Re: Wolfram, meet Stefan and Boltzmann
- To: mathgroup at smc.vnet.net
- Subject: [mg117451] Re: Wolfram, meet Stefan and Boltzmann
- From: Curtis Osterhoudt <cfo at lanl.gov>
- Date: Sat, 19 Mar 2011 05:17:54 -0500 (EST)
Perhaps.
In[1]:= 1 + 1
Out[1]= 2
In[2]:= $Version
Out[2]= "7.0 for Linux x86 (32-bit) (November 11, 2008)"
In[3]:= AbsoluteTiming[Integrate[x^3/(Exp[x] - 1), {x, 0, Infinity}]]
Out[3]= {15.64583`7.645943600598422, \[Pi]^4/15}
On Friday, March 18, 2011 05:01:46 SigmundV wrote:
> Very weird timings AES is getting. On my cheap Dell Inspiron 1545 (4
> GB RAM, Intel Pentium Dual Core CPU) running Ubuntu 10.10 64-bit I get
>
> In[1]:= AbsoluteTiming[Integrate[x^3/(Exp[x] - 1), {x, 0, Infinity}]]
> Out[1]= {2.659503, \[Pi]^4/15}
>
> with
>
> In[2]:= $Version
> Out[2]= "8.0 for Linux x86 (64-bit) (November 7, 2010)"
>
> This is on the high end of timings posted here, but no surprise given
> my low-end hardware. The timings posted by others using similar
> hardware to AES's points to his specific MacBook Pro being at fault.
>
> It also astonished me that AES is not familiar with the term
> 'antiderivative'. The derivative of the antiderivative is the function
> itself.
>
> /Sigmund
>
>
>
> On Mar 17, 11:30 am, AES <sieg... at stanford.edu> wrote:
> > David Lichtbau notes:
> >
> > > (3) Find the antiderivative. It is
> >
> > > In[22]:= InputForm[Integrate[x^3/(Exp[x] - 1), x]]
> >
> > > Out[22]//InputForm=
> > > -x^4/4 + x^3*Log[1 - E^x] + 3*x^2*PolyLog[2, E^x] -
> > > 6*x*PolyLog[3, E^x] + 6*PolyLog[4, E^x]
> >
> > The original integral shown above is used to evaluate the
> > Stefan-Boltzmann constant sigma in the expression P/A = sigma T^4 fo=
> r
> > total power per unit area radiated by a blackbody surface at temperature
> > T. Physically it represents integrating over all the frequencies or
> > wavelengths coming from the blackbody radiator.
> >
> > I'm not personally familiar with the term "antiderivative" (nor the term
> > "PolyLog" for that matter); but if it means in essence "indefinite
> > integral" then this antiderivative might be used to evaluate the total
> > P/A coming from a _frequency_ or _spectrally_ filtered blackbody source
> > using a flat-topped passband filter.
> >
> > Hmm -- wonder if the thermodynamic or blackbody-radiation communities
> > know about that?
>
>
>