From Eudora
Zips
07/30/2006 04:45p
exp(-alpha x)
07/30/2006 04:42p
modifications.
07/30/2006 04:45p
different exponential.
07/30/2006 08:38a
produce aigau.
11,804 tfun.zip – The final codes described in #Decay as
2,389 tests.zip – test.f from steve and test.for with minimal
1,711 tglagu.zip – code for integrating a polynomial and a
2,931 orig.zip – copy of code for integrating a Gaussian to
I modified the Gauss Laguerre code to calculate
an integral whose value I know (in this case
1/(x*x+100)^1.5 from 2 to infinity. The exact
result is 0.00803884 but the numerical results
are off. Here's the code. There is no input.
Steve Alexander
test.f I compiled this using Watcom
wfl386 test.f
test
0.0079439532626720
0.0065542826199757
Two different results, neither equal to the actual answer See discussion in #The test
function. I changed to the best alpha – bob’s changes to test.f are in lower case - found
in the section #Decay as exp(-alpha x) and saved this as test.for
Wfl386 test.for
test
0.0080365285494317
0.0080100288655011
The results below are more accurate due to the fact that they have 16 digit values
of the Gauss Laguerre quadrature points, but the primary problem at this point is that the
integrand falls off as x3 which is quite different from the exponential decay assumed by
Gauss Laguerre. We can subtract a 1/x3 and integrate it analytically, but that goes
beyond the notion of easy quadrature.
Bob
http://www.phys.ufl.edu/~coldwell/aigau/aiglz/Welcome.htm
“TGLAGU.FOR is set up to test Gauss Laguerre integration on an analytic function.
obsolete
will need revision to be used.”
I started in
C:\public_html\aigau\aiglz>
All fortran and ide files were copied to the present folder aiglz.wpj
The main code is AiGlz.for
IMPLICIT REAL*8 (A-H,O-Z)
DATA SQRPI/1.772453850905516D0/
DATA N/500/
OPEN(1,FILE='BRACK.OUT')
OPEN(2,FILE='AIGAUSS.OUT')
OPEN(3,FILE='ERR.OUT')
DO I=1,N
Z=-7+(I-1)*6/(1D0*N)
BR=BRGLAG(Z,ERR)
AT2=-EXP(-Z*Z)*BR/(2*Z*SQRPI)
WRITE(1,'(3G24.16)')Z,BR,ERR
WRITE(2,'(3G24.16)')Z,AT2,ERR
WRITE(3,'(2G15.6)')Z,ABS(ERR)
ENDDO
END
C$INCLUDE BRGLAGU
C$INCLUDE GLAGU
The BR refers to AiGauss( z )  exp   z 2  Brack  z  in
C:\public_html\aigau\aiglz\Welcome.doc

 t2  
exp

t
exp
    2 dt 

 4 z   (1)
Brack  z    0


2z 




Note that z < 0 in this accounting for the – in the definition of AT2 above. The file
BRGLAGU.FOR
FUNCTION BRGLAG(Z,ERT)
IMPLICIT REAL*8 (A-H,O-Z)
COMMON/PASS/ZP
EXTERNAL AIGI
ZP=Z
IF(Z.EQ.0)THEN
BRGLAG=.5D0
ERT=0
RETURN
ENDIF
BRGLAG=GLAGU1(AIGI,DIFF)
ERT=DIFF/BRGLAG
END
FUNCTION AIGI(T)
IMPLICIT REAL*8 (A-H,O-Z)
COMMON/PASS/Z
ARG=T/(2*Z)
AIGI=EXP(-ARG*ARG)
C NOTE THAT THE EXP(-T) IS PART OF GLAG INTEGRAND
RETURN
END
This code uses the function exp  t 2 / 4 z 2  so that the full integral is

N
0
i 1
2
2
2
2
 exp  t  exp  t / 4 z  dt   wi exp   xi / 4 z  (2)
This is evaluated in GLAGU.FOR
2
3
4
5
6
7
8
9
A
B
C
D
E
2
3
4
5
6
7
8
9
A
B
C
D
E
FUNCTION GLAGU1(FUN,DIFF)
IMPLICIT REAL*8 (A-H,O-Z)
DIMENSION XI1(19),AI1(19),XI2(20),AI2(20)
DATA (XI1(I),AI1(I),I=1,19)/7.415878375720509D-2,
.1767684749159125D0,.3912686133199946D0,.3004781436072544D0,
.9639573439979580D0,.2675995470381750D0,1.796175582068328D0,
.1599133721355802D0,2.893651381873784D0,.6824937997614911D-1,
4.264215539627766D0,.2123930760654432D-1,5.918141561644049D0,
.4841627351148396D-2,7.868618915334734D0,.8049127473813668D-3,
10.13242371681527D0,.9652472093153502D-4,12.73088146384240D0,
.8207305258051031D-5,15.69127833983589D0,.4830566724730773D-6,
19.04899320982355D0,.1904991361123286D-7,22.85084976082948D0,
.4816684630928062D-9,27.16066932741145D0,.7348258839551144D-11,
32.06912225186224D0,0.6202275387572616D-13,37.71290580121965D0,
.2541430843015423D-15,44.31736279583150D0,.4078861296825712D-18,
52.31290245740438D0,.1707750187593837D-21,62.80242315350038D0,
.6715064649908190D-26/
DATA (XI2(I),AI2(I),I=1,20)/.7053988969198875D-1,
.1687468018511139D0,.3721268180016114D0,.2912543620060683D0,
.9165821024832736D0,.2666861028670013D0,1.707306531028344D0,
.1660024532695068D0,2.749199255309432D0,.7482606466879237D-1,
4.048925313850887D0,.2496441730928322D-1,5.615174970861617D0,
.6202550844572237D-2,7.459017453671063D0,.1144962386476908D-2,
9.594392869581097D0,.1557417730278120D-3,12.03880254696432D0,
.1540144086522492D-4,14.81429344263074D0,.1086486366517982D-5,
17.94889552051938D0,.5330120909556715D-7,21.47878824028501D0,
.1757981179050582D-8,25.45170279318691D0,.3725502402512321D-10,
29.93255463170061D0,.4767529251578191D-12,35.01343424047900D0,
.3372844243362438D-14,40.83305705672857D0,.1155014339500399D-16,
47.61999404734650D0,.1539522140582344D-19,55.81079575006390D0,
.5286442725569158D-23,66.52441652561575D0,.1656456612499023D-27/
GLAGU0=0
DO I=1,19
GLAGU0=GLAGU0+AI1(I)*FUN(XI1(I))
ENDDO
GLAGU1=0
DO I=1,20
GLAGU1=GLAGU1+AI2(I)*FUN(XI2(I))
ENDDO
DIFF=GLAGU1-GLAGU0
RETURN
END
These were checked before, I zip them into orig.zip
Test power of x
C:\public_html\class2K\integration\GaussLag.doc
The code TGLAGU.FOR is set up to test the integrals of powers of x. The relevant
portions are
5
IMPLICIT REAL*8 (A-H,O-Z)
COMMON /PPOWOF/NPOW
COMMON /EXPON/ALPHA
EXTERNAL POWOFX,EXPONF
PRINT*,' ENTER THE POWER OF X'
READ(*,*)NPOW
ANFAC=1
DO I=2,NPOW
ANFAC=ANFAC*I
ENDDO
IF(NPOW.EQ.-10)GOTO 10
TEST=GLAGU(POWOFX,DIFF)
PRINT*,' TEST=',TEST
PRINT*,' NFAC=',ANFAC
PRINT*,' DIFF=',DIFF
GOTO 5
…
FUNCTION POWOFX(X)
IMPLICIT REAL*8 (A-H,O-Z)
COMMON /PPOWOF/NPOW
POWOFX=X**NPOW
RETURN
END
Glagu was changed so that the function is GLAGU rather than GLAGU1 as for the error
function integral. Note that the integral of xN is N!.
run
ENTER THE POWER OF X
0
TEST=
1.000000000000000
NFAC=
1.000000000000000
DIFF= -1.110223024625160D-016
ENTER THE POWER OF X
5
TEST=
120.000000000000000
NFAC=
120.000000000000000
DIFF= 1.421085471520200D-014
ENTER THE POWER OF X
10
TEST= 3628800.000000000000000
NFAC= 3628800.000000000000000
DIFF=
0.000000000000000
ENTER THE POWER OF X
15
TEST= 1.307674368000000D+012
NFAC= 1.307674368000000D+012
DIFF=
0.000732421875000
ENTER THE POWER OF X
The first difference is due to the truncation error for 16 digits
ENTER THE POWER OF X
20
TEST= 2.432902008176640D+018
NFAC= 2.432902008176640D+018
DIFF=
4608.000000000000000
ENTER THE POWER OF X
25
TEST= 1.551121004333100D+025
NFAC= 1.551121004333100D+025
DIFF= 3.435973836800000D+010
ENTER THE POWER OF X
30
TEST= 2.652528598121910D+032
NFAC= 2.652528598121910D+032
DIFF= 5.404319552844600D+017
ENTER THE POWER OF X
35
TEST= 1.033314796638610D+040
NFAC= 1.033314796638610D+040
DIFF= 1.088033237653170D+025
ENTER THE POWER OF X
Note that these last values are still exact, even though the differences indicate only 15
digits.
They should be exact until about 36 for 19 points and 38 for 20 points.
40
TEST= 8.159152832419790D+047
NFAC= 8.159152832478980D+047
DIFF= 4.499026736539390D+039 – 8 digits
ENTER THE POWER OF X
45
TEST= 1.196221877898960D+056
NFAC= 1.196222208654800D+056
DIFF= 1.159802700720010D+051 – 5 digits
ENTER THE POWER OF X
50
TEST= 3.041191034335690D+064
NFAC= 3.041409320171340D+064
DIFF= 2.275593042063180D+061 – 3 digits
ENTER THE POWER OF X
Test alpha
10
PRINT*,' ENTER ALPHA'
READ(*,*)ALPHA
IF(ALPHA.GT.2D0)STOP
TEST=GLAGU(EXPONF,DIFF)
PRINT*,' TEST=',TEST
PRINT*,' ANAL=',-1/ALPHA
PRINT*,' DIFF=',DIFF
GOTO 10
END
…
FUNCTION EXPONF(X)
IMPLICIT REAL*8 (A-H,O-Z)
COMMON /EXPON/ALPHA
ALPHAP=ALPHA+1
ARG=ALPHAP*X
IF(X.GT.90D0)THEN
EXPONF=1.22D39
RETURN
ENDIF
IF(X.LT.-90.D0)THEN
EXPONF=0D0
RETURN
ENDIF
EXPONF=EXP(ARG)
RETURN
END


e x dx 
1

  e  x e 1 x dx
0

Note that alpha must be less than 1 for the integral to converge.
0
ENTER THE POWER OF X
-10
ENTER ALPHA
-1
TEST=
1.000000000000000
ANAL=
1.000000000000000
DIFF= -1.110223024625160D-016
ENTER ALPHA
-.5
TEST=
2.000000000000000
ANAL=
2.000000000000000
DIFF= 2.220446049250310D-016
ENTER ALPHA
The above values are better than the old table due to the increased precision in
GLAGU.FOR
-.2
TEST=
4.999998821388100
ANAL=
5.000000000000000
DIFF= 1.427392587238790D-006
ENTER ALPHA
-.1
TEST=
9.994306877313890
ANAL=
10.000000000000000
DIFF=
0.002717118659016
ENTER ALPHA
ENTER ALPHA
-.05
TEST=
19.506829422016500
ANAL=
20.000000000000000
DIFF=
0.105475627052240
ENTER ALPHA
-.025
TEST=
33.671550728920300
ANAL=
40.000000000000000
DIFF=
0.642007601415472
ENTER ALPHA
Very negative value of alpha are also of interest
ENTER ALPHA
-4
TEST=
0.2499999966161755
ANAL=
0.2500000000000000
DIFF= 5.7768075045760980D-009
ENTER ALPHA
-8
TEST=
0.1249529377004137
ANAL=
0.1250000000000000
DIFF= 2.8712992098584020D-005
ENTER ALPHA
-16
TEST=
0.0596715658622538
ANAL=
0.0625000000000000
DIFF=
0.0007053403985482
ENTER ALPHA
-32
TEST=
0.0189501719947952
ANAL=
0.0312500000000000
DIFF=
0.0012068368532991
ENTER ALPHA
-64
TEST=
0.0019825548409274
ANAL=
0.0156250000000000
DIFF= 3.2914875963343910D-004
The good range with 8+ digit accuracy is -4<<-.5
The test function
I copied tglagu to tfun.for and made tfun.wpj. The function above is
1
integrated from 2 to  = 0.00803884. Write this as
3/ 2
2
x

100




2
1
x
2
 100 
3/ 2
dx  0.00803884
Let x’=x-2, and multiply by the needed exponential

exp  x 
dx  0.00803884
3/ 2
0 exp   x 
2
 x  2   100

TEST=
Steve Claims
DIFF=

(3)
(4)
0.0079532072818606
0.00803884
0.0000092540550930986930
0.00008563272
The problem is that 1/x3 is not even close to exponential in its decay. Thus there is more
long range part than one would expect. Assume that the form is
Decay as exp(-alpha x)
exp  x 

 exp   x 
0
 x  2
2
 100

3/ 2
dx  0.00803884
(5)
Note that this is not the same alpha as that in the section #Test alpha. Now let x’=x





 1

exp  x '
exp

x
'
dx '  0.00803884
(6)



3/ 2 
0
2

   x '


    2   100  

 
   
With alpha at out disposal. In particular the slow convergence of x-3 implies that we
might want alpha in (5) to be on the order of 0.01 or so. This places the points at x/
much further out than before. tfun.for tfun.wpj
ENTER A POSITIVE VALUE FOR ALPHA
1
TEST=
0.0079532072818606
Steve Claims 0.00803884
DIFF= 9.2540550930986930D-006
ENTER A POSITIVE VALUE FOR ALPHA
.5
TEST=
0.0080166513768168
Steve Claims 0.00803884
DIFF= 2.4569700125334770D-006
ENTER A POSITIVE VALUE FOR ALPHA
.25
TEST=
0.0080332038327021
Steve Claims 0.00803884
DIFF= 6.3010351074137270D-007
ENTER A POSITIVE VALUE FOR ALPHA
.125
TEST=
0.0080374194522396
Steve Claims 0.00803884
Diff
0.000000149
Actual diff
0.000001420
DIFF= 1.4932011877660800D-007  The estimate with the smallest error
ENTER A POSITIVE VALUE FOR ALPHA
.0625
TEST=
0.0080373109389459
Steve Claims 0.00803884
DIFF= 7.8236501858258700D-007
ENTER A POSITIVE VALUE FOR ALPHA
Further thoughts



a
1
s
2


2 3/ 2
1
 s2   2 
tfun.wpj
ENTER A POSITIVE VALUE FOR ALPHA
.05
TEST=
0.0080328627579463
3/ 2
ds 
ds 
s
 2  s2   2 
1/ 2
1
2

(7)
a
 2  a2   2 
1/ 2
Steve Claims 0.00803884
Analytic =
0.0080388386486182
DIFF= 1.2883721617444610D-006
By subtracting out the analytic integral, I can get to any desired accuracy.
The actual integral of interest is

 3

v  p    3  ds exp  p 2 /  s 2   2   Re 1  exp z  Rs ', pi  w  jz  Rs ', pi    / s '3 / 2 (8)
0
 j 1

 




s '  s2   2
(9)
z  Rs ' jpx / 2 s '
The exponential of p2/s2 rapidly goes to 1, while the 1-exp()w is plotted in 1-w.htm and
also rapidly becomes 1. The 1/s3/2 is analytic for all ranges. Thus from a to infinity


 3
 
3
v  a, p     ds  exp   p 2 / s 2   Re 1  exp z  Rs, pi  w  jz  Rs, pi     1 / s 3 / 2
a
 j 1
 





1
a

3  2 
2
2
2 1/ 2 

 a    

(10)


