teoria de errores nuevo

8/17/2019 Teoria de Errores Nuevo

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1

METODOS NUMERICOS I

Ing. William Chauca Nolasco

2016-I


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2

REPRESENTACIÓN BINARIA


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htt!""num#$icalm#tho%s.#ng.us&.#%u'

Cómo se representa un número decimal

21012 10610710710510276.257 −−

×+×+×+×+×=


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(

)as# 2

1875.11

)21212020(

)21212021()0011.1011(

104321

0123

2

=

×+×+×+×+

×+×+×+×=

−−−−


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*

Table 1 Con+#$ting a ,as#-10 int#g#$ to ,ina$ $#$#s#ntation.

Cociente Resto

11"2 *

*"2 2

2"2 1

1"2 0

o$ lo tanto

CONERSION !E "N ENTERO EN BASE 1# ABINARIA

01 a=11 a=

20 a=

31 a=

2

2012310

)1011(

)()11(

=

= aaaa


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6

Sta$t

Inut /N10

i 0

Di+i%# N , 2 to g#t uoti#nt 3 4$#main%#$ R

ai R

Is 3 05

n i/N

10 /a

n. . .a

0

2

STO

Int#g#$ N to ,# con+#$t#% to,ina$ &o$mat

ii17N3

No

8#s


7/939

N$%ERO !ECI%A& 'RACCIONARIO ABINARIO

Num,#$Num,#$%#su#s%#cimal

Num,#$ ant#s%#cimal

0.'9* 0.'9*

0.9* 0.9*

1.* 0.*

1.0 0.0

Table (. Con+#$ting a ,as#-10 &$action to ,ina$ $#$#s#ntation.

o$ lo tanto

10 −= a

20 −= a

31 −= a

41 −= a

2

2432110

)0011.0(

)()1875.0(

=

=−−−−

aaaa

21875.0 ×

2375.0 ×

275.0 ×

25.0 ×


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Sta$t

Inut /;10

Multil ; , 2 to g#tnum,#$ ,#&o$# %#cimal7 San% a&t#$ %#cimal7 T

ai R

Is T 05

n i/;

10 /a

-1. . .a

-n

2

STO

;$action ; to ,# con+#$t#%to ,ina$ &o$mat

No

8#s

T =−= F1,ii

1i −=


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N$%ERO !ECI%A& A BINARIO

t#n#mos

Entonc#s

( ) ( )210

?.?1875.11 =

210 )1011()11( =

210 )0011.0()1875.0( =

210 )0011.1011()1875.11( =


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Todos los números decimales )raccionarios no sepueden representar e*actamente

NumberNumber a)ter

decimal

Number be)ore!ecimal

0.6 0.6

1.2 0.2

0.( 0.(

0.: 0.:

1.6 0.6

Table +, Con-ertin. a base/1# )raction to appro*imate binar0representation,

23.0 ×26.0 ×22.0 ×24.0 ×28.0 ×

10 −= a

21 −= a

3

0−

= a

40 −= a51 −= a

28125.0)01001.0()()3.0( 225432110 ==≈ −−−−− aaaaa


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Otra manera de -er lacon-ersión

Con+#$t to ,as# 2( )101875.11

( )

( ) 2

0123

013

13

3

10

1011

21212021

222

122

3211

=×+×+×+×=

++=

++=+=


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( )

( ) 2

4321

43

3

10

0011.

21212020

22

0625.021875.0

=

×+×+×+×=

+=

+=

−−−−

−−

−

( ) ( )210

0011.10111875.11 =


13/931'


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1(

REPRESENTACIÓN !E& P"NTO'&OTANTE


15/931*

Punto fotante decimal: FormaCientíca

2

3

2

10.56782aswrittenis78.256



×−−

×+

×+−


16/9316

E=#mlo

E=#mlo! a$a

>a &o$ma #s!

o

exponent10mantissasign ××em 10××σ

2105678.2 ×−

2

5678.2

1

==−=

e

m

σ


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'ormato de la coma otante para losnúmeros binarios

1 no s# almac#na mi#nt$as u# s# %a si#m$# a$a s#$ 1.

( )

( ) ( )[ ] mm

m y e

22 101mantissa

ve-or1ve,or0n!m"ero sign

2


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E2emploPalabra hipotetica de 9 bits

th# ?$st ,it is us#% &o$ th# sign o& th# num,#$7 th# s#con% ,it &o$ th# sign o& th# #@on#nt7 th# n#@t &ou$ ,its &o$ th# mantissa7 an% th# n#@t th$## ,its &o$ th# #@on#nt

3e 4a-e t4e representation as

0 0 1 0 1 1 1 0 1

Sign o& th#num,#$

mantissaSign o& th##@on#nt

#@on#nt

( ) ( ) ( )

( ) ( )22

5

2210

1011011.1

21011011.111.11011075.54

×≅

×==


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#psi$on %e $a &a'!ina

D#?ni%o como la m#%i%a %# #@actitu%

calcula%o o$ %i$#ncia #nt$# 1 #l nAm#$osigui#nt# u# u#%#n s#$ $#$#s#nta%os


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E@aml#

Di#B ,it o$%

Signo %#l num#$oSigno %#l #@on#nt#

Sigui#nt# cuat$o ,its a$a #l #@on#nt#Sigui#nt# cuat$o ,its a$a la mantissa

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 1N#@tnum,#$

4210625.1 −=−=∈mach

( )101=

( ) ( )102 0625.10001.1 ==


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Error relati-o 0 epsilon de la ma5uina

E@aml#

1# bit 6ord 7si.n8 si.n o) e*ponent8 9 )or e*ponent8 9 )or mantissa:

0 1 0 1 1 0 1 1 0 0

Sign o& th#num,#$

mantissa

Sign o& th##@on#nt

#@on#nt

El error verdadero relativo absoluto en la representación

de un número será menor entonces el épsilon de lamáuina

( ) ( )

( ) ( )20110

2

5

210

21100.1

21100.102832.0

−

−

×=

×≅

( ) ( )

0625.02034472.0

02832.0

0274375.002832.0

0274375.021100.1

4

0110

22

=


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IEEE ;


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IEEE-9*( ;loating oint

Stan%a$%Estandari>a la representación de los números depunto otante en di-ersas computadoras ensimple 0 doble precisión

Estandari>a la representación de las operacionesde punto otante en di-ersas computadoras,


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RE'ERENCIA I%PORTANTE

Ca%a $osional #n ci#ncia # ing#ni#$a /#incluso si ust#% no lo #s %#,# sa,#$ so,$#a$itmtica %# unto Fotant#G


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2*

IEEE-9*( ;o$mat Singl#$#cision

+( bits )or sin.le precision

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Sign/s

)ias#%E@on#nt /#H

Mantissa /m

( ) 1272 21)1(a$!e . −××−= e s m


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26

E@aml#11 1 0 1 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Sign/s

)ias#%E@on#nt /#H

Mantissa /m

( ) ( ) 1272 2.11a$!e −××−= e s m

( ) ( ) 127)10100010(21

2210100000.11 −××−=

( ) ( ) 127162

2625.11

−

××−= ( ) ( ) 1035 105834.52625.11 ×−=××−=

E l 2


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29

E@aml#2

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

Sign/s

)ias#%E@on#nt /#H

Mantissa /m

R#$#s#nt -*.*:'(@1010 as a singl#$#cision Foating oint num,#$.

( ) ( ) ?110 2?.11105834.5 ±××−=×−


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2:

E*ponent )or +( Bit IEEE/;


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Casos especiales de e#ponentes

Kctual $ang# o&

an%

are reserved $or special numbers

Kctual $ang# o&

e′2541 ≤′≤ e

0=′e 255=′e

127126 ≤≤− e

e


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S#cial E@on#nts an% Num,#$s

all B#$os

all on#s

s m Represents

0 all zeros all zeros 0

1 all zeros all zeros -0

0 all ones all zeros

1 all ones all zeros

0 or 1 all ones non-zero NaN

0=′e255=′e

e′

∞∞−


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'1

IEEE-9*( ;o$mat

Th# la$g#st num,#$ , magnitu%#

Th# small#st num,#$ , magnitu%#

Machin# #silon

( ) 381272 1040.321........1.1 ×=×

( ) 381262 1018.220......00.1 −− ×=×

723 101*.12 −− ×==machε


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'2


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''

'uentes de ERROR


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'(

%os son las Fuentes de errores en

calculo numérico

1 Roun% oL #$$o$

2 T$uncation #$$o$


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'*

Roun% oL E$$o$

Causa%o o$ $#$#s#nta$a$o@ima%am#nt# un num#$o.

333333.03

1≅

...4142.12 ≅


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'6

$o,l#mas c$#a%os o$ #$$o$ %#$#%on%#o

(? americanos son asesinados en )ebrero el (<de 1@@1 por un misil Scud ira5u en !4a4ran8Arabia saudi,

El sistema de de$ensa P&'()*' no pudo rastreare interceptar el +cud, Por ué-


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'9

$o,l#ma con #l Misil at$iotEl ciclo de relo2 de 11# de se.undos )uerepresentado en re.istro de punto 2o (9/bitcreó un error de @8< * 1# /? se.undos,

>a ,at#$a #$a #nc#n%i%o o$ 100 ho$ascons#cuti+as7 as causan%o una in#@actitu%%#

1+r

3600s100+r 0.1s

s10*.5 8

×××= −

s342.0=


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':

El $ango calcula%o #n #l sist#ma u# s#alcanBa,a #l misil #$a 6:9 m#t$os

El ,lanco consi%#$a%o #$a a$a una %istanciamao$ %# 1'9 m#t$os


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(0

Th#$mal E@ansion Co#ci#nt +s T#m#$atu$#

/(

o

F) , (inin

o

F)

-340 2.45

-300 3.07

-220 4.08

-160 4.72

-80 5.43

0 6.00

40 6.24

80 6.47

T D D ∆=∆ α


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(1

R#g$#ssing Data in E@c#l/g#n#$al &o$mat

α = -1E-05T2 + 0.0062T + 6.0234


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(2

O,s#$+#% an% $#%ict#% alu#s

/(oF) , (ininoF)iven

, (ininoF)re%ite%

-340 2.45 2.76

-300 3.07 3.26

-220 4.08 4.18

-160 4.72 4.78

-80 5.43 5.46

0 6.00 6.02

40 6.24 6.26

80 6.47 6.46

α = -1E-05T2 + 0.0062T + 6.0234


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('

R#g$#ssing Data in E@c#l

/sci#nti?c &o$mat

α = -1.2360E-05T2 + 6.2714E-03T + 6.0234


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((

Obser-ed and Predicted alues

/(oF) , (ininoF)iven

, (ininoF)re%ite%

-340 2.45 2.46

-300 3.07 3.03

-220 4.08 4.05

-160 4.72 4.70

-80 5.43 5.44

0 6.00 6.02

40 6.24 6.25

80 6.47 6.45

α = -1.2360E-05T2 + 6.2714E-03T + 6.0234


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(*

Obser-ed and Predicted alues

T(oF) , (in/in/oF)

Given

, (in/in/oF)Predicted

, (in/in/oF)

Predicted

-340 !4" !4# !$#

-300 3!0$ 3!03 3!#

-0 4!0% 4!0" 4!1%

-1#0 4!$ 4!$0 4!$%

-%0 "!43 "!44 "!4#

0 #!00 #!0 #!0

40 #!4 #!" #!#

%0 #!4$ #!4" #!4#

α = -1.2360E-05T2 + 6.2714E-03T + 6.0234

α = -1E-05T2 + 0.0062T + 6.0234


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(6

Error de Truncamieno


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(9

Truncation errorError Causado por Truncamiento o apro*imadoen un procedimiento matem=tico,


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(:

E*ample o) Truncation Error

TaPing onl a t#$ms o& a Maclau$in s#$i#s to

a$o@imat#

I& onl ' t#$ms a$# us#%7

....................32

132

++++= x x

xe x

xe

++−=

21

2

x xe Error Truncation x


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(<

Anot4er E*ample o) Truncation Error

Using a ?nit# to a$o@imat#

P

Q

secant line

tangent line

'i.ure 1, K$o@imat# %#$i+ati+# using ?nit# Q@

x∆ )( x f ′

x

x f x x f x f

∆−∆+

≈′ )()(

)(


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*0

Anot4er E*ample o) Truncation

Error"sin. nite rectan.les to appro*imate an inte.ral,


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*1

Example 1 —Maclaurin seriesCalculat# th# +alu# o& ith an a,solut#

$#lati+# a$o@imat# #$$o$ o& l#ss than 1.

n

1 1

2 2.2 1.2 54.545

3 2.*2 0.72 24.658

4 3.208 0.288 8.*776

5 3.2*44 0.0864 2.6226

6 3.3151 0.020736 0.62550

6 t#$ms a$# $#ui$#%. o man a$# $#ui$#% to g#t at l#ast 1 signi?cant%igit co$$#ct in ou$ ans#$5

2.1e

...................3

2.1

2

2.12.11

322.1

++++=e

a E a∈2.1e


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*2

E*ample ( D!ierentiation

;in% &o$ using

an%

Th# actual +alu# is

T$uncation #$$o$ is th#n7

Can ou ?n% th# t$uncation #$$o$ ith

)3( f ′ 2)( x x f = x

x f x x f x f

∆

−∆+≈′

)()()(

2.0=∆ x

2.0

)3()2.03()3(

f f f

−+=

2.0

)3()2.3( f f

−= 2.0 32.3 22

−= 2.0*24.10 −

= 2.024.1

= 2.6=

,2)( x x f = 632)3( =×= f

2.02.66 −=−

1.0=∆ x


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*'

E*ample + D Inte.rationUs# to $#ctangl#s o& #ual i%th to a$o@imat# th# a$#a un%#$ th#cu$+# &o$

o+#$ th# int#$+al2)( x x f = *,39

∫ *

3

2dx x


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*(

Choosing a i%th o& '7 # ha+#

Kctual +alu# is gi+#n ,

T$uncation #$$o$ is th#n

Can ou ?n% th# t$uncation #$$o$ ith ( $#ctangl#s5

)6*()()36()(6

2

3

2

*

3

2−+−=

==∫ x x

x xdx x

3)6(3)3( 22 +=

13510827 =+=

∫

*

3

2dx x

*

3

3

3

= x 234

3

3* 33=

−=

**135234 =−


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**

%edicion de Errores

P*(./E 0E%)(


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*6

P*(./E 0E%)(E((*(E+-

Para determinar la e*actitud del resultadosnumFrico,

Para desarrollar los criterios de al.oritmositerati-os,


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*9

ERROR ER!A!ERO

D#?ni%o como la %i$#ncia #nt$# #l +alo$ +#$%a%#$oo,t#ni%o #n un clculo #l +alo$ a$o@ima%o#ncont$a%o con un mto%o num$ico #tc.

True Error G True alue H Appro*imate alue


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*:

E@aml#T$u# E$$o$

Th# %#$i+ati+#7 o& a &unction

can ,# a$o@imat#% , th# #uation7

I& an%

a ;in% th# a$o@imat# +alu# o&

, T$u# +alu# o&

c T$u# #$$o$ &o$ a$t /a

)( x f ′ )( x f

h

x f h x f x f

)()()(

−+≈

xe x f 5.07)( = 3.0=h

)2( f

)2( f

Solution


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*<

Solution!

a ;o$ an%2= x 3.0=h

3.0

)2()3.02()2(

f f f

−+≈

3.0

)2()3.2( f f −=

3.0

77 )2(5.0)3.2(5.0 ee −=

3.0

028.1*107.22 −= 263.10=


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60

Solution!

, Th# #@act +alu# o& can ,# &oun% , using

ou$ Pnol#%g# o& %iL#$#ntial calculus.

So th# t$u# +alu# o& is

T$u# #$$o$ is calculat#% as

T$u# alu# V K$o@imat# alu#

)2( f

xe x f 5.07)( =

xe x f 5.05.07)( ××=

xe 5.05.3=

)2(5.05.3)2( e f =

)2( f

5140.*=

=t E 722.0263.105140.* −=−=


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61

Error (elativo 1erdadero

D#?ni%o como #l coci#nt# #nt$# #l #$$o$ +#$%a%#$o7 #l +alo$+#$%a%#$o.

.

.r

Error Verdadero

Valor Verdaderoε =


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62

E*ampleDRelati-e True Error

;olloing &$om th# $#+ious #@aml# &o$ t$u# #$$o$7

?n% th# $#lati+# t$u# #$$o$ &o$

at ith

;$om th# $#+ious #@aml#7

R#lati+# T$u# E$$o$ is %#?n#% as

as a #$c#ntag#7

xe x f 5.07)( =)2( f 3.0=h

722.0−=t E

a$!e/r!e

#rror /r!e=∈t

5140.*

722.0−= 075888.0−=

100075888.0 ×−=∈t 5888.7−=

E$$o$ %# a$o imacion


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6'

E$$o$ %# a$o@imacion 3u o%#mos hac#$ si los +alo$#s +#$%a%#$os no s#

sa,#n o son mu %i&cil %# o,t#n#$5 El #$$o$ a$o@ima%o s# %#?n# como la %i$#ncia

#nt$# la a$o@imacion actual la a$o@imacinant#$io$.

. .a E Aprox Actual Aprox Anterior = −

E l A i t E


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6(

E*ampleDAppro*imate Error

;o$ at ?n% th# &olloing7

a using

, using

c a$o@imat# #$$o$ &o$ th# +alu# o& &o$ a$t ,

Solution!

a ;o$ an%

xe x f 5.07)( = 2= x

)2( f ′ 3.0=h

)2( f ′ 15.0=h

)2( f ′

h

x f h x f x f

)()()(

−+≈

2= x 3.0=h

3.0

)2()3.02()2(

f f f

−+≈

/cont )2()32( ff


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6*

/cont.

, ;o$ an%

3.0

)2()3.2( f f −=

3.0

77 )2(5.0)3.2(5.0 ee −=

3.0

028.1*107.22 −= 263.10=

2= x 15.0=h

15.0

)2()15.02()2(

f f f

−+≈

15.0

)2()15.2( f f −=

/cont.


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66

/cont.

c So th# a$o@imat# #$$o$7 is

Present &ppro#imation 2 Previous &ppro#imation

15.0

77 )2(5.0)15.2(5.0 ee −=

15.0

028.1*50.20 −= 8800.*=

a E

=a E

263.108800.* −=

38300.0−=

E$$o$ R#lati+o a$o@ima%o


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69

E$$o$ R#lati+o a$o@ima%o

D#?ni%o como #l coci#nt# #nt$# #l #$$o$

a$o@ima%o la actual a$o@imacin.

.

.a

Error Aproximado

Aproximacion acual ε =

E l R l ti K i t E


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6:

E@aml#R#lati+# K$o@imat# E$$o$

;o$ at 7 ?n% th# $#lati+# a$o@imat#

#$$o$ using +alu#s &$om an%

Solution!

;$om E@aml# '7 th# a$o@imat# +alu# o&

usingan%

using

$#s#nt K$o@imation V $#+ious K$o@imation

xe x f 5.07)( = 2= x

3.0=h 15.0=h

263.10)2( =′ f

3.0=h 8800.*)2( =′ f 15.0=h

=a E

263.108800.* −=

38300.0−=

cont.


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6<

K$o@imat# E$$o$

$#s#nt K$o@imation

as a #$c#ntag#7

K,solut# $#lati+# a$o@imat# #$$o$s ma also n##% to ,#calculat#%7

=∈a

8800.*

38300.0−= 038765.0−=

8765.3100038765.0 −=×−=∈a

:038765.0: −=∈a 8765.3or 038765.0=

Cómo utili3ar el error relativo absoluto como


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90

I& h#$# is a $#-s#ci?#% tol#$anc#7 th#n

no &u$th#$ it#$ations a$# n#c#ssa$ an% th# $oc#ss is sto#%.

I& at l#ast m signi?cant %igits a$# $#ui$#% to ,# co$$#ct in th# ?nalans#$7 th#n

Cómo utili3ar el error relativo absoluto comocriterio de paro-

sa ∈≤∈ :: s∈

105.0:: 2 ma−×≤∈

Ta,l# o& alu#s


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91

Ta,l# o& alu#s

;o$ at ith +a$ing st# siB#7

0.' 10.26' N"K 0

0.1*


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92


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9'

REPASO !E &A SERIE !ETA&OR

WIWIDKTK INX.


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9(

JuF es la serie de Ta0lorKAl.unos de los e2emplos de la serie de Ta0lor 5ue "d, debe4aber -isto

WIWIDKTK INX.

+−+−=

642

1)os(642 x x x

x

+−+−=753

)sin(753 x x x

x x

++++=32

132 x x

xe x

X l T l S i


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9*

X#n#$al Talo$ S#$i#s Th# g#n#$al &o$m o& th# Talo$ s#$i#s is gi+#n ,

Con la con%icin %# u# to%as sus %#$i+a%as %# &/@ son continuas #@ista #n #l int#$+alo Y@7@hZ

Como 4abra dic4o Ar5umedes8 L dame el valor de la $unción enun solo punto4 5 al valor de todos los 6primero4 se7undo4etcétera8 de sus derivados en ese solo punto4 puedo darle el

valor de la $unción en cualuier otro punto

WIWIDKTK INX.

( ) ( ) ( ) ( ) ( )

+′′′

+′′

+′+=+ 3232

h x f

h x f

h x f x f h x f

E@aml#Talo$ S#$i#s


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96

a # a o S# #s

;in% th# +alu# o& gi+#n that

an% all oth#$ high#$ o$%#$ %#$i+ati+#s

o& at a$# B#$o.

Solution!

WIWIDKTK INX.

( )6 f ( ) ,1254 = f ( ) ,744 =′ f

( ) ,304 =′′ f ( ) 64 =′′′ f ( ) x f 4= x

( ) ( ) ( ) ( ) ( ) +′′′+′′+′+=+32

32 h x f

h x f h x f x f h x f

4= x

246 =−=h

Solution! /cont.


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99

Sinc# th# high#$ o$%#$ %#$i+ati+#s a$# B#$o7

Not# that to ?n% #@actl7 # onl n##% th# +alu#

o& th# &unction an% all its %#$i+ati+#s at som# oth#$ oint7 in this cas#

WIWIDKTK INX.

( ) ( ) ( ) ( ) ( ) 3

2

42

2

424424

32

f f f f f ′′′+′′+′+=+

( ) ( )

+

++=

3

26

2

2302741256

32

f

860148125 +++= 341=( )6 f

4= x

!eri-ation )or %aclaurin Series )or


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9:

e*

D#$i+# th# Maclau$in s#$i#s

Th# Maclau$in s#$i#s is siml th# Talo$ s#$i#s a,out th# oint@0

WIWIDKTK INX.

++++= 321

32 x x

xe x

( ) ( ) ( ) ( ) ( ) ( ) ( ) +′′′′′+′′′′+′′′+′′+′+=+5432

5432 h x f h x f h x f h x f h x f x f h x f

( ) ( ) ( ) ( ) ( ) ( ) ( ) +′′′′′+′′′′+′′′+′′+′+=+

5

0

4

0

3

0

2

00005432 h

f h

f h

f h

f h f f h f

Sinc# an% xn x x x e x f e x f e x f e x f ==′′=′= )(,...,)( ,)(,)( 1)0( 0 == e f n


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9<

th# Maclau$in s#$i#s is th#n

So7

WIWIDKTK INX.

ffff )(,,)(,)(,)( )(f

...53

)(

52

)()()()( 3

02

000 h

eh

eheeh f +++=

...3

1

2

11 32 hhh +++=

...32

1)(32

++++= x x

x x f

E$$o$ in Talo$ S#$i#s


80/93

:0

E$$o$ in Talo$ S#$i#s

h#$# th# $#main%#$ is gi+#n ,

h#$#

that is7 c is som# oint in th# %omain Y@7@hZ

Th# Talo$ olnomial o& o$%#$ n o& a &unction f(x) ith

(n+1) continuous %#$i+ati+#s in th# %omain Y x,x+h] isgi+#n ,

WIWIDKTK INX.

( ) ( ) ( ) ( ) ( ) ( ) ( ) x Rn

h x f

h x f h x f x f h x f n

nn ++++′+=+

2

2

( ) ( ) ( ) ( )c f

n

h x x R n

n

n

1

1

)1(

++

+−=

h xc x +


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:1

Th# Talo$ s#$i#s &o$ at oint is gi+#n ,

WIWIDKTK INX.

!e la serie cuantos mas tFrminos se tome el error disminu0e 0

por lo tanto se tendr= una me2or estimación de la )unción a sercalculada,

Cu=ntos tFrminos se re5ueriran para conse.uir unaapro*imación de e1 dentro de una ma.nitud del error -erdaderode menos de 1#/M,

xe 0= x

++++++=5432

15432 x x x x

xe x

Solution!


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:2

Using t#$ms o& Talo$ s#$i#s gi+#s #$$o$ ,oun% o&

Sinc#

WIWIDKTK INX.

( )1+n

( )

( )

( )( )

( )c f n

h x

x R n

n

n

1

1

1

++

+

−

=

xe x f h x === )(,1,0

( ) ( )

( )( ) ( )c f

n R n

n

n

1

1

1

100 +

+

+

−=

( )( )

cn

en 1

1 1

+−=

+

h xc x +


83/93

:'

So i& # ant to ?n% out ho man t#$ms it oul%

$#ui$# to g#t an a$o@imation o& ithin a

magnitu%# o& t$u# #$$o$ o& l#ss than 7

So < t#$ms o$ mo$# a$# n##%#% to g#t a t$u# #$$o$

l#ss than

WIWIDKTK INX.

1e

610−

610)1(

−<+ne

en 6

10)1( >+

310)1( 6 ×>+n

*≥n

610−


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:(


85/93

:*

PROPAACION !E

ERRORES


86/93

:6

Propa.acion de errores

En mFtodos numFricos8 los c=lculos nose 4acen con números e*actos, Cómoestas ine*actitudes se propa.an en losc=lculosK

E@aml# 1!


87/93

:9

Encontrar los limites de Propagacion en la suma de dos numeros. Ejemplo si

uno esta calculando X +Y donde

X = 1.5 ± 0.05Y = 3.4 ± 0.04

Solution

Maximum possile !alue o" X = 1.55 # Y = 3.44

Maximum possile !alue o" X +Y =1.55+3.44=4.$$

Minimum possile !alue o" X =1.45 # Y = 3.3%.

Minimum possile !alue o" X +Y =1.45+3.3%=4.&1

Por lo tanto

4.&1 ' ( + ) '4.$$.

Propa.ación de errores en


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::

p .)órmulas

SI $ #s una &uncin %# +a$ias +a$ia,l#s @17 @27 @'7 [.@n-17@n

#ntonc#s #l +alo$ m@imo osi,l# %#l #$$o$ #n $ #sta%a%o!

n

n

n

n

X X

f X

X

f X

X

f X

X

f f ∆

∂∂

+∆∂

∂++∆

∂∂

+∆∂∂

≈∆ −−

1

1

2

2

1

1

.......

E@aml# 2!


89/93

:<

>a t#nsin #n un mi#m,$o a@ial %# una s#ccin$#$#s#ntati+a cua%$a%a #sta %a%o!

Encu#nt$# #l #$$o$ m@imo osi,l# m@imo #n la

m#%i%a %# lala t#nsion.

E h

F 2

∈=

;*.072 ±= F

mm1.04 ±=h

a5.170 ±= E

Solucion


90/93


91/93


92/93

SolutionS#a!

Entonc#s!

El cam,io $#lati+o #sta %a%o o$!

a su,st$accin %# los nAm#$os u# son mu c#$canos o casi igual#s c$#ain#@actitu%#s in%#s#a%as. Usa$#mos la &$mula a$a la $oagacin %##$$o$7 #l cual s# %#mu#st$a u# #sto #s +#$%a%.

y x z −=

y y

z x

x

z z ∆

∂∂

+∆∂∂

=∆

y x ∆−+∆= )1()1(

y x ∆+∆=

y x

y x

z

z

−

∆+∆=

∆

o$ #=#mlo si!


93/93

=

0.6667

66.67

001.02 ±= x

001.0003.2 ±= y

:003.22:

001.0001.0

−

+

=

∆ z

z

teoria de errores nuevo

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