Dear Oliver
This is a reply to your previous email.
Q1.
> 3. About /which is *correct* and should always be expected:
> /Unfortunately, it is not correct.
> Why is not it correct? To illustrate, assume we are going to analyze the
> stability of a system such as a satellite, airplane, etc. using this
> kind of over-estimated approach, we may obtain that the system is
> unstable whereas the system is not actually unstable and just over
> estimation is blame.
Oliver told:
If you conclude that the system is unstable, your conclusion is wrong.
When you have over-estimated results and they contain stable solutions
as well as unstable solutions, you cannot conclude whether the system is
unstable.
Please notice that, in control theory, that the dynamical system is stable or not is very crucial when an expert is designing a controller for the system. When we have over-estimated results and they contain stable solutions as well as unstable solutions, means that it is possible that the system is unstable. As a result, the decision maker/the expert has to consider this bad situation and has to design a controller for stabilizing the system before doing anything else. This issue leads to so many problems and efforts which are not in fact needed. Because over-estimated results contain pieces of information which are not possible, and don't happen in fact. These pieces of information are misleading the analyzer and that expert.
Q2.
> /I can certainly find formulas which will bring RDM beyond its limits too.
> /
> You are very welcome, and I am all ears.
Oliver told:
a1 = F1 / m
a2 = F2 / m
Both forces are uncertain, but equal with inverse direction
F1 = -F2 = -1 + 2α₁
m = 3
Now we compute a1+a2 (which should equal zero). Since we have
intermediate results, we will probably have to do outward rounding for
the computation of a1 and a2 (1/3 and 2/3 cannot be stored in a finite
binary number format). Then the sum will not equal zero, but be some
kind of uncertainty around zero.
Would you say that Newton laws are violated by the computation?
A) Your calculation is not correct. Please see carefully that paper you found and mentioned before concerning RDM. The solution is as follows:
F1 = -1 + 2α₁ , F2 = 1 - 2α₁
F1+F2=(-1 + 2α₁)+(1 - 2α₁)=0
which is exactly zero.
so no laws are violated.
Q3. Very important
1. In that software is mul(x,y) equal to sqr(x) whenever x=y?
Oliver told:
The sqr (square) function comes from IEEE Std 1788-2015. You can use it
to compute x². mul(x, x) will give a different result, because both
parameters are considered independent of each other, which may lead to
overestimation.
You told that symbolic computation in Octave gives better results, because it is a tighter enclosure of the exact values. Yes you're right, but please bear in mind that symbolic computation is not the same as RDM, in general. That's, one can obtain two different solution based on RDM and symbolic computation, indeed, RDM gives us better solution. see the following example.
A) Please consider the following example which shows based on IEEE Std.1788 we have over-estimated result, and based on symbolic computation (your suggestion) we have a result different from RDM.
let f(x)=x(10-x), where x=[-1 1];
IEEE Std.1788 reads
f=[-11 11]
if we expand f(x) as f(x)=10x-x^2, then based on IEEE Std.1788 one can also get the following: f=[-10 10]-[0 1]=[-11 10]. However, the exact result is [-11 9].
If someone would like to use Octave, then they have to do one of the following: