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Subject: Re: [boost] boost interval arithmetic
From: Daniel Duffy (dduffy_at_[hidden])
Date: 2015-02-05 06:53:03

This is my second e-mail ever to Boost, so please accept my apologies if it is not following Boost protocol.

Anyways, I have posted ticket #5498 (a show-stopper for sure) (I am a numerical analyst) 4 years ago. The contact replied with a question after a long time..

Long story short .. cannot take transcendental functions of intervals

Interval Arithmetic is well-defined (read Moore). I even had a chapter in it in Volume II of my Boost C++ books.

It's a big pity about these flames and the reactions. It would seem that Thijs vd Bergh has a normal question regarding this library? It should be possible to give an answer?

Is it not possible to discuss this without flames? BTW some interval operations return a bool while in some cases it is neither true nor false.

Best regards

Daniel J. Duffy

From: Boost [boost-bounces_at_[hidden]] on behalf of ðÁ×ÅÌ ëÕÄÁÎ [coodan_at_[hidden]]
Sent: 15 January 2015 13:45
To: boost_at_[hidden]
Subject: Re: [boost] boost interval arithmetic

Thu, 15 Jan 2015 13:16:19 +0100 "Thijs (M.A.) van den Berg" <thijs_at_[hidden]>:
>> OK, but that is enough to conclude that [-inf, inf] IS( [-inf, -1] U [1, inf]) is not true and this operator returns not correct result, no matter is the way it corrupts own result documented or not.
>No, you are wrong here again, you claim so much but demonstrate nothing. You seem to lack the ability to reason precisely and mix up set theory with numerical interval computations.
>You should stick to pen and paper math, learn about logical inference
>(A and NOT(A))->proves the earth is flat
>š,lean about numerical representation theory (like floats) and don't touch a computer or post in forums until you do that!
>What in your opinion is the value of "d" in the following statement?
>double d=1/3;
>Unsubscribe & other changes:

I would like you say me first that you really think that [-inf, inf] IS EMPTY, as I asked first.

But, still, OK, double d=1/3 will contain a result of int operator/ as values you are dividing are int. So what? Next question you will ask will be a result of 'double d=1%3;' ???!

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