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25-Diamond I
July 16, 2014
Solved

Bug with symbolic solve?

  • July 16, 2014
  • 3 replies
  • 2125 views

Has anybody an explanation why the symbolic solve produces that many wrong solution if used without the modifer "fully"?

May we call it a bug? Are there other examples?

16.07.png

Best answer by AlanStevens

I think it's clear enough where the extra solutions come from; Mathcad must do something like the following on solving the equation.

oddsolns.PNG

Alan

3 replies

19-Tanzanite
July 16, 2014

May we call it a bug?

. Is there something else we might call it?

Has anybody an explanation why the symbolic solve produces that many wrong solution if used without the modifer "fully"?

Because there's a bug

Werner_E25-Diamond IAuthor
25-Diamond I
July 16, 2014

Richard Jackson wrote:

May we call it a bug?

. Is there something else we might call it?

Hmm, its PTC - maybe its a feature!? PTC marketing sure can make that possible.

Has anybody an explanation why the symbolic solve produces that many wrong solution if used without the modifer "fully"?

Because there's a bug

Yes, and obviously "fully" makes the symbolics to think about its results a second time

Alan's explanation make it clear whats happening - false solutions coming from taking the equation to the power of 4.

24-Ruby IV
July 16, 2014

and too few solutions:

root.png

24-Ruby IV
July 16, 2014

I use others tools in some cases:

wolfram.png

19-Tanzanite
July 16, 2014

I think it's clear enough where the extra solutions come from; Mathcad must do something like the following on solving the equation.

oddsolns.PNG

Alan

Werner_E25-Diamond IAuthor
25-Diamond I
July 16, 2014

AlanStevens wrote:

I think it's clear enough where the extra solutions come from; Mathcad must do something like the following on solving the equation.

Thank! Yes, at least with your explanation its clear what happens.

Nevertheless a bug which should not happen.

Also intersting the different behaviour when using the fourth root instead.

I know the difference between (-1)^(1/3) and the third root of -1 (only the latter is -1) but not sure about the fourth root and how the symbolics sees it.

Werner