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Symbolic evaluation precision

SOLVED
CamilleLaflamme
Regular Member

Symbolic evaluation precision

Hi, I would like to simplify the solution, but I want the rest of the program to use all of the numbers. 

The Float command doesnt not allow that request. It uses the new numerical value.

Also, In my case, I can't change the display precision. It is not available while using symbolic evaluation. 

 

Thank you

Camille

 

CamilleLaflamme_0-1595352187563.pngCamilleLaflamme_1-1595352206005.png

 

1 ACCEPTED SOLUTION

Accepted Solutions

Re: Symbolic evaluation precision

As epsilon.B seems to be undefined, you can't use numeric evaluation whits its ability to format the displayed result the way you'd like it being shown.

 

Unfortunately the symbolic evaluation does not know any commands or modifiers to format the result.

The "float" modifier affects the precision of the whole calculation, not just the displayed result.

 

In case you don't mind the red error which stems from the numeric as epsilon.B is not defined, you may do it as shown below:

Werner_E_0-1595356817921.png

 

As you can see, epsilon.B still has full precision. Don't be tempted to use an inline evaluation with "float", as this will limit the precision of epsilon.b:

Werner_E_1-1595356921056.png

 

Another approach comes to my mind but it involves kind of cheating and I am not sure if you can use it in your work. You may define epsilon.A:=1 somewhere out of sight (to the right of the sheet or in a collapsed region at the top) and then use epsilon.A similar to a unit in a numerical evaluation which you can format as you like.
You should also add clear.sym(epsilon.A) so epsilon.A still shows in symbolic evaluations.

Werner_E_0-1595358162768.png

 

The drawback of course is, that you have to add epsilon.A (or epsilon.A^2 in the last region) manually and this sure is failure prone. And it gets worse if more unknown variables are added (like A, E in your sheet).

 

Another approach which may not look so clean, though, would be to set up all as function of the unknowns

epsilon.B(epsilon.A):=....

which are evaluated numerically only when all unknowns are specified.

 

View solution in original post

1 REPLY 1

Re: Symbolic evaluation precision

As epsilon.B seems to be undefined, you can't use numeric evaluation whits its ability to format the displayed result the way you'd like it being shown.

 

Unfortunately the symbolic evaluation does not know any commands or modifiers to format the result.

The "float" modifier affects the precision of the whole calculation, not just the displayed result.

 

In case you don't mind the red error which stems from the numeric as epsilon.B is not defined, you may do it as shown below:

Werner_E_0-1595356817921.png

 

As you can see, epsilon.B still has full precision. Don't be tempted to use an inline evaluation with "float", as this will limit the precision of epsilon.b:

Werner_E_1-1595356921056.png

 

Another approach comes to my mind but it involves kind of cheating and I am not sure if you can use it in your work. You may define epsilon.A:=1 somewhere out of sight (to the right of the sheet or in a collapsed region at the top) and then use epsilon.A similar to a unit in a numerical evaluation which you can format as you like.
You should also add clear.sym(epsilon.A) so epsilon.A still shows in symbolic evaluations.

Werner_E_0-1595358162768.png

 

The drawback of course is, that you have to add epsilon.A (or epsilon.A^2 in the last region) manually and this sure is failure prone. And it gets worse if more unknown variables are added (like A, E in your sheet).

 

Another approach which may not look so clean, though, would be to set up all as function of the unknowns

epsilon.B(epsilon.A):=....

which are evaluated numerically only when all unknowns are specified.

 

View solution in original post

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