points (G Melquiond)Etant donnes 3 points du plan p q et r On veut savoir si pqr sont alignes dans le sens horaire ou dans le sens anti-horaire

float det = ( qx - px ) ( r y - py )1048576 - ( qy - py ) ( r x - px ) i f ( det gt 0) return POSITIVE i f ( det lt 0) return NEGATIVE return ZERO

p

r

q

pyrypyqy

pxrxpxqxsignerqporient

)(2

Premiere guerre du Golfe - 100h plus tard28 GI morts et 98 blesses

iraquiens US

antiminusmissile Patriotmissile Scud

Premiere guerre du Golfe - explicationLanti-missile Patriot est prevu pour fonctionner pendant quelques heuresMais ca marchait tellement bien quils lont laisse brancheLhorloge interne ajoute 01s a chaque tic Mais 01 nest pas exact en binaire rarr une petite erreur a chaque ticrarr au bout de 100h une erreur susante pour rater le missile

Mon banquier ma propose cet investissement vous me donnez e=271828 currenlannee suivante je prends 1curren de frais et je multiplie par 1lannee suivante je prends 1curren de frais et je multiplie par 2lannee suivante je prends 1curren de frais et je multiplie par 3 apres n ans je prends 1curren de frais et je multiplie par nPour recuperer mon argent il y a 1curren de fraisDans 50 ans pour ma retraite combien dargent aurai-je

6500127190887506580397803840893860464266619976132998982702001269694679091261802058160555785871702272129904252549070848000000000

Matlab 或 C(format double)

x = 27182818284590452354for annee = 1 50 x = (x - 1) anneeend x = -4396803930182069e+048

00200

x

Fori 1 i 51 i x x 1ix SimplifyxNx 1 10

MATLAB1718281828459045e+0001436563656918090e+0001309690970754271e+0001238763883017082e+0001193819415085411e+0001162916490512465e+0001140415433587258e+0001123323468698061e+0001109911218282548e+0001099112182825479e+0001090234011080270e+0001082808132963237e+0001076505728522079e+0001071080199309108e+0001066202989636622e+0001059247834185953e+0001007213181161205e+0001298372609016951e-001-1653309204286779e+001-3506618408573559e+002-7384898658004473e+003

1718281828459045

1436563656918090

1309690970754271

1238763883017086

1193819415085428

1162916490512569

1140415433587986

1123323468703890

1109911218335008

1099112183350075

1090234016850830

1082808202209954

1076506628729406

1071092802211678

1066392033175171

1062272530802742

1058633023646621

1055394425639172

1052494087144259

1049881742885176

1047516600588701

Mathematica

includeltmathhgtmain()int n=1double x=10 h=10 deriv=cos(x) diffquoerror f(x)=sin(x) printf(deriv=136enderiv)printf(h diffquo abs(deriv-diffquo) n)

while(nlt=20)h=h10diffquo=(sin(x+h)-sin(x))herror=fabs(deriv-diffquo)printf(51e 136e 136e n h diffquo error)n++

h

xfhxfxf

)()()(

Approximating a derivative by a difference quotient

Error (absolute value of derivative minus difference quotient) as a function of h (Log-Log scale)

The truncation error of the finite difference approximation is bounded by Mh2 Assuming the error in function values is bounded by ε the rounding error is bounded by 2 εh Total computational error is

h

Mh 22

8102 Mh

Where M is a bound on |frdquo(t)| for t near x There is a tradeoff between truncation error and rounding error in choosing the step size h It is minimized when

x=ones(351)x(1)=112x(2)=6111for i=335 x(i)=111-(1130-3000x(i-2))x(i-1)endx(35)

Consider the recurrence

In exact arithmetic the xk form a monotonically increasing sequence that converges to 6 Implement the recurrence and compare the computed x34 with the true value 5998 (to four correct signicant figures)

1161211

)30001130(111

10

11

xx

xxx kkk

1161211

)30001130(111

10

11

xx

xxx kkk

30001130111)( 23 xxxxp

)100()55(3025251130111)( 223 xxxxxxq

The mathematical convergence depends on the initial conditions x0 x1convergence to 100 for almost all (x0 x1)convergence to 6 x0(11- x1)=30convergence to 5 x0=x1=5 (exactly in machine representation)

fplot(x^3-111x^2+1130x-3000[-400400])

fplot(x^3-111x^2+1130x-3000[47]) fplot(x^3-111x^2+1130x-3000[-10110])

syms x1 x2 x3x1=sym(11)sym(2)x2=sym(6111)for i=335 x3=111-(1130-3000x1)x2 x1=x2 x2=x3endx3x3=1721981182794095961389986301287093876567205105910375321asymp599797252168491

Array[x35]x[1]=112x[2]=6111Do[x[i+2]=111-(1130-3000x[i])x[i+1]i133](For[i=1ilt=33i++x[i+2]=111-(1130-3000x[i])x[i+1]])(While[ilt=33x[i+2]=111-(1130-3000x[i])x[i+1]i++] )x[35]

037532167205105912870938765

8998630179409596131721981182

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

82982839)( 23 xxxxP

82982839)( 23 xxxxP

Orientation de 3 points (G Melquiond)Etant donnes 3 points du plan p q et r On veut savoir si pqr sont alignes dans le sens horaire ou dans le sens anti-horaire

float det = ( qx - px ) ( r y - py )1048576 - ( qy - py ) ( r x - px ) i f ( det gt 0) return POSITIVE i f ( det lt 0) return NEGATIVE return ZERO

p

r

q

pyrypyqy

pxrxpxqxsignerqporient

)(2

Premiere guerre du Golfe - 100h plus tard28 GI morts et 98 blesses

iraquiens US

antiminusmissile Patriotmissile Scud

Premiere guerre du Golfe - explicationLanti-missile Patriot est prevu pour fonctionner pendant quelques heuresMais ca marchait tellement bien quils lont laisse brancheLhorloge interne ajoute 01s a chaque tic Mais 01 nest pas exact en binaire rarr une petite erreur a chaque ticrarr au bout de 100h une erreur susante pour rater le missile

Mon banquier ma propose cet investissement vous me donnez e=271828 currenlannee suivante je prends 1curren de frais et je multiplie par 1lannee suivante je prends 1curren de frais et je multiplie par 2lannee suivante je prends 1curren de frais et je multiplie par 3 apres n ans je prends 1curren de frais et je multiplie par nPour recuperer mon argent il y a 1curren de fraisDans 50 ans pour ma retraite combien dargent aurai-je

6500127190887506580397803840893860464266619976132998982702001269694679091261802058160555785871702272129904252549070848000000000

Matlab 或 C(format double)

x = 27182818284590452354for annee = 1 50 x = (x - 1) anneeend x = -4396803930182069e+048

00200

x

Fori 1 i 51 i x x 1ix SimplifyxNx 1 10

MATLAB1718281828459045e+0001436563656918090e+0001309690970754271e+0001238763883017082e+0001193819415085411e+0001162916490512465e+0001140415433587258e+0001123323468698061e+0001109911218282548e+0001099112182825479e+0001090234011080270e+0001082808132963237e+0001076505728522079e+0001071080199309108e+0001066202989636622e+0001059247834185953e+0001007213181161205e+0001298372609016951e-001-1653309204286779e+001-3506618408573559e+002-7384898658004473e+003

1718281828459045

1436563656918090

1309690970754271

1238763883017086

1193819415085428

1162916490512569

1140415433587986

1123323468703890

1109911218335008

1099112183350075

1090234016850830

1082808202209954

1076506628729406

1071092802211678

1066392033175171

1062272530802742

1058633023646621

1055394425639172

1052494087144259

1049881742885176

1047516600588701

Mathematica

includeltmathhgtmain()int n=1double x=10 h=10 deriv=cos(x) diffquoerror f(x)=sin(x) printf(deriv=136enderiv)printf(h diffquo abs(deriv-diffquo) n)

while(nlt=20)h=h10diffquo=(sin(x+h)-sin(x))herror=fabs(deriv-diffquo)printf(51e 136e 136e n h diffquo error)n++

h

xfhxfxf

)()()(

Approximating a derivative by a difference quotient

Error (absolute value of derivative minus difference quotient) as a function of h (Log-Log scale)

The truncation error of the finite difference approximation is bounded by Mh2 Assuming the error in function values is bounded by ε the rounding error is bounded by 2 εh Total computational error is

h

Mh 22

8102 Mh

Where M is a bound on |frdquo(t)| for t near x There is a tradeoff between truncation error and rounding error in choosing the step size h It is minimized when

x=ones(351)x(1)=112x(2)=6111for i=335 x(i)=111-(1130-3000x(i-2))x(i-1)endx(35)

Consider the recurrence

In exact arithmetic the xk form a monotonically increasing sequence that converges to 6 Implement the recurrence and compare the computed x34 with the true value 5998 (to four correct signicant figures)

1161211

)30001130(111

10

11

xx

xxx kkk

1161211

)30001130(111

10

11

xx

xxx kkk

30001130111)( 23 xxxxp

)100()55(3025251130111)( 223 xxxxxxq

The mathematical convergence depends on the initial conditions x0 x1convergence to 100 for almost all (x0 x1)convergence to 6 x0(11- x1)=30convergence to 5 x0=x1=5 (exactly in machine representation)

fplot(x^3-111x^2+1130x-3000[-400400])

fplot(x^3-111x^2+1130x-3000[47]) fplot(x^3-111x^2+1130x-3000[-10110])

syms x1 x2 x3x1=sym(11)sym(2)x2=sym(6111)for i=335 x3=111-(1130-3000x1)x2 x1=x2 x2=x3endx3x3=1721981182794095961389986301287093876567205105910375321asymp599797252168491

Array[x35]x[1]=112x[2]=6111Do[x[i+2]=111-(1130-3000x[i])x[i+1]i133](For[i=1ilt=33i++x[i+2]=111-(1130-3000x[i])x[i+1]])(While[ilt=33x[i+2]=111-(1130-3000x[i])x[i+1]i++] )x[35]

037532167205105912870938765

8998630179409596131721981182

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

82982839)( 23 xxxxP

Orientation de 3 points (G Melquiond)Etant donnes 3 points du plan p q et r On veut savoir si pqr sont alignes dans le sens horaire ou dans le sens anti-horaire

float det = ( qx - px ) ( r y - py )1048576 - ( qy - py ) ( r x - px ) i f ( det gt 0) return POSITIVE i f ( det lt 0) return NEGATIVE return ZERO

p

r

q

pyrypyqy

pxrxpxqxsignerqporient

)(2

Premiere guerre du Golfe - 100h plus tard28 GI morts et 98 blesses

iraquiens US

antiminusmissile Patriotmissile Scud

Premiere guerre du Golfe - explicationLanti-missile Patriot est prevu pour fonctionner pendant quelques heuresMais ca marchait tellement bien quils lont laisse brancheLhorloge interne ajoute 01s a chaque tic Mais 01 nest pas exact en binaire rarr une petite erreur a chaque ticrarr au bout de 100h une erreur susante pour rater le missile

Mon banquier ma propose cet investissement vous me donnez e=271828 currenlannee suivante je prends 1curren de frais et je multiplie par 1lannee suivante je prends 1curren de frais et je multiplie par 2lannee suivante je prends 1curren de frais et je multiplie par 3 apres n ans je prends 1curren de frais et je multiplie par nPour recuperer mon argent il y a 1curren de fraisDans 50 ans pour ma retraite combien dargent aurai-je

6500127190887506580397803840893860464266619976132998982702001269694679091261802058160555785871702272129904252549070848000000000

Matlab 或 C(format double)

x = 27182818284590452354for annee = 1 50 x = (x - 1) anneeend x = -4396803930182069e+048

00200

x

Fori 1 i 51 i x x 1ix SimplifyxNx 1 10

MATLAB1718281828459045e+0001436563656918090e+0001309690970754271e+0001238763883017082e+0001193819415085411e+0001162916490512465e+0001140415433587258e+0001123323468698061e+0001109911218282548e+0001099112182825479e+0001090234011080270e+0001082808132963237e+0001076505728522079e+0001071080199309108e+0001066202989636622e+0001059247834185953e+0001007213181161205e+0001298372609016951e-001-1653309204286779e+001-3506618408573559e+002-7384898658004473e+003

1718281828459045

1436563656918090

1309690970754271

1238763883017086

1193819415085428

1162916490512569

1140415433587986

1123323468703890

1109911218335008

1099112183350075

1090234016850830

1082808202209954

1076506628729406

1071092802211678

1066392033175171

1062272530802742

1058633023646621

1055394425639172

1052494087144259

1049881742885176

1047516600588701

Mathematica

includeltmathhgtmain()int n=1double x=10 h=10 deriv=cos(x) diffquoerror f(x)=sin(x) printf(deriv=136enderiv)printf(h diffquo abs(deriv-diffquo) n)

while(nlt=20)h=h10diffquo=(sin(x+h)-sin(x))herror=fabs(deriv-diffquo)printf(51e 136e 136e n h diffquo error)n++

h

xfhxfxf

)()()(

Approximating a derivative by a difference quotient

Error (absolute value of derivative minus difference quotient) as a function of h (Log-Log scale)

The truncation error of the finite difference approximation is bounded by Mh2 Assuming the error in function values is bounded by ε the rounding error is bounded by 2 εh Total computational error is

h

Mh 22

8102 Mh

Where M is a bound on |frdquo(t)| for t near x There is a tradeoff between truncation error and rounding error in choosing the step size h It is minimized when

x=ones(351)x(1)=112x(2)=6111for i=335 x(i)=111-(1130-3000x(i-2))x(i-1)endx(35)

Consider the recurrence

In exact arithmetic the xk form a monotonically increasing sequence that converges to 6 Implement the recurrence and compare the computed x34 with the true value 5998 (to four correct signicant figures)

1161211

)30001130(111

10

11

xx

xxx kkk

1161211

)30001130(111

10

11

xx

xxx kkk

30001130111)( 23 xxxxp

)100()55(3025251130111)( 223 xxxxxxq

The mathematical convergence depends on the initial conditions x0 x1convergence to 100 for almost all (x0 x1)convergence to 6 x0(11- x1)=30convergence to 5 x0=x1=5 (exactly in machine representation)

fplot(x^3-111x^2+1130x-3000[-400400])

fplot(x^3-111x^2+1130x-3000[47]) fplot(x^3-111x^2+1130x-3000[-10110])

syms x1 x2 x3x1=sym(11)sym(2)x2=sym(6111)for i=335 x3=111-(1130-3000x1)x2 x1=x2 x2=x3endx3x3=1721981182794095961389986301287093876567205105910375321asymp599797252168491

Array[x35]x[1]=112x[2]=6111Do[x[i+2]=111-(1130-3000x[i])x[i+1]i133](For[i=1ilt=33i++x[i+2]=111-(1130-3000x[i])x[i+1]])(While[ilt=33x[i+2]=111-(1130-3000x[i])x[i+1]i++] )x[35]

037532167205105912870938765

8998630179409596131721981182

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

Orientation de 3 points (G Melquiond)Etant donnes 3 points du plan p q et r On veut savoir si pqr sont alignes dans le sens horaire ou dans le sens anti-horaire

float det = ( qx - px ) ( r y - py )1048576 - ( qy - py ) ( r x - px ) i f ( det gt 0) return POSITIVE i f ( det lt 0) return NEGATIVE return ZERO

p

r

q

pyrypyqy

pxrxpxqxsignerqporient

)(2

Premiere guerre du Golfe - 100h plus tard28 GI morts et 98 blesses

iraquiens US

antiminusmissile Patriotmissile Scud

Premiere guerre du Golfe - explicationLanti-missile Patriot est prevu pour fonctionner pendant quelques heuresMais ca marchait tellement bien quils lont laisse brancheLhorloge interne ajoute 01s a chaque tic Mais 01 nest pas exact en binaire rarr une petite erreur a chaque ticrarr au bout de 100h une erreur susante pour rater le missile

Mon banquier ma propose cet investissement vous me donnez e=271828 currenlannee suivante je prends 1curren de frais et je multiplie par 1lannee suivante je prends 1curren de frais et je multiplie par 2lannee suivante je prends 1curren de frais et je multiplie par 3 apres n ans je prends 1curren de frais et je multiplie par nPour recuperer mon argent il y a 1curren de fraisDans 50 ans pour ma retraite combien dargent aurai-je

6500127190887506580397803840893860464266619976132998982702001269694679091261802058160555785871702272129904252549070848000000000

Matlab 或 C(format double)

x = 27182818284590452354for annee = 1 50 x = (x - 1) anneeend x = -4396803930182069e+048

00200

x

Fori 1 i 51 i x x 1ix SimplifyxNx 1 10

MATLAB1718281828459045e+0001436563656918090e+0001309690970754271e+0001238763883017082e+0001193819415085411e+0001162916490512465e+0001140415433587258e+0001123323468698061e+0001109911218282548e+0001099112182825479e+0001090234011080270e+0001082808132963237e+0001076505728522079e+0001071080199309108e+0001066202989636622e+0001059247834185953e+0001007213181161205e+0001298372609016951e-001-1653309204286779e+001-3506618408573559e+002-7384898658004473e+003

1718281828459045

1436563656918090

1309690970754271

1238763883017086

1193819415085428

1162916490512569

1140415433587986

1123323468703890

1109911218335008

1099112183350075

1090234016850830

1082808202209954

1076506628729406

1071092802211678

1066392033175171

1062272530802742

1058633023646621

1055394425639172

1052494087144259

1049881742885176

1047516600588701

Mathematica

includeltmathhgtmain()int n=1double x=10 h=10 deriv=cos(x) diffquoerror f(x)=sin(x) printf(deriv=136enderiv)printf(h diffquo abs(deriv-diffquo) n)

while(nlt=20)h=h10diffquo=(sin(x+h)-sin(x))herror=fabs(deriv-diffquo)printf(51e 136e 136e n h diffquo error)n++

h

xfhxfxf

)()()(

Approximating a derivative by a difference quotient

Error (absolute value of derivative minus difference quotient) as a function of h (Log-Log scale)

The truncation error of the finite difference approximation is bounded by Mh2 Assuming the error in function values is bounded by ε the rounding error is bounded by 2 εh Total computational error is

h

Mh 22

8102 Mh

Where M is a bound on |frdquo(t)| for t near x There is a tradeoff between truncation error and rounding error in choosing the step size h It is minimized when

x=ones(351)x(1)=112x(2)=6111for i=335 x(i)=111-(1130-3000x(i-2))x(i-1)endx(35)

Consider the recurrence

In exact arithmetic the xk form a monotonically increasing sequence that converges to 6 Implement the recurrence and compare the computed x34 with the true value 5998 (to four correct signicant figures)

1161211

)30001130(111

10

11

xx

xxx kkk

1161211

)30001130(111

10

11

xx

xxx kkk

30001130111)( 23 xxxxp

)100()55(3025251130111)( 223 xxxxxxq

The mathematical convergence depends on the initial conditions x0 x1convergence to 100 for almost all (x0 x1)convergence to 6 x0(11- x1)=30convergence to 5 x0=x1=5 (exactly in machine representation)

fplot(x^3-111x^2+1130x-3000[-400400])

fplot(x^3-111x^2+1130x-3000[47]) fplot(x^3-111x^2+1130x-3000[-10110])

syms x1 x2 x3x1=sym(11)sym(2)x2=sym(6111)for i=335 x3=111-(1130-3000x1)x2 x1=x2 x2=x3endx3x3=1721981182794095961389986301287093876567205105910375321asymp599797252168491

Array[x35]x[1]=112x[2]=6111Do[x[i+2]=111-(1130-3000x[i])x[i+1]i133](For[i=1ilt=33i++x[i+2]=111-(1130-3000x[i])x[i+1]])(While[ilt=33x[i+2]=111-(1130-3000x[i])x[i+1]i++] )x[35]

037532167205105912870938765

8998630179409596131721981182

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

Premiere guerre du Golfe - 100h plus tard28 GI morts et 98 blesses

iraquiens US

antiminusmissile Patriotmissile Scud

Premiere guerre du Golfe - explicationLanti-missile Patriot est prevu pour fonctionner pendant quelques heuresMais ca marchait tellement bien quils lont laisse brancheLhorloge interne ajoute 01s a chaque tic Mais 01 nest pas exact en binaire rarr une petite erreur a chaque ticrarr au bout de 100h une erreur susante pour rater le missile

Mon banquier ma propose cet investissement vous me donnez e=271828 currenlannee suivante je prends 1curren de frais et je multiplie par 1lannee suivante je prends 1curren de frais et je multiplie par 2lannee suivante je prends 1curren de frais et je multiplie par 3 apres n ans je prends 1curren de frais et je multiplie par nPour recuperer mon argent il y a 1curren de fraisDans 50 ans pour ma retraite combien dargent aurai-je

6500127190887506580397803840893860464266619976132998982702001269694679091261802058160555785871702272129904252549070848000000000

Matlab 或 C(format double)

x = 27182818284590452354for annee = 1 50 x = (x - 1) anneeend x = -4396803930182069e+048

00200

x

Fori 1 i 51 i x x 1ix SimplifyxNx 1 10

MATLAB1718281828459045e+0001436563656918090e+0001309690970754271e+0001238763883017082e+0001193819415085411e+0001162916490512465e+0001140415433587258e+0001123323468698061e+0001109911218282548e+0001099112182825479e+0001090234011080270e+0001082808132963237e+0001076505728522079e+0001071080199309108e+0001066202989636622e+0001059247834185953e+0001007213181161205e+0001298372609016951e-001-1653309204286779e+001-3506618408573559e+002-7384898658004473e+003

1718281828459045

1436563656918090

1309690970754271

1238763883017086

1193819415085428

1162916490512569

1140415433587986

1123323468703890

1109911218335008

1099112183350075

1090234016850830

1082808202209954

1076506628729406

1071092802211678

1066392033175171

1062272530802742

1058633023646621

1055394425639172

1052494087144259

1049881742885176

1047516600588701

Mathematica

includeltmathhgtmain()int n=1double x=10 h=10 deriv=cos(x) diffquoerror f(x)=sin(x) printf(deriv=136enderiv)printf(h diffquo abs(deriv-diffquo) n)

while(nlt=20)h=h10diffquo=(sin(x+h)-sin(x))herror=fabs(deriv-diffquo)printf(51e 136e 136e n h diffquo error)n++

h

xfhxfxf

)()()(

Approximating a derivative by a difference quotient

Error (absolute value of derivative minus difference quotient) as a function of h (Log-Log scale)

The truncation error of the finite difference approximation is bounded by Mh2 Assuming the error in function values is bounded by ε the rounding error is bounded by 2 εh Total computational error is

h

Mh 22

8102 Mh

Where M is a bound on |frdquo(t)| for t near x There is a tradeoff between truncation error and rounding error in choosing the step size h It is minimized when

x=ones(351)x(1)=112x(2)=6111for i=335 x(i)=111-(1130-3000x(i-2))x(i-1)endx(35)

Consider the recurrence

In exact arithmetic the xk form a monotonically increasing sequence that converges to 6 Implement the recurrence and compare the computed x34 with the true value 5998 (to four correct signicant figures)

1161211

)30001130(111

10

11

xx

xxx kkk

1161211

)30001130(111

10

11

xx

xxx kkk

30001130111)( 23 xxxxp

)100()55(3025251130111)( 223 xxxxxxq

The mathematical convergence depends on the initial conditions x0 x1convergence to 100 for almost all (x0 x1)convergence to 6 x0(11- x1)=30convergence to 5 x0=x1=5 (exactly in machine representation)

fplot(x^3-111x^2+1130x-3000[-400400])

fplot(x^3-111x^2+1130x-3000[47]) fplot(x^3-111x^2+1130x-3000[-10110])

syms x1 x2 x3x1=sym(11)sym(2)x2=sym(6111)for i=335 x3=111-(1130-3000x1)x2 x1=x2 x2=x3endx3x3=1721981182794095961389986301287093876567205105910375321asymp599797252168491

Array[x35]x[1]=112x[2]=6111Do[x[i+2]=111-(1130-3000x[i])x[i+1]i133](For[i=1ilt=33i++x[i+2]=111-(1130-3000x[i])x[i+1]])(While[ilt=33x[i+2]=111-(1130-3000x[i])x[i+1]i++] )x[35]

037532167205105912870938765

8998630179409596131721981182

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

Mon banquier ma propose cet investissement vous me donnez e=271828 currenlannee suivante je prends 1curren de frais et je multiplie par 1lannee suivante je prends 1curren de frais et je multiplie par 2lannee suivante je prends 1curren de frais et je multiplie par 3 apres n ans je prends 1curren de frais et je multiplie par nPour recuperer mon argent il y a 1curren de fraisDans 50 ans pour ma retraite combien dargent aurai-je

6500127190887506580397803840893860464266619976132998982702001269694679091261802058160555785871702272129904252549070848000000000

Matlab 或 C(format double)

x = 27182818284590452354for annee = 1 50 x = (x - 1) anneeend x = -4396803930182069e+048

00200

x

Fori 1 i 51 i x x 1ix SimplifyxNx 1 10

MATLAB1718281828459045e+0001436563656918090e+0001309690970754271e+0001238763883017082e+0001193819415085411e+0001162916490512465e+0001140415433587258e+0001123323468698061e+0001109911218282548e+0001099112182825479e+0001090234011080270e+0001082808132963237e+0001076505728522079e+0001071080199309108e+0001066202989636622e+0001059247834185953e+0001007213181161205e+0001298372609016951e-001-1653309204286779e+001-3506618408573559e+002-7384898658004473e+003

1718281828459045

1436563656918090

1309690970754271

1238763883017086

1193819415085428

1162916490512569

1140415433587986

1123323468703890

1109911218335008

1099112183350075

1090234016850830

1082808202209954

1076506628729406

1071092802211678

1066392033175171

1062272530802742

1058633023646621

1055394425639172

1052494087144259

1049881742885176

1047516600588701

Mathematica

includeltmathhgtmain()int n=1double x=10 h=10 deriv=cos(x) diffquoerror f(x)=sin(x) printf(deriv=136enderiv)printf(h diffquo abs(deriv-diffquo) n)

while(nlt=20)h=h10diffquo=(sin(x+h)-sin(x))herror=fabs(deriv-diffquo)printf(51e 136e 136e n h diffquo error)n++

h

xfhxfxf

)()()(

Approximating a derivative by a difference quotient

Error (absolute value of derivative minus difference quotient) as a function of h (Log-Log scale)

The truncation error of the finite difference approximation is bounded by Mh2 Assuming the error in function values is bounded by ε the rounding error is bounded by 2 εh Total computational error is

h

Mh 22

8102 Mh

Where M is a bound on |frdquo(t)| for t near x There is a tradeoff between truncation error and rounding error in choosing the step size h It is minimized when

x=ones(351)x(1)=112x(2)=6111for i=335 x(i)=111-(1130-3000x(i-2))x(i-1)endx(35)

Consider the recurrence

In exact arithmetic the xk form a monotonically increasing sequence that converges to 6 Implement the recurrence and compare the computed x34 with the true value 5998 (to four correct signicant figures)

1161211

)30001130(111

10

11

xx

xxx kkk

1161211

)30001130(111

10

11

xx

xxx kkk

30001130111)( 23 xxxxp

)100()55(3025251130111)( 223 xxxxxxq

The mathematical convergence depends on the initial conditions x0 x1convergence to 100 for almost all (x0 x1)convergence to 6 x0(11- x1)=30convergence to 5 x0=x1=5 (exactly in machine representation)

fplot(x^3-111x^2+1130x-3000[-400400])

fplot(x^3-111x^2+1130x-3000[47]) fplot(x^3-111x^2+1130x-3000[-10110])

syms x1 x2 x3x1=sym(11)sym(2)x2=sym(6111)for i=335 x3=111-(1130-3000x1)x2 x1=x2 x2=x3endx3x3=1721981182794095961389986301287093876567205105910375321asymp599797252168491

Array[x35]x[1]=112x[2]=6111Do[x[i+2]=111-(1130-3000x[i])x[i+1]i133](For[i=1ilt=33i++x[i+2]=111-(1130-3000x[i])x[i+1]])(While[ilt=33x[i+2]=111-(1130-3000x[i])x[i+1]i++] )x[35]

037532167205105912870938765

8998630179409596131721981182

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

6500127190887506580397803840893860464266619976132998982702001269694679091261802058160555785871702272129904252549070848000000000

Matlab 或 C(format double)

x = 27182818284590452354for annee = 1 50 x = (x - 1) anneeend x = -4396803930182069e+048

00200

x

Fori 1 i 51 i x x 1ix SimplifyxNx 1 10

MATLAB1718281828459045e+0001436563656918090e+0001309690970754271e+0001238763883017082e+0001193819415085411e+0001162916490512465e+0001140415433587258e+0001123323468698061e+0001109911218282548e+0001099112182825479e+0001090234011080270e+0001082808132963237e+0001076505728522079e+0001071080199309108e+0001066202989636622e+0001059247834185953e+0001007213181161205e+0001298372609016951e-001-1653309204286779e+001-3506618408573559e+002-7384898658004473e+003

1718281828459045

1436563656918090

1309690970754271

1238763883017086

1193819415085428

1162916490512569

1140415433587986

1123323468703890

1109911218335008

1099112183350075

1090234016850830

1082808202209954

1076506628729406

1071092802211678

1066392033175171

1062272530802742

1058633023646621

1055394425639172

1052494087144259

1049881742885176

1047516600588701

Mathematica

includeltmathhgtmain()int n=1double x=10 h=10 deriv=cos(x) diffquoerror f(x)=sin(x) printf(deriv=136enderiv)printf(h diffquo abs(deriv-diffquo) n)

while(nlt=20)h=h10diffquo=(sin(x+h)-sin(x))herror=fabs(deriv-diffquo)printf(51e 136e 136e n h diffquo error)n++

h

xfhxfxf

)()()(

Approximating a derivative by a difference quotient

Error (absolute value of derivative minus difference quotient) as a function of h (Log-Log scale)

The truncation error of the finite difference approximation is bounded by Mh2 Assuming the error in function values is bounded by ε the rounding error is bounded by 2 εh Total computational error is

h

Mh 22

8102 Mh

Where M is a bound on |frdquo(t)| for t near x There is a tradeoff between truncation error and rounding error in choosing the step size h It is minimized when

x=ones(351)x(1)=112x(2)=6111for i=335 x(i)=111-(1130-3000x(i-2))x(i-1)endx(35)

Consider the recurrence

In exact arithmetic the xk form a monotonically increasing sequence that converges to 6 Implement the recurrence and compare the computed x34 with the true value 5998 (to four correct signicant figures)

1161211

)30001130(111

10

11

xx

xxx kkk

1161211

)30001130(111

10

11

xx

xxx kkk

30001130111)( 23 xxxxp

)100()55(3025251130111)( 223 xxxxxxq

The mathematical convergence depends on the initial conditions x0 x1convergence to 100 for almost all (x0 x1)convergence to 6 x0(11- x1)=30convergence to 5 x0=x1=5 (exactly in machine representation)

fplot(x^3-111x^2+1130x-3000[-400400])

fplot(x^3-111x^2+1130x-3000[47]) fplot(x^3-111x^2+1130x-3000[-10110])

syms x1 x2 x3x1=sym(11)sym(2)x2=sym(6111)for i=335 x3=111-(1130-3000x1)x2 x1=x2 x2=x3endx3x3=1721981182794095961389986301287093876567205105910375321asymp599797252168491

Array[x35]x[1]=112x[2]=6111Do[x[i+2]=111-(1130-3000x[i])x[i+1]i133](For[i=1ilt=33i++x[i+2]=111-(1130-3000x[i])x[i+1]])(While[ilt=33x[i+2]=111-(1130-3000x[i])x[i+1]i++] )x[35]

037532167205105912870938765

8998630179409596131721981182

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

MATLAB1718281828459045e+0001436563656918090e+0001309690970754271e+0001238763883017082e+0001193819415085411e+0001162916490512465e+0001140415433587258e+0001123323468698061e+0001109911218282548e+0001099112182825479e+0001090234011080270e+0001082808132963237e+0001076505728522079e+0001071080199309108e+0001066202989636622e+0001059247834185953e+0001007213181161205e+0001298372609016951e-001-1653309204286779e+001-3506618408573559e+002-7384898658004473e+003

1718281828459045

1436563656918090

1309690970754271

1238763883017086

1193819415085428

1162916490512569

1140415433587986

1123323468703890

1109911218335008

1099112183350075

1090234016850830

1082808202209954

1076506628729406

1071092802211678

1066392033175171

1062272530802742

1058633023646621

1055394425639172

1052494087144259

1049881742885176

1047516600588701

Mathematica

includeltmathhgtmain()int n=1double x=10 h=10 deriv=cos(x) diffquoerror f(x)=sin(x) printf(deriv=136enderiv)printf(h diffquo abs(deriv-diffquo) n)

while(nlt=20)h=h10diffquo=(sin(x+h)-sin(x))herror=fabs(deriv-diffquo)printf(51e 136e 136e n h diffquo error)n++

h

xfhxfxf

)()()(

Approximating a derivative by a difference quotient

Error (absolute value of derivative minus difference quotient) as a function of h (Log-Log scale)

The truncation error of the finite difference approximation is bounded by Mh2 Assuming the error in function values is bounded by ε the rounding error is bounded by 2 εh Total computational error is

h

Mh 22

8102 Mh

Where M is a bound on |frdquo(t)| for t near x There is a tradeoff between truncation error and rounding error in choosing the step size h It is minimized when

x=ones(351)x(1)=112x(2)=6111for i=335 x(i)=111-(1130-3000x(i-2))x(i-1)endx(35)

Consider the recurrence

In exact arithmetic the xk form a monotonically increasing sequence that converges to 6 Implement the recurrence and compare the computed x34 with the true value 5998 (to four correct signicant figures)

1161211

)30001130(111

10

11

xx

xxx kkk

1161211

)30001130(111

10

11

xx

xxx kkk

30001130111)( 23 xxxxp

)100()55(3025251130111)( 223 xxxxxxq

The mathematical convergence depends on the initial conditions x0 x1convergence to 100 for almost all (x0 x1)convergence to 6 x0(11- x1)=30convergence to 5 x0=x1=5 (exactly in machine representation)

fplot(x^3-111x^2+1130x-3000[-400400])

fplot(x^3-111x^2+1130x-3000[47]) fplot(x^3-111x^2+1130x-3000[-10110])

syms x1 x2 x3x1=sym(11)sym(2)x2=sym(6111)for i=335 x3=111-(1130-3000x1)x2 x1=x2 x2=x3endx3x3=1721981182794095961389986301287093876567205105910375321asymp599797252168491

Array[x35]x[1]=112x[2]=6111Do[x[i+2]=111-(1130-3000x[i])x[i+1]i133](For[i=1ilt=33i++x[i+2]=111-(1130-3000x[i])x[i+1]])(While[ilt=33x[i+2]=111-(1130-3000x[i])x[i+1]i++] )x[35]

037532167205105912870938765

8998630179409596131721981182

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

includeltmathhgtmain()int n=1double x=10 h=10 deriv=cos(x) diffquoerror f(x)=sin(x) printf(deriv=136enderiv)printf(h diffquo abs(deriv-diffquo) n)

while(nlt=20)h=h10diffquo=(sin(x+h)-sin(x))herror=fabs(deriv-diffquo)printf(51e 136e 136e n h diffquo error)n++

h

xfhxfxf

)()()(

Approximating a derivative by a difference quotient

Error (absolute value of derivative minus difference quotient) as a function of h (Log-Log scale)

The truncation error of the finite difference approximation is bounded by Mh2 Assuming the error in function values is bounded by ε the rounding error is bounded by 2 εh Total computational error is

h

Mh 22

8102 Mh

Where M is a bound on |frdquo(t)| for t near x There is a tradeoff between truncation error and rounding error in choosing the step size h It is minimized when

x=ones(351)x(1)=112x(2)=6111for i=335 x(i)=111-(1130-3000x(i-2))x(i-1)endx(35)

Consider the recurrence

In exact arithmetic the xk form a monotonically increasing sequence that converges to 6 Implement the recurrence and compare the computed x34 with the true value 5998 (to four correct signicant figures)

1161211

)30001130(111

10

11

xx

xxx kkk

1161211

)30001130(111

10

11

xx

xxx kkk

30001130111)( 23 xxxxp

)100()55(3025251130111)( 223 xxxxxxq

The mathematical convergence depends on the initial conditions x0 x1convergence to 100 for almost all (x0 x1)convergence to 6 x0(11- x1)=30convergence to 5 x0=x1=5 (exactly in machine representation)

fplot(x^3-111x^2+1130x-3000[-400400])

fplot(x^3-111x^2+1130x-3000[47]) fplot(x^3-111x^2+1130x-3000[-10110])

syms x1 x2 x3x1=sym(11)sym(2)x2=sym(6111)for i=335 x3=111-(1130-3000x1)x2 x1=x2 x2=x3endx3x3=1721981182794095961389986301287093876567205105910375321asymp599797252168491

Array[x35]x[1]=112x[2]=6111Do[x[i+2]=111-(1130-3000x[i])x[i+1]i133](For[i=1ilt=33i++x[i+2]=111-(1130-3000x[i])x[i+1]])(While[ilt=33x[i+2]=111-(1130-3000x[i])x[i+1]i++] )x[35]

037532167205105912870938765

8998630179409596131721981182

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

Error (absolute value of derivative minus difference quotient) as a function of h (Log-Log scale)

The truncation error of the finite difference approximation is bounded by Mh2 Assuming the error in function values is bounded by ε the rounding error is bounded by 2 εh Total computational error is

h

Mh 22

8102 Mh

Where M is a bound on |frdquo(t)| for t near x There is a tradeoff between truncation error and rounding error in choosing the step size h It is minimized when

x=ones(351)x(1)=112x(2)=6111for i=335 x(i)=111-(1130-3000x(i-2))x(i-1)endx(35)

Consider the recurrence

In exact arithmetic the xk form a monotonically increasing sequence that converges to 6 Implement the recurrence and compare the computed x34 with the true value 5998 (to four correct signicant figures)

1161211

)30001130(111

10

11

xx

xxx kkk

1161211

)30001130(111

10

11

xx

xxx kkk

30001130111)( 23 xxxxp

)100()55(3025251130111)( 223 xxxxxxq

The mathematical convergence depends on the initial conditions x0 x1convergence to 100 for almost all (x0 x1)convergence to 6 x0(11- x1)=30convergence to 5 x0=x1=5 (exactly in machine representation)

fplot(x^3-111x^2+1130x-3000[-400400])

fplot(x^3-111x^2+1130x-3000[47]) fplot(x^3-111x^2+1130x-3000[-10110])

syms x1 x2 x3x1=sym(11)sym(2)x2=sym(6111)for i=335 x3=111-(1130-3000x1)x2 x1=x2 x2=x3endx3x3=1721981182794095961389986301287093876567205105910375321asymp599797252168491

Array[x35]x[1]=112x[2]=6111Do[x[i+2]=111-(1130-3000x[i])x[i+1]i133](For[i=1ilt=33i++x[i+2]=111-(1130-3000x[i])x[i+1]])(While[ilt=33x[i+2]=111-(1130-3000x[i])x[i+1]i++] )x[35]

037532167205105912870938765

8998630179409596131721981182

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

The truncation error of the finite difference approximation is bounded by Mh2 Assuming the error in function values is bounded by ε the rounding error is bounded by 2 εh Total computational error is

h

Mh 22

8102 Mh

Where M is a bound on |frdquo(t)| for t near x There is a tradeoff between truncation error and rounding error in choosing the step size h It is minimized when

x=ones(351)x(1)=112x(2)=6111for i=335 x(i)=111-(1130-3000x(i-2))x(i-1)endx(35)

Consider the recurrence

In exact arithmetic the xk form a monotonically increasing sequence that converges to 6 Implement the recurrence and compare the computed x34 with the true value 5998 (to four correct signicant figures)

1161211

)30001130(111

10

11

xx

xxx kkk

1161211

)30001130(111

10

11

xx

xxx kkk

30001130111)( 23 xxxxp

)100()55(3025251130111)( 223 xxxxxxq

The mathematical convergence depends on the initial conditions x0 x1convergence to 100 for almost all (x0 x1)convergence to 6 x0(11- x1)=30convergence to 5 x0=x1=5 (exactly in machine representation)

fplot(x^3-111x^2+1130x-3000[-400400])

fplot(x^3-111x^2+1130x-3000[47]) fplot(x^3-111x^2+1130x-3000[-10110])

syms x1 x2 x3x1=sym(11)sym(2)x2=sym(6111)for i=335 x3=111-(1130-3000x1)x2 x1=x2 x2=x3endx3x3=1721981182794095961389986301287093876567205105910375321asymp599797252168491

Array[x35]x[1]=112x[2]=6111Do[x[i+2]=111-(1130-3000x[i])x[i+1]i133](For[i=1ilt=33i++x[i+2]=111-(1130-3000x[i])x[i+1]])(While[ilt=33x[i+2]=111-(1130-3000x[i])x[i+1]i++] )x[35]

037532167205105912870938765

8998630179409596131721981182

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

x=ones(351)x(1)=112x(2)=6111for i=335 x(i)=111-(1130-3000x(i-2))x(i-1)endx(35)

Consider the recurrence

In exact arithmetic the xk form a monotonically increasing sequence that converges to 6 Implement the recurrence and compare the computed x34 with the true value 5998 (to four correct signicant figures)

1161211

)30001130(111

10

11

xx

xxx kkk

1161211

)30001130(111

10

11

xx

xxx kkk

30001130111)( 23 xxxxp

)100()55(3025251130111)( 223 xxxxxxq

The mathematical convergence depends on the initial conditions x0 x1convergence to 100 for almost all (x0 x1)convergence to 6 x0(11- x1)=30convergence to 5 x0=x1=5 (exactly in machine representation)

fplot(x^3-111x^2+1130x-3000[-400400])

fplot(x^3-111x^2+1130x-3000[47]) fplot(x^3-111x^2+1130x-3000[-10110])

syms x1 x2 x3x1=sym(11)sym(2)x2=sym(6111)for i=335 x3=111-(1130-3000x1)x2 x1=x2 x2=x3endx3x3=1721981182794095961389986301287093876567205105910375321asymp599797252168491

Array[x35]x[1]=112x[2]=6111Do[x[i+2]=111-(1130-3000x[i])x[i+1]i133](For[i=1ilt=33i++x[i+2]=111-(1130-3000x[i])x[i+1]])(While[ilt=33x[i+2]=111-(1130-3000x[i])x[i+1]i++] )x[35]

037532167205105912870938765

8998630179409596131721981182

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

1161211

)30001130(111

10

11

xx

xxx kkk

30001130111)( 23 xxxxp

)100()55(3025251130111)( 223 xxxxxxq

The mathematical convergence depends on the initial conditions x0 x1convergence to 100 for almost all (x0 x1)convergence to 6 x0(11- x1)=30convergence to 5 x0=x1=5 (exactly in machine representation)

fplot(x^3-111x^2+1130x-3000[-400400])

fplot(x^3-111x^2+1130x-3000[47]) fplot(x^3-111x^2+1130x-3000[-10110])

syms x1 x2 x3x1=sym(11)sym(2)x2=sym(6111)for i=335 x3=111-(1130-3000x1)x2 x1=x2 x2=x3endx3x3=1721981182794095961389986301287093876567205105910375321asymp599797252168491

Array[x35]x[1]=112x[2]=6111Do[x[i+2]=111-(1130-3000x[i])x[i+1]i133](For[i=1ilt=33i++x[i+2]=111-(1130-3000x[i])x[i+1]])(While[ilt=33x[i+2]=111-(1130-3000x[i])x[i+1]i++] )x[35]

037532167205105912870938765

8998630179409596131721981182

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

fplot(x^3-111x^2+1130x-3000[47]) fplot(x^3-111x^2+1130x-3000[-10110])

syms x1 x2 x3x1=sym(11)sym(2)x2=sym(6111)for i=335 x3=111-(1130-3000x1)x2 x1=x2 x2=x3endx3x3=1721981182794095961389986301287093876567205105910375321asymp599797252168491

Array[x35]x[1]=112x[2]=6111Do[x[i+2]=111-(1130-3000x[i])x[i+1]i133](For[i=1ilt=33i++x[i+2]=111-(1130-3000x[i])x[i+1]])(While[ilt=33x[i+2]=111-(1130-3000x[i])x[i+1]i++] )x[35]

037532167205105912870938765

8998630179409596131721981182

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

syms x1 x2 x3x1=sym(11)sym(2)x2=sym(6111)for i=335 x3=111-(1130-3000x1)x2 x1=x2 x2=x3endx3x3=1721981182794095961389986301287093876567205105910375321asymp599797252168491

Array[x35]x[1]=112x[2]=6111Do[x[i+2]=111-(1130-3000x[i])x[i+1]i133](For[i=1ilt=33i++x[i+2]=111-(1130-3000x[i])x[i+1]])(While[ilt=33x[i+2]=111-(1130-3000x[i])x[i+1]i++] )x[35]

037532167205105912870938765

8998630179409596131721981182

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

Array[x35]x[1]=112x[2]=6111Do[x[i+2]=111-(1130-3000x[i])x[i+1]i133](For[i=1ilt=33i++x[i+2]=111-(1130-3000x[i])x[i+1]])(While[ilt=33x[i+2]=111-(1130-3000x[i])x[i+1]i++] )x[35]

037532167205105912870938765

8998630179409596131721981182

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

Cancellationgtgtx=12e-8 (1-cos(x))x^2ans=07710

保留 32位有效数字得049999999999999999400001004443762gtgt

2

2

)2sin(

2

1

x

x

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

acb 42

acxx

a

acbbsignbx

2

4)(

21

2

1

))()(( csbsassA

))())(())(((4

1cbabaccbaA

Solving a Quadratic EquationIf

Rename a b c so that agebgec (Kahan)

The parentheses are essential

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

n

iin xx

ns

1

22

1

1

n

iixn

x1

1

2

11

22 1

1

1 n

ii

n

iin xn

xn

s

Computing the Sample Variance

(requires two passes through the data)

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

)1exp(e

n

n ne

11lim

Accumulation of Rounding ErrorsApproximate

by taking finite n in the definition

e=exp(1)for i=115 n=10^i (1+1n)^n-eend

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

11 10

1

1

01

11

1

A

11)1( fl

10

00

0

1

1

01

11

1ˆˆ11

ULA

Instability Without Cancellation(1) The Need for Pivoting

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

(2) An Innocuous Calculation

for i=160 x=sqrt(x)endfor i=160 x=x^2end

gtgtx=10ans=1

gtgtx=05ans=0

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

(3) An Infinite Sum

48644934066816

1 2

12

n n

Sum in the opposite order from smallest to largest

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

||||

xxx

Axx

504010

407030

206040

A

Rounding Errors Can Be BeneficialPower method

Consider the matrix

Which has eigenvalues 0 04394 1161 and an eigenvector [111]T

corresponding to the eigenvalue zero

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

Gauss

Givens

Neumann

Goldstine

Turing

Wilkinson

Backward error analysis

Condition number

)( Afx A

AA

)(~ AAfx

le 向前误差 条件数 X 向后误差

数据 解

(A nearby problem)

Condition number backward error

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

数值运算的简单原则 运算次序 求和时从小到大相加避免相近的数相减避免小分母 分母小会造成浮点溢出避免大数吃小数简化计算步骤选用稳定的算法

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

References1 C W Ueberhuber Numerical Computation 1 Springer 19972 Sylvie Boldo Pourquoi mon ordinateur calcule faux CR INRIA-Equip

e-projet ProVal 18 janvier 20083 M T Heath Scientific Computing An Introductory Survey McGraw-Hil

l Companies Inc 20024 M L Overton Numerical Computing with IEEE Floating Point Arithmet

ic SIAM 20015 N Higham Accuracy ans Stability of Numerical Algorithms 2nd edition

SIAM 2002

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36

1 用 For 循环计算

2 用 While 循环找到最小的正整数 k 使得

3 计算函数值

4 取 n=10^15 计算

5 下一题复习用 Matlab 做图并为下次内容做准备做出下面二次型对应的曲面

提示[X Y]=meshgrid(-6056)Z=05(3X^2 + 4XY+6Y^2)-2X+8Y再用 surf(XYZ)

100000

12

1

n n

)1021()cos(1

)( 82

fx

xxf

n

nnf

11)(

8

2

62

23

2

1)(

bAxbAxxxf TT

124

14

10590

1

k

n n

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36