On Tue, Nov 25, 2014 at 11:30 AM, Bill Page <bill.p...@newsynthesis.org> wrote:
> On 25 November 2014 at 01:11, Ondřej Čertík <ondrej.cer...@gmail.com> wrote:
>> On Mon, Nov 24, 2014 at 10:23 PM, Bill Page <bill.p...@newsynthesis.org>
>> wrote:
>>> ...
>>> I am not very interested in real numbers. I am interested in the
>>> algebra. Would you say that
>>>
>>> sqrt(x^2).diff(x) = sqrt(x^2)/x
>>>
>>> is OK?
>>
>> I think so, using the following calculation:
>>
>> sqrt(x^2).diff(x) = exp(1/2*log(x^2)).diff(x) = exp(1/2*log(x^2)) *
>> 1/2 * 1/x^2 * 2*x = sqrt(x^2)/x
>>
>> The function exp(1/2*log(x^2)) that we differentiate is analytic, so I
>> don't see any issue here.
>>
>
> I did not ask whether it was technically correct or not. What I meant
> was is this expression what you would expect given the rest of the
> machinery of differentiation in any given computer algebra system?
Ah ok. I would actually expect to get x/sqrt(x^2), which is equivalent.
>
>>
>>> But I don't want to be forced to make a choice of branch until
>>> I actually need to evaluate an expression numerically.
>>
>> I understand that's what you want. I am just trying to understand how
>> exactly this works.
>>
>
> OK.
>
>>> ...
>>> I think what you are trying to say is
>>>
>>> (A) log(exp(z)) = { z + 2*pi*i*n | for all n in Integer}
>>
>> Exactly, that's what I meant.
>> ...
>>>
>>> Although it may seem simple in this case, in general implementing
>>> sets with comprehension like this requires logic and takes us
>>> outside of algebra as such into the realm of theorem proving.
>>
>> Sure. But that's what you want, correct?
>>
>
> No, not at all. I want this to be "algebraic", not some theorem of
> predicate calculus. That is what I meant by taking
>
> x + conjugate(x)
>
> as the definition of a real valued variable.
Ok.
>
>> ...
>>>
>>>>
>>>> This is precisely the part that I don't understand with the approach
>>>> (A). log(a*b), log(a) and log(b) are all multivalued, so you would
>>>> naively think, that log(a*b)-log(a)-log(b) = 0 + 2*pi*i*n, for all
>>>> "n". But I think this is not the case, I think the "n" in log(a*b) is
>>>> coupled to the implicit "n" in log(a) and log(b) in such a way, that
>>>> the result is exactly 0. Can you clarify exactly how this works?
>>>
>>> Try it this way:
>>>
>>> a*b = exp(?1)
>>> a = exp(?2)
>>> b = exp(?3)
>>>
>>> I think 'normalize' is saying that there is a solution that makes
>>>
>>> ?1 - ?2 - ?3 = 0.
>>
>> Ok, but why wouldn't normalize return 2*pi*i instead? Or 4*pi*i?
>
> These are equivalent in the sense of having the same number of
> algebraically independent transcendental kernels, i.e. none.
I don't understand that. Is the result of normalize() multivalued?
Or how else could 0 be equivalent to 2*pi*i or 4*pi*i?
>
>>
>> In other words, how exactly are the operations on the multivalued
>> sets log(x) defined?
>
> FriCAS does not perform operations on multivalued sets to determine the above.
Ok. Though my question stands, how are the operations defined in your approach?
>
>>>
>>> I meant that I did not understand what you are proposing for how to
>>> represent the value of 'log(z)' symbolically, i.e. when the value of z
>>> is unknown.
>>
>> Ah ok. I would represent it by the approach (B). But then, as we
>> talked about, it's not true that conjugate(log(z)) = log(conjugate(z)).
>> Since you want this property to hold, then the approach (B) does
>> not work for you, obviously. So I am trying to understand how
>> exactly are all the operations defined in your approach. You said
>> your approach is not (A) exactly. So I am just trying to understand.
>>
>
> OK.
>
>>
>> This discussion is about how a CAS should handle (complex)
>> differentiation. Since it started here, I would finish it here, so
>> that the whole thread is in one mailinglist for future reference.
>>
>
> OK. It would be nice to know if other sage-devel subscribers actually
> remain interested...
>
> Let's return to differentiation for a moment. Using your definitions
> what would you say is the correct result for
>
> log(exp(z-conjugate(z))).diff(z)
>
> My patched version of FriCAS based your definition in this thread
> currently returns 0. Do you get the same result?
No, the derivative is most definitely not zero:
log(exp(z-conjugate(z))).diff(z) =
exp(z-conjugate(z))/exp(z-conjugate(z)) * [1 - 1*exp(-2*i*theta)] = 1
- exp(-2*i*theta)
In other words, the two Wirtinger derivatives are 1 and -1. You can
easily check numerically that this formula is correct for all complex
"x" and angles theta, I've done it here:
https://github.com/certik/theoretical-physics/blob/f9406a02ef8e04b2daa669f444148186b6b892e8/src/math/code/test_complex_diff.py#L118
and it works.
>
> Since the derivative is 0 would we want to say therefore that
>
> log(exp(z-conjugate(z)))
>
> is a constant?
If you got 0, then I think you can say that the function is constant.
We didn't get 0, so the function is not constant.
> If not, isn't this an argument for needing another
> derivative?
In some of your previous emails you wrote that this theta factor
"still looks ugly to you". Maybe it's ugly, but it's correct, as you
fell into this trap yourself: if you omit theta and implicitly assume
theta=0, then you don't know if what you got is analytic or not.
Perhaps this comment in a sympy issue might help:
https://github.com/sympy/sympy/issues/8502#issuecomment-64415017
Essentially the formula with theta is equivalent to just returning a
tuple of the two Wirtinger derivatives. So what holds for one approach
holds for the other one.
> The result of this test currently causes a problem during
> manipulations of expressions of this form. Check the two Wirtinger
> derivatives for this case. If we have both derivatives we can avoid
> this problem quite easily as my previous version of the patch showed.
My current best solution is to define a function `diff(x, theta=0)`,
where the theta argument is 0 by default, but you can pass any angle
into it, or a symbol theta if you want. That way you won't get the
theta factors by default, but if in doubt, you can always get them.
Let me know if you have a better proposal.
Ondrej
>
> Bill.
>
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