You might want to take a look at the book by Garret Sobczyk
New Foundations in Mathematics
The Geometric Concept of Number
Birkhauser
On 12/13/2014 09:08 AM, Joachim Durchholz wrote:
Am 13.12.2014 um 06:27 schrieb Richard Fateman:
On Thursday, December 11, 2014 8:16:09 AM UTC-8, Joachim Durchholz
wrote:
Am 11.12.2014 um 00:40 schrieb Richard Fateman:
1994 paper by Adam Dingle and Richard Fateman
Branch Cuts in Computer Algebra, (ISSAC '94 proceedings. also search
online).
That paper assumes that everything can be refactored to logarithms plus
arithmetic.
Does that assumption hold? I could imagine that expressions containing
irreducible integrals might not be normalizable in that fashion.
Repeating the unanswered questions:
Does the paper assume that everything is normalizable to log/exp?
If yes, does this assumption hold in general?
If the assumption does not hold in general, how large is the class of
problems where it does hold, as opposed to all problems that might be
relevant to somebody doing symbolic math?
When you say things about sqrt(), does it generalize to cuberoot?
If it
does not, you are in trouble, or will be down the road.
What is the principal value of (1)^(1/6) ?
Well, there are 6 possible values.
Heh. I didn't write the question, and I knew the answer :-)
> You could pick (same as sqrt rule?) the positive one, namely 1.
That doesn't strike me as a good choice,
I think picking *any* principal root would be problematic, because one
would miss equalities (think x^12 = x^15).
But that wasn't my point anyway. I was thinking about calculating for
all principal roots in parallel, and not choosing at all until it
turns out that some choice is inconsistent with other assumptions.
> since powers of 1 do not generate
the other roots.
What's the definition of "generate"? That you can get all principal
roots by taking a power of the chosen value?
Not that I think that property is *that* important, you can always use
polar representation and generate additively (that's more in line with
the approach in your paper anyway).
You think you can use sqrt because the quadratic formula says
-b+-sqrt(b^2....)
etc.
So you think you know what sqrt means.
Not sure what assumptions you assume.
No assumptions whatsoever.
You're making assumptions about my knowledge of math if you say "so
you think you know xxx".
I was wondering about these.
But in that formula you can switch the values +- and the formula
is still valid.
Not sure what you mean with that - switching the signs means I still
get
the same set of expressions.
That's why the formula works. switching the sign on sqrt just
exchanges the
roots.
The point now being...?
How many other formulas do you expect to fiddle with where that is
true?
Certainly not this one: sqrt(y^2) = abs(y).
That could still be handled by doing case distinctions.
If you plot abs(y) you get a V-shaped curve, with a singularity at 0.
neither square root of y^2 has such a plot.
Yeah, you need both plots.
And you need to match them with abs() using a case distinction. Which
happen to be the same as the one abs() is making (otherwise the
equation wouldn't hold).
This is an interesting case though, since it tells us that we need a
quantor: Are we interested in "there exists a branch where the
expression holds" or in "the expression holds in all branches"?
I think we need the former when determining all solutions to an
equation, and the latter when verifying an assumption.
I'm not familiar with the term quantor,
Oh right, it's "quantifier" in English.
> but it makes sense to me to
have 2 separate questions here.
Agreeing on that one then.
So if you go off and do the wrong thing, it is probably prudent to
understand that you are doing the wrong thing.
Actually the "wrong thing" could be the right thing in specific
circumstances (experimental physics is full of this kind of stuff),
perhaps that is why some people advise against learning
math from a physicist.
Heh. I can understand that sentiment.
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