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System does not have a finite number of solutions. Why?

Cornel
18-Opal

System does not have a finite number of solutions. Why?

Hello,

Cornel_0-1693995886136.png

 

Cornel_4-1693996122318.png

 


Why? Because results for a, b, and c should give:

Cornel_3-1693996085217.png

 

MC Prime 9 fille attached.

5 REPLIES 5
LucMeekes
23-Emerald III
(To:Cornel)

The error message is ridiculous, considering that:

LucMeekes_0-1694000101045.png

also does not have a finite amount of solutions. As it is, Prime will give:

LucMeekes_1-1694000164559.png

and you can have:

LucMeekes_3-1694000508428.png

 

I'd say, report it as a bug. Especially considering that Mathcad 11 gives:

LucMeekes_2-1694000392547.png

 

Success!
Luc

 

Werner_E
24-Ruby V
(To:Cornel)

Well, not every innovation is an improvement! This is especially true for Prime.
Here are the results of Prime 6 with the old muPad engine still available there, compared to your disappointing results with the new FriCas/Axiom symbol kernel.

Werner_E_0-1694000919818.pngWerner_E_1-1694000931951.png

Of course the "solution" a=any number b=c=0 is wrong, as a,b and c can't be chosen to be zero (same for x2)!

 

As Luc said - if you feel like doing so and think it may help, you may consider reporting your results to PTC support as a bug in the new symbolic engine.

 

@LucMeekes 

 

I found one workaround...

Cornel_0-1694001785214.png

 

For example, from the first equation we can solve for b:

Cornel_1-1694001812125.png

Then with that b:

Cornel_2-1694001882459.png 

Cornel_0-1694002210874.png

 

And then:

Cornel_1-1694002218008.png

 

 

LucMeekes
23-Emerald III
(To:Werner_E)

"Of course the "solution" a=any number b=c=0 is wrong, as a,b and c can't be chosen to be zero (same for x2)!"

There is something to say for it: the equations can/could also have been written as:

x2*x1*a*c = b

x3*x1*a*c = c-b

x4*b*(c-b) = x1*a*c

Now if c and b are 0, then a (and x1...x4) can be any number:

LucMeekes_2-1694005146098.png

But the fact that b and c appear dominantly in the denominator of a division prevents them from being 0.

However, that is still no excuse to flag an error for too many solutions.

 

Success!
Luc


However, that is still no excuse to flag an error for too many solutions.

Sure!

 

For a fully correct solution it would be necessary to say that the result is only valid if x1, x2, x3, x3 and x2+x3 are not zero, otherwise there is no solution (given the original equations).

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