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MathGroup Archive 2005

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Re: Computation speeds: numerical vs symbolic

  • To: mathgroup at smc.vnet.net
  • Subject: [mg55879] Re: Computation speeds: numerical vs symbolic
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Fri, 8 Apr 2005 01:37:15 -0400 (EDT)
  • Organization: The University of Western Australia
  • References: <d2tfah$ql7$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

In article <d2tfah$ql7$1 at smc.vnet.net>, AES <siegman at stanford.edu> 
wrote:

> No actual question, or at least no immediately useful question in this 
> post.  I'm just expressing surprise and/or asking for enlightenment on 
> what seems like a vast disparity in speeds of symbolic vs numerical 
> calculations in the latest version of Mathematica.
> 
> To be specific, take an expression like
> 
>   v = Exp[ -a (s+y)] ( Exp[2 a y] Cos[s0 + q (s-y)] - Cos[s0 + q (s+y)] )
> 
> which happens to have come up in a problem I'm just working on (y and s 
> are the independent variables of interest, dimensionless surrogates for 
> space and time; a, q and s0 are parameters; q happens to be an integer 
> in the end).  I then want to evaluate the integral
> 
>    w = Integrate[v^2, {y, 0 , Pi}]
> 
> (symbolically) and make some plots of w versus s, with values of a, s0 
> and q being set by rules for each plot.
> 
> The integration involved here (nothing but products of exponentials and 
> cosines)  I would describe as trivial but tedious,  that is, I could do 
> it by hand, but it would be a (very) dull job.  Yet on my Mac iBook G4 
> under Mathematica 5.1, it takes 15 seconds to do this particular 
> integration.  

It is trivial -- and you can speed things up considerably using 
indefinite integration and substituting in the limits (as you would do 
by hand). 

  int = Integrate[v^2, y];

  Timing[Simplify[(int /. y -> Pi) - int /. y -> 0), Element[q,Integers]]

You can justify this approach for this example.

> If I follow the above cell with a second cell putting in 
> the "simplifying" assumption
> 
>    w = Integrate[v^2, {y, 0 , Pi}, Assumptions->Element[q,Integers]]
> 
> this calculation takes 27 seconds.  
> 
> Oddly, if I immediate re-evaluate either of the cells it only takes 2.7 
> seconds or 5 seconds, respectively, for the results to come up the 
> second time,  but having evaluated the first cell (without the 
> Assumption) doesn't seem to speed up the second cell (with the 
> Assumption), so it's not like some integration rules have gotten loaded 
> and cached in evaluating the first cell and used in evaluating the 
> second cell. So, I don't see how I can use this speedup in any practical 
> way (???).

Indeed, you cannot.
 
> The speed contrast comes when I then make a single DisplayPlot 
> containing Plots of the w function from above and two very similar 
> functions wp and wpp versus s for 10 instances of rules specifying 
> values of a, s0 and q,  in other words, thirty decaying-sinusoid-like 
> curves in all, each containing 5 or 6 complete cycles.  This plot 
> appears on the screen faster than I can release the Enter key to create 
> it.
> 
> So, I'm just curious, how can a calculate and plot command like this 
> be so stunningly fast?  and the trivially simple symbolic integration 
> above be so slow?

I'm suprised that you think this. In general, direct numerical 
computation is _much_ easier and faster than symbolic computation, 
especially definite symbolic integration.

Cheers,
Paul

-- 
Paul Abbott                                   Phone: +61 8 6488 2734
School of Physics, M013                         Fax: +61 8 6488 1014
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Crawley WA 6009                      mailto:paul at physics.uwa.edu.au 
AUSTRALIA                            http://physics.uwa.edu.au/~paul


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