196 reputation
5
bio website
location Krakow, Poland
age 25
visits member for 2 years, 2 months
seen Oct 16 at 9:30

Nov
2
accepted How to draw this kind of plot?
Nov
2
comment How to draw this kind of plot?
Thank you! That's exactly what I needed.
Nov
2
accepted Why does Mathematica not seem to know that $i^x$ cannot be equal to $0$?
Nov
2
asked How to draw this kind of plot?
Oct
31
comment Why does Mathematica not seem to know that $i^x$ cannot be equal to $0$?
@Nasser: I'm using V 8.0. I didn't realise this would differ between versions.
Oct
31
awarded  Commentator
Oct
31
comment Why does Mathematica not seem to know that $i^x$ cannot be equal to $0$?
Well, I think in your example the two problems are mathematically different. As $x$ goes to $-\infty$, $i^x$ certainly takes the values $1$ and $-1$ infinitely often (for $x$ congruent to $4$ and $2$ modulo $4$, respectively). After replacing $I$ by $2I$ in your example, the limit comes out as $0$.
Oct
31
comment Why does Mathematica not seem to know that $i^x$ cannot be equal to $0$?
Thanks! It works as I was hoping it would.
Oct
30
asked Why does Mathematica not seem to know that $i^x$ cannot be equal to $0$?
Jun
23
comment Finite precision - how to get rid of “near zeros”
Chop is just what I was looking for. Thank you!
Jun
23
comment Finite precision - how to get rid of “near zeros”
@Kuba: Precisely what I needed! Thanks.
Jun
23
asked Finite precision - how to get rid of “near zeros”
Jun
23
accepted Set theoretic operations on sets of real numbers
May
29
comment Set theoretic operations on sets of real numbers
@J.M. I did not, thank you for pointing this out. I will try this and report if it worked.
May
29
asked Set theoretic operations on sets of real numbers
May
7
accepted Why is arithmetic faster for inexact arithmetic?
May
7
comment Why is arithmetic faster for inexact arithmetic?
Thank you! This clarifies a lot.
May
7
asked Why is arithmetic faster for inexact arithmetic?
Apr
29
comment Exploiting self-adjointness when changing basis
Thank you, that was just what I needed. And yes, I did mean $Q' = P^{-1} Q P$; just corrected it for increased readibility.
Apr
29
awarded  Scholar