ggrafendorfer wrote:
> Hi Robert,
> this was not a misunderstanding, there is an "n" missing :-), I
> corrected it:
> I wanted to write
>
> then i^2 should also be of type CDF, but
>
> rather then
>
> the i^2 should also ....
>
> as an answer to your statement, namely that i^2 is getting turned into
> CDF(i)^CDF(2)
>
> I hope this is clear now
>
>
>> Generally in Sage we prefer exact arithmetic to numeric
>> approximations, because once you start doing numeric things one has
>> to deal with rounding errors, etc. and there's no going back.
>
> and that's exactly the reason why I don't understand why "i^2" is not
> be treated like an exact symbolic expression
>
See the bottom of the message for my suggestion, which basically is to
use the python complex i if you really want speed (i.e., at the top of
your program, do "i = complex(I)"
I'll take a try at explaining this. This may be more for the interested
reader than for you specifically, so sorry if it's rehashing ground that
has already been covered.
In Sage, as you know, we have numerous ways of representing i. The
default way is to represent it using a symbolic expression:
sage: type(i)
<class 'sage.functions.constants.I_class'>
sage: type(i^2)
<class 'sage.calculus.calculus.SymbolicArithmetic'>
Of course, we can also treat i as a numeric quantity (basically, we
treat complex numbers as double precision floating point numbers; one
double for the real part, one for the imaginary part). That's using CDF:
sage: cdf_i=CDF.0
sage: cdf_i
1.0*I
sage: cdf_i^2
-1.0 + 1.22460635382e-16*I
sage: cdf_i*cdf_i
-1.0
Unfortunately, as you see above, this representation is prone to
roundoff error and gives slightly inaccurate results when squaring.
That's what happens when you work with a numerical i (that's why the
default "i" in Sage is symbolic).
You could also declare i as the root of the polynomial x^2+1; that's
what Robert did earlier, and is still fast, and what's more, it is exact.
sage: K.<ii> = QuadraticField(-1)
sage: ii
ii
sage: ii^2
-1
In fact, using the quadratic field is the fastest Sage solution from
what we have above:
sage: #Symbolic I
sage: timeit('i^2')
625 loops, best of 3: 86.7 µs per loop
sage: # CDF I
sage: timeit('cdf_i^2')
625 loops, best of 3: 41.3 µs per loop
sage: # Exact I as a root of a polynomial
sage: timeit('ii^2')
625 loops, best of 3: 5.65 µs per loop
However, for your purposes, maybe using the builtin python complex type
is best; it is by far the fastest and it has the property that you want
(namely, i^2=-1)
sage: python_i = complex(I)
sage: python_i
1j
sage: python_i^2
(-1+0j)
sage: timeit('python_i^2')
625 loops, best of 3: 2.72 µs per loop
Jason
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