Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2003
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2003

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Re: A question on interval arithmetic

  • To: mathgroup at smc.vnet.net
  • Subject: [mg43759] Re: [mg43735] Re: A question on interval arithmetic
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Fri, 3 Oct 2003 02:28:46 -0400 (EDT)
  • References: <blcq5l$p5f$1@smc.vnet.net> <200310020651.CAA15238@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Steve Luttrell wrote:
> 
> You need to average  (r1*r2)/(r1+r2) over the required interval.
> 
> Here is how I did it evaluating over the 2-dimensional tolerance region
> around r1=r10 and r2=r20. I assume each resistor independently has a
> tolerance +/-a which is uniformly distributed over an interval of length 2a,
> which gives a probability density 1/(4a^2) over the 2-dimensional tolerance
> region for the pair of resistors. I had to evaluate indefinite integrals and
> then substitute in limits because definite integration seemed to lock
> Mathematica up (I have version 5 for Windows).
> 
> Integrate[((r1*r2)/(r1 + r2))*(1/(4*a^2)), r1]
> Simplify[(% /. {r1 -> r10 + a}) - (% /. {r1 -> r10 - a})]
> Integrate[%, r2]
> Simplify[(% /. {r2 -> r20 + a}) - (% /. {r2 -> r20 - a})]
> 
> --
> Steve Luttrell
> West Malvern, UK

Addressing specifically the definite Integrate issue, it is slow due to
processing of parameters. It will help considerably if you pass in
assumptions. For example:

In[5]:= InputForm[Timing[Integrate[((r1*r2)/(r1 + r2))*(1/(4*a^2)),
  {r2,r20-a,r20+a}, {r1,r10-a,r10+a},
  Assumptions->{r10>0,r20>0,a>0,a<r10,a<r20}]]]
        
Out[5]//InputForm= 
{8.55*Second, 
 -(-8*a^2*r10 - 8*a^2*r20 - 2*(r10^3 + r20^3 + 3*a^2*(r10 + r20))*
     Log[r10 + r20] + (-2*a^3 + r10^3 + r20^3 + 3*a^2*(r10 + r20) - 
      3*a*(r10^2 + r20^2))*Log[-2*a + r10 + r20] + 
    2*a^3*Log[2*a + r10 + r20] + 3*a^2*r10*Log[2*a + r10 + r20] + 
    3*a*r10^2*Log[2*a + r10 + r20] + r10^3*Log[2*a + r10 + r20] + 
    3*a^2*r20*Log[2*a + r10 + r20] + 3*a*r20^2*Log[2*a + r10 + r20] + 
    r20^3*Log[2*a + r10 + r20])/(12*a^2)}


Another possibility is to suppress generation of conditions involving
the parameters. This tends to speed the computation, at the risk of
obtaining a result that may be incorrect for the actual parameter value
regions you have in mind.

In[1]:= InputForm[Timing[Integrate[((r1*r2)/(r1 + r2))*(1/(4*a^2)),
  {r2,r20-a,r20+a}, {r1,r10-a,r10+a}, GenerateConditions->False]]]
        
Out[1]//InputForm= 
{2.94*Second, 
 -(-8*a^2*r10 - 8*a^2*r20 - 2*(r10^3 + r20^3 + 3*a^2*(r10 + r20))*
     Log[r10 + r20] + (-2*a^3 + r10^3 + r20^3 + 3*a^2*(r10 + r20) - 
      3*a*(r10^2 + r20^2))*Log[-2*a + r10 + r20] + 
    2*a^3*Log[2*a + r10 + r20] + 3*a^2*r10*Log[2*a + r10 + r20] + 
    3*a*r10^2*Log[2*a + r10 + r20] + r10^3*Log[2*a + r10 + r20] + 
    3*a^2*r20*Log[2*a + r10 + r20] + 3*a*r20^2*Log[2*a + r10 + r20] + 
    r20^3*Log[2*a + r10 + r20])/(12*a^2)}


Daniel Lichtblau
Wolfram Research


  • Prev by Date: Automatic execution of a notebook
  • Next by Date: simple vactor magitude question
  • Previous by thread: Re: A question on interval arithmetic
  • Next by thread: Re: A question on interval arithmetic