Smooth functions
anthony burke
any assistance.
Is it correct that a smooth function is a function that has a
derivative of all orders, that it can be diferentiated an infinite
number of times, and for a function of several real variables, the
definition is that it has all its partial derivatives of all orders?
Assuming the above is the case, and I have a function and
differentiate it to get the zero function, e.g. f(x) = 1, f'(x)=0 am
I right in thinking that the derivative of this is zero and so on,
i.e. f''(x)=0, f'''(x) = 0...?
A smooth function has all (partial) derivatives defined, but can these
be defined as zero, like in the above case? Is f(x) = x^2 a smooth
function, i.e. f'(x) = 2x, f''(x) = 2, f'''(x) = 0, and all other
derivatives are zero? Or are smooth functions restricted to functions
like trig and exponentials etc?
I am just unsure as to whether the derivatives being zero equates to
the derivatives not being defined.
Thanks in advance,
Anthony
I still think I struggle with maths more than Einstein, I suspect he
was just being modest...
Dirk Van de moortel
"anthony burke" <anthon...@hotmail.com> wrote in message news:8a9174dc.02091...@posting.google.com...
> Hello, this may seem like a trivial post, but I would be grateful for
> any assistance.
>
> Is it correct that a smooth function is a function that has a
> derivative of all orders, that it can be diferentiated an infinite
> number of times, and for a function of several real variables, the
> definition is that it has all its partial derivatives of all orders?
I guess most of the time it goes like you outlined.
But if I'm not mistaken, *sometimes* a function is also called
"smooth" if the first order derivative exists and is continuous.
>
> Assuming the above is the case, and I have a function and
> differentiate it to get the zero function, e.g. f(x) = 1, f'(x)=0 am
> I right in thinking that the derivative of this is zero and so on,
> i.e. f''(x)=0, f'''(x) = 0...?
yes.
>
> A smooth function has all (partial) derivatives defined, but can these
> be defined as zero, like in the above case? Is f(x) = x^2 a smooth
> function, i.e. f'(x) = 2x, f''(x) = 2, f'''(x) = 0, and all other
> derivatives are zero? Or are smooth functions restricted to functions
> like trig and exponentials etc?
No, a constant function like f(x) = k is so smooth that it
almost hurts. They don't come any smoother.
>
> I am just unsure as to whether the derivatives being zero equates to
> the derivatives not being defined.
Being zero is perfectly defined, zero being not just
another number, but still a perfectly acceptable one ;-)
> Thanks in advance,
>
> Anthony
>
>
> I still think I struggle with maths more than Einstein, I suspect he
> was just being modest...
I think he really was.
Dirk Vdm
Virgil
anthon...@hotmail.com (anthony burke) wrote:
> Hello, this may seem like a trivial post, but I would be grateful for
> any assistance.
>
> Is it correct that a smooth function is a function that has a
> derivative of all orders, that it can be diferentiated an infinite
> number of times, and for a function of several real variables, the
> definition is that it has all its partial derivatives of all orders?
It will usually mean that, but it may, ocasionally, only mean it has
continuous first derivatives. Which meaning is usually clear from
context
>
> Assuming the above is the case, and I have a function and
> differentiate it to get the zero function, e.g. f(x) = 1, f'(x)=0 am
> I right in thinking that the derivative of this is zero and so on,
> i.e. f''(x)=0, f'''(x) = 0...?
Once you get a zero function, all further derivatives are
automatically zero functions too.
>
> A smooth function has all (partial) derivatives defined, but can these
> be defined as zero, like in the above case? Is f(x) = x^2 a smooth
> function, i.e. f'(x) = 2x, f''(x) = 2, f'''(x) = 0, and all other
> derivatives are zero? Or are smooth functions restricted to functions
> like trig and exponentials etc?
Smooth functions include all polynomial functions (in any number of
variables) and rational functions with no poles (quotients of
polynomials whose divisor polynomials are never zero).
>
> I am just unsure as to whether the derivatives being zero equates to
> the derivatives not being defined.
Zero is perfectly well defined real number, and a function whose
value is always zero is just a special case of a constant function.
Such functions are exremely smooth, since their derivatives of all
orders exist and are always zero functions.
David C. Ullrich
burke) wrote:
>Hello, this may seem like a trivial post, but I would be grateful for
>any assistance.
>
>Is it correct that a smooth function is a function that has a
>derivative of all orders, that it can be diferentiated an infinite
>number of times, and for a function of several real variables, the
>definition is that it has all its partial derivatives of all orders?
The word "smooth" is used with various meanings (check out what
"smooth" means in Zygmund "Trigonometric Series" for example!)
but this is one of the most common meanings, yes.
>Assuming the above is the case, and I have a function and
>differentiate it to get the zero function, e.g. f(x) = 1, f'(x)=0 am
>I right in thinking that the derivative of this is zero and so on,
>i.e. f''(x)=0, f'''(x) = 0...?
Of course.
>A smooth function has all (partial) derivatives defined, but can these
>be defined as zero, like in the above case? Is f(x) = x^2 a smooth
>function, i.e. f'(x) = 2x, f''(x) = 2, f'''(x) = 0, and all other
>derivatives are zero?
Yes.
>Or are smooth functions restricted to functions
>like trig and exponentials etc?
No.
>I am just unsure as to whether the derivatives being zero equates to
>the derivatives not being defined.
No. If f' = 0 then f has a derivative.
Kevin Foltinek
> Is it correct that a smooth function is a function that has a
> derivative of all orders, that it can be diferentiated an infinite
> number of times, and for a function of several real variables, the
> definition is that it has all its partial derivatives of all orders?
As others have pointed out, "smooth" can mean several things,
depending on context; however, the above definition of smooth is not
quite correct (in all the times I have seen the word used). Instead,
there is a subtly different definition: that all of the partial
derivatives of all orders exist *and* are continuous. (For functions
of one variable, the existence of derivatives implies continuity [by
the existence of the next derivative and the fact that
differentiability implies continuity]; however, it makes a big
difference for multivariate functions.)
> I am just unsure as to whether the derivatives being zero equates to
> the derivatives not being defined.
Zero exists just like any other number; the function that is
identically zero is a perfectly acceptable function. If derivative=0
then derivative exists.
Kevin.
David C. Ullrich
<folt...@math.utexas.edu> wrote:
>anthon...@hotmail.com (anthony burke) writes:
>
>> Is it correct that a smooth function is a function that has a
>> derivative of all orders, that it can be diferentiated an infinite
>> number of times, and for a function of several real variables, the
>> definition is that it has all its partial derivatives of all orders?
>
>As others have pointed out, "smooth" can mean several things,
>depending on context; however, the above definition of smooth is not
>quite correct (in all the times I have seen the word used). Instead,
>there is a subtly different definition: that all of the partial
>derivatives of all orders exist *and* are continuous. (For functions
>of one variable, the existence of derivatives implies continuity [by
>the existence of the next derivative and the fact that
>differentiability implies continuity]; however, it makes a big
>difference for multivariate functions.)
Holy cow, what's-his-name is right, I should be fired. I'm willing
to admit the following only because I've recently read that nobody
of any consequence reads sci.math:
Duh. An hour ago if you'd asked me I would have said that while
of course the existence of, say, first-order partials does not
imply continuity, but if you assume _enough_ partials exist that
_does_ imply continuity, so if all the partials exist then they
are all continuous. But that's nonsense. For the benefit of
anyone who knows a little about what sort of things a smooth
function can do, but who nonethless has the same gaping hole
in his understanding that I did (if there is such a person
other than me):
It's easy to give an example. For example, say
V = {(x,y) : |x| <= |y|/2} union {(x,y) : |y| <= |x|/2},
and let
D = {(x,y) : (x,y) <> (0,0) and |x| = |y|}.
It's clear that there exists an f which is smooth
except at the origin and which equals 0 at every
point of V and 1 at every point of D. And now the
fact that f vanishes on V shows that every single
friggin partial of f at the origin exists (and
equals 0); this is clear by induction on the order
of the partial, because all the partials of f vanish
at every point (x,0) with x <> 0 and every point
(0,y) with y <> 0.
What an idiot I've been all my life. Just read a book
"Everything You Know is Wrong" - I think I should send
this to the editor for inclusion in the next edition.
Thanks for clearing _that_ up. Maybe there's some hope -
at least it didn't take me long to decide that your
obviously absurd statement was in fact correct. Oh
well. Remind me not to believe _anything_ that's
obviously true unless I actually know how to prove
it. (And please don't show this to the department
head here at OSU.) I wonder what other idiocies
are lurking in this sorry substitute for a brain.
Let's see, the fundamental theorem of calculus, no
I can prove that...
>> I am just unsure as to whether the derivatives being zero equates to
>> the derivatives not being defined.
>
>Zero exists just like any other number; the function that is
>identically zero is a perfectly acceptable function. If derivative=0
>then derivative exists.
>
>Kevin.
David C. Ullrich
V.Gopal
The process of natural radioactive is described as follows: Number of
atoms decaying at any instant of time (or dN) is equal to the number
of atoms (N) (balance) in the mass or dN=F(N). Although the process
natural radioactive decay is described in terms of number of atoms, as
if 'mass' is atomistic, 'number' that describes the process, is never
used in any calculation. The initial number of atoms is replaced by
'mass'(=1) so that we do not have to restrict number of half lives
(decay continues indefinitely) as if mass is continuous, not
atomistic. The process of natural radioactive decay is represented by
a smooth function - any number of successive differentiation of this
function will not give a constant.
1/2+1/2^2+1/2^3------------1/2^n------up to INFINITY is not a smooth
function. The sum of this series doesnot constitute a continuous mass.
The moment we say that 'duration' (space or time) between consecutive
terms of the above series is constant (constant half-life), the sum of
the series
or the 'mass' becomes continuous and does not remain atomistic. The
process represents an exponential function which is 'time' both, base
and exponent. Every smooth function must be a function of time.
V.Gopal
I do not think 'smooth' can mean several things. It is associated with
our sense of touch of, say, an edge or a surface. Therefore we have to
associate
'smooth' with CONTINUOUS lines and surfaces of geometry. A continuous
straight line or a plain flat surface (a plane in geometry) is smooth.
A continuous curve (open or closed) is smooth. A parabola or sphere is
smooth. But any discontinuous line or surface is not 'smooth' although
it may be possible to CONSTRUCT a 'smooth' line or surface by putting
them together. But the process
of 'putting things together' is in expressible in symbolic
mathematics. This is the reason why we cannot consider a hyperbola
expressed as XY=1 is smooth. Here we assign values to X and calculate
the value of Y. By this process we can never fill all the points (they
will be infinite in number) and make the curve 'smooth'. The values of
TanA (A= 0 to 90 degrees) cannot become continuous by joining (by
putting together) different values of TanA by a line. If we want to
make values of TanA continuous then we have to express TanA as the
function of time. TanA as the function of time would allow us to draw
all the open curves but not any closed curve, like a circle or an
ellipse. The formula for all possible closed curve
is(dA/dL)*(dL/dT)=constant (here dA is differential of 'angle' and
dA/dL is instantaneous curvature). The expression represents Keppler's
law. The 'smooth'is an integral constituted of a continuous variable
in space-time (we can never know when a pleasent warmth has become a
painful heat) or something always constant in time (sleep), both are
inexpressible.
Kevin Foltinek
> Just read a book "Everything You Know is Wrong"
You sure about that?
Kevin.
David C. Ullrich
<folt...@math.utexas.edu> wrote:
Yes. If you assume enough (mixed) partials exist
then it's easy to concoct an argument showing that
f is continuous, and it follows that I just read
that book.
>Kevin.
David C. Ullrich
David C. Ullrich
wrote:
>Kevin Foltinek <folt...@math.utexas.edu> wrote in message news:<wo6n0qf...@linux60.ma.utexas.edu>...
>> anthon...@hotmail.com (anthony burke) writes:
>>
>> > Is it correct that a smooth function is a function that has a
>> > derivative of all orders, that it can be diferentiated an infinite
>> > number of times, and for a function of several real variables, the
>> > definition is that it has all its partial derivatives of all orders?
>>
>> As others have pointed out, "smooth" can mean several things,
>> depending on context; however, the above definition of smooth is not
>> quite correct (in all the times I have seen the word used). Instead,
>> there is a subtly different definition: that all of the partial
>> derivatives of all orders exist *and* are continuous. (For functions
>> of one variable, the existence of derivatives implies continuity [by
>> the existence of the next derivative and the fact that
>> differentiability implies continuity]; however, it makes a big
>> difference for multivariate functions.)
>>
>> > I am just unsure as to whether the derivatives being zero equates to
>> > the derivatives not being defined.
>>
>> Zero exists just like any other number; the function that is
>> identically zero is a perfectly acceptable function. If derivative=0
>> then derivative exists.
>>
>> Kevin.
>
>I do not think 'smooth' can mean several things.
Doesn't matter whether you think that or not. In mathematics
words mean _exactly_ what their definitions say they mean,
no more and no less. And the word "smooth" is given various
different definitions in different mathematical contexts.
David C. Ullrich
Bob Pease
"V.Gopal" <vgop...@rediffmail.com> wrote in message
news:38af3945.02091...@posting.google.com...
According to Wolfram
Smooth Function A smooth function is a function that has continuous
second-order derivatives over some domain. A function can therefore be said
to be smooth over a restricted interval such as (a, b) or [a, b].
http://mathworld.wolfram.com/SmoothFunction.html,
Unfortunately a math definition cannot agree with everyone's unique vision
of the context of the word.
RJ Pease
Kevin Foltinek
> On 19 Sep 2002 11:21:38 -0500, Kevin Foltinek
> <folt...@math.utexas.edu> wrote:
>
> >David C. Ullrich <ull...@math.okstate.edu> writes:
> >
> >> Just read a book "Everything You Know is Wrong"
> >
> >You sure about that?
>
> Yes. If you assume enough (mixed) partials exist
> then it's easy to concoct an argument showing that
> f is continuous, and it follows that I just read
> that book.
Aren't you using the Axiom of Choice in that argument?
Kevin.
Kevin Foltinek
> I do not think 'smooth' can mean several things.
Your (evidently uninformed) opinion is irrelevant to the consensus
definitions used by the mathematical community.
> It is associated with
> our sense of touch of, say, an edge or a surface. Therefore we have to
> associate
> 'smooth' with CONTINUOUS lines and surfaces of geometry.
Sandpaper is continuous. Go rub 60 grit paper for an hour and then
tell us if continuous things are smooth.
> [snip decreasingly meaningful sequence of words]
Kevin.
Dirk Van de moortel
"Kevin Foltinek" <folt...@math.utexas.edu> wrote in message news:wo6hegl...@linux60.ma.utexas.edu...
Absolute value abs(x) is not smooth but it looks a *lot*
smoother than y(x) = (1000000*x)^2 which *is* smooth.
I'll sit on the abs peak, you can take the other one.
Dirk Vdm
Dave Rusin
Bob Pease <bobp...@concentric.net> wrote:
>According to Wolfram
>Smooth Function A smooth function is a function that has continuous
>second-order derivatives over some domain. A function can therefore be said
>to be smooth over a restricted interval such as (a, b) or [a, b].
>http://mathworld.wolfram.com/SmoothFunction.html,
>
>Unfortunately a math definition cannot agree with everyone's unique vision
>of the context of the word.
Nor, apparently, need it agree with mathematicians' common "visions" of
the word. Which is the mathematical community which would naturally take
"smooth" to mean C^2? I would have expected C^1 and C^\infty to be more
natural interpretations. I guess the family of C^2 functions is
technically convenient for some tasks, but there are others which
work better in other settings.
I like the MathWorld website a lot, but it's sort of unfortunate that
some people believe it represents a practicing mathematician's
understanding of the discipline. Sometimes yes, sometimes no.
Ours is at least as specialized discipline as, say, law; but while
some people have gotten into the habit of writing "IANAL but...", they
seem less willing to write "I am not a mathematician but...".
dave
Kevin Foltinek
> Absolute value abs(x) is not smooth but it looks a *lot*
> smoother than y(x) = (1000000*x)^2 which *is* smooth.
Not when I looked at it. Of course, I chose my axis scaling
intelligently.
(Hint: the smoothness property is invariant under scaling of either
axis.)
Kevin.
Dirk Van de moortel
"Kevin Foltinek" <folt...@math.utexas.edu> wrote in message news:wo6admd...@linux60.ma.utexas.edu...
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> writes:
>
> > Absolute value abs(x) is not smooth but it looks a *lot*
> > smoother than y(x) = (1000000*x)^2 which *is* smooth.
>
> Not when I looked at it. Of course, I chose my axis scaling
> intelligently.
Normal bottoms prefer abs(x).
Nanobottoms prefer (1000000*x)^2.
>
> (Hint: the smoothness property is invariant under scaling of either
> axis.)
>
> Kevin.
Perhaps I should have added a ";-)" in my previous post.
I will add one now.
Look carefully:
;-)
Dirk Vdm
Bob Pease
"Dave Rusin" <ru...@vesuvius.math.niu.edu> wrote in message
news:amdh1g$3g0$1...@news.math.niu.edu...
> In article <amdd66$1...@dispatch.concentric.net>,
> Bob Pease <bobp...@concentric.net> wrote:
>
> >According to Wolfram
> >Smooth Function A smooth function is a function that has continuous
> >second-order derivatives over some domain. A function can therefore be
said
> >to be smooth over a restricted interval such as (a, b) or [a, b].
> >http://mathworld.wolfram.com/SmoothFunction.html,
> >
> >Unfortunately a math definition cannot agree with everyone's unique
vision
> >of the context of the word.
>
> Nor, apparently, need it agree with mathematicians' common "visions" of
> the word. Which is the mathematical community which would naturally take
> "smooth" to mean C^2? I would have expected C^1 and C^\infty to be more
> natural interpretations. I guess the family of C^2 functions is
> technically convenient for some tasks, but there are others which
> work better in other settings.
To reprise myself,
Ya gotta form a consensus, or at least a truce.
James and James gives a working definition that is more precise than
mathworld.
It also might be less useful to a person studying CalcIII who just needs
some guideline.
> I like the MathWorld website a lot, but it's sort of unfortunate that
> some people believe it represents a practicing mathematician's
> understanding of the discipline. Sometimes yes, sometimes no.
> Ours is at least as specialized discipline as, say, law; but while
> some people have gotten into the habit of writing "IANAL but...", they
> seem less willing to write "I am not a mathematician but...".
I don't know that we have a disagreement here.
Sometimes, however, flexibility is a need when communicating among varying
levels of specialization or , indeed, among different levels of expertise.
RJ Pease
David C. Ullrich
<folt...@math.utexas.edu> wrote:
>David C. Ullrich <ull...@math.okstate.edu> writes:
>
>> On 19 Sep 2002 11:21:38 -0500, Kevin Foltinek
>> <folt...@math.utexas.edu> wrote:
>>
>> >David C. Ullrich <ull...@math.okstate.edu> writes:
>> >
>> >> Just read a book "Everything You Know is Wrong"
>> >
>> >You sure about that?
>>
>> Yes. If you assume enough (mixed) partials exist
>> then it's easy to concoct an argument showing that
>> f is continuous, and it follows that I just read
>> that book.
>
>Aren't you using the Axiom of Choice in that argument?
It's clear that AC _follows_ from the result that if
f has partials of all orders then f is continuous.
>Kevin.
David C. Ullrich
David C. Ullrich
wrote:
>In article <amdd66$1...@dispatch.concentric.net>,
>Bob Pease <bobp...@concentric.net> wrote:
>
>>According to Wolfram
>>Smooth Function A smooth function is a function that has continuous
>>second-order derivatives over some domain. A function can therefore be said
>>to be smooth over a restricted interval such as (a, b) or [a, b].
>>http://mathworld.wolfram.com/SmoothFunction.html,
>>
>>Unfortunately a math definition cannot agree with everyone's unique vision
>>of the context of the word.
>
>Nor, apparently, need it agree with mathematicians' common "visions" of
>the word. Which is the mathematical community which would naturally take
>"smooth" to mean C^2?
My favorite is Zygmund "Trigonometric Functions", where smooth
functions are functions in what everyone else calls the
"little-oh Zugmund class" (f is continuous and
(f(x+h) - 2 f(x) + f(x-h))/h -> 0 as h -> 0.
No typo, it's h, not h^2. This is more than continuous but
much less than C^1; just happens to be something with convenient
characterizations in terms of Fourier series.)
>I would have expected C^1 and C^\infty to be more
>natural interpretations. I guess the family of C^2 functions is
>technically convenient for some tasks, but there are others which
>work better in other settings.
>
>I like the MathWorld website a lot, but it's sort of unfortunate that
>some people believe it represents a practicing mathematician's
>understanding of the discipline. Sometimes yes, sometimes no.
>Ours is at least as specialized discipline as, say, law; but while
>some people have gotten into the habit of writing "IANAL but...", they
>seem less willing to write "I am not a mathematician but...".
>
>dave
David C. Ullrich
Jon Miller
>On 17 Sep 2002 14:26:20 -0700, anthon...@hotmail.com (anthony
>burke) wrote:
>
>>Assuming the above is the case, and I have a function and
>>differentiate it to get the zero function, e.g. f(x) = 1, f'(x)=0 am
>>I right in thinking that the derivative of this is zero and so on,
>>i.e. f''(x)=0, f'''(x) = 0...?
>>
>
>Of course.
>
Assuming the OP meant what he wrote, and that the derivatives are the 0
function, and not that the value of the derivative at x just happens to
be 0.
Jon Miller
Kevin Foltinek
> Perhaps I should have added a ";-)" in my previous post.
No need for the smiley; I was being deliberately over-serious.
Kevin.
Kevin Foltinek
> On 19 Sep 2002 15:47:07 -0500, Kevin Foltinek
> <folt...@math.utexas.edu> wrote:
>
> >David C. Ullrich <ull...@math.okstate.edu> writes:
> >
> >> Yes. If you assume enough (mixed) partials exist
> >> then it's easy to concoct an argument showing that
> >> f is continuous, and it follows that I just read
> >> that book.
> >
> >Aren't you using the Axiom of Choice in that argument?
>
> It's clear that AC _follows_ from the result that if
> f has partials of all orders then f is continuous.
I'm pretty sure that the above is true only if you assume that you did
in fact read the book.
It seems, then, that AC is equivalent to you having read the book
"Everything You Know is Wrong".
In retrospect, this should not have been at all surprising; the title
of the book suggests explicitly assigning a truth value to the set of
things you know, but implicit is the set of all consequences of things
you know, and according to my copy of the book "Choice for Complete
Idiots", discussing the consequences involves making some choices.
Also, choices have consequences. QED.
Kevin.
David C. Ullrich
<folt...@math.utexas.edu> wrote:
Hard to argue with logic like that.
>Kevin.
David C. Ullrich
V.Gopal
continuous second order derivatives over some domain. What I insist is
that the function that you are talking about - the function with
CONTINUOUS second order DERIVATIVE - cannot have any description
within a RESTRICTED INTERVAL - it cannot be expressed or conveyed by
any symbolic expression. A continuous line
of finite length cannot have a numerical/symbolic expression. I cannot
understand whether such a description of 'Continuous Function' makes
any sense.
Bob Pease
take it up with the author of the definition.
>
>What I insist is
> that the function that you are talking about - the function with
> CONTINUOUS second order DERIVATIVE - cannot have any description
> within a RESTRICTED INTERVAL - it cannot be expressed or conveyed by
> any symbolic expression. A continuous line
> of finite length cannot have a numerical/symbolic expression. I cannot
> understand whether such a description of 'Continuous Function' makes
> any sense.
Please refer to James and James for a more precise definition.
RJ P
V.Gopal
"Sand paper is continuous. Go, rub 60 grit paper for an hour and then
tell us if continuous things are smooth".
You are 'measuring' smoothness in the scale of your size. Einstein
introduced the idea of 'Thought Experiment'. Use it and think yourself
riding on a wheel of radius equal to the size of a nucleus of an atom
and you move at infinitely low speed (you must always be in touch with
what makes YOUR sand-paper continuous. IF the sand paper IS continuous
like any continuous line in geometry
then the sand paper is definitely smooth.
This helps us to define theoretical 'smoothness'. 'Smoothness' is a
quality created or measured by a point in motion. A Continuously
moving 'heavy' point-mass gennerates a smooth line. A line is not
smooth if and only if the point STOPS MOVING at least for a
moment(comes to a state of rest in perfect agreement with Newton's
law) then changes its direction (suddenly?) and begins to move again
(according to Newton's law). Even if speed and direction change
continuously the line remains 'smooth'and always has a curvature
without an angle. Any angle would destroy 'smoothness'.(Mass has
momentum and does not allow a sudden change in direction).
Instantaneous velocity is dL/dT=TanA. Therefore TanA, expressed as a
function of time generates a 'smooth function'. Here the 'curve' is
without any angle, it always has a curvature that makes it smooth. Tan
A, as a function of time is a straight line even if speed changes but
direction does not change and even when both change.
L vs T 'graph' for motion under gravity is is a logarithmic scale
which we cannot describe.
When 'rate of change' of velocity or frequency or temperature is
'uniform' then
we can never know EXACTLY at what point of space or time motion
bbecomes aceleration becomes frightening or painful.
I have started discussions on "Quantification in Physics and its
consequences" "Does Zero exist?" and "Zeno's paradox - why it is
wrong?". I request all serious mathematicians to join these
discussions.
Bob Pease
What makes you think that serious people would join you at all?
Clearly the Popester is not usually serious, except when it gets so deep
that you need a shovel and hip boots!!!
Rj Pease
Curator
Dobbs University
Kevin Foltinek
> This helps us to define theoretical 'smoothness'.
> [...] Continuously moving 'heavy' point-mass [...] Newton's law
> [...] Mass [...] momentum [...] time [...] gravity [...]
> velocity or frequency or temperature [...] space or time [...]
All of this belongs in the subject of the use of mathematics in the
modelling of our physical universe, and is irrelevant to the
mathematical definitions themselves.
Kevin.