[Maxima-bugs] [ maxima-Bugs-3547315 ] Problems with bfloat

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Bugs item #3547315, was opened at 2012-07-22 09:34
Message generated for change (Comment added) made by chrisrein
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Category: Lisp Core - Floating point
Group: None
Status: Open
Resolution: None
Priority: 5
Private: No
Submitted By: christoph reineke (chrisrein)
Assigned to: Nobody/Anonymous (nobody)
Summary: Problems with bfloat

Initial Comment:
Enter:

(%i2) fpprec:30;
(%o2) 30
(%i3) bfloat((45.12*34.78923)^2);
(%o3) 2.46392687692829128354787826538b6
           2.46392687692829131776b6 is the correct value.

build_info("5.27.0","2012-05-08 11:27:57","i686-pc-mingw32","GNU Common Lisp (GCL)","GCL 2.6.8")

Regards

Chris

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>Comment By: christoph reineke (chrisrein)
Date: 2012-07-23 02:12

Message:
Sorry, but I don’t understand you.
<Can you explain why you think the value you give is the correct value?>
Which value?

(45.12*34.78923)^2=2.46392687692829131776b6=the correct value.
                                     
Since I expect that the correct value has more than 16 digits after the
comma,
I use bfloat and fpprec=30.
I quote from the documentation:
“bfloat (expr) Function
Converts all numbers and functions of numbers in expr to bigfloat numbers.
The
number of significant digits in the resulting bigfloats is specified by the
global 
variable fpprec.” …all numbers…!!

Now Maxima returns 30 digits, but the result is wrong. That’s all.

<Thus, the simplifier has simplified the float expression…>
Which simplifier?
If a CAS computes a simple multiplication each digit of the displayed
floating point number should be correct (apart from the last), no matter
how
many digits the user wants. Even if I erroneously “simplify” something,
Maxima
should never display a wrong result. That’s my opinion.

How do I get the correct result?

Thanks for your help!


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Comment By: Raymond Toy (rtoy)
Date: 2012-07-22 13:07

Message:
Can you explain why you think the value you give is the correct value?

Consider this transcript:

(%i1) fpprec:30;
(%o1)                                 30
(%i2) trace(bfloat);
(%o2)                              [bfloat]
(%i3) bfloat((45.12*34.78923)^2);
1 Enter bfloat [2463926.876928291]
1 Exit  bfloat 2.46392687692829128354787826538b6
(%o3)                  2.46392687692829128354787826538b6

Thus, the simplifier has simplified the float expression before bfloat gets
a chance to look at it.


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