-- Wrong limits do not commute
Han de Bruijn
and an answer. Some teachers think that they are smart by asking wrong
questions and nevertheless expect right answers to those questions.
But, of course, the "right" answer to a wrong question is still wrong.
Especially with limits, wrong questions are imposed upon students, in
order to demonstrate, for example, that: limits do not always commute.
Example 1, by Horand Gassmann, from "Why does everyone do it?":
http://groups.google.nl/group/sci.math/browse_frm/thread/bfa9f4f2c780e6f
f_n(x) = 1 if n <= x < n+1
= 0 otherwise
Then int_{-infinity}^{+infinity} f_n(x)dx = 1, and hence
lim{n->infinity}[int...] = 1, as well, while
lim{n->infinity} f_n(x) = 0 for every x, and so
int_{-infinity}^{+infinity} [lim...] = 0.
Let's analyse this. We shall first unravel the meaning of these limits:
int_{-infinity}^{+infinity} f_n(x)dx = lim int_(-L)^(+L) f_n(x)dx
L->oo
Therefore what's really going on is the following (n natural, L real):
lim lim int_(-L)^(+L) f_n(x)dx ( = lim 1 = 1 )
n->oo L->oo n->oo
lim lim int_(-L)^(+L) f_n(x)dx ( = lim 0 = 0 )
L->oo n->oo L->oo
Where the answer between parentheses is the one expected by your maths
teacher. Now Horand himself has published a picture that is associated
with this little problem:
--------
: :
---------------------^----------- ^ ------^---------------- f_n
L n L n+1 L
Apart from the limits, as long as we are in the finitary domain, then
it's obvious that there exist the following possibilities for L and n:
L < n ; L = n ; n < L < n+1 ; n+1 = L ; n+1 < L
Now Horand says that it's possible to "fix" n and let L go to infinity
or that it's possible to fix L and let n go to infinity. _I_ say that
such a "fixed" real L or natural n is an undefined concept. A real is
just a real and there's nothing "fixed" or "variable" with it. We have
to live wit the fact that _both_ n and L approach limiting values and
in _whatever_ fashion. So what is the proper way to express this idea?
With other words: what would have been the _right question_ belonging
to this excercise ? The following is my proposal:
lim int_(-L)^(+L) f_n(x)dx
min(n,L)->oo
And now the _proper_ the answer is that this limit: does not exist. Or
rather that it has a value between 0 and 1, not by coincidence the two
extremes found by Horand's students (at least those who got an 'A' for
their exam). In this case two limits do not commute and the associated
right questioned limit does not exist. Hmm, would that be systematical?
Example 2, by Horand Gassmann, from "Why does everyone do it?":
http://groups.google.nl/group/sci.math/browse_frm/thread/bfa9f4f2c780e6f
Let g_n(x) = 2.n^2.x if 0 <= x < 1/(2n)
= 2n - 2.n^2.x if 1/(2n) <= x < 1/n
= 0 everywhere else.
These functions are triangular, and they all disappear outside of
[0,1], so I can compute int_0^1{g_n(x) dx} = 1 for every n.
And again, for every x, lim{n->infinity g_n(x) = 0.
So again, the limit and the integration can't be interchanged.
Sure, g_n(x) = 0 , sure. It's not the first time I've seen infinities
suddenly disappear by mainstream mathematician's "ingenuity":
http://hdebruijn.soo.dto.tudelft.nl/www/grondig/natural.htm#bv
http://hdebruijn.soo.dto.tudelft.nl/jaar2006/ballen.jpg
As Mr. Gassmann has admitted, this is a proper picture of the problem:
http://hdebruijn.soo.dto.tudelft.nl/jaar2008/gassmann.jpg
We see that the function g_n(x) becomes a sharp peak at x=0 for n->oo
and that, geometrically, it certainly not disappears or becomes zero.
The vemin is again in the false doctrine that it would be possible to
"fix" x and let n go to infinity. Again, HdB says that such a "fixed"
real x is an _undefined_ concept. A real is just a real and there's
nothing "fixed" or "variable" with it. Such is even more obvious with
the present function g_n(x) , where x and n are clearly intermixed,
by definition. So we _can_ have: 0 <= x < 1/(2n) , 1/(2n) <= x < 1/n ,
1/n <= x , and nothing prevents us from groing to limits _while_ this
is the case. If we just do this, then what we get is what physicists
know as a delta function:
delta(x) = 0 for x <> 0
= any for x = 0 ; integral_(-oo)^(+oo) delta(x) dx = 1
It's easy to see that: lim g_n(x) = delta(x)
n->oo
Whatever definition might be the "right" one, a delta function roughly
speaking is just a very large peak around x = 0 with area normed to 1.
With the finitary domain in mind, there is no problem in understanding
what delta functions actually are (even if they are a bit beyond x=0).
Note that, in this case, it's _not_ even the wrong question. It's just
that professors expect a wrong answer from their students. The second
answer is wrong, namely, and the first one is right:
lim [ int_0^1 g_n(x) dx ] = 1 = int_0^1 [ lim g_n(x) ] dx
n->oo n->oo
Example 3, from a (hypothetical, I've made it up) textbook.
x^2 - y^2
Let F(x,y) = --------- .
x^2 + y^2
To avoid trouble with the mainstreamers, we could define F at (x,y) =
(0,0) as the mean value F(0,0) = 0 . As far as I'm concerned myself,
I'd rather define F(0,0) as 0/0 = anything. And in view of the sequel
that anything is restricted to values between -1 and +1. Proceeding:
x^2 - y^2 x^2/y^2 - 1
lim ( lim --------- ) = lim ( lim ----------- ) = - 1
x->oo y->oo x^2 + y^2 x->oo y->oo x^2/y^2 + 1
x^2 - y^2 1 - y^2/x^2
lim ( lim --------- ) = lim ( lim ----------- ) = + 1
y->oo x->oo x^2 + y^2 y->oo x->oo 1 + y^2/x^2
So again, according to the textbook, the limits do not commute.
x^2 - y^2 x^2
lim ( lim --------- ) = lim ( --- ) = + 1
x->0 y->0 x^2 + y^2 x->0 x^2
x^2 - y^2 - y^2
lim ( lim --------- ) = lim ( ----- ) = - 1
y->0 x->0 x^2 + y^2 y->0 + y^2
So again, according to the textbook, the limits do not commute.
Unravel the limits. We will do so by introducing polar coordinates:
x = r.cos(phi) ; y = r.sin(phi)
x^2 - y^2 r^2 ( cos^2(phi) - sin^2(phi) )
Giving --------- = ------------------------------- = cos(2.phi)
x^2 + y^2 r^2 ( cos^2(phi) + sin^2(phi) )
Now we can understand immediately why the professor's limits for x and
y _indeed_ do not commute ! At (x,y) = 0 , this function is _singular_
and it assumes any value between -1 and +1. And the latter is the case
everywhere else in the (x,y)-plane. So unless there is a clear path to
infinity, any value between -1 and +1 would be good enough for an 'A'.
See? The function is a circular wave, or rather the wave perpendicular
to it and has period Pi. Here is an (ASCII) contour map of it. Notice
the singularity at the origin.
0 -1 0
\ | /
\ | /
\ | /
\ | /
\ | /
\|/
+1 ------------------- +1
/|\
/ | \
/ | \
/ | \
/ | \
/ | \
0 -1 0
Why not ask then _proper_ questions belonging to answers ? Here comes:
x^2 - y^2
lim ( --------- ) = undefined (between -1 and +1)
|(x,y)|->0 x^2 + y^2
x^2 - y^2
lim ( --------- ) = undefined (between -1 and +1)
|(x,y)|->oo x^2 + y^2
With other words: let the absolute value of the vector (x,y) approach
zero, or infinity, and see what happens. (Neighbourhoods and such ..)
Excercise 4.
x.y
Let G(x,y) = --------- . I'm sure you can do this one by yourself.
x^2 + y^2
Wrong limits do not commute. Almost dare to say now that non commuting
limits are simply .. wrong. Any counter examples are quite welcome.
Guess it's not so much superstition but merely a sense of reality that
so many people let believe that: properly posed limits _do_ commute.
Han de Bruijn
David C. Ullrich
Fine. So what? When people talk about L being "fixed" that
doesn't mean that L is some special sort of real number, a
fixed one as opposed to other reals that are maybe not so
fixed, it's just meant to say that the _notation_ "L"
refers to _one_ real number. It's a fact that for every real
L the limit as n tends to infinity is one thing and for every
natural number n the limit as L tends to infinity is
something else. There you are, exactly the same fact
expressed without the word "fixed".
>We have
>to live wit the fact that _both_ n and L approach limiting values and
>in _whatever_ fashion. So what is the proper way to express this idea?
>With other words: what would have been the _right question_ belonging
>to this excercise ? The following is my proposal:
>
> lim int_(-L)^(+L) f_n(x)dx
> min(n,L)->oo
>
>And now the _proper_ the answer is that this limit: does not exist. Or
>rather that it has a value between 0 and 1,
No, the limit does not exist.
>not by coincidence the two
>extremes found by Horand's students (at least those who got an 'A' for
>their exam). In this case two limits do not commute and the associated
>right questioned limit does not exist. Hmm, would that be systematical?
Sounds like you've uncovered some deep conspiracy here.
The idea that iterated limits are "wrong" and the limit as above
is "right" is very strange. It _is_ a fact that if I(n, L) is a
function of two variables and the two iterated limits are not the
same then the limit as min(n,K) tends to infinity cannot exist.
This is easy to prove, and very well known.
>Example 2, by Horand Gassmann, from "Why does everyone do it?":
>http://groups.google.nl/group/sci.math/browse_frm/thread/bfa9f4f2c780e6f
>
> Let g_n(x) = 2.n^2.x if 0 <= x < 1/(2n)
> = 2n - 2.n^2.x if 1/(2n) <= x < 1/n
> = 0 everywhere else.
>
> These functions are triangular, and they all disappear outside of
> [0,1], so I can compute int_0^1{g_n(x) dx} = 1 for every n.
> And again, for every x, lim{n->infinity g_n(x) = 0.
> So again, the limit and the integration can't be interchanged.
>
>Sure, g_n(x) = 0 , sure. It's not the first time I've seen infinities
>suddenly disappear by mainstream mathematician's "ingenuity":
>
>http://hdebruijn.soo.dto.tudelft.nl/www/grondig/natural.htm#bv
>http://hdebruijn.soo.dto.tudelft.nl/jaar2006/ballen.jpg
>
>As Mr. Gassmann has admitted, this is a proper picture of the problem:
>
>http://hdebruijn.soo.dto.tudelft.nl/jaar2008/gassmann.jpg
>
>We see that the function g_n(x) becomes a sharp peak at x=0 for n->oo
>and that, geometrically, it certainly not disappears or becomes zero.
>The vemin is again in the false doctrine that it would be possible to
>"fix" x and let n go to infinity. Again, HdB says that such a "fixed"
>real x is an _undefined_ concept. A real is just a real and there's
>nothing "fixed" or "variable" with it.
What doesn't exist is the point you're trying to make.
Again, the fact is just that for every x, the limit as n tends
to infinity is 0. When people say "for every fixed x"
they're just trying to make it more clear exactly what
is meant.
>Such is even more obvious with
>the present function g_n(x) , where x and n are clearly intermixed,
>by definition. So we _can_ have: 0 <= x < 1/(2n) , 1/(2n) <= x < 1/n ,
>1/n <= x , and nothing prevents us from groing to limits _while_ this
>is the case. If we just do this, then what we get is what physicists
>know as a delta function:
>
> delta(x) = 0 for x <> 0
> = any for x = 0 ; integral_(-oo)^(+oo) delta(x) dx = 1
>
>It's easy to see that: lim g_n(x) = delta(x)
> n->oo
>
>Whatever definition might be the "right" one, a delta function roughly
>speaking is just a very large peak around x = 0 with area normed to 1.
>With the finitary domain in mind, there is no problem in understanding
>what delta functions actually are (even if they are a bit beyond x=0).
There certainly is a notion of limit for which the limit of these g_n
is a "delta function". So what? That's simply a different sort of
limit.
>Note that, in this case, it's _not_ even the wrong question. It's just
>that professors expect a wrong answer from their students. The second
>answer is wrong, namely, and the first one is right:
>
> lim [ int_0^1 g_n(x) dx ] = 1 = int_0^1 [ lim g_n(x) ] dx
> n->oo n->oo
Your notion of "right" and "wrong" is wrong. It's like
those nasty professors are saying things about apples,
and you're saying they're wrong. You agree that the
things they say about apples are correct, but nonetheless
they're wrong because they should be talking about oranges.
It's a very strange sort of "right" and "wrong" - if I say
something that's true it's wrong because _you_ say I
should be talking about something else.
You should really learn a little more about what those evil professors
say.
The limit of that function as (x,y) -> (0,0) does not exist (that's
the correct way to express the same fact that you're expressing
by saying the function is singular or that the limit is "anything".
That limit does not exist).
_Proving_ that that limit does not exist is a standard exercise
that those nasty profesors ask their students to do. And the
simplest proof that that limit does not exist is simply to
note that the two iterated limits are not equal.
Many of the mathematical points you raise in all this
are correct, although many are expressed in very unfortunate
language. But the things you say about what those professors
say versus what they should say are hilarious - none of the
correct mathematical points you raise are news to thos
profesors, and they're not something that they've been
trying to hide from their students.
It's like you tell us that professors tell their students
that cats are not dogs, and that it's wrong for profesors
to tell their students this because dogs bark!
Honest, it _is_ like that. Yes, dogs bark. But that's
not news, and it's not something that professors
have been misinforming their students about when
they tell them that cats are not dogs.
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
Denis Feldmann
[many things, correct of course, about nasty professors and
non-commutating limits]
I understand why you would like to remind the innocent reader of the
reality. But you lose your precious time feeding this troll, and I am
sure no serious student is in real danger of confusion over those
things. So, lets do math for a change. Same way as the g_n sequence
does converge in some sense towards the "delta function" (which is not
a function, but never mind), or that the sequence (1,-1,1,-1,...)
converges towards 0 in some sense (Cesaro, say), is there a way, for
instance, to get a "limit" to (x^2-y^2)/(x^2+y^2) (probably 0, of
course) and, more generally, is there some general setting where all
those pesky non-commutative limits at last commute?
William Hughes
> Wrong limits do not commute. Almost dare to say now that non commuting
> limits are simply .. wrong. Any counter examples are quite welcome.
>
Given
A transistor dissipates power propotional to its linear dimension.
a) for a fixed chip size what is the power dissipated by the chip
as the transistor size gets very small?
b) for a fixed transistor size what is the power disapated by the chip
as the chip size gets very small?
The limits in a) and b) are hardly wrong.
However,
c) what is the power dissipated by the chip as the transistor size
and the chip size get very small?
Has no answer.
The related question,
d) what is the dissipated power density as the transistor size and the
chip size get very small?
has an answer.
At best you could say "In properly posed problems limits _do_
commute".
So now all we need to do is define "properly posed problem. Oh I
know,
a properly posed problem is one in which limits commute.
- William Hughes
Jesse F. Hughes
the rest of us, it never occurred to them that (1) they could be
nested and (2) limits don't necessarily commute.
Hence (?) the mathematicians are wrong.
--
Jesse F. Hughes
"If this novel could be compared to sculpture I'd have to compare it
to the Sistine Cieling."
-- An insightful review of Ryskamp's "Nature Studies"
Jesse F. Hughes
> Synopsis: All this time Han claimed to understand limits better than
> the rest of us, it never occurred to them that (1) they could be
^^^^ him
> nested and (2) limits don't necessarily commute.
>
> Hence (?) the mathematicians are wrong.
Oops.
--
Jesse F. Hughes
"Every country has its stupid people. It just so happens that
America's stupid people are louder." -- Ling Cheung, Sociologist
The World Wide Wade
Han de Bruijn <Han.de...@DTO.TUDelft.NL> wrote:
LOL. Look, the word "fixed" is there to help you; if you find it
confusing, then just abandon it. For each x in [0,1] you have a
sequence of numbers (g_n(x)), and for each n in N, you have a function
g_n on [0,1]. That's all that is being said.
The World Wide Wade
<aderamey.addw-F0D...@newsgroups.comcast.net>,
By the way, did you run screaming out of calculus class when the
professor said "fix x and take the derivative with respect to y"? Are
partial derivatives also "wrong limits"?
Han de Bruijn
We would be on speaking terms here, iff the following would be the case:
lim int_(-L)^(+L) f_n(x)dx = 0 with _one_ L (such that L < n = oo)
n->oo
lim int_(-L)^(+L) f_n(x)dx = 1 with _one_ n (such that n+1 < L = oo)
L->oo
The augmentations between parentheses are not necessary then. But this
is _not_ what our teacher is doing. Because:
lim lim int_(-L)^(+L) f_n(x)dx
n->oo L->oo
lim lim int_(-L)^(+L) f_n(x)dx
L->oo n->oo
So now the n that has been declared fixed suddenly becomes _variable_,
i.e. more than one real number, and the L that has been declared fixed
suddenly becomes variable, i.e. more than one real number. Which is not
what we've agreed upon.
Han de Bruijn
Han de Bruijn
> David C. Ullrich wrote (answering Han de Bruinj):
>
> [many things, correct of course, about nasty professors and
> non-commutating limits]
>
> I understand why you would like to remind the innocent reader of the
> reality. But you lose your precious time feeding this troll, and I am
> sure no serious student is in real danger of confusion over those
> things.
I wish you were right .. Worse ! Serious mathematics is in real danger
of confusion over those things. Well, not really over things as simple
as these limits, but the generalizations of these things. (Take a look
at e.g. point set topology) I became aware of these things at the time
I studied Lie Groups at the Eindhoven University of Technology. <quote>
My absolute favorite is a book I borrowed from the library many years
ago and never succeeded in finding it back again. It's written by one
of Lie's pupils: Georg Scheffers (1866–1945). It's written in German &
titled "Differentialgleichungen von Sophus Lie" or some such like it.
I still remember quite well that Georg Scheffers' book is abundant with
lucid geometrical interpretations and very much readable. Not this book,
but several others by Georg Scheffers about Lie groups can be found at:
http://concise.britannica.com/ebc/article-9048172/Sophus-Lie
http://0-www.search.eb.com.library.uor.edu/eb/article-9048172?tocId=9...
Or try Google("Sophus Lie" "Georg Scheffers" "differential equations")
</quote> The dark side of this story is: _topology_. Got stuck on it.
Got *PLONK*ed by my superiors, at last. And this is my sweet revenge.
> So, lets do math for a change. Same way as the g_n sequence
> does converge in some sense towards the "delta function" (which is not a
> function, but never mind), or that the sequence (1,-1,1,-1,...)
> converges towards 0 in some sense (Cesaro, say), is there a way, for
> instance, to get a "limit" to (x^2-y^2)/(x^2+y^2) (probably 0, of
> course) and, more generally, is there some general setting where all
> those pesky non-commutative limits at last commute?
You've clearly not understood that _wrong questions_ can be formulated
within mathematics. The answer is: no.
Han de Bruijn
Han de Bruijn
> Synopsis: All this time Han claimed to understand limits better than
> the rest of us, it never occurred to them that (1) they could be
> nested and (2) limits don't necessarily commute.
>
> Hence (?) the mathematicians are wrong.
Let's consider again what a (double) limit really is.
lim lim f(x,y) = F
x->oo y->oo
For any real number epsilon > 0 there exists real numbers M , N such
that if x > L and y > M then | f(x,y) - F | < epsilon . Right ?
This is a perfectly finitary definition. From this definition alone, it
follows that the double limit is (1) not "nested" (?) and (2) commutes.
Han de Bruijn
Han de Bruijn
>>LOL. Look, the word "fixed" is there to help you; if you find it
>>confusing, then just abandon it. For each x in [0,1] you have a
>>sequence of numbers (g_n(x)), and for each n in N, you have a function
>>g_n on [0,1]. That's all that is being said.
>
> By the way, did you run screaming out of calculus class when the
> professor said "fix x and take the derivative with respect to y"? Are
> partial derivatives also "wrong limits"?
Please read my response to David C. Ullrich. There is no problem at all
when taking only _one_ limit with a function of two variables. IMO there
is only a problem when taking a _double_ limit with such functions. Uhm,
you know about the rule d^2f/(dx.dy) = d^2f/(dy.dx) , for those partial
derivatives, which is always valid in physics, but not in mathematics.
Han de Bruijn
Mariano Suárez-Alvarez
> Jesse F. Hughes wrote:
> > Synopsis: All this time Han claimed to understand limits better than
> > the rest of us, it never occurred to them that (1) they could be
> > nested and (2) limits don't necessarily commute.
>
> > Hence (?) the mathematicians are wrong.
>
> Let's consider again what a (double) limit really is.
>
> lim lim f(x,y) = F
> x->oo y->oo
>
> For any real number epsilon > 0 there exists real numbers M , N such
> that if x > L and y > M then | f(x,y) - F | < epsilon . Right ?
Not really.
Consider the trivial example: it should be obvious that
lim_{x -> +infty} lim_{y -> +infty} x/y = 0
yet the function x/y is not even bounded on any
set of the form { (x,y) : x > M, y > N }.
-- m
Jesse F. Hughes
> Jesse F. Hughes wrote:
>
>> Synopsis: All this time Han claimed to understand limits better than
>> the rest of us, it never occurred to them that (1) they could be
>> nested and (2) limits don't necessarily commute.
>>
>> Hence (?) the mathematicians are wrong.
>
> Let's consider again what a (double) limit really is.
>
> lim lim f(x,y) = F
> x->oo y->oo
>
> For any real number epsilon > 0 there exists real numbers M , N such
> that if x > L and y > M then | f(x,y) - F | < epsilon . Right ?
Wrong.
First, we say that
lim_{y -> oo} f(x,y) = g(x)
iff for all e > 0, there is an M > 0 such that for all y > M and for
all x,
| g(x) - f(x,y) | < e.
Notice in particular that lim_{y -> oo} f(x,y), if it exists, is a
function R -> R.
Thus,
lim lim f(x,y) = F
x->oo y->oo
iff there is a g(x) such that lim_{y -> oo} f(x,y) = g(x) and for all
e > 0, there is an M such that for all x > M,
| g(x) - F | < e.
I don't know why you thought otherwise. Don't you know the definition
of limit?
> This is a perfectly finitary definition. From this definition alone, it
> follows that the double limit is (1) not "nested" (?) and (2)
> commutes.
Yes, but that is not the definition of
lim lim f(x,y) = F
x->oo y->oo
But no fret! Now you know the definition. From this, it should be
easy to see that the two limits do not always commute.
--
"The math doesn't care about their mortgages. It doesn't care about
their political needs. [...] Today's mathematicians have to hate
mathematics because mathematics doesn't look out for them. It doesn't
pay attention to their needs." --- JSH analyzes mathematicians
Virgil
I do not know how physicists think about it, but to mathematicians. In
lim lim int_(-L)^(+L) f_n(x)dx
n->oo L->oo
n must be regarded as constant while determining the limit when L->oo.
And after taking that limit while L -> oo, there is no L remaining in
g(n) = lim_{L -> oo} int_(-L)^(+L) f_n(x)dx, if such a g(n) exists,
to affect the value when considering the limit as n -> oo.
Virgil
Han de Bruijn <Han.de...@DTO.TUDelft.NL> wrote:
But it is quite possible for both iterated limits to exist without the
double limit existing. See, e.g., "Counterexamples in Analysis".
So, among mathematicians, there is distinction between
lim_{x->a} lim_{y->b} f(x)
and
lim_{y->b} lim_{x->a} f(x)
and
lim_{(x,y)->(a,b)} f(x),
particularly in the case when the last of these does not exist but one
or both of first two do exist.
The World Wide Wade
Han de Bruijn <Han.de...@DTO.TUDelft.NL> wrote:
But you wrote "A real is just a real and there's nothing "fixed" or
"variable" with it." I'm not sure how one could muster the courage to
take a partial derivative after that.
Interesting that you mention d^2f/(dxdy), d^2f/(dydx). If f is twice
continuously differentiable, those two quantities agree. This is a
very nice result, don't you think? Apparently wrong limits sometimes
become right.
Similarly, there are nice results about when lim_n->oo int g_n(x) dx =
int [lim_n->oo g_n(x)] dx; for example, there is the Lebesgue
dominated convergence theorem.
It would seem to me that even with your unusual belief system, you
would have to admit it would be a good idea to study "when wrong
limits become right".
The World Wide Wade
Han de Bruijn <Han.de...@DTO.TUDelft.NL> wrote:
So how come
int_[a,b] int_[c,d] f(x,y) dx dy
= int_[c,d] int_[c,d] f(x,y) dy dx
for any continuous f on [a,b] x [c,d]? Aren't those double limits?
Are they wrong limits?
Virgil
<aderamey.addw-1B9...@newsgroups.comcast.net>,
The World Wide Wade <aderam...@comcast.net> wrote:
I can think of lots of situations in which
int_[a,b] int_[c,d] f(x,y) dx dy
= int_[c,d] int_[a,b] f(x,y) dy dx
But,unless a = c and b = d, I don't really expect that
int_[a,b] int_[c,d] f(x,y) dx dy
= int_[c,d] int_[c,d] f(x,y) dy dx
very often.
Han de Bruijn
So you find my definition: not right. Instead, what you do is first take
the (single) limit lim_{y -> +infty} x/y (keeping x "fixed") resulting
in 0 and then you take the (single) limit lim_{x -> +infty} 0 giving
(of course) 0 . That's what you did, right?
Han de Bruijn
Han de Bruijn
> So, among mathematicians, there is distinction between
>
> lim_{x->a} lim_{y->b} f(x)
> and
> lim_{y->b} lim_{x->a} f(x)
> and
> lim_{(x,y)->(a,b)} f(x),
>
> particularly in the case when the last of these does not exist but one
> or both of first two do exist.
Okay. Thus, if I write
> For any real number epsilon > 0 there exists real numbers M , N such
> that if x > L and y > M then | f(x,y) - F | < epsilon .
then the correct notation for this is: lim_{(x,y)->(a,b)} f(x,y) = F .
Am I right this time?
Hand de Bruijn
Virgil
Han de Bruijn <Han.de...@DTO.TUDelft.NL> wrote:
Han is again wrong mathematically , to define
lim lim f(x,y) = F
x->oo y->oo
to mean the same as
lim f(x,y) = F
(x,y)->(oo,oo)
Mathematically, at least, they are different, as the example
lim_{x -> +infty} lim_{y -> +infty} x/y = 0
quite properly shows.
Virgil
Han de Bruijn <Han.de...@DTO.TUDelft.NL> wrote:
Not quite. Try lim_{(x,y)->(oo,oo)} f(x,y) = F
Han de Bruijn
> In article <9ea0a$48c77903$82a1e228$31...@news1.tudelft.nl>,
> Han de Bruijn <Han.de...@DTO.TUDelft.NL> wrote:
>
>>Virgil wrote:
>>
>>>So, among mathematicians, there is distinction between
>>>
>>>lim_{x->a} lim_{y->b} f(x)
>>>and
>>>lim_{y->b} lim_{x->a} f(x)
>>>and
>>>lim_{(x,y)->(a,b)} f(x),
>>>
>>>particularly in the case when the last of these does not exist but one
>>>or both of first two do exist.
>>
>>Okay. Thus, if I write
>>
>>>For any real number epsilon > 0 there exists real numbers M , N such
>>>that if x > L and y > M then | f(x,y) - F | < epsilon .
Huh ? Did I say this ? Must be:
For any real number epsilon > 0 there exists real numbers L , M such
that if x > L and y > M then | f(x,y) - F | < epsilon .
>>then the correct notation for this is: lim_{(x,y)->(a,b)} f(x,y) = F .
>>
>>Am I right this time?
>
> Not quite. Try lim_{(x,y)->(oo,oo)} f(x,y) = F
Sorry. That's what I meant, of course. The other one is:
For any real number epsilon > 0 there exists real numbers L , M such
that if | x - a | < L and | y - b | < M then | f(x,y) - F | < epsilon
Han de Bruijn
Tonico
wrote:
*************************************************************
A waste of time...again... **Sigh**
Han, buddy: why do you insist over and over again to show to everybody
in this forum that you suck, and big time, in pretty basic, elementary
stuff in mathematics?
Why this craving for notoriety in being wrong in mathematics in a
mathematics forum, and even discussing and arguing nonsenses over and
over again...with mathematicians!?
This time is not only Cantor and set theory: now you're pissed off
about limits, how they're defined (something that seems to be way
above your comprehension), how to work with them, etc.
So, in some rather amazing, surprising way, you seem to be trying to
expand your field of cranking from set theory to other realms of
mathematics...why?? Why won't you first master thoroughly in depth all
the crankhood details in set theory and THEN pass over to crank around
in calculus, geometric algebra, number theory, etc?
Hey, who knows: perhaps after some time you'll be able, just as JHS
has, to achieve heights like proving FLT with the unique and only aid
of an abacus and prayers to Zoroaster, or even showing RH to be true
by using matches and dancing the cherokee dance rain.
Stay in what you're good, Han: Set Theory and Cantor cranking.
Regards
Tonio
Denis Feldmann
No, youre wrong. He is not such a good crank. Otoh, he is quite good at
trolling...
>
> Regards
> Tonio
>
David C. Ullrich
<Han.de...@DTO.TUDelft.NL> wrote:
You're not making any sense. Read what I wrote again - there is
nothing "agreed upen" when we say that the limit of g(n,L)
for a "fixed L" as n tends to infinity is 0; that simply says that
for every L the limit as n tends to infinity is 0. And it follows
from that that the limit as L tends to infinity of the limit as
n tends to infinity is 0.
David C. Ullrich
<Han.de...@DTO.TUDelft.NL> wrote:
>Jesse F. Hughes wrote:
>
>> Synopsis: All this time Han claimed to understand limits better than
>> the rest of us, it never occurred to them that (1) they could be
>> nested and (2) limits don't necessarily commute.
>>
>> Hence (?) the mathematicians are wrong.
>
>Let's consider again what a (double) limit really is.
>
> lim lim f(x,y) = F
> x->oo y->oo
>
>For any real number epsilon > 0 there exists real numbers M , N such
>that if x > L and y > M then | f(x,y) - F | < epsilon . Right ?
Wrong. The above says that for every epsilon > 0 there exists
N such that |F - lim_{y->oo} f(x,y)} < epsilon for every x > N.
What you give is the definition of
lim_{(x,y) -> (oo, oo)} f(x,y) = F.
That's something _different_. If you think the two
are the same then of course you're going to deduce
that there are problems somewhere. But that's
not a problem with mathematics, it's just a problem
with your misunderstanding of mathematical notation.
>This is a perfectly finitary definition. From this definition alone, it
>follows that the double limit is (1) not "nested" (?) and (2) commutes.
>
>Han de Bruijn
David C. Ullrich
Han de Bruijn
> *************************************************************
>
> A waste of time...again... **Sigh**
>
> Han, buddy: why do you insist over and over again to show to everybody
> in this forum that you suck, and big time, in pretty basic, elementary
> stuff in mathematics?
That "pretty basic, elementary stuff in mathematics" is pretty basic in
our debates about _foundational_ issues. Look up e.g. in the WM threads
(WM = Wolfgang Mueckenheim) how often the counter argument is used (by
meanstreamers) that WM's limits do not commute, while WM thinks they do.
> Why this craving for notoriety in being wrong in mathematics in a
> mathematics forum, and even discussing and arguing nonsenses over and
> over again...with mathematicians!?
Uhm, if you find it disturbing, please leave this Internet coffeeshop
and go to the censored 'sci.math.research', where you certainly can meet
people of your own taste. (BTW. It's so incredibly boring there .. )
> This time is not only Cantor and set theory: now you're pissed off
> about limits, how they're defined (something that seems to be way
> above your comprehension), how to work with them, etc.
>
> So, in some rather amazing, surprising way, you seem to be trying to
> expand your field of cranking from set theory to other realms of
> mathematics...why?? Why won't you first master thoroughly in depth all
> the crankhood details in set theory and THEN pass over to crank around
> in calculus, geometric algebra, number theory, etc?
>
> Hey, who knows: perhaps after some time you'll be able, just as JHS
> has, to achieve heights like proving FLT with the unique and only aid
> of an abacus and prayers to Zoroaster, or even showing RH to be true
> by using matches and dancing the cherokee dance rain.
>
> Stay in what you're good, Han: Set Theory and Cantor cranking.
I'm not going to (im)prove on Fermat's Last Theorem, not going to prove
Goldbach Conjecture or prove the Riemann Hypothesis. Maybe I'm a crank,
but not _that_ much. (Actually, I've done research on these topics, but
never published it, for good reasons. Oh well, everybody did an attempt
on these issues .. Not ? )
Han de Bruijn
Han de Bruijn
> On Tue, 09 Sep 2008 16:00:41 +0200, Han de Bruijn
> <Han.de...@DTO.TUDelft.NL> wrote:
>
>>Jesse F. Hughes wrote:
>>
>>>Synopsis: All this time Han claimed to understand limits better than
>>>the rest of us, it never occurred to them that (1) they could be
>>>nested and (2) limits don't necessarily commute.
>>>
>>>Hence (?) the mathematicians are wrong.
>>
>>Let's consider again what a (double) limit really is.
>>
>> lim lim f(x,y) = F
>> x->oo y->oo
>>
>>For any real number epsilon > 0 there exists real numbers M , N such
>>that if x > L and y > M then | f(x,y) - F | < epsilon . Right ?
>
> Wrong. The above says that for every epsilon > 0 there exists
> N such that |F - lim_{y->oo} f(x,y)} < epsilon for every x > N.
>
> What you give is the definition of
>
> lim_{(x,y) -> (oo, oo)} f(x,y) = F.
>
> That's something _different_. If you think the two
> are the same then of course you're going to deduce
> that there are problems somewhere. But that's
> not a problem with mathematics, it's just a problem
> with your misunderstanding of mathematical notation.
Yes. I've learned that now. Thanks.
Han de Bruijn
Jesse F. Hughes
> No, youre wrong. He is not such a good crank. Otoh, he is quite good at
> trolling...
He's a *great* crank. You take that back.
--
"I believe that my job is to go out and explain to the people what's on
my mind. That's why I'm having this press conference, see? I'm telling
you what's on my mind. And what's on my mind is winning the war on
terror." --- George W. Bush
William Hughes
> Mariano Suárez-Alvarez wrote:
>
> > Consider the trivial example: it should be obvious that
>
> > lim_{x -> +infty} lim_{y -> +infty} x/y = 0
>
> > yet the function x/y is not even bounded on any
> > set of the form { (x,y) : x > M, y > N }.
>
> So you find my definition: not right. Instead, what you do is first take
> the (single) limit lim_{y -> +infty} x/y (keeping x "fixed") resulting
> in 0 and then you take the (single) limit lim_{x -> +infty} 0 giving
> (of course) 0 . That's what you did, right?
Since that is what the notation means, yes.
Do not confuse the iterated limits
A: lim lim f(x,y)
x->oo y->oo
(first vary y , then vary x)
With the double limit
B: lim f(x,y)
(x,y)->(oo,oo)
(vary x and y in either order or both together)
The two are related, (e.g. if the limit in B exists then
the limits in A commute), however even if the limit in B does
not exist the limits
lim f(x,y) and lim f(x,y)
x->oo y->oo
are not "wrong".
- William Hughes
Han de Bruijn
In a sense that they are not wrong in _common_ mathematics. But I think
you can know quite well, meanwhile, what my objective is: the _meaning_
behind all this, in a physical sense - (or if you prefer: _geometrical_
sense; the latter is advantageous in that you stay within mathematics).
I can "see" quite well how - what's the name of it - the following limit
works out in this sense:
B: lim f(x,y)
(x,y)->(oo,oo)
I can imagine that one devises a path, and travelling along that path
leads to infinity. It may happen that the limit is independent of the
path and then we have this limit. As e.g with: f(x,y) = 1/(x^2 + y^2)
But I still find the meaning of the following limit at least confusing:
A: lim lim f(x,y)
x->oo y->oo
x.y
When applied to e.g. --------- = sin(2.phi)/2 with x = r.cos(phi)
x^2 + y^2 y = r.sin(phi)
Then with limit A we first go to infinity along a line parallel to the
y-axis, giving 0 because the angle phi approaches Pi/2 . Right ?
But then we have to shift that parallel line towards infinity and _that_
stumps me, because now I can't "see" anymore what's happening. (This is
merely a consequence of _both_ my brain halves working. I can understand
it with the "formal" half but not with the "visual" half). Consequently,
I still find that the limit A is indeed "wrong". Flame shield ... well,
think I can take it. Do we have to live with "counter intuition" or can
we refuse and let the other brain half speak ?
Han de Bruijn
Denis Feldmann
> Denis Feldmann <denis.feldm...@neuf.fr> writes:
>
>> No, youre wrong. He is not such a good crank. Otoh, he is quite good at
>> trolling...
>
> He's a *great* crank. You take that back.
>
time). You wanna bet ? (but I admit he is a *great* troll, if you like)
William Hughes
Even in a physical sense, meaningful limits may or may not commute.
(Hint, check posts that you ignored for an example)
So limits that do not commute are not "wrong" in
in either the mathematical or the physical sense.
- William Hughes
Han de Bruijn
I did not ignore it, but rather thought it was a confirmation instead of
a refutation. (Suppose you mean the story about transistors and chips)
Han de Bruijn
Jesse F. Hughes
I think I'd take that bet, if we restrict it to his mathematical
claims. But how will we settle it?
--
"We want a single platform. We're trying to get there using the
carrot, or blackmail, or rewards, or whatever you call it."
-- Madison, WI, superintendent Rainwater grasps subtlety in the
operating system wars.
Jesse F. Hughes
> But then we have to shift that parallel line towards infinity and _that_
> stumps me, because now I can't "see" anymore what's happening. (This is
> merely a consequence of _both_ my brain halves working. I can understand
> it with the "formal" half but not with the "visual" half). Consequently,
> I still find that the limit A is indeed "wrong". Flame shield ... well,
> think I can take it. Do we have to live with "counter intuition" or can
> we refuse and let the other brain half speak ?
I don't know why you have problems with it.
Take a function f(x,y). A graph of that function is a surface. If we
consider the limit as x -> oo of f(x,y), we get a curve: a function
g(y). Now, you can visualize what it means to take lim_{y->oo} g(y),
so there you go.
--
Jesse F. Hughes
"I get to make things move just by saying a few things. When I post
now the math world has to tremble, even if it does so quietly, hoping
that no one else notices." -- James S. Harris has the power.
victor_me...@yahoo.co.uk
> That "pretty basic, elementary stuff in mathematics" is pretty basic in
> our debates about _foundational_ issues. Look up e.g. in the WM threads
> (WM = Wolfgang Mueckenheim) how often the counter argument is used (by
> meanstreamers) that WM's limits do not commute, while WM thinks they do.
He's an eejit, just like you.
Victor Meldrew
"I don't believe it!"
victor_me...@yahoo.co.uk
> But then we have to shift that parallel line towards infinity and _that_
> stumps me, because now I can't "see" anymore what's happening. (This is
> merely a consequence of _both_ my brain halves working.
Or of _neither_ of your brain halves working?
William Hughes
wrote:
> William Hughes wrote:
Nope. Two meaningful limits that do not commute. A clear refutation
to the idea that limits which do not commute are "wrong".
- William Hughes
Tonico
> But I still find the meaning of the following limit at least confusing:
>
> A: lim lim f(x,y)
> x->oo y->oo
> x.y
> When applied to e.g. --------- = sin(2.phi)/2 with x = r.cos(phi)
> x^2 + y^2 y = r.sin(phi)
>
> Then with limit A we first go to infinity along a line parallel to the
> y-axis, giving 0 because the angle phi approaches Pi/2 . Right ?
>
> But then we have to shift that parallel line towards infinity and _that_
> stumps me, because now I can't "see" anymore what's happening. (This is
> merely a consequence of _both_ my brain halves working. I can understand
> it with the "formal" half but not with the "visual" half). Consequently,
> I still find that the limit A is indeed "wrong". Flame shield ... well,
> think I can take it. Do we have to live with "counter intuition" or can
> we refuse and let the other brain half speak ?
>
> Han de Bruijn-
************************************************************
So you can't "see" anymore what's happening, and instead of putting
the blame on your lack of mathematical education and/or on your own
stupidity, you'd rather put the blame on mathematics....aha.
Some rather comfortable way to put the blame on others rather that in
you.
And Han: your several brain halves (how many do we already have there:
six, seven halves...?) speaking to you might be a sign for you to
search for some professional help: you know, those little voices in
your head...tsk,tsk,tsk. Not to mention that those voices seem to be
as lost in basic mathematics as you are.
Regards
Tonio
Mariano Suárez-Alvarez
> In article <9ea0a$48c77903$82a1e228$31...@news1.tudelft.nl>,
>
>
>
> > Virgil wrote:
>
> > > So, among mathematicians, there is distinction between
>
> > > lim_{x->a} lim_{y->b} f(x)
> > > and
> > > lim_{y->b} lim_{x->a} f(x)
> > > and
> > > lim_{(x,y)->(a,b)} f(x),
>
> > > particularly in the case when the last of these does not exist but one
> > > or both of first two do exist.
>
> > Okay. Thus, if I write
>
> > > For any real number epsilon > 0 there exists real numbers M , N such
> > > that if x > L and y > M then | f(x,y) - F | < epsilon .
>
> > then the correct notation for this is: lim_{(x,y)->(a,b)} f(x,y) = F .
>
> > Am I right this time?
>
> > Hand de Bruijn
>
> Not quite. Try lim_{(x,y)->(oo,oo)} f(x,y) = F
While this notation is probably correct, it is
rather unusual, I think.
The idea of limit that Han is entertaining is
the one corresponding to the convergence of
nets phi : R x R --> R where R x R, the cartesian
square of the real numbers, is ordered with the
product order of the usual orders in the factors.
This kind of limit rarely shows up in calculus,
and while your notation is quite indcative of
the specific kind of limit, I do not think
it is standard.
-- m
Virgil
Han de Bruijn <Han.de...@DTO.TUDelft.NL> wrote:
Note that, in mathematics, lim_{(x,y)->(oo,oo)} x*y/(x^2 + y^2)
does not exist, since, for example, along the path y = m*x for m > 0 the
limit would be m/(1+m^2), which depends on m.
However both iterated limits
lim_{x->oo} lim_{y->oo} x*y/(x^2 + y^2) and
lim_{y->oo} lim_{x->oo} x*y/(x^2 + y^2)
exist and are zero.
If physicists chose to restrict their attentions to functions for which
this sort of behavior does not occur, that does not mean that there are
no such functions.
The World Wide Wade
Virgil <Vir...@gmale.com> wrote:
> In article
> <aderamey.addw-1B9...@newsgroups.comcast.net>,
> The World Wide Wade <aderam...@comcast.net> wrote:
>
> > In article <72600$48c68383$82a1e228$20...@news1.tudelft.nl>,
> > Han de Bruijn <Han.de...@DTO.TUDelft.NL> wrote:
> >
> > > The World Wide Wade wrote:
> > >
> > > >>LOL. Look, the word "fixed" is there to help you; if you find it
> > > >>confusing, then just abandon it. For each x in [0,1] you have a
> > > >>sequence of numbers (g_n(x)), and for each n in N, you have a function
> > > >>g_n on [0,1]. That's all that is being said.
> > > >
> > > > By the way, did you run screaming out of calculus class when the
> > > > professor said "fix x and take the derivative with respect to y"? Are
> > > > partial derivatives also "wrong limits"?
> > >
> > > Please read my response to David C. Ullrich. There is no problem at all
> > > when taking only _one_ limit with a function of two variables. IMO there
> > > is only a problem when taking a _double_ limit with such functions. Uhm,
> > > you know about the rule d^2f/(dx.dy) = d^2f/(dy.dx) , for those partial
> > > derivatives, which is always valid in physics, but not in mathematics.
> > >
> > > Han de Bruijn
> >
> > So how come
> >
> >
> > int_[a,b] int_[c,d] f(x,y) dx dy
> >
> > = int_[c,d] int_[c,d] f(x,y) dy dx
> >
> > for any continuous f on [a,b] x [c,d]? Aren't those double limits?
> > Are they wrong limits?
>
> I can think of lots of situations in which
>
> int_[a,b] int_[c,d] f(x,y) dx dy
>
> = int_[c,d] int_[a,b] f(x,y) dy dx
>
> But,unless a = c and b = d, I don't really expect that
>
> int_[a,b] int_[c,d] f(x,y) dx dy
>
> = int_[c,d] int_[c,d] f(x,y) dy dx
>
> very often.
Virgil, it was a typo, kind of like yours here
<Virgil-AE1B6A....@comcast.dca.giganews.com>
Stop being a dork.
Han de Bruijn
> Han de Bruijn <Han.de...@DTO.TUDelft.NL> writes:
>
>>But then we have to shift that parallel line towards infinity and _that_
>>stumps me, because now I can't "see" anymore what's happening. (This is
>>merely a consequence of _both_ my brain halves working. I can understand
>>it with the "formal" half but not with the "visual" half). Consequently,
>>I still find that the limit A is indeed "wrong". Flame shield ... well,
>>think I can take it. Do we have to live with "counter intuition" or can
>>we refuse and let the other brain half speak ?
>
> I don't know why you have problems with it.
>
> Take a function f(x,y). A graph of that function is a surface. If we
> consider the limit as x -> oo of f(x,y), we get a curve: a function
> g(y). Now, you can visualize what it means to take lim_{y->oo} g(y),
> so there you go.
(Seems that you have two brain halves working as well here.) Okay, I can
see that for _some_ functions. But how about f(x,y) = x.y/(x^2 + y^2) ,
being equal to sin(2.phi)/2 in polar coordinates. Quite a messy thing
for x -> oo . Here are a few isolines (a few due to ASCII limitations):
-1 0 +1
\ | /
\ | /
\ | /
\ | /
\ | /
\|/
0 ------------------- 0
/|\
/ | \
/ | \
/ | \
/ | \
/ | \
+1 0 -1
Now you can think that for x -> oo the function becomes zero. Allright.
But then for y -> oo you go upwards, and my geometrical brain half says
that you will meet the isoline y = x somewhere. Resulting in a value
+ 1 (and others)eventually. But no, the official (i.e. formal) result
is 0 . Confused ..
Han de Bruijn
Han de Bruijn
I'll take it. Back to homework.
Han de Bruijn
Han de Bruijn
> On Sep 10, 4:36 pm, Han de Bruijn <Han.deBru...@DTO.TUDelft.NL> wrote:
>
>>But I still find the meaning of the following limit at least confusing:
>>
>> A: lim lim f(x,y)
>> x->oo y->oo
>> x.y
>>When applied to e.g. --------- = sin(2.phi)/2 with x = r.cos(phi)
>> x^2 + y^2 y = r.sin(phi)
>>
>>Then with limit A we first go to infinity along a line parallel to the
>>y-axis, giving 0 because the angle phi approaches Pi/2 . Right ?
>>
>>But then we have to shift that parallel line towards infinity and _that_
>>stumps me, because now I can't "see" anymore what's happening. (This is
>>merely a consequence of _both_ my brain halves working. I can understand
>>it with the "formal" half but not with the "visual" half). Consequently,
>>I still find that the limit A is indeed "wrong". Flame shield ... well,
>>think I can take it. Do we have to live with "counter intuition" or can
>>we refuse and let the other brain half speak ?
>
> ************************************************************
>
> So you can't "see" anymore what's happening, and instead of putting
> the blame on your lack of mathematical education and/or on your own
> stupidity, you'd rather put the blame on mathematics....aha.
> Some rather comfortable way to put the blame on others rather that in
> you.
>
> And Han: your several brain halves (how many do we already have there:
> six, seven halves...?) speaking to you might be a sign for you to
> search for some professional help: you know, those little voices in
> your head...tsk,tsk,tsk. Not to mention that those voices seem to be
> as lost in basic mathematics as you are.
As long as there is music in one's head as well, one shall not fear.
Han de Bruijn
Han de Bruijn
> On Sep 10, 4:45 am, Virgil <Vir...@gmale.com> wrote:
>
>>In article <9ea0a$48c77903$82a1e228$31...@news1.tudelft.nl>,
>> Han de Bruijn <Han.deBru...@DTO.TUDelft.NL> wrote:
>>
>>>Virgil wrote:
>>
>>>>So, among mathematicians, there is distinction between
>>
>>>>lim_{x->a} lim_{y->b} f(x)
>>>>and
>>>>lim_{y->b} lim_{x->a} f(x)
>>>>and
>>>>lim_{(x,y)->(a,b)} f(x),
>>
>>>>particularly in the case when the last of these does not exist but one
>>>>or both of first two do exist.
>>
>>>Okay. Thus, if I write
>>
>>>>For any real number epsilon > 0 there exists real numbers M , N such
>>>>that if x > L and y > M then | f(x,y) - F | < epsilon .
>>
>>>then the correct notation for this is: lim_{(x,y)->(a,b)} f(x,y) = F .
>>
>>>Am I right this time?
>>
>>Not quite. Try lim_{(x,y)->(oo,oo)} f(x,y) = F
>
> While this notation is probably correct, it is
> rather unusual, I think.
>
> The idea of limit that Han is entertaining is
> the one corresponding to the convergence of
> nets phi : R x R --> R where R x R, the cartesian
> square of the real numbers, is ordered with the
> product order of the usual orders in the factors.
> This kind of limit rarely shows up in calculus,
> and while your notation is quite indcative of
> the specific kind of limit, I do not think
> it is standard.
Please educate us about limit notations thet _are_ standard.
Han de Bruijn
Han de Bruijn
> In article <Virgil-FC7481....@comcast.dca.giganews.com>,
> Virgil <Vir...@gmale.com> wrote:
>
>>In article
>><aderamey.addw-1B9...@newsgroups.comcast.net>,
>> The World Wide Wade <aderam...@comcast.net> wrote:
>>
>>>In article <72600$48c68383$82a1e228$20...@news1.tudelft.nl>,
>>> Han de Bruijn <Han.de...@DTO.TUDelft.NL> wrote:
>>>
>>>>The World Wide Wade wrote:
>>>>
>>>>>>LOL. Look, the word "fixed" is there to help you; if you find it
>>>>>>confusing, then just abandon it. For each x in [0,1] you have a
>>>>>>sequence of numbers (g_n(x)), and for each n in N, you have a function
>>>>>>g_n on [0,1]. That's all that is being said.
>>>>>
>>>>>By the way, did you run screaming out of calculus class when the
>>>>>professor said "fix x and take the derivative with respect to y"? Are
>>>>>partial derivatives also "wrong limits"?
>>>>
>>>>Please read my response to David C. Ullrich. There is no problem at all
>>>>when taking only _one_ limit with a function of two variables. IMO there
>>>>is only a problem when taking a _double_ limit with such functions. Uhm,
>>>>you know about the rule d^2f/(dx.dy) = d^2f/(dy.dx) , for those partial
>>>>derivatives, which is always valid in physics, but not in mathematics.
>>>
>>>So how come
>>>
>>> int_[a,b] int_[c,d] f(x,y) dx dy
>>>
>>> = int_[c,d] int_[c,d] f(x,y) dy dx
>>>
>>>for any continuous f on [a,b] x [c,d]? Aren't those double limits?
>>>Are they wrong limits?
>>
>>I can think of lots of situations in which
>>
>> int_[a,b] int_[c,d] f(x,y) dx dy
>>
>> = int_[c,d] int_[a,b] f(x,y) dy dx
>>
>>But,unless a = c and b = d, I don't really expect that
>>
>> int_[a,b] int_[c,d] f(x,y) dx dy
>>
>> = int_[c,d] int_[c,d] f(x,y) dy dx
>>
>>very often.
>
> Virgil, it was a typo, kind of like yours here
>
> <Virgil-AE1B6A....@comcast.dca.giganews.com>
>
> Stop being a dork.
Maybe, but, thanks to the typo, I don't understand it either.
Han de Bruijn
Virgil
Han de Bruijn <Han.de...@DTO.TUDelft.NL> wrote:
Your geometric brain should have told you that once you have gone to the
limit indicated by x -> oo while y remains fixed and finite, that you
are on a line through origin with slope equal to 0, and must remain
there.
Han de Bruijn
Take "thet", make it "that" and I would really appreciate if you can not
only educate us about the notation but about limits in higher dimensions
in general. Indeed, I haven't seen much of it as well. I know that, with
complex analysis for example, there are things like circles with radius
approaching infinity, for the purpose of calculating integrals (which in
the end turn out to be useful for analysis with _real_ numbers: nice !).
Han de Bruijn
Han de Bruijn
And then we do y -> oo , so I must go up through a line with a slope of
90 degrees, while having arrived at an infinite horizontal distance from
the origin. How can we do that ?
Han de Bruijn
Alois Steindl
Hello,
it would make some (pedagogical) fun to explain these issues to a
first year student.
The two-level operation
lim lim f(x, y)
y->oo x->oo
simply behaves this way: First you keep y fixed and let x go to
infinity, that might give you some value g(y). In your example you
simply obtain g(y) = 0 for all values of y.
Next you observe the behaviour of this function g(y), as y approaches
infinity.
Since you obtain different results when you approach (oo, oo)
differently, e.g. along a ray with a fixed slope, you know, that the
function doesn't have an unique limit for (oo, oo).
There is nothing wrong about that, you simple have to be aware that
taking limits needs proper care.
Alois
Han de Bruijn
> Han de Bruijn <Han.de...@DTO.TUDelft.NL> writes:
>
>>And then we do y -> oo , so I must go up through a line with a slope of
>>90 degrees, while having arrived at an infinite horizontal distance from
>>the origin. How can we do that ?
>
> Hello,
> it would make some (pedagogical) fun to explain these issues to a
> first year student.
>
> The two-level operation
> lim lim f(x, y)
> y->oo x->oo
>
> simply behaves this way: First you keep y fixed and let x go to
> infinity, that might give you some value g(y). In your example you
> simply obtain g(y) = 0 for all values of y.
> Next you observe the behaviour of this function g(y), as y approaches
> infinity.
> Since you obtain different results when you approach (oo, oo)
> differently, e.g. along a ray with a fixed slope, you know, that the
> function doesn't have an unique limit for (oo, oo).
> There is nothing wrong about that, you simple have to be aware that
> taking limits needs proper care.
Yes I can understand that, formally. But now we try the other way around
and substitute (VERY LARGE, VERY LARGE) for (oo, oo). Because that's the
_physical_ (or, if you prefer: the geometrical) _meaning_ of the limits.
And then we see that the answer = 0 is kind of nonsense. I've decided to
simply call it "wrong". (Why show mercy: _they_ have so often said it to
this author).
Han de Bruijn
Alois Steindl
Hello,
it seems that that's the source of your confusion: You assume that
both entries in (VERY LARGE, VERY LARGE) are of the same size.
Its the same kind of misconception, when beginners look at
0/0 and think that it should be 1, because x/x is 1.
But 0/0 could also occur in combinations like x^2/x^3 or x^7/x^2 (and
of course, much more complicated ones.)
You seem to believe, that "the point" (oo, oo) is defined as
lim (R, R)
R->oo
but there are much more possibilities like (R^2, R), (log R, R),
(log(log(R)), exp(exp(R))), ...
Along these different curves to infinity the limit behaves
differently. All the "endpoints" of the curves are representations of
(oo, oo), but the limits are different. This f isn't just so nice to
give the same limit in any case.
Forget your imagination, that oo is some kind of point, especially in
more than 1 dimension, it is more a concept.
Again: All the "endpoints" of curves (x(s), y(s)) with
lim x(s) = oo, lim = oo
s->oo y->oo
are perfect candidates for the "geometrical meaning" of (oo, oo).
But as I already told you, that topic is appropriate for young
students. Since I assume, that you are much older, you should have
understood these concepts long ago.
Alois
--
Alois Steindl, Tel.: +43 (1) 58801 / 32558
Inst. for Mechanics and Mechatronics Fax.: +43 (1) 58801 / 32598
Vienna University of Technology, A-1040 Wiedner Hauptstr. 8-10
Han de Bruijn
> Your notion of "right" and "wrong" is wrong. It's like
> those nasty professors are saying things about apples,
> and you're saying they're wrong. You agree that the
> things they say about apples are correct, but nonetheless
> they're wrong because they should be talking about oranges.
> It's a very strange sort of "right" and "wrong" - if I say
> something that's true it's wrong because _you_ say I
> should be talking about something else.
Yes. And this brings me to a general point. I've said the following in
another thread, namely in response to Tonio in the article
http://groups.google.nl/group/sci.math/msg/f51ee837a2fc0ee1
> Then you'd also have trouble with this. The power of mathematics is not
> its generality, but its ability to deal with special cases.
In short: mathematics is TOO GENERAL. Mathematicians are so affraid of
missing something that the precautions they take are close to paranoia.
Now I will be the last one to say that one should'nt carefully consider
anything possible, but it should be a great relief if some kind of, say,
mathematics BY DEFAULT could be available. And my _great dream_ is that
such mathematics should reflect .. sense of reality. I could mention a
lot of examples, but just one may be sufficient for our purpose: my old
friend the 'scinc' function.
http://groups.google.nl/group/sci.math/msg/c83b58832ecdff3f
Any definition other than with sinc(0) = 1 will never have applications
outside mathematics. That's why default mathematics should only contain
_this_ definition and no others.
How can I be so sure of this ? I'm so sure because the _continuous_ and
the _discrete_, in _science_, are just two ways of looking at the _same_
thing. For example, an ideal gas is _continuous_ in Computational Fluid
Dynamics, but it is _discrete_ in Statistical Mechanics. _Singularities_
arise in continuous functions where the discrete substrate can no longer
be hidden. I became aware of this due to a discovery called "Fluid Tube
Continuum", while I was working on apparatus for nuclear reactors, quite
a while ago. With such a rough real world continuum, the place where the
model breaks down, and the discrete becomes _visible_ can be pinpointed.
It leads to the rather surprising result that such breakdown will happen
if the "primary" mass flow within the apparatus exceeds a well defined,
though approximately calculated value. See my website for more details:
http://hdebruijn.soo.dto.tudelft.nl/QED/index.htm#ft
I mean, before somebody says that default ("realistic") mathematics is
a hopeless excercise, the _whole_ picture must be taken into account.
In the somewhat desperate hope that I can convince someone, somehow ..
Han de Bruijn
Han de Bruijn
> Denis Feldmann <denis.feldm...@neuf.fr> writes:
>
>>Jesse F. Hughes a écrit :
>>
>>>Denis Feldmann <denis.feldm...@neuf.fr> writes:
>>>
>>>>No, youre wrong. He is not such a good crank. Otoh, he is quite good at
>>>>trolling...
>>>
>>>He's a *great* crank. You take that back.
>>
>>Never. That would implies he really believes what he says (most of the
>>time). You wanna bet ? (but I admit he is a *great* troll, if you like)
>
> I think I'd take that bet, if we restrict it to his mathematical
> claims. But how will we settle it?
I'm so sorry to jump in here and disturb your bad, bad game ..
What exactly is the difference between a crank and a troll ?
(Can't find a proper translation for "trolling" in my dictionary)
Iff I have to suffer from name calling, maybe I'd like to to suffer
from the _right_ name calling, huh :-(
Han de Bruijn
Han de Bruijn
> Han de Bruijn <Han.de...@DTO.TUDelft.NL> writes:
>
>>Yes I can understand that, formally. But now we try the other way around
>>and substitute (VERY LARGE, VERY LARGE) for (oo, oo). Because that's the
>>_physical_ (or, if you prefer: the geometrical) _meaning_ of the limits.
>>And then we see that the answer = 0 is kind of nonsense. I've decided to
>>simply call it "wrong". (Why show mercy: _they_ have so often said it to
>>this author).
>
> Hello,
> it seems that that's the source of your confusion: You assume that
> both entries in (VERY LARGE, VERY LARGE) are of the same size.
> Its the same kind of misconception, when beginners look at
> 0/0 and think that it should be 1, because x/x is 1.
Nope. I'm not _that_ inexperienced. I think that 0/0 = anything .
> But 0/0 could also occur in combinations like x^2/x^3 or x^7/x^2 (and
> of course, much more complicated ones.)
> You seem to believe, that "the point" (oo, oo) is defined as
> lim (R, R)
> R->oo
> but there are much more possibilities like (R^2, R), (log R, R),
> (log(log(R)), exp(exp(R))), ...
> Along these different curves to infinity the limit behaves
> differently. All the "endpoints" of the curves are representations of
> (oo, oo), but the limits are different. This f isn't just so nice to
> give the same limit in any case.
Exactly.
> Forget your imagination, that oo is some kind of point, especially in
> more than 1 dimension, it is more a concept.
>
> Again: All the "endpoints" of curves (x(s), y(s)) with
> lim x(s) = oo, lim = oo
> s->oo y->oo
> are perfect candidates for the "geometrical meaning" of (oo, oo).
>
> But as I already told you, that topic is appropriate for young
> students. Since I assume, that you are much older, you should have
> understood these concepts long ago.
I do. Please look up my Internet profile, if you want. You will see then
that my objectives are always evolving around the same theme, which has
become kind of an obsession: the _meaning_ of mathematics in a _science_
environment. The whole thread fits precisely in this perspective, and it
is not a surprise for the older 'sci.math' debaters that it does. Maybe
I'm a bit dishonest in suggesting that I don't really know things, but I
am never dishonest in trying to fulfill my destination here.
Han de Bruijn
Jesse F. Hughes
> Jesse F. Hughes wrote:
>
>> Han de Bruijn <Han.de...@DTO.TUDelft.NL> writes:
>>
>>>But then we have to shift that parallel line towards infinity and _that_
>>>stumps me, because now I can't "see" anymore what's happening. (This is
>>>merely a consequence of _both_ my brain halves working. I can understand
>>>it with the "formal" half but not with the "visual" half). Consequently,
>>>I still find that the limit A is indeed "wrong". Flame shield ... well,
>>>think I can take it. Do we have to live with "counter intuition" or can
>>>we refuse and let the other brain half speak ?
>>
>> I don't know why you have problems with it.
>>
>> Take a function f(x,y). A graph of that function is a surface. If we
>> consider the limit as x -> oo of f(x,y), we get a curve: a function
>> g(y). Now, you can visualize what it means to take lim_{y->oo} g(y),
>> so there you go.
>
> (Seems that you have two brain halves working as well here.)
Nonsense. Imagination and visualization is a pretty common tool in
understanding mathematics, but I don't regard it as "two brain
halves". *Most* folks trying to understand calculus rely on visual
imagery to some extent, I'd wager.
> Okay, I can see that for _some_ functions. But how about f(x,y) =
> x.y/(x^2 + y^2) , being equal to sin(2.phi)/2 in polar
> coordinates. Quite a messy thing for x -> oo . Here are a few
> isolines (a few due to ASCII limitations):
>
> -1 0 +1
> \ | /
> \ | /
> \ | /
> \ | /
> \ | /
> \|/
> 0 ------------------- 0
> /|\
> / | \
> / | \
> / | \
> / | \
> / | \
> +1 0 -1
>
> Now you can think that for x -> oo the function becomes zero. Allright.
> But then for y -> oo you go upwards, and my geometrical brain half says
> that you will meet the isoline y = x somewhere. Resulting in a value
> + 1 (and others)eventually. But no, the official (i.e. formal) result
> is 0 . Confused ..
Again, I don't see *why* you're confused, really.
lim lim f(x,y)
y -> oo x -> oo
means: take the limit with respect to x first. The result is a
function of y. Take the limit with respect to y now.
Your "geometrical brain half" seems determined to forget what it's
doing here. Or perhaps you think that the function
lim f(x,y)
x -> oo
really is a cross-section of f(x,y) found at some point x = a. It
isn't.
So, what is
lim f(x,y) ?
x -> oo
As you've said, it's the function g(y) = 0. Now, surely, you have no
doubt what the limit of that function is as y -> oo.
--
Jesse F. Hughes
"Besides, discoverers are too proud to kiss butt. Indiana Jones would
never kiss some academic's ass to get published, and neither will I."
--James Harris
Jesse F. Hughes
You're pulling this requirement plumb out of your ass. It just isn't
so.
It's the exact issue here: Taking lim_y lim_x is not the same thing
as taking a limit along the line y = x. If part of your brain is
saying that it is, perhaps you should stop listening to that part,
because that part hasn't a clue about what lim_y lim_x *means*.
--
Jesse F. Hughes
"That's cool for us in Alabama. 'Cause you know, it's either this or
the monster truck rally." -- An Alabaman expresses appreciation for
local repertory theater on NPR
Jesse F. Hughes
> Yes I can understand that, formally. But now we try the other way around
> and substitute (VERY LARGE, VERY LARGE) for (oo, oo). Because that's the
> _physical_ (or, if you prefer: the geometrical) _meaning_ of the
> limits.
That is *not* the physical or geometrical meaning of the nested limit
lim_y lim_x f(x,y).
You really are bound and determined to get this wrong.
--
"So I speak before a crowd of the damned, cursed to be unloved
throughout time, with only their hatred and bile to comfort them now,
having betrayed what should have been their one true lover:
Mathematics." -- James Harris reaches a bit
Han de Bruijn
> Han de Bruijn <Han.de...@DTO.TUDelft.NL> writes:
>
>>Jesse F. Hughes wrote:
>>
>>>Han de Bruijn <Han.de...@DTO.TUDelft.NL> writes:
>>>
>>>>But then we have to shift that parallel line towards infinity and _that_
>>>>stumps me, because now I can't "see" anymore what's happening. (This is
>>>>merely a consequence of _both_ my brain halves working. I can understand
>>>>it with the "formal" half but not with the "visual" half). Consequently,
>>>>I still find that the limit A is indeed "wrong". Flame shield ... well,
>>>>think I can take it. Do we have to live with "counter intuition" or can
>>>>we refuse and let the other brain half speak ?
>>>
>>>I don't know why you have problems with it.
>>>
>>>Take a function f(x,y). A graph of that function is a surface. If we
>>>consider the limit as x -> oo of f(x,y), we get a curve: a function
>>>g(y). Now, you can visualize what it means to take lim_{y->oo} g(y),
>>>so there you go.
>>
>>(Seems that you have two brain halves working as well here.)
>
> Nonsense. Imagination and visualization is a pretty common tool in
> understanding mathematics, but I don't regard it as "two brain
> halves". *Most* folks trying to understand calculus rely on visual
> imagery to some extent, I'd wager.
Which doesn't mean that both abilities are located in the same brain
center.
Surely. And I have no doubt that IT MAKES NO SENSE (literally) as well.
Oh, you know what _senses_ are ?
Han de Bruijn
Jesse F. Hughes
> Jesse F. Hughes wrote:
>
>> Denis Feldmann <denis.feldm...@neuf.fr> writes:
>>
>>>Jesse F. Hughes a écrit :
>>>
>>>>Denis Feldmann <denis.feldm...@neuf.fr> writes:
>>>>
>>>>>No, youre wrong. He is not such a good crank. Otoh, he is quite good at
>>>>>trolling...
>>>>
>>>>He's a *great* crank. You take that back.
>>>
>>>Never. That would implies he really believes what he says (most of the
>>>time). You wanna bet ? (but I admit he is a *great* troll, if you like)
>>
>> I think I'd take that bet, if we restrict it to his mathematical
>> claims. But how will we settle it?
>
> I'm so sorry to jump in here and disturb your bad, bad game ..
>
> What exactly is the difference between a crank and a troll ?
> (Can't find a proper translation for "trolling" in my dictionary)
A crank is a person who believes that he has a revolutionary or
momentous mathematical/scientific discovery, but this discovery is
without merit (or with trivial merit). He exaggerates his own
importance. Cranks predate the internet and formerly attracted
attention by self-published pamphlets and books and by letters to
various academics.
Trolling may be a new phenomena. In its early form, one would say
something off the wall (on Usenet) just to see how many folks missed
the joke and reacted as if it were serious. It was a fairly
lighthearted jest, but nowadays, we think of folks whose main
contributions are trolls without any evident humor. A troll posts
whatever will get the most reaction from others, regardless of his own
personal assessment of merits. He posts just to get attention,
whereas a crank posts to get attention for ideas he thinks are
revolutionary.
Thus, a troll is fundamentally dishonest (at least not motivated by
honesty), while a crank is deluded.
>
> Iff I have to suffer from name calling, maybe I'd like to to suffer
> from the _right_ name calling, huh :-(
Delusion is better. Be a crank.
And not to worry: I think you are!
--
"So now, The Hammer is here, and with it, the end of days. The world will be
destroyed, and then remade, as foretold. You will be lost, with your
children, and then there will be others, and one day they will be tested, and
will pass, but that is another story." --James S Harris gets a bit excited.
Jesse F. Hughes
I think that your feigned ignorance of taking nested limits is indeed
dishonest. In this respect, you're crossing the line from crank to
troll. Why not stick to what you really dispute and stop pretending
that this evil brain half keeps muddling your calculus?
--
Jesse F. Hughes
"Readers should remember that being able to post on Usenet does not mean
a person actually has expertise in a particular area or even knows
ANYTHING significant in that area." -- James S. Harris
Han de Bruijn
> On Sep 8, 6:53 am, Han de Bruijn <Han.deBru...@DTO.TUDelft.NL> wrote:
>
>>Wrong limits do not commute. Almost dare to say now that non commuting
>>limits are simply .. wrong. Any counter examples are quite welcome.
>
> Given
>
> A transistor dissipates power propotional to its linear dimension.
Let T = linear dimension of transistor, C = linear dimension of chip,
P_T = power dissipated by transistor, P_T = power dissipated by chip,
k = certain constant, N = number of transistors on chip. Then we have:
P_T = k.T^3 , N = C/T^2 (approximately), P_C = N.P_T = k.C.T . Right?
I hope so. My knowledge of electronics is a bit rusty.
> a) for a fixed chip size what is the power dissipated by the chip
> as the transistor size gets very small?
lim P_C = lim k.C.T = 0
T->0 T->0
> b) for a fixed transistor size what is the power dissipated by the chip
> as the chip size gets very small?
lim P_C = lim k.C.T = 0
C->0 C->0
> The limits in a) and b) are hardly wrong.
> However,
>
> c) what is the power dissipated by the chip as the transistor size
> and the chip size get very small?
lim P_C = lim k.C.T = 0
C->0,T->0 C->0,T->0
> Has no answer.
Yes? (Suppose I don't have the right expression for small chip sizes ..)
> The related question,
>
> d) what is the dissipated power density as the transistor size and the
> chip size get very small?
>
> has an answer.
lim P_C/C = lim k.T = 0
C->0,T->0 C->0,T->0
> At best you could say "In properly posed problems limits _do_
> commute".
> So now all we need to do is define "properly posed problem. Oh I
> know, a properly posed problem is one in which limits commute.
??
Han de Bruijn
Han de Bruijn
Uhm, maybe the best of both worlds .. I'm immune, I'm immune ..
Han de Bruijn
Han de Bruijn
> Han de Bruijn <Han.de...@DTO.TUDelft.NL> writes:
>
>>I do. Please look up my Internet profile, if you want. You will see then
>>that my objectives are always evolving around the same theme, which has
>>become kind of an obsession: the _meaning_ of mathematics in a _science_
>>environment. The whole thread fits precisely in this perspective, and it
>>is not a surprise for the older 'sci.math' debaters that it does. Maybe
>>I'm a bit dishonest in suggesting that I don't really know things, but I
>>am never dishonest in trying to fulfill my destination here.
>
> I think that your feigned ignorance of taking nested limits is indeed
> dishonest. In this respect, you're crossing the line from crank to
> troll. Why not stick to what you really dispute and stop pretending
> that this evil brain half keeps muddling your calculus?
I've made quite clear what I really want to dispute: the significance of
those nested limits with respect to other issues than pure formalistics.
When talking about honesty, I didn't expect _you_ would sink into this:
http://phiwumbda.org/~jesse/teaching/BusEthics-F08/syll.html
And hey ! To be honest, mind the supermarket paradigm in your lessons !
Han de Bruijn
William Hughes
> William Hughes wrote:
> > On Sep 8, 6:53 am, Han de Bruijn <Han.deBru...@DTO.TUDelft.NL> wrote:
>
> >>Wrong limits do not commute. Almost dare to say now that non commuting
> >>limits are simply .. wrong. Any counter examples are quite welcome.
>
> > Given
>
> > A transistor dissipates power propotional to its linear dimension.
>
> Let T = linear dimension of transistor, C = linear dimension of chip,
> P_T = power dissipated by transistor, P_T = power dissipated by chip,
> k = certain constant, N = number of transistors on chip. Then we have:
> P_T = k.T^3 ,
nope
P_T = (power per transistor) * (number of transistors)
= (k * T) * (area of chip / area of transistor)
= (k * T) * ( C^2 / T^2)
= kC^2/T
> N = C/T^2 (approximately), P_C = N.P_T = k.C.T . Right?
> I hope so. My knowledge of electronics is a bit rusty.
>
> > a) for a fixed chip size what is the power dissipated by the chip
> > as the transistor size gets very small?
>
> lim P_C = lim k.C.T = 0
> T->0 T->0
>
lim P_T = lim k.C^2/T = oo
T->0 T->0
> > b) for a fixed transistor size what is the power dissipated by the chip
> > as the chip size gets very small?
>
> lim P_C = lim k.C.T = 0
> C->0 C->0
lim P_C = lim k.C^2/T = 0
T->0 C->0
> > The limits in a) and b) are hardly wrong.
> > However,
>
> > c) what is the power dissipated by the chip as the transistor size
> > and the chip size get very small?
>
lim P_T = lim kC^2/T
C->0,T->0 C->0,T->0
Has no answer.
So the limits are meaninful (for a fixed chip size as the transistor
size gets small the power dissipated by the chip gets very large;
for a fixed transistor size as the chip size gets small the power
dissipated by the chip gets very small) but they do not commute.
- William Hughes
Han de Bruijn
> On Sep 11, 8:13 am, Han de Bruijn <Han.deBru...@DTO.TUDelft.NL> wrote:
>
>>William Hughes wrote:
>>
>>>On Sep 8, 6:53 am, Han de Bruijn <Han.deBru...@DTO.TUDelft.NL> wrote:
>>
>>>>Wrong limits do not commute. Almost dare to say now that non commuting
>>>>limits are simply .. wrong. Any counter examples are quite welcome.
>>
>>>Given
>>
>>>A transistor dissipates power propotional to its linear dimension.
>>
>>Let T = linear dimension of transistor, C = linear dimension of chip,
>>P_T = power dissipated by transistor, P_T = power dissipated by chip,
>>k = certain constant, N = number of transistors on chip. Then we have:
>>P_T = k.T^3 ,
Typo, must be: P_C = power dissipated by chip. I'm too hasty ..
> nope
Whew ! Sorry ! Ooops ! How could I misread the question so much:
> P_T = (power per transistor) * (number of transistors)
>
> = (k * T) * (area of chip / area of transistor)
>
> = (k * T) * ( C^2 / T^2)
>
> = kC^2/T
Yeah, you're quite right, of course, if what you mean is:
P_T = k.T and P_C = k.C^2/T .
>>N = C/T^2 (approximately), P_C = N.P_T = k.C.T . Right?
>>I hope so. My knowledge of electronics is a bit rusty.
>>
>>>a) for a fixed chip size what is the power dissipated by the chip
>>> as the transistor size gets very small?
>>
>>lim P_C = lim k.C.T = 0
>>T->0 T->0
>
> lim P_T = lim k.C^2/T = oo
> T->0 T->0
lim P_C = lim k.C^2/T = oo
T->0 T->0
So that limit _does not exist_. Right ?
>>>b) for a fixed transistor size what is the power dissipated by the chip
>>> as the chip size gets very small?
>>
>>lim P_C = lim k.C.T = 0
>>C->0 C->0
>
> lim P_C = lim k.C^2/T = 0
> T->0 C->0
lim P_C = lim k.C^2/T = 0
C->0 C->0
>>>The limits in a) and b) are hardly wrong.
Apart from the fact that the first one (a) does _not_ exist.
>>>However,
>>
>>>c) what is the power dissipated by the chip as the transistor size
>>> and the chip size get very small?
>>
> lim P_T = lim kC^2/T
> C->0,T->0 C->0,T->0
>
> Has no answer.
Correct: 0/0 is anything. But now we divide the power by the chip area,
being C^2 , and get zero for the limit of the power _density_, right ?
> So the limits are meaninful (for a fixed chip size as the transistor
> size gets small the power dissipated by the chip gets very large;
> for a fixed transistor size as the chip size gets small the power
> dissipated by the chip gets very small) but they do not commute.
Not at all. Your "meaningful" limit (a) does not exist, namely.
Han de Bruijn
Jesse F. Hughes
Sure.
And yet, I do not come to the same conclusion as you. Take the limit
in terms of x. You get a function g(y). Take its limit.
Neither my senses (which report nothing at all about this) nor my
imagination leads me to any conclusion but that the latter limit is 0.
--
"But you people are scum of the earth who pretend to be something that
is clearly beyond you--real mathematicians. I wouldn't be having
these problems with Gauss or Euler. I wouldn't be having these
problems with Fermat or Archimedes." -- James S. Harris on pretending
Jesse F. Hughes
> Jesse F. Hughes wrote:
>
>> Han de Bruijn <Han.de...@DTO.TUDelft.NL> writes:
>>
>>>I do. Please look up my Internet profile, if you want. You will see then
>>>that my objectives are always evolving around the same theme, which has
>>>become kind of an obsession: the _meaning_ of mathematics in a _science_
>>>environment. The whole thread fits precisely in this perspective, and it
>>>is not a surprise for the older 'sci.math' debaters that it does. Maybe
>>>I'm a bit dishonest in suggesting that I don't really know things, but I
>>>am never dishonest in trying to fulfill my destination here.
>>
>> I think that your feigned ignorance of taking nested limits is indeed
>> dishonest. In this respect, you're crossing the line from crank to
>> troll. Why not stick to what you really dispute and stop pretending
>> that this evil brain half keeps muddling your calculus?
>
> I've made quite clear what I really want to dispute: the
> significance of those nested limits with respect to other issues
> than pure formalistics.
Perhaps you want to claim: every nested limit in physics (and other
sciences) is a limit that commutes.
Perhaps this is so, perhaps not. I'll leave that discussion to
others.
But that claim has nothing to do with your feigned confusion over the
limit
lim_y->oo lim_x->oo xy/(x^2 + y^2).
> When talking about honesty, I didn't expect _you_ would sink into this:
>
> http://phiwumbda.org/~jesse/teaching/BusEthics-F08/syll.html
>
> And hey ! To be honest, mind the supermarket paradigm in your
> lessons !
Teaching business ethics is shameful? News to me.
--
"It's my belief that when religion and pseudoscience achieve an
official status within a culture [...], then genocide, war,
oppression, injustice, and economic stagnation are sure to follow."
-- David Petry, on why |X| < |P(X)| is bad, bad, bad.
Virgil
For one thing, 90 degrees is not a slope at all.
Do you mean a slope of tan(90 degrees)?
For another, how does one get off of a line with slope zero at a point
for which y is still finite but x has already "gone to infinity"?
Virgil
Han de Bruijn <Han.de...@DTO.TUDelft.NL> wrote:
Since the value of x*y/(x^2-y^2) varies between 0 (acheived) and oo
(only approachable) depending on the path towards (oo,oo) one follows,
there can be no single limit., and "the limit" cannot exist.
However each iterated limit exists and is zero.
At least as long as one goes by the formal definitions and does not try
to impose one's own geometric intuition on top of the mathematics.
> And then we see that the answer = 0 is kind of nonsense. I've decided to
> simply call it "wrong". (Why show mercy: _they_ have so often said it to
> this author).
And here is another occasion on which that author is spouting nonsense,
at least regarding the mathematics involved.
Virgil
Han de Bruijn <Han.de...@DTO.TUDelft.NL> wrote:
> Any definition other than with sinc(0) = 1 will never have applications
> outside mathematics. That's why default mathematics should only contain
> _this_ definition and no others.
You are quite free to define your own "default mathematics", so long as
you do not insist that anyone else be bound by it.
But the you must equally allow anyone else to define his or her own
"default mathematics" without let or hindrance as long as they do not
bind you to it.
But there is a non-default standard for mathematics in general, and in
that standard the definition of a double limit (two variables
simultaneously) is different from the corresponding definitions for
iterated limits (in which the limiting for one variable precedes the
limiting of the other).
Virgil
Han de Bruijn <Han.de...@DTO.TUDelft.NL> wrote:
> I'm a bit dishonest in suggesting that I don't really know things, but I
> am never dishonest in trying to fulfill my destination here.
May I suggest that your "destination", which is ultimately the grave,
may be unrelated to what you see as your destiny.
Virgil
Han de Bruijn <Han.de...@DTO.TUDelft.NL> wrote:
> > So, what is
> >
> > lim f(x,y) ?
> > x -> oo
> >
> > As you've said, it's the function g(y) = 0. Now, surely, you have no
> > doubt what the limit of that function is as y -> oo.
>
> Surely. And I have no doubt that IT MAKES NO SENSE (literally) as well.
What you probably mean is that it does not have any physical
interpretation that you are comfortable with, but that there be any
physical interpretation is a hang up almost exclusively limited to
physicists like you.
In mathematics itself, there is no need for any such physical
interpretation. And to suggest that there ought to be is a common sin of
physicists.
Virgil
Han de Bruijn <Han.de...@DTO.TUDelft.NL> wrote:
> I've made quite clear what I really want to dispute: the significance of
> those nested limits with respect to other issues than pure formalistics.
In other words, the significance of mathematics outside of mathematics.
Within mathematics, definitions rule.
Without mathematics, what rules?
William Hughes
Nope. The limit is oo (e.g. as the size of the transisor gets
small the power dissipated by the chip gets large, and you can make
this as large as you want). If you insist on finite limits
take
P_C = min(M, k.C^2/T)
where M is some maximum power.
- William Hughes
umu...@gmail.com
> Neither my senses (which report nothing at all about this) nor my
> imagination leads me to any conclusion but that the latter limit is 0.
It's typical that your senses report nothing at all about mathematical
truth. Be not surprised then that the mathematics you come up with may
turn out to be be non-sense.
Han de Bruijn
umu...@gmail.com
>
> Teaching business ethics is shameful? News to me.
Capitalism is compatible with ethics? News to me.
Han de Bruijn
umu...@gmail.com
> At least as long as one goes by the formal definitions and does not try
> to impose one's own geometric intuition on top of the mathematics.
I've always thought that geometric intuition is _part of_ mathematics.
Han de Bruijn
umu...@gmail.com
[ whatever ]
Thanks for the formula describing the power consumption P_C of a chip
with linear dimension C containing transistors with linear dimension
T.
The formula is (apart from a constant): P_C = C^2/T .
Here the quotient N = C^2/T^2 is the number of transistors on the
chip.
One thing should be noted: N is a natural. Meaning that William
Hughes'
formula is only valid for C = sqrt(N).T : straight lines through the
origin in the (C,T) plane. And since the minimal value of N is 1, only
the following are relevant: C = T , C = sqrt(2).T , C = sqrt(3).T , ..
Consequently also T > 0 , T > 0 , C > T (last one greater or equal).
C>0 C>T
| /
| /
| /
| /
| / forbidden
|/
--------------------- T>0
|
|
|
The search for a MEANINGFUL limit could end in the following: consider
the number of transistors on chip as fixed (therefore travel along one
of the lines C = sqrt(N).T , for fixed N). What is the limit of P_C
if the transistor size approaches zero then ? Answer: lim (N.T) = 0 .
T->0
Han de Bruijn
Jesse F. Hughes
It's true! It's true! I've never seen the curve
x y
f(x,y) = ---------
x^2 + y^2
I've seen two-dimensional projections of part of that curve, and
that's pretty close. But I only knew that the projection represented
the curve by reasoning about the behavior of the curve.
And such projections were bounded. They didn't show me what happened
arbitrarily far out. So, I reiterate: My senses did not lead me to
any conclusion about the limit at all.
--
"No sane person actually believes that religion mumbo-jumbo[...] Of
course, very few people [...] would ever admit that they don't
actually believe any of it. Of course I can't prove this, so don't
ask. But you know it's true as well as I do." -- Mensanator
Jesse F. Hughes
Rather off-topic, but I see nothing inherently immoral about trading
goods freely.
In any case, I don't really give a shit what you think about
(a) capitalism or (b) the courses I teach.
--
"I liked the world a lot better over ten years ago. I believed in a
lot more things. Hell, most people believed in a lot more things.
Back then the United States was still, well, known as most people used
to know the United States." -- James S. Harris in a nostalgic mood
Virgil
"Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> umu...@gmail.com writes:
>
> > On 11 sep, 17:26, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >
> >> Neither my senses (which report nothing at all about this) nor my
> >> imagination leads me to any conclusion but that the latter limit is 0.
> >
> > It's typical that your senses report nothing at all about mathematical
> > truth. Be not surprised then that the mathematics you come up with may
> > turn out to be be non-sense.
>
> It's true! It's true! I've never seen the curve
>
> x y
> f(x,y) = ---------
> x^2 + y^2
How is it a "curve" at all?
A surface, maybe, if interpreted as {(x,y,z) in R^3: z = f(x,y)}.
Jesse F. Hughes
You're right, of course. I meant "surface" or "function" or something
similar.
--
Jesse F. Hughes
"The American people would have been incredibly proud of watching our
military folks dispense with basic health care needs to people who
needed help." --George W. Bush, March 13, 2007
Virgil
"Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> Virgil <Vir...@gmale.com> writes:
>
> > In article <87ljxxa...@phiwumbda.org>,
> > "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >
> >> umu...@gmail.com writes:
> >>
> >> > On 11 sep, 17:26, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> >
> >> >> Neither my senses (which report nothing at all about this) nor my
> >> >> imagination leads me to any conclusion but that the latter limit is 0.
> >> >
> >> > It's typical that your senses report nothing at all about mathematical
> >> > truth. Be not surprised then that the mathematics you come up with may
> >> > turn out to be be non-sense.
> >>
> >> It's true! It's true! I've never seen the curve
> >>
> >> x y
> >> f(x,y) = ---------
> >> x^2 + y^2
> >
> >
> > How is it a "curve" at all?
> >
> > A surface, maybe, if interpreted as {(x,y,z) in R^3: z = f(x,y)}.
>
> You're right, of course. I meant "surface" or "function" or something
> similar.
I've made worse errors.
William Hughes
> [ whatever ]
Ok, we will go back to accepting oo as a valid limit.
>
> Thanks for the formula describing the power consumption P_C of a chip
> with linear dimension C containing transistors with linear dimension
> T.
>
> The formula is (apart from a constant): P_C = C^2/T .
>
> Here the quotient N = C^2/T^2 is the number of transistors on the
> chip.
> One thing should be noted: N is a natural. Meaning that William
> Hughes'
> formula is only valid for C = sqrt(N).T
Nope. If you insist that N be a whole number the correct number of
transistors is
N = floor(C^2/T^2)
Of course the minumum number of transistors on a chip is not 1 but 0.
So the limits
i: lim P_C = lim k* floor(C^2/T) = oo
T->0 T->0
and
ii: lim P_C = lim k* floor(C^2/T) = 0
C->0 C->0
are meaningful and do not commute.
<snip discussion of another limit>
Whether another limit is meaningful or not has nothing
to do with whether i and ii are meaningful.
- William Hughes
Tonico
*************************************************************
Big...huge...humongous mistake, Han: intuition, ANY intuition, is no
part of ANY part of mathematics, and hasn't been ever, and it looks
like it never will.
Of course, you can say YOU use your intuition, geometric or whatever,
to work out this or that problem in maths in this or that fashion, but
it still isn't part of mathematics.
This sheds a little more light on your rather thinking that much of
mathematics HAS to be wrong simply because it doesn't fit some
intuitive idea you have.
I'm beginning to arrive to the enlighting conclusion that you treat
mathematics as you'd treat one of your kids: "don't do that, behave
like this and that, don't open your mouth while eating", etc., but it
only sounds more like "don't say this and that parts in limits theory
correct since they MUST be wrong because I can't see how this can do
this or that thing, don't define stuff like that because one half of
brain doesn't understand that, behave properly and don't talk about
infinite stuff since I can't see how I will use it in physcis or while
buying stuff in the market, etc.
You are pissed off at maths because it doesn't behave as you have
decided, the hell knows why, it should behave...amazing!
Of course, it could also be that you're pissed off at mathematics
because it is stuff way above your head, and thus you feel it "must"
be wrong, otherwise such an ashtonishing developed intelligence like
yours should grasp it...but it doesn't, so it is wrong or else it all
is nonsense!
Oh, well....
Regards
Tonio
galathaea
how do you negotiate repeatable symbol manipulation?
do you think it is definitions all the way down?
why do mathematicians always fall back to some "naive" theory
like a metalogic for formation or application?
do you think language can form
without some faculty
(no matter how faulty)
for establishing bisimulation?
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
galathaea: prankster, fablist, magician, liar
umu...@gmail.com
> umum...@gmail.com writes:
> > On 11 sep, 17:29, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> >> Teaching business ethics is shameful? News to me.
>
> > Capitalism is compatible with ethics? News to me.
>
> Rather off-topic, but I see nothing inherently immoral about trading
> goods freely.
Neither do I. Define: freely.
> In any case, I don't really give a shit what you think about
> (a) capitalism or (b) the courses I teach.
Han de Bruijn
umu...@gmail.com
> So the limits
>
> i: lim P_C = lim k* floor(C^2/T) = oo
> T->0 T->0
> and
>
> ii: lim P_C = lim k* floor(C^2/T) = 0
> C->0 C->0
>
> are meaningful and do not commute.
I rest my case.
Han de Bruijn
William Hughes
Good. Note that you now agree that your
statement "properly posed limits _do_ commute"
is nonsense.
- William Hughes
umu...@gmail.com
Might be wrong about the meaning of "I rest my case".
Anyway, these are your words, not mine.
Han de Bruijn
Virgil
<342f8b43-fd68-48b0...@e53g2000hsa.googlegroups.com>,
umu...@gmail.com wrote:
You just lost your case.
William Hughes
I assume it means that you do not indend to present
any more arguments. As you have not presented any argument
that the limits are not meaningful, I assume you accept the example.
- William Hughes
Jesse F. Hughes
> On 12 sep, 22:51, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> umum...@gmail.com writes:
>> > On 11 sep, 17:29, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>>
>> >> Teaching business ethics is shameful? News to me.
>>
>> > Capitalism is compatible with ethics? News to me.
>>
>> Rather off-topic, but I see nothing inherently immoral about trading
>> goods freely.
>
> Neither do I. Define: freely.
As I said, I don't see any reason to debate this point with you,
especially not on sci.math. I don't really care about your opinion.
>> In any case, I don't really give a shit what you think about
>> (a) capitalism or (b) the courses I teach.
--
Jesse F. Hughes
"It doesn't even pay to be brilliant any more. Every goddamn moron
with an attitude has a computer and can chatter you to the point of
wishing suicide." --- James S. Harris, one or the other.
Han de Bruijn
I accept the example as a confirmation of the fact that wrong limits do
not commute. And the reverse is true as well.
Han de Bruijn
William Hughes
> I accept the example as a confirmation of the fact that wrong limits do
> not commute.
Since the limits in the example are not
"wrong" this is nonsensical.
- William Hughes