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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



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