Re: using N function
- To: mathgroup at smc.vnet.net
- Subject: [mg76336] Re: using N function
- From: dimitris <dimmechan at yahoo.com>
- Date: Sun, 20 May 2007 02:21:37 -0400 (EDT)
- References: <f2me6h$lu4$1@smc.vnet.net>
Watch the differences. In[34]:= o = Cos[31^(1/3)]; (*a number very close to -1; see http://mathworld.wolfram.com/AlmostInteger.html) Then In[35]:= N@o Out[35]= -1. In[36]:= N@o//InputForm Out[36]//InputForm= -0.999999977527746 In[37]:= o//N[#,MachinePrecision]&//InputForm Out[37]//InputForm= -0.999999977527746 But In[38]:= o//N[#,$MachinePrecision]&//InputForm Out[38]//InputForm= -0.9999999775277460225993505331`15.954589770191003 Almost 16 digits of precision. Also execute the following command In[50]:= FrontEndExecute[{HelpBrowserLookup["MainBook", "3.1.4", "4.14"]}] You will be gone to a relevant section of Mathematica information with enough information for understanding their differences! Note that I don't know if the the last command works in version 6! Dimitris / kkwweett : > Good morning Mathematica users, > > > Can anyone tell me, please, which of a),b) or c) is wrong : > > a) > <MathematicaHelp> > <quote> > > N[expr] is equivalent to N[expr, MachinePrecision]. > > </quote> > </MathematicaHelp> > > b) > In[1]:=N[Exp[Sqrt[163 ]Pi] - 6403203] > Out[1]=256. > > > c) > In[2]:=N[Exp[Sqrt[163 ]Pi] - 6403203,$MachinePrecision] > Out[2]=743.9999999999993 > > ? > > Thank you