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

Cornel
19-Tanzanite

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
25-Diamond I
(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.

 

Cornel
19-Tanzanite
(To:Werner_E)

@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

Werner_E
25-Diamond I
(To:LucMeekes)


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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