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From: Ostap S. B. M. Bender Jr. on 17 Jan 2010 02:20
On Jan 13, 4:06 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
> Hi Mr. Eastham.
>
> On Jan 13, 3:21 pm, Chip Eastham <hardm...(a)gmail.com> wrote:
>
>
>
> > On Jan 13, 12:45 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
>
> > > To Dr. Lewis et al,
>
> > > On Jan 13, 8:12 am, "Robert H. Lewis" <rle...(a)fordham.edu> wrote:
>
> > > > > > >${-oo,+oo} F(x) (dx)^.5   (1)
>
> > > > > > Where did you run across this? The notation makes very little
> > > > > > sense.
> > > > > Jays doing just fine, David I think you need to take a basic
> > > > > calculus course, let me explain (using Jay's example), that I
> > > > > frequently encounter.
>
> > > > > V = dX/dT = Velocity.
> > > > > sqrt(V) = sqrt (dX/dT) , sqrt(dX) = sqrt(V)*sqrt(dT).
> > > > > That's SOP in calculus, from that a bit of algebraic massage
> > > > > produces *generally* a means to solve.
> > > > > Regards
> > > > > Ken S. Tucker
>
> > > >   SOP in Calculus?  Elementary course?   No way!
>
> > > >  I have been teaching calculus at all levels for close to 30 years, and I have never seen \sqrt(dx).  It is absurd on the face of it.  Perhaps some physicists have come up with this and found a coherent usage?
>
> > > Dr. Lewis, I too have taught and refined Calculus and very much
> > > respect the reasoning and notation conventions.
> > > I think it's reasonable to examine the 'differential coefficents',
> > > apart from the ratio, let's start here,
>
> > > V=dX/dT, sqrt(V)= sqrt(dX/dT)
>
> > > that's an expression of V by differential coefficients,
> > > The apparent confusion results from writing the above as,
>
> > > sqrt(V) = sqrt(dX)/sqrt(dT)   , THE STEP.
>
> > > Here we ask, does THE STEP represent a clear definition
> > > of symbolic logic?
>
> Above, I'm working on getting a handle on 'what is the
> problem'.
>
> > No, it is not clear by any definition of
> > derivative.
>
> I'll need to clarify, that we're a bit deeper than
> derivatives, specifically 'differential co-efficents'
> then a bit beyond that, a respected mathematician
> Herman Weyl was fond of.
>
> > The casual treatment of the
> > derivative dX/dT as a ratio of two values
> > mixed with taking a square root in the
> > numerator and denominator is lacking in
> > logical justification.
>
> Good enough!
> Allow me to move to integration ($) and then pose,
>
> dU = dV*dV , dV = sqrt(dU)
>

What is 'dV*dV'? A function? A number? A bird? A mammal? A mineral?

>
> and then moving to integration, this is expected to
> be a logical question, can we solve,
>
> $ dU = $$ dV dV  ,   THE 2nd STEP.
>
> I'm stop pending inbounds.
>
> > I'd be happy of course to see such a
> > framework, but it falls outside the
> > experiences my life as a simple country
> > mathematician have afforded.
> > regards, chip
>
> Bet I'm a simpler country mathafella, care to
> discuss probability.
> Regards
> Ken S. Tucker

From: Ken S. Tucker on 17 Jan 2010 16:02
Hello Fredrick, nice to meet you.

On Jan 15, 5:55 pm, Frederick Williams <frederick.willia...(a)tesco.net>
wrote:
> "Ken S. Tucker" wrote:
>
> > [...]
>
> Work an example for us. What's
> the integral from 0 to 1 of x sqrt(dx)

Very reasonable question.

Let x=U^2 then d(x=U^2) => (dx = dU^2), then sub to Fredricks
example,

1
$ U^2 dU = 1/3.
0

Regards
Ken S. Tucker
From: Ken S. Tucker on 17 Jan 2010 17:16
On Jan 16, 11:20 pm, "Ostap S. B. M. Bender Jr."
<ostap_bender_1...(a)hotmail.com> wrote:
> On Jan 13, 4:06 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
>
>
>
> > Hi Mr. Eastham.
>
> > On Jan 13, 3:21 pm, Chip Eastham <hardm...(a)gmail.com> wrote:
>
> > > On Jan 13, 12:45 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
>
> > > > To Dr. Lewis et al,
>
> > > > On Jan 13, 8:12 am, "Robert H. Lewis" <rle...(a)fordham.edu> wrote:
>
> > > > > > > >${-oo,+oo} F(x) (dx)^.5 (1)
>
> > > > > > > Where did you run across this? The notation makes very little
> > > > > > > sense.
> > > > > > Jays doing just fine, David I think you need to take a basic
> > > > > > calculus course, let me explain (using Jay's example), that I
> > > > > > frequently encounter.
>
> > > > > > V = dX/dT = Velocity.
> > > > > > sqrt(V) = sqrt (dX/dT) , sqrt(dX) = sqrt(V)*sqrt(dT).
> > > > > > That's SOP in calculus, from that a bit of algebraic massage
> > > > > > produces *generally* a means to solve.
> > > > > > Regards
> > > > > > Ken S. Tucker
>
> > > > > SOP in Calculus? Elementary course? No way!
>
> > > > > I have been teaching calculus at all levels for close to 30 years, and I have never seen \sqrt(dx). It is absurd on the face of it. Perhaps some physicists have come up with this and found a coherent usage?
>
> > > > Dr. Lewis, I too have taught and refined Calculus and very much
> > > > respect the reasoning and notation conventions.
> > > > I think it's reasonable to examine the 'differential coefficents',
> > > > apart from the ratio, let's start here,
>
> > > > V=dX/dT, sqrt(V)= sqrt(dX/dT)
>
> > > > that's an expression of V by differential coefficients,
> > > > The apparent confusion results from writing the above as,
>
> > > > sqrt(V) = sqrt(dX)/sqrt(dT) , THE STEP.
>
> > > > Here we ask, does THE STEP represent a clear definition
> > > > of symbolic logic?
>
> > Above, I'm working on getting a handle on 'what is the
> > problem'.
>
> > > No, it is not clear by any definition of
> > > derivative.
>
> > I'll need to clarify, that we're a bit deeper than
> > derivatives, specifically 'differential co-efficents'
> > then a bit beyond that, a respected mathematician
> > Herman Weyl was fond of.
>
> > > The casual treatment of the
> > > derivative dX/dT as a ratio of two values
> > > mixed with taking a square root in the
> > > numerator and denominator is lacking in
> > > logical justification.
>
> > Good enough!
> > Allow me to move to integration ($) and then pose,
>
> > dU = dV*dV , dV = sqrt(dU)
>
> What is 'dV*dV'? A function? A number? A bird? A mammal? A mineral?

The easiest I know is to begin with an Area=Length^2, I'll
write as A=L^2, with A=L=0 to start, then

d(Area) = d(Length)^2 == dA = dL^2.

In the world of differentials, only change counts, absolute
values are arbituary.

In General Relativity oft used is

ds^2 = g_uv dx^u dx^v = 0,

(describes the light path)
then some authors go to dA = ds^2.

It looks ok to me, the integral

$ dA = $ ds^2 = the same constant of integration,

and the same proper integral.
Ken
PS:Just got dizzy, washed a naked girls back.
From: Ostap S. B. M. Bender Jr. on 17 Jan 2010 22:57
On Jan 17, 2:16 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
> On Jan 16, 11:20 pm, "Ostap S. B. M. Bender Jr."
>
>
>
> <ostap_bender_1...(a)hotmail.com> wrote:
> > On Jan 13, 4:06 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
>
> > > Hi Mr. Eastham.
>
> > > On Jan 13, 3:21 pm, Chip Eastham <hardm...(a)gmail.com> wrote:
>
> > > > On Jan 13, 12:45 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
>
> > > > > To Dr. Lewis et al,
>
> > > > > On Jan 13, 8:12 am, "Robert H. Lewis" <rle...(a)fordham.edu> wrote:
>
> > > > > > > > >${-oo,+oo} F(x) (dx)^.5   (1)
>
> > > > > > > > Where did you run across this? The notation makes very little
> > > > > > > > sense.
> > > > > > > Jays doing just fine, David I think you need to take a basic
> > > > > > > calculus course, let me explain (using Jay's example), that I
> > > > > > > frequently encounter.
>
> > > > > > > V = dX/dT = Velocity.
> > > > > > > sqrt(V) = sqrt (dX/dT) , sqrt(dX) = sqrt(V)*sqrt(dT).
> > > > > > > That's SOP in calculus, from that a bit of algebraic massage
> > > > > > > produces *generally* a means to solve.
> > > > > > > Regards
> > > > > > > Ken S. Tucker
>
> > > > > >   SOP in Calculus?  Elementary course?   No way!
>
> > > > > >  I have been teaching calculus at all levels for close to 30 years, and I have never seen \sqrt(dx).  It is absurd on the face of it.  Perhaps some physicists have come up with this and found a coherent usage?
>
> > > > > Dr. Lewis, I too have taught and refined Calculus and very much
> > > > > respect the reasoning and notation conventions.
> > > > > I think it's reasonable to examine the 'differential coefficents',
> > > > > apart from the ratio, let's start here,
>
> > > > > V=dX/dT, sqrt(V)= sqrt(dX/dT)
>
> > > > > that's an expression of V by differential coefficients,
> > > > > The apparent confusion results from writing the above as,
>
> > > > > sqrt(V) = sqrt(dX)/sqrt(dT)   , THE STEP.
>
> > > > > Here we ask, does THE STEP represent a clear definition
> > > > > of symbolic logic?
>
> > > Above, I'm working on getting a handle on 'what is the
> > > problem'.
>
> > > > No, it is not clear by any definition of
> > > > derivative.
>
> > > I'll need to clarify, that we're a bit deeper than
> > > derivatives, specifically 'differential co-efficents'
> > > then a bit beyond that, a respected mathematician
> > > Herman Weyl was fond of.
>
> > > > The casual treatment of the
> > > > derivative dX/dT as a ratio of two values
> > > > mixed with taking a square root in the
> > > > numerator and denominator is lacking in
> > > > logical justification.
>
> > > Good enough!
> > > Allow me to move to integration ($) and then pose,
>
> > > dU = dV*dV , dV = sqrt(dU)
>
> > What is 'dV*dV'? A function? A number? A bird? A mammal? A mineral?
>
> The easiest I know is to begin with an Area=Length^2, I'll
> write as A=L^2, with A=L=0 to start, then
>
> d(Area) = d(Length)^2   ==    dA = dL^2.
>

You mean the area of a square? OK. Indeed, you can write:

dA(L) = d(L^2)

But how do you get form this to (dL)^2 ?



>
> In the world of differentials, only change counts, absolute
> values are arbituary.
>
> In General Relativity oft used is
>
> ds^2 = g_uv dx^u dx^v = 0,
>
> (describes the light path)
> then some authors go to dA = ds^2.
>
> It looks ok to me, the integral
>
> $ dA = $ ds^2 = the same constant of integration,
>

Sure, an expression like $ d(s^2) looks fine. But what does the
expression $ (ds)^2 mean?

>
> PS:Just got dizzy, washed a naked girls back.
>

Are you a teenager?
From: Ken S. Tucker on 18 Jan 2010 04:05
On Jan 17, 7:57 pm, "Ostap S. B. M. Bender Jr."
<ostap_bender_1...(a)hotmail.com> wrote:
> On Jan 17, 2:16 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
>
>
>
> > On Jan 16, 11:20 pm, "Ostap S. B. M. Bender Jr."
>
> > <ostap_bender_1...(a)hotmail.com> wrote:
> > > On Jan 13, 4:06 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
>
> > > > Hi Mr. Eastham.
>
> > > > On Jan 13, 3:21 pm, Chip Eastham <hardm...(a)gmail.com> wrote:
>
> > > > > On Jan 13, 12:45 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
>
> > > > > > To Dr. Lewis et al,
>
> > > > > > On Jan 13, 8:12 am, "Robert H. Lewis" <rle...(a)fordham.edu> wrote:
>
> > > > > > > > > >${-oo,+oo} F(x) (dx)^.5 (1)
>
> > > > > > > > > Where did you run across this? The notation makes very little
> > > > > > > > > sense.
> > > > > > > > Jays doing just fine, David I think you need to take a basic
> > > > > > > > calculus course, let me explain (using Jay's example), that I
> > > > > > > > frequently encounter.
>
> > > > > > > > V = dX/dT = Velocity.
> > > > > > > > sqrt(V) = sqrt (dX/dT) , sqrt(dX) = sqrt(V)*sqrt(dT).
> > > > > > > > That's SOP in calculus, from that a bit of algebraic massage
> > > > > > > > produces *generally* a means to solve.
> > > > > > > > Regards
> > > > > > > > Ken S. Tucker
>
> > > > > > > SOP in Calculus? Elementary course? No way!
>
> > > > > > > I have been teaching calculus at all levels for close to 30 years, and I have never seen \sqrt(dx). It is absurd on the face of it. Perhaps some physicists have come up with this and found a coherent usage?
>
> > > > > > Dr. Lewis, I too have taught and refined Calculus and very much
> > > > > > respect the reasoning and notation conventions.
> > > > > > I think it's reasonable to examine the 'differential coefficents',
> > > > > > apart from the ratio, let's start here,
>
> > > > > > V=dX/dT, sqrt(V)= sqrt(dX/dT)
>
> > > > > > that's an expression of V by differential coefficients,
> > > > > > The apparent confusion results from writing the above as,
>
> > > > > > sqrt(V) = sqrt(dX)/sqrt(dT) , THE STEP.
>
> > > > > > Here we ask, does THE STEP represent a clear definition
> > > > > > of symbolic logic?
>
> > > > Above, I'm working on getting a handle on 'what is the
> > > > problem'.
>
> > > > > No, it is not clear by any definition of
> > > > > derivative.
>
> > > > I'll need to clarify, that we're a bit deeper than
> > > > derivatives, specifically 'differential co-efficents'
> > > > then a bit beyond that, a respected mathematician
> > > > Herman Weyl was fond of.
>
> > > > > The casual treatment of the
> > > > > derivative dX/dT as a ratio of two values
> > > > > mixed with taking a square root in the
> > > > > numerator and denominator is lacking in
> > > > > logical justification.
>
> > > > Good enough!
> > > > Allow me to move to integration ($) and then pose,
>
> > > > dU = dV*dV , dV = sqrt(dU)
>
> > > What is 'dV*dV'? A function? A number? A bird? A mammal? A mineral?
>
> > The easiest I know is to begin with an Area=Length^2, I'll
> > write as A=L^2, with A=L=0 to start, then
>
> > d(Area) = d(Length)^2 == dA = dL^2.
>
> You mean the area of a square? OK. Indeed, you can write:
>
> dA(L) = d(L^2)
>
> But how do you get form this to (dL)^2 ?

Well in a science experiment shining light through an aperature,
we may wish to calculate the area of projection as we move the
screen the image is projected upon, thinking in terms of a movie
projector.

We can use, (A=Area, r=distance from bulb, t=time)

dA/dt = (dr/dt)^2 = dr^2 / dt^2,

Plug in a number like dr/dt=2 and the rate of change of the
projected Area is dA/dt=4, so the projected Area of the image
increased 4x as the distance from the bulb doubled.

> > In the world of differentials, only change counts, absolute
> > values are arbituary.
>
> > In General Relativity oft used is
>
> > ds^2 = g_uv dx^u dx^v = 0,
>
> > (describes the light path)
> > then some authors go to dA = ds^2.
>
> > It looks ok to me, the integral
>
> > $ dA = $ ds^2 = the same constant of integration,
>
> Sure, an expression like $ d(s^2) looks fine. But what does the
> expression $ (ds)^2 mean?

Ostap, the subtle notation change is respected.

In GR ds=0 so we have $ ds ds and that gives,

$ 0 ds = constant of integration.

In physics I understand that constant to be number a "N"
that is a scalar (invariant).

> > PS:Just got dizzy, washed a naked girls back.
>
> Are you a teenager?

Betcha when I see 90 I'll still get dizzy bathing gals,
wifey called me to scrub her back, it's tough work but
gives me an excuse to stay alive ;-).
Ken
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