Re: evaluate the floor s to yield a number

*To*: mathgroup at smc.vnet.net*Subject*: [mg109017] Re: evaluate the floor s to yield a number*From*: Bill Rowe <readnews at sbcglobal.net>*Date*: Sat, 10 Apr 2010 06:54:48 -0400 (EDT)

On 4/9/10 at 3:34 AM, ertlejack at sbcglobal.net (John Ertle Jr.) wrote: >In[82]:= J = 367*y + floor (7*(y + floor ((m + 9)/12))/4) + int >(275*m/9) + da + >1721013.5 >Out[82]= 2.44254*10^6 + 7/4 floor (1966 + (17 floor)/12) + (2200 >int)/ >9 >How do you evaluate the floor s in the output in this example to >yield a number for the Julian Date. You will have a lot more success if you use valid Mathematica syntax and spend some time reading the documentation. All built-in Mathematica functions start with an upper case letter and function arguments should be surrounded by square brackets, "[" and "]" not "(" and ")". That is In[34]:= 367*y + Floor[7*(y + Floor [(m + 9)/12])/4] + IntegerPart [275*m/9] + da + 1721013.5 /. {y -> 2010, m -> 4, da -> 15} Out[34]= 2.46234*10^6