Re: [Maxima-discuss] e, pi, gamma, log(2)

[Volker v. N. <vol...@gm...>]

Am 22.09.2014 18:27, schrieb Raymond Toy:
> 
>     Volker> fpprec:30$
>     Volker> x:0.5b0$
>     Volker> asin(x);            --> 5.23598775598298873077107230547b−1
>     Volker> bs_asin(x);         --> 5.23598775598298873077107230547b−1
>     Volker> sin(asin(x));       --> 5.00000000000000000000000000001b−1
>     Volker> bs_sin(bs_asin(x)); --> 5.00000000000000000023968760547b−1
> 
> Interesting. I'm sure this can be fixed, but it seems like an odd
> artifact.
> 

I fixed it. Now it seems that the bs_ functions are more accurate.

fpprec:32$ x:0.5b0$

asin(x);    5.2359877559829887307710723054659b−1 wrong last digit
bs_asin(x); 5.2359877559829887307710723054658b−1 OK

sin(x);     4.7942553860420300027328793521557b−1 OK
bs_sin(x);  4.7942553860420300027328793521557b−1 OK

sin(asin(x));       5.0000000000000000000000000000001b−1
bs_sin(bs_asin(x)); 5.0b−1
asin(sin(x));       5.0000000000000000000000000000001b−1
bs_asin(bs_sin(x)); 5.0b−1


> 
>     Volker> My functions have a bug where I need help. In case the return value is a
>     Volker> power of 2 and fpround has rounded from below I have the problem to fix
>     Volker> the exponent. It seems that I do not understand the mechanism of fpround
>     Volker> and *m in float.lisp. Can someone help?
> 
> Can you give an example of the problem? I have not looked at fpround
> in a very long time and don't really remember how it works anymore.
> 

fpprec:30$ x:0.5b0$

asin(sin(x));        5.0b−1
bs_asin(bs_sin(x));  2.5b−1  BUG


The essential lines of code:

(defun bs-asincos (x fn)
  (let ((phi
         (let* ((fpprec (+ fpprec 12))

I leave out some lines here.
phi is now computed with 12 extra bits and
the following line is the correct result but with 12 bits extra.

           (print (bs-carg1 re-z im-z fpprec)) )))

After returning from the block of higher precision I want to round back
to the original precision:

    (list (print (fpround (car phi))) (print (cadr phi))) )) ; BUG


print 1 : (20769187434139310514121985316880257 -1)  OK
print 2 : 2535301200456458802993406410752  OK
print 3 : -1  WRONG

The exponent must be 0 instead of -1. The binary output shows the
reason. fpround rounds from below.

print 1 :
(111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110000001
 -1)
print 2 :
100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

print 3 : -1

How do I get noticed when to change the exponent?
How can *m help here?


Thanks in advance
Volker


> Ray
> 
> 
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