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From: A N Niel on 13 Jan 2010 11:43

> It is absurd on the face of it.

That never stopped the physicists!
From: Ken S. Tucker on 13 Jan 2010 12:45
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?

> Robert H. Lewis
> Mathematics Department
> Fordham University

Regards
Ken S. Tucker
From: achille on 13 Jan 2010 18:08
On Jan 14, 12:43 am, A N Niel <ann...(a)nym.alias.net.invalid> wrote:
> > It is absurd on the face of it.
>
> That never stopped the physicists!

[Put on my ex-physicist hat]
Don't blame physicists on this! We haven't come up
with sqrt(dx) yet. Jay is simply confused about the
two possible use of dq in the context of path integral.
[Take off my ex-physicist hat]
From: Chip Eastham on 13 Jan 2010 18:21
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?

No, it is not clear by any definition of
derivative. 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.

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


From: Ken S. Tucker on 13 Jan 2010 19:06
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)

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
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