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2018-03-16 13:39:00 +0200 | asked a question | Assumptions on symbolic functions I have a symbolic function f = function('f')(x). Is there a way to assume it is real and get abs(exp(I*f)) = 1? |
2018-02-08 13:42:16 +0200 | commented answer | Sage 8.1 eats memory until system freeze Thanks for the code optimization. But the problem remains: a working code in 7.5.1 makes 8.1 to use more and more memory. I think this is a regression. Shall I open a bug report? |
2018-02-07 11:16:09 +0200 | asked a question | Sage 8.1 eats memory until system freeze Hi, I have a quite long code (which uses arbitrary precision real numbers) which runs perfectly on Sage 7.5.1 (ppa for Mint 17.3 - Ubuntu 14.04). On Sage 8.1 (sage-8.1-Ubuntu_14.04-x86_64.tar.bz2) it starts to eat the memory until it freezes the system. I would like to help in debugging. Is there something that I can try/run/test? I can also upload the code, if necessary. Finally, I have a minimal working code If I run this in 7.5.1, I see (in top) the memory percentage stable around 2.5. If I run in 8.1, it grows up to 7.3 before code termination. If I increase the length of the loops, memory usage continues to grow. |
2017-08-29 13:40:52 +0200 | commented answer | Pade approximation Hi, is it possible to go back to a power series? If I give I get the error: AttributeError: 'FractionFieldElement_1poly_field' object has no attribute 'taylor' |
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2016-07-29 14:09:33 +0200 | asked a question | Pass arguments to an octave call Hi, given, say, a = [1,2,3,4], is it possible to pass this vector to an octave call, like octave('a+1')? Of course, this last does not work. |
2016-07-29 14:04:51 +0200 | answered a question | More digits from octave Hi, I found myself a solution, for instance: |
2016-07-22 14:56:39 +0200 | commented answer | More digits from octave Sorry, try this instead a=octave('0.123456789'). I chose pi just to be short. I get a=0.123457, lost two digits. |
2016-07-22 12:39:08 +0200 | asked a question | More digits from octave Hi, how can I get all the 15/16 digits from a command like pi=octave('pi')? |
2016-07-22 12:37:53 +0200 | asked a question | More digits from octave Hi, how can I get all the 15/16 digits from a command like pi = octave('pi')? |
2014-07-14 10:34:12 +0200 | commented answer | taylor expansion with arbritary precision numbers Thanks, it works, And in order to get the coefficients? f.series(x,5).coefficients() seems to me in double precision. |
2014-07-11 08:58:26 +0200 | asked a question | taylor expansion with arbritary precision numbers Hi, if I define a function with arbritary precision numbers (e.g., f=0.123456789123456789*log(1+x)) and then compute its Taylor expansion (f.taylor(x,0,5)) it seems to me that the coefficients are given in double precision, whereas if I compute them (e.g., by derivative(f,x,5)(x=0)/factorial(5)) they are in original precision. First of all, am I right or is it only a visualisation difference? If I'm right, is it possible to compute the Taylor expansion with the original precision? Cheers, Marco |
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2013-11-07 03:50:53 +0200 | commented answer | Continued fraction of pi by hand Thank you for the explanation. I see that this question was tagged "confirmed_bug". Is there a place where I can follow what happens to this bug? Any maintainer working on it? |
2013-11-06 07:38:43 +0200 | answered a question | Continued fraction of pi by hand Thanks for the quick answer. I extended the precision but now a[28] is wrong (compared to continued_fraction). So 100 bits is not enough. How I discover it? |
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2013-11-06 04:41:54 +0200 | asked a question | Continued fraction of pi by hand Hello, I'm a newbie in Sage. I tried to compute (sagenb.org) the continued fraction of pi by hand with the following commands the answer (14 terms) is correct, but then I get Am I doing something wrong? Cheers |