Doubt about formula transcription
in [Fortran]
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From: Leonardo Marques on 7 Apr 2008 15:36 hey guys, im with a little problem when im transcripting a math formula from maple to fortran, because when i put a solution on the equation, i got a result diferent from zero. The formula is: 2*arctan(sin(a)*cos(a)/(.8-cos(a)^2))-4/9*Pi ; I've transcripted to fortran as: 2*atan((sin(a)*cos(a))/(0.8- cos(a)**2))-((pi)*4/9) There's correct?! a piece of my code: ! usando o metodo da zoada kkkkk program missel !constantes e variaveis real FUNCAO ! programa write(*,*)'Valor de f:',FUNCAO(1.023762140); end ! funcoes, para facilitar, tava foda ficar repetindo... real function FUNCAO(a); real pi,a; parameter(k=0.8,pi=3.141592653); FUNCAO=dble(2*atan((sin(a)*cos(a))/(k-cos(a)**2))-((pi)*4/9)) return end
From: Michael Metcalf on 7 Apr 2008 16:09 "Leonardo Marques" <surf3r0(a)gmail.com> wrote in message news:9c4bdee7-084e-4644-8b90-74cbef524878(a)u3g2000hsc.googlegroups.com... > hey guys, > > im with a little problem when im transcripting a math formula from > maple to fortran, because when i put a solution on the equation, i got > a result diferent from zero. > > The formula is: 2*arctan(sin(a)*cos(a)/(.8-cos(a)^2))-4/9*Pi ; > I've transcripted to fortran as: 2*atan((sin(a)*cos(a))/(0.8- > cos(a)**2))-((pi)*4/9) > I don't know Maple, but in Fortran 4/9 gives zero. Try 4.0/9.0. Regards, Mike Metcalf
From: fj on 7 Apr 2008 16:34 On 7 avr, 22:09, "Michael Metcalf" <michaelmetc...(a)compuserve.com> wrote: > "Leonardo Marques" <surf...(a)gmail.com> wrote in message > > news:9c4bdee7-084e-4644-8b90-74cbef524878(a)u3g2000hsc.googlegroups.com... > > > hey guys, > > > im with a little problem when im transcripting a math formula from > > maple to fortran, because when i put a solution on the equation, i got > > a result diferent from zero. > > > The formula is: 2*arctan(sin(a)*cos(a)/(.8-cos(a)^2))-4/9*Pi ; > > I've transcripted to fortran as: 2*atan((sin(a)*cos(a))/(0.8- > > cos(a)**2))-((pi)*4/9) > > I don't know Maple, but in Fortran 4/9 gives zero. Try 4.0/9.0. > > Regards, > > Mike Metcalf I don't agree : (pi)*4/9 is computed as ((pi)*4)/9) and this is OK. But the "parameter(k=0.8" is strange, k being no declared REAL !
From: e p chandler on 7 Apr 2008 16:36 On Apr 7, 4:09 pm, "Michael Metcalf" <michaelmetc...(a)compuserve.com> wrote: > "Leonardo Marques" <surf...(a)gmail.com> wrote in message > > news:9c4bdee7-084e-4644-8b90-74cbef524878(a)u3g2000hsc.googlegroups.com... > > > hey guys, > > > im with a little problem when im transcripting a math formula from > > maple to fortran, because when i put a solution on the equation, i got > > a result diferent from zero. > > > The formula is: 2*arctan(sin(a)*cos(a)/(.8-cos(a)^2))-4/9*Pi ; > > I've transcripted to fortran as: 2*atan((sin(a)*cos(a))/(0.8- > > cos(a)**2))-((pi)*4/9) > > I don't know Maple, but in Fortran 4/9 gives zero. Try 4.0/9.0. > > Regards, > > Mike Metcalf The underlying mess has masked the implicit typing of K as INTEGER. k=0.8 sets K to 0. :-(.
From: Gordon Sande on 7 Apr 2008 16:38
On 2008-04-07 16:36:05 -0300, Leonardo Marques <surf3r0(a)gmail.com> said: > hey guys, > > im with a little problem when im transcripting a math formula from > maple to fortran, because when i put a solution on the equation, i got > a result diferent from zero. > > The formula is: 2*arctan(sin(a)*cos(a)/(.8-cos(a)^2))-4/9*Pi ; > I've transcripted to fortran as: 2*atan((sin(a)*cos(a))/(0.8- > cos(a)**2))-((pi)*4/9) > > There's correct?! > a piece of my code: > > ! usando o metodo da zoada kkkkk > program missel > !constantes e variaveis > real FUNCAO > ! programa > write(*,*)'Valor de f:',FUNCAO(1.023762140); > end > ! funcoes, para facilitar, tava foda ficar repetindo... > real function FUNCAO(a); > real pi,a; parameter(k=0.8,pi=3.141592653); > FUNCAO=dble(2*atan((sin(a)*cos(a))/(k-cos(a)**2))-((pi)*4/9)) > return > end real function FUNCAO(a); real pi,a real k parameter(k=0.8,pi=3.141592653) FUNCAO = 2.0 * atan( (sin(a)*cos(a)) / (k-cos(a)**2) )- pi*4.0/9.0 return end The DBLE serves no obvious purpose as it argument will be single and the result is then assigned to a single. The 2 would be converted by default but easier to make it real. k is a default integer so needs an explicit decalaration. In general all should be declared and IMPLICIT NONE used. The value is simple enough that the use of a PARAMETER is just an extra source of error. As writen it will be 0 because the value will be truncated. The exponent **2 is fine. The form **2.0 may not work if applied to a negative value and may be less accurate as this form general powering with log and exp rather than just squaring. (Some compilers will notice that it is a special case and use squaring.) 4/9 evaluates to 0 as noted elsewhere because it is all integer. All the brackets are not really needed once the literals become real. In Fortran 90 form this would be real function FUNCAO ( a ) implicit none real :: a real, parameter :: pi=3.141592653 FUNCAO = 2.0 * atan( (sin(a)*cos(a)) / (0.8-cos(a)**2) )- pi*4.0/9.0 return end as the readable form of PARAMETER has evolved since F77. If you really meant to use double precision then ALL the real literals will need a D0 to promote then to double. It is not important for 4.0 and 9.0 but matters a lot for PI. |