On 24 November 2014 at 17:43, Ondřej Čertík <ondrej.cer...@gmail.com> wrote:
> On Mon, Nov 24, 2014 at 1:57 PM, Bill Page <bill.p...@newsynthesis.org> wrote:
> ...
>>
>> In FriCAS 'abs' is already a kernel function and it implemented the
>> derivative of 'abs' even before my proposed patch but I think the
>> current definition is wrong:
>>
>> (14) -> D(abs(x),x)
>>
>> abs(x)
>> (14) ------
>> x
>> Type: Expression(Integer)
>
> I think that's correct for real numbers, i.e. x/abs(x) = abs(x) / x.
>
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?
>> Rather, I think the correct definition of 'log(z)' is the solution of
>>
>> z = exp(?)
>>
>> So we can write z=exp(log(z)) by definition.
>
> Indeed, exp(log(z))=z always,
Not just "true always". It is the definition of 'log(z)'.
>
> In both (A) and (B),
> it is true that z = exp(log(z)). However, these are different:
>
> (A) log(exp(z)) = z + 2*pi*i*n
> (B) log(exp(z)) = z + 2*pi*i*floor((pi-Im z) / 2*pi)
>
> In (B), you get a single value, but in (A) you get multiple values,
> one for each "n".
>
Better to say (A) is true "\forall n \in Integer"
> I am very familiar with the approach (B) and I think I understand
> exactly what follows from what and how to derive all the formulas.
(B) is about choosing a particular "branch" of 'log' - the principal
value. But I don't want to be forced to make a choice of branch until
I actually need to evaluate an expression numerically.
> But I am not 100% sure with (A), I was hoping you can help, since
> that's the approach that you want to use in FriCAS.
(A) is not quite the approach I want to use for 'Expression' in
FriCAS. I want 'log(exp(z))' to be the solution of
exp(z) = exp(?)
The best way to say that algebraically (symbolically) is just
'log(exp(z))', i.e. without evaluation. This is what FriCAS already
does in the case of
(1) -> z:Expression Complex Integer
Type: Void
(2) -> exp(log(z))
(2) z
Type: Expression(Complex(Integer))
(3) -> log(exp(z))
z
(3) log(%e )
Type: Expression(Complex(Integer))
but unfortunately not in the case of 'Expression Integer'.
(4) -> z:Expression Integer
Type: Void
(5) -> log(exp(z))
(5) z
Type: Expression(Integer)
I think that what it currently does for 'Expression Integer' should be
considered a bug.
> I think what I wrote is correct for (A), but please correct me if I am wrong.
>
>> This is exactly analogous
>> to the treatment of 'sqrt(z)' as the solution to
>>
>> z = ? * ?
>>
We have
sqrt(z)^2 = z
but
sqrt(z^2) = sqrt(z^2)
>>> However, this definition quickly becomes impractical, because you
>>> need to be able to numerically evaluate symbolic expressions, and
>>> you would need to carry the symbolic term 2*pi*i*n around.
>>
>> We do not need an extra term. We only need axioms for the correct
>> behavior of the expression 'log(z)'. But 'log(z)' does not denote a
>> function in the sense of a many-to-one mapping. The inverse of a
>> function is only a function (possibly partial) if the function is
>> injective (one-to-one).
>
> Sure, that's why you have "n" in the formula for log(z) in (A), and
> the function is multivalued over all "n".
>
No. As you have written it, it is a function of two variables with
one value for each z and n.
f(z,n) = z + 2*pi*i*n
I think what you are trying to say is
(A) log(exp(z)) = { z + 2*pi*i*n | for all n in Integer}
and
sqrt(z) = { x | x^2 =z }
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.
>>
>> The Riemann surface is an important tool in complex analysis but I
>> have yet to see it used explicitly in any computer algebra system as
>> a representation of complex functions.
>
> I thought the Riemann surface gives you a way to couple "n" in
> log(a*b), log(a) and log(b) in such a way that you get exactly
> log(a*b)-log(a)-log(b)=0. I.e. that the Riemann surface is (A).
> I agree that I haven't seen it used in CAS. But I think it must be
> implicitly used in FriCAS somehow. Maybe you can clarify that.
I don't think it is used in FriCAS at all. A Riemann surface is a
manifold. At each point in the manifold we need to evaluate a
function, but how would you propose to represent this surface
algebraically? For example, the surface of a sphere is a manifold.
For that we need a co-ordinate system that labels each point on the
surface. In a sense the important thing about a Riemann surface is
its global topology and this is different for each complex holomorphic
function. For 'exp' it might make sense to include 'n' as one of
these co-ordinates. But this is not what FriCAS does.
>> ...
>> In order to numerically evaluate a symbolic expression it is indeed
>> necessary to choose a branch in the case of "multi-valued functions",
>> i.e. expressions like 'sqrt(x)' and 'log(x)'. But this choice should
>> not effect the axiomatic properties of these expressions.
>
> I see. I think in the definition:
>
> (A) log(z) = log|z| + i*arg(z) + 2*pi*i*n
>
> the result is multivalued, and you obtain all the numerical values by
> plugging different integers for "n".
>
Sure but "plugging different integers for n" makes n a parameter.
>>
>> In FrCAS we have
>>
>> (2) -> normalize(log(a*b)-log(a)-log(b))
>>
>> (2) 0
>> Type: Expression(Integer)
>>
>> and with my proposed patch we also have 'conjugate(log(z)) =
>> log(conjugate(z))' by definition. So this is like your (A).
>
> What does Expression(Integer) mean? Does it mean that "a" and "b"
>are integers? I.e. does the above hold if a=b=-1?
No. Integer is the domain of the coefficients of the rational
function. In other words Expression Integer is a ratio of two
polynomials with coefficients from Integer and variables from Symbol
or a kernels over Expression Integer (recursively).
>
> 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.
Maybe there is a way to think of this as a kind of coupling of n's if
that was the way that log was represented but I am quite sure that
this is not what 'normalize' is doing.
>>
>> Now I am a little unclear on what you are proposing. Are you
>> suggesting that symbolic computations (such as those using
>> 'Expression' in FriCAS) should somehow introduce an independent
>> 'argument' function instead of defining it as
>>
>> argument(x) = log(x/abs(x))/%i
> ...
>>
>> or that 'abs' is somehow an important part of these equalities?
>
> I am not sure what you mean by this question.
>
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.
>> Note
>> that 'normalize' does not have to introduce these sort of terms to
>> in order to return 0 in result (2) above.
>
> See my questions above about this.
>
'normalize' re-writes Expression using the minimum number of
algebraically independent transcendental kernels. In the case above
that is just 0.
--
Let's take a step back: Is this discussion going any place useful for
you? We seem to be talking mostly about FriCAS. Does this discussion
really belong here on the sage-devel mailing list?
Bill.
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