The following functions/activities/qualities/processess are smooth: 1) Continuous unchanging information - duration of the state of equilibrium, uniform velocity, a thing expressed directly as the function of time, when a thing is an element of itself, a thing expressed as the function of itself e.g. dX=f(X)and X=F(dX), the idea of a fluid like eather, 'space' in geometry, time, SLEEP without any dream, silence of the mind like meditation. All have internal contiguity and have no parts (smooth is continuous and partless). Anything whose first derivative is constant or zero is smooth. Any number can be used to specify its physical (space-time) dimension. The smooth cannot have a description of its own. We cannot associate reason or cause with any of these. The following also are smooth - Continuously and (and orderly) change in information involving a finite space-time: It involves continuous change in reference and continuous forgetting of the past. State of change, a thing/activity expressed as the function of itself, the formula for making a zero-error prediction, thinking in order to find the means of achieving a goal, MEDITATION, dX=F(X) but X is not F(dX), elongation, explosion, implosion, a thing/activity expressed indirectly as the continuous function of time like radioactive decay, chain reaction, creation, motion under gravity, uniform acceleration, a field of influence, PV=RT. We ahve to associate cause and effect with these functions and activities. These smooth functions and activities are incommunicable. Smoothness is is lost if there is discontinuity, if there are space-corpuscles in space and time corpuscles in time, if state of change is replaced by series of equilibrium states. PV=constant is not smooth (but PV=RT is smooth) A smooth function/activity has to be understood by psychophysical parallelism, intellectual sympathy, introspection and meditation. Everything in nature is smooth. No language is smooth. Nature is meant to enhance our capacity for sympathy, introspection and meditation. We can realize the exictence of God only by understanding 'The smooth'.
> A smooth function/activity has to be understood by psychophysical > parallelism, intellectual sympathy, introspection and meditation. > Everything in nature is smooth. No language is smooth.
[snip]
Learn some math.
Cargo cultism - whether it is some idiot native constructing a thatched hut to emulate an airfield control tower or some idiot New Age wombat screaming "look at meeeee!" - is a null set. The universe doesn't care what you think, (heck, neither do we). OTOH, if you do something clever with geometry, algebra follows in lockstop via Euler's equation.
You aren't clever. Shannonizing technical vocabulary in hopes of a hit has the same statistics as Saddamn Hussein lighting up Bagdhad skies with anti-aircraft fire during the first night of Desert Storm. The real guardians of civilization flew in, bombed the living shit out of anything they desired, and flew out with a single casualty. No Stealth aircraft was hit in 1271 missions. Does Allah have execrable aim? Test of faith!
-- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) "Quis custodiet ipsos custodes?" The Net!
In article <38af3945.0209301116.77a62...@posting.google.com>,
V.Gopal <vgopa...@rediffmail.com> wrote: >A smooth function/activity has to be understood by psychophysical >parallelism, intellectual sympathy, introspection and meditation.
df/dx exists along the domain of the function.
-- "A nice adaptation of conditions will make almost any hypothesis agree with the phenomena. This will please the imagination but does not advance our knowledge." -- J. Black, 1803.
> The following functions/activities/qualities/processess are smooth: > 1) Continuous unchanging information - duration of the state of > equilibrium, uniform velocity, a thing expressed directly as the > function of time, when a thing is an element of itself, a thing > expressed as the function of itself e.g. dX=f(X)and X=F(dX), the idea > of a fluid like eather, 'space' in geometry, time, SLEEP without any > dream, silence of the mind like meditation. All have internal > contiguity and have no parts (smooth is continuous and partless). > Anything whose first derivative is constant or zero is smooth. Any > number can be used to specify its physical (space-time) dimension. The > smooth cannot have a description of its own. We cannot associate > reason or cause with any of these. > The following also are smooth - Continuously and (and orderly) change > in information involving a finite space-time: It involves continuous > change in reference and continuous forgetting of the past. > State of change, a thing/activity expressed as the function of itself, > the formula for making a zero-error prediction, thinking in order to > find the means of achieving a goal, MEDITATION, dX=F(X) but X is not > F(dX), elongation, explosion, implosion, a thing/activity expressed > indirectly as the continuous function of time like radioactive decay, > chain reaction, creation, motion under gravity, uniform acceleration, > a field of influence, PV=RT. We ahve to associate cause and effect > with these functions and activities. These smooth functions and > activities are incommunicable. > Smoothness is is lost if there is discontinuity, if there are > space-corpuscles in space and time corpuscles in time, if state of > change is replaced by series of equilibrium states. PV=constant is not > smooth (but PV=RT is smooth) > A smooth function/activity has to be understood by psychophysical > parallelism, intellectual sympathy, introspection and meditation. > Everything in nature is smooth. No language is smooth. Nature is meant > to enhance our capacity for sympathy, introspection and meditation. We > can realize the exictence of God only by understanding 'The smooth'.
You have a mild attack of verbal diarrhoea. It is not fatal, but it stinks.
Bryan Reed wrote: > In article <RN3m9.140174$8o4.20...@afrodite.telenet-ops.be>, > Dirk Van de moortel <dirkvandemoor...@ThankS-NO-SperM.hotmail.com> wrote: >>"V.Gopal" <vgopa...@rediffmail.com> wrote in message >>> A smooth function >> is a function with a continuous derivative.
> In the math classes I took, it was a function where all derivatives are > continuous. But of course usage varies with context.
It's interesting that C-infinity functions are often described as "infinitely many times continuously differentiable" when just "infinitely many times differentiable" is sufficient (the existence of the n+1-st derivative implies the continuity of the n-th derivative). Not that there's anything wrong with that. It just seems that, to an already awkward phrase, you wouldn't want to add words you don't have to add.
"Bryan Reed" <bwr...@u.washington.edu> wrote in message news:analvd$22u2$1@nntp6.u.washington.edu... > In article <RN3m9.140174$8o4.20...@afrodite.telenet-ops.be>, > Dirk Van de moortel <dirkvandemoor...@ThankS-NO-SperM.hotmail.com> wrote:
On Mon, 30 Sep 2002 19:45:44 GMT, Uncle Al <Uncle...@hate.spam.net> wrote in <3D98A9DC.698E8...@hate.spam.net> that:
>Cargo cultism -
I've seen you more than once uttering this. The notion itself says you're stupid.
You misunderstood that "cargo" thing. You misinterpreted their intention and incentives. You did the same mistake (together with that jackass "Feynman") that your archaeologists do everyday describing why there are ornaments and even food left with the dead of 5000 years back, etc.
There are some aspects of humanities that Westerners do not qualify to bother with. Don't exhibit that void then all the time, it doesn't look good.
On Mon, 30 Sep 2002 19:29:43 -0700, Mike Oliver <oli...@math.ucla.edu> wrote:
>Bryan Reed wrote: >> In article <RN3m9.140174$8o4.20...@afrodite.telenet-ops.be>, >> Dirk Van de moortel <dirkvandemoor...@ThankS-NO-SperM.hotmail.com> wrote: >>>"V.Gopal" <vgopa...@rediffmail.com> wrote in message >>>> A smooth function >>> is a function with a continuous derivative.
>> In the math classes I took, it was a function where all derivatives are >> continuous. But of course usage varies with context.
>It's interesting that C-infinity functions are often described >as "infinitely many times continuously differentiable" when >just "infinitely many times differentiable" is sufficient >(the existence of the n+1-st derivative implies the continuity >of the n-th derivative).
I found out recently that I've been an idiot about exactly this point all my life. In several variables the fact that f has partial derivatives of all orders at every point does not imply that f is continuous.
(Simple example: It's not hard to see that there exists a function in R^2 which is smooth (all partials continuous) except at the origin, which vanishes at the origin and in a neighborhood of (coordinate axes minus the origin) and which equals 1 on (diagonal minus the origin). It follows that all the partials of f at the origin equal 0 although f is not continuous.)
>Not that there's anything wrong >with that. It just seems that, to an already awkward phrase, >you wouldn't want to add words you don't have to add.
>In the math classes I took, it was a function where all derivatives are >continuous. But of course usage varies with context.
>Bryan
If just the first derivative is continuous, that's enough to ensure there are no corners in the function.
-- "A nice adaptation of conditions will make almost any hypothesis agree with the phenomena. This will please the imagination but does not advance our knowledge." -- J. Black, 1803.
> Well, I know some people think smooth functions are more sexy then ones that > aren't...
If this posting has made so many angry and go out of track, I am sure I make sense. In any case let me know whether PV=RT and PV=ocnstant both are smooth OR only PV=RT is smooth and PV=constant (series of equilibrium states) not smooth. If you make a distinction how did you make it. I hope I will get an answer and not angry words.
Mike Oliver <oli...@math.ucla.edu> writes: > It's interesting that C-infinity functions are often described > as "infinitely many times continuously differentiable" when > just "infinitely many times differentiable" is sufficient > (the existence of the n+1-st derivative implies the continuity > of the n-th derivative). Not that there's anything wrong > with that. It just seems that, to an already awkward phrase, > you wouldn't want to add words you don't have to add.
The C^infty functions are described either as "infinitely many times differentiable" or "having infinitely many continuous partial derivatives". Differentiability implies existence of partial derivatives, but the converse is false; however, the existence of continuous partial derivatives implies differentiability, so the latter description implies the former (and, as you note, the former implies continuous differentiability which implies the latter).
> There are some aspects of humanities that Westerners do > not qualify to bother with.
Oh, sure, if you want it that way. I'll keep my nose out of feces smeared, rat infested, half starved coddling non-western business, if they will just never put their smelly, dirty fingernailed, disease ridden bodies anywhere that has anything to do with western things. Like, say, computer networks. Or modern metallurgy. Or modern medicine. Or airplanes. Or automobiles. Or television. Socks
"David C. Ullrich" wrote: > On Mon, 30 Sep 2002 19:29:43 -0700, Mike Oliver <oli...@math.ucla.edu> > wrote: >> It's interesting that C-infinity functions are often described >> as "infinitely many times continuously differentiable" when >> just "infinitely many times differentiable" is sufficient >> (the existence of the n+1-st derivative implies the continuity >> of the n-th derivative).
> I found out recently that I've been an idiot about exactly > this point all my life. In several variables the fact that > f has partial derivatives of all orders at every point > does not imply that f is continuous.
That hadn't occurred to me; I was thinking in terms of functions of a single variable. For a function of several variables I suppose the analogous thing would not be just that all partial derivatives exist but that all directional derivatives do.
Mike Oliver <oli...@math.ucla.edu> writes: >"David C. Ullrich" wrote:
>> I found out recently that I've been an idiot about exactly >> this point all my life. In several variables the fact that >> f has partial derivatives of all orders at every point >> does not imply that f is continuous.
>That hadn't occurred to me; I was thinking in terms of functions >of a single variable. For a function of several variables I suppose >the analogous thing would not be just that all partial derivatives >exist but that all directional derivatives do.
I believe you suppose wrong (in dimension 2 and up). I further believe (with better reason) that the correct analogous thing would be that the "total derivative" exist, and that *its* total derivative exist, and so on ad infinitum. "Partial derivatives" belong where you have put them, in the same context as "several variables"--but the context of a single, Banach-space (say) variable is a better context, and there (total) derivatives are much more natural than "partial" (or even "directional") derivatives.
Lee Rudolph wrote: > Mike Oliver <oli...@math.ucla.edu> writes: >> "David C. Ullrich" wrote: >>> I found out recently that I've been an idiot about exactly >>> this point all my life. In several variables the fact that >>> f has partial derivatives of all orders at every point >>> does not imply that f is continuous.
>> That hadn't occurred to me; I was thinking in terms of functions >> of a single variable. For a function of several variables I suppose >> the analogous thing would not be just that all partial derivatives >> exist but that all directional derivatives do.
> I believe you suppose wrong (in dimension 2 and up). I further > believe (with better reason) that the correct analogous thing > would be that the "total derivative" exist, and that *its* > total derivative exist, and so on ad infinitum. "Partial > derivatives" belong where you have put them, in the same > context as "several variables"--but the context of a single, > Banach-space (say) variable is a better context, and there > (total) derivatives are much more natural than "partial" > (or even "directional") derivatives.
What's a total derivative? Basically the gradient, generalized to tensor fields?
>Lee Rudolph wrote: >> Mike Oliver <oli...@math.ucla.edu> writes: >>> "David C. Ullrich" wrote: >>>> I found out recently that I've been an idiot about exactly >>>> this point all my life. In several variables the fact that >>>> f has partial derivatives of all orders at every point >>>> does not imply that f is continuous.
>>> That hadn't occurred to me; I was thinking in terms of functions >>> of a single variable. For a function of several variables I suppose >>> the analogous thing would not be just that all partial derivatives >>> exist but that all directional derivatives do.
>> I believe you suppose wrong (in dimension 2 and up). I further >> believe (with better reason) that the correct analogous thing >> would be that the "total derivative" exist, and that *its* >> total derivative exist, and so on ad infinitum. "Partial >> derivatives" belong where you have put them, in the same >> context as "several variables"--but the context of a single, >> Banach-space (say) variable is a better context, and there >> (total) derivatives are much more natural than "partial" >> (or even "directional") derivatives.
>What's a total derivative? Basically the gradient, generalized >to tensor fields?
It's clear that Lee is right about the total derivatives being the correct analogue. Also seems very likely that he's right about directional derivatives not being enough - I'm not sure whether I'll come up with a counterexample before hitting the Send button.
If f'(x) = a in one variable that says that
f(x + h) = f(x) + ah + o(h).
The total derivative is precisely analogous to the derivative in one variable, _when_ you look at it that way. Say f : R^n -> R^m. f is differentiable at x, with derivative D, if there exists a linear map D : R^n -> R^m such that
f(x + h) = f(x) + Dh + o(h).
Making it clear that for example, a differentiable function must be continuous. If f has continuous partial derivatives then it is differentiable; otoh if f is differentiable at x then it has partials at x; the partials are the same as the entries in the matrix representing the derivative.
Ok. It's not immediately obvious to me how to give an example of a function that has directional derivatives at every point, and such that the directional derivatives all have directional derivatives, etc, without being continuous. It's "clear" that such examples exist (I hope I'm wrong about that, would give me an excuse for the idiocy mentioned above.)
But it's easy to give an example of a function on R^2 that has directional derivatives at every point although it's not continuous. The function is going to be smooth everywhere except at the origin and satisfy 0 <= f <= 1 everywhere.
Say C is (union of two circles minus the origin), where the circles have radius 1 and centers at (0,1) and (0,-1). We're going to have f(x,0) = 0 for all x. The plane minus the x-axis is the union of rays from the origin, each of which intersects C in exactly one point. On each ray f is smooth; f = 1 at the point of the ray that intersects C, and f -> 0 "fast" as you approach the origin along each ray.
On each ray there's a little bump at the point the ray intersects C, and f = 0 on that ray except near the bump. The bumps tend to the origin as the ray approaches the x-axis, so f remains smooth on the x-axis. f is smooth everywhere except at the origin, and f has directional derivatives at the origin since the restriction to each ray is smooth, but f is not continuous at the origin, since f(0,0) = 0 but f = 1 on C.
"David C. Ullrich" wrote: > Say C is (union of two circles minus the origin), > where the circles have radius 1 and centers at > (0,1) and (0,-1). We're going to have f(x,0) = 0 > for all x. The plane minus the x-axis is the > union of rays from the origin, each of which > intersects C in exactly one point. On each ray > f is smooth; f = 1 at the point of the ray that > intersects C, and f -> 0 "fast" as you approach > the origin along each ray.
> On each ray there's a little bump at the point > the ray intersects C, and f = 0 on that ray > except near the bump. The bumps tend to the > origin as the ray approaches the x-axis, so > f remains smooth on the x-axis. f is smooth > everywhere except at the origin, and f has > directional derivatives at the origin since > the restriction to each ray is smooth, but > f is not continuous at the origin, since > f(0,0) = 0 but f = 1 on C.
It looks pretty good, but I'm missing just what's happening on the y-axis.
>> Say C is (union of two circles minus the origin), >> where the circles have radius 1 and centers at >> (0,1) and (0,-1). We're going to have f(x,0) = 0 >> for all x. The plane minus the x-axis is the >> union of rays from the origin, each of which >> intersects C in exactly one point. On each ray >> f is smooth; f = 1 at the point of the ray that >> intersects C, and f -> 0 "fast" as you approach >> the origin along each ray.
>> On each ray there's a little bump at the point >> the ray intersects C, and f = 0 on that ray >> except near the bump. The bumps tend to the >> origin as the ray approaches the x-axis, so >> f remains smooth on the x-axis. f is smooth >> everywhere except at the origin, and f has >> directional derivatives at the origin since >> the restriction to each ray is smooth, but >> f is not continuous at the origin, since >> f(0,0) = 0 but f = 1 on C.
>It looks pretty good, but I'm missing just what's >happening on the y-axis.
The positive half of the y-axis is a ray just like all the others (except the ones on the x-axis): It intersects C in one point (0,2) and the restriction of f to that ray has a bump at that point and vanishes except close to that point.
"David C. Ullrich" wrote: > The positive half of the y-axis is a ray just like > all the others (except the ones on the x-axis): > It intersects C in one point (0,2) and the restriction > of f to that ray has a bump at that point and > vanishes except close to that point.
Oh, I see. I had the circles set horizontally instead of vertically.
> A smooth function/activity has to be understood by psychophysical > parallelism, intellectual sympathy, introspection and meditation. > Everything in nature is smooth. No language is smooth. Nature is meant > to enhance our capacity for sympathy, introspection and meditation. We > can realize the exictence of God only by understanding 'The smooth'.
> In article <38af3945.0209301116.77a62...@posting.google.com>, > V.Gopal <vgopa...@rediffmail.com> wrote:
> >A smooth function/activity has to be understood by psychophysical > >parallelism, intellectual sympathy, introspection and meditation.
> df/dx exists along the domain of the function.
And the domain doesn't have any holes.
(Searching for a quibble).
Otherwise you could have
y = 0 (x < 0) y = x (x > 0) y undefined (x = 0)
One might also quible about functions like y = sin(1/x)/x
The function is continuous everywhere and the derivative exists everywhere, but if you rode the function into x = 0, you would take an infinitely nauseating roller coster ride.
Edward Green <nullde...@aol.com> wrote: >glhan...@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message ><news:anaag8$avk$3@rainier.uits.indiana.edu>... >> In article <38af3945.0209301116.77a62...@posting.google.com>, >> V.Gopal <vgopa...@rediffmail.com> wrote:
>> >A smooth function/activity has to be understood by psychophysical >> >parallelism, intellectual sympathy, introspection and meditation.
I do like "derivative is continous" better, that's probably the real definition of smooth. I was figuring that at a sharp corner, like x=0 for y=|x|, the derivative is ambiguous. Approach it from the right and it's 1, approach it from the left and it's -1, but right at x=0 there is no derivative. Same thing, I suppose, unless you carefully define
f(x) = -x for x >= 0 = x for x < 0
"Greater than or equals" removes the ambiguity.
>One might also quible about functions like y = sin(1/x)/x
>The function is continuous everywhere and the derivative exists >everywhere, but if you rode the function into x = 0, you would take an >infinitely nauseating roller coster ride.
-- "A nice adaptation of conditions will make almost any hypothesis agree with the phenomena. This will please the imagination but does not advance our knowledge." -- J. Black, 1803.
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