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 June 26th, 2014, 03:43 AM #1 Newbie   Joined: Jun 2014 From: India Posts: 1 Thanks: 0 Mathematical Models No Mathematical Model is 100% accurate. Am I saying it wrong? If yes, why?
June 26th, 2014, 03:52 AM   #2
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Quote:
 Am I saying it wrong?
Don't know. I'd like to see a proof of the statement if it's correct.

I guess that means I believe that more than likely, if that statement really has an answer, it's no, but if that is what I believe, doesn't mean I'm right

 June 26th, 2014, 05:22 AM #3 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,150 Thanks: 730 Math Focus: Physics, mathematical modelling, numerical and computational solutions If you are making a mathematical model of a real physical process then the answer is no because you can never measure a value with zero error. If you are making a model that replicates the behaviour of another model then you could in principle make an exact model which is an exact replica of an existing model and therefore get exact calculations. However, in practise, even computer algorithms involve error every time you use floating point arithmetic. As an example of a model replicating another, I could choose: $\displaystyle y = x+1$ with $\displaystyle x={1,2,3}$ or $\displaystyle y = x+2$ with $\displaystyle x={0,1,2}$. The models are technically "different" but give identical results.
 June 26th, 2014, 05:52 AM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms I think that discrete models can be completely accurate. For example, the associative property says that rather than sum the votes for an election directly we can break them into precincts and count each one and then total: a + b + c + d = (a + b) + (c + d)
June 26th, 2014, 06:01 AM   #5
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Quote:
 Originally Posted by Benit13 However, in practise, even computer algorithms involve error every time you use floating point arithmetic.
Although, in principle, calculations with real numbers (using only addition, subtraction, multiplication, and division) can be done exactly by the Tarski-Seidenberg theorem. There are limits to what you can do, though -- Richardson's theorem shows that adding sin, abs, and exp along with log(2) and pi is too much (and I think this has been improved over the years).

June 26th, 2014, 06:21 AM   #6
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Quote:
 Originally Posted by CRGreathouse (and I think this has been improved over the years).
I am not familiar with any other development on the constant problem in recent days.

June 26th, 2014, 10:04 AM   #7
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 Originally Posted by mathbalarka I am not familiar with any other development on the constant problem in recent days.
Not since Richardson?!? Hooboy, you have a lot of reading to do.

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