Why Miles Mathis is wrong!

[David]

David, You have taken the quote out of context.

Since I don’t own a Mathis decoder ring (or know the secret handshake, for that matter), I can only evaluate his material as written:

Calculus is worked upon the curve equation. You must have a curve equation in order to find a derivative.

If above statement requires translation from Mathis-speak to common English, then that would be your area of expertise, not mine.

[David]

The generalized derivative equation is

y' = nxn-1

http://milesmathis.com/expon.html

Wrong! That equation is commonly known as the “power rule”. It is just one of the many rules of differentiation (power rule, product rule, chain rule, et cetera). It is not a “generalized derivative equation” as Mathis claims. It is a differentiation "rule" that applies to polynomials only.

This is the generalized derivative equation:

[Chromium6]

The generalized derivative equation is

y' = nxn-1

http://milesmathis.com/expon.html

Wrong! That equation is commonly known as the “power rule”. It is just one of the many rules of differentiation (power rule, product rule, chain rule, et cetera). It is not a “generalized derivative equation” as Mathis claims. It is a differentiation "rule" that applies to polynomials only.

This is the generalized derivative equation:

David, simply for the sake of clarity, please include quotes like these in that paper:
---


All of this is very ugly, as I think most people can see. We are told that we cannot find the tangent directly, even 300 years after Newton. This must be ridiculous after the publication of my long paper (http://milesmathis.com/are.html), because there I am able to find the derivative directly and precisely. I am able to derive the basic derivative equation straight from a table of differentials (and he's not the only one to do so AFAIK--CSR), and my generalized equation is exact and complete. It is not an estimate or an approximation, since we never go to zero or to an infinitesimal. Readers complain that I don't prove the equation for all numbers, only integers, but I don't need to prove it for all numbers. All numbers are defined by integers, so any extension of the basic equation is true by definition. Yes, I prove my basic and general equation from a table of integers, but the solution is not limited to integers, and this should be clear to anyone awake. Any such proof that is proved for integers is proved for all numbers, since the number line is defined by integers. Exponential notation is defined by integers. Likewise, logarithms are defined by integers and by exponential notation, so that anything proved for integers must be proved for exponents and logarithms. There is no such thing as an exponent or integer or log that is not defined by the number line, and since the number line is defined by the integers, my proof is generalized automatically. All we have to do is make a simple table of differentials, using the same method I used for integers (as I will do again below).

[LongtimeAirman]

David wrote:

Miles Mathis wrote:
The generalized derivative equation is

y' = nxn-1

http://milesmathis.com/expon.html

Wrong! That equation is commonly known as the “power rule”. It is just one of the many rules of differentiation (power rule, product rule, chain rule, et cetera). It is not a “generalized derivative equation” as Mathis claims. It is a differentiation "rule" that applies to polynomials only.

This is the generalized derivative equation:

23b7112ec7aa5d19157cf84bd3b392e8.png (1.29 KiB) Viewed 13323 times

In http://www.thunderbolts.info/wp/forum/phpB ... 165#p98625

David wrote:

You can see the obvious similarities between Lagrange's approach to differential calculus, and the Mathis approach. Both are based on algebra (finding derivatives algebraically), without need for infinitesimals or limits. But Lagrange takes it much further than Mathis went, with actual applications (differential geometry).

However, there is one big drawback; the technique is limited to polynomials only. That’s the same problem Mathis ran into. You can only go so far with algebra, and then you hit a brick wall. Once you enter the realm of non-polynomials, you have no choice but to use infinitesimals and limits; it’s just the nature of the beast.

And

“Rational trigonometry avoids direct use of transcendental functions like sine and cosine by substituting their squared equivalents. Rational trigonometry requires fewer steps to solve typical problems and avoids logical inconsistencies associated with classical trigonometry.”

But the theory has its critics (citing Wikipedia):

“Paul J. Campbell, writing in Mathematics Magazine: "the author claims that this new theory will take 'less than half the usual time to learn'; but I doubt it, and it would still have to be interfaced with the traditional concepts and notation.

“New Scientist's Gefter described the approach of Wildberger as an example of finitism.”

I don’t know enough about his theory to offer an opinion. But if Wildberger's goal is to purge mathematics of infinite-series, infinitesimals, limits, transcendental numbers, et cetera, then we are definitely in opposite camps.

David, You know very well that Miles would redefine calculus without points or infinitesimals, but in all your attacks, you are comfortable misrepresenting him and misleading others.

The derivative, y' = nxn-1, is the “generalized derivative equation” as found in Miles' original calculus paper: "A Redefinition Of The Derivative", http://milesmathis.com/are.html, using constant differentials.

You say Miles is “Wrong” simply because he does not agree with standard calculus and is therefore in the “opposite camp”.

You said, “Once you enter the realm of non-polynomials, you have no choice but to use infinitesimals and limits; it’s just the nature of the beast”. I cannot agree. That is probably the thinking that let standard calculus depart from reality in the first place.

REMCB

[David]

David, You know very well that Miles would redefine calculus without points or infinitesimals, but in all your attacks, you are comfortable misrepresenting him and misleading others.

The derivative, y' = nxn-1, is the “generalized derivative equation” as found in Miles' original calculus paper: "A Redefinition Of The Derivative", http://milesmathis.com/are.html, using constant differentials.

That equation is the “power rule”. It was discovered by Cavalieri in the 1630’s, before the advent of calculus. It is not a “generalized derivative equation”, as Mathis claims; it is just a rule that applies *only* to polynomials.

You say Miles is “Wrong” simply because he does not agree with standard calculus and is therefore in the “opposite camp”.

No! Mathis is wrong because his tables are based entirely on the “power rule”, which means they will only work for a polynomial. When Mathis attempts to find the derivative of a non-polynomial (for instance, “Trig Derivatives without Calculus”, http://milesmathis.com/trig.html), his method fails.

So it’s not a matter of agreeing or disagreeing, the Mathis method of differentiation simply doesn't work. Mathis can bitch and moan about limits and infinitesimals, but his proposed alternative is flawed.

You said, “Once you enter the realm of non-polynomials, you have no choice but to use infinitesimals and limits; it’s just the nature of the beast”. I cannot agree. That is probably the thinking that let standard calculus depart from reality in the first place.

Fair enough. If you disagree, then give us a demonstration. Show us how to find derivatives of non-polynomials (for instance, the derivative of sin(x)), without invoking limits or infinitesimals. So far, no one has been able to do it; but if you can I'd love to see it.

[David]

Acceleration is said to be the derivative of the velocity, but it isn't. The derivative is the rate of change of the curve, and a velocity isn't a curve. The derivative of any straight line is a constant, since the rate of change of any straight line is a constant. You can't really differentiate a velocity, since there isn't any variation.

http://milesmathis.com/expon.html

Just a few comments back I gave an example where calculus was used to find the derivative of a straight line; not a curve, a straight line. So Mathis is flat out wrong. The velocity, without any doubt whatsoever, has a derivative; even when the velocity is constant (unchanging), it still has a derivative.

Wikipedia tells us that “the derivative of velocity with respect to time is the acceleration,” but that is simply false. In fact, it is upside down. You can't differentiate a velocity into an acceleration, since a velocity has a constant rate of change.

Oh yes I can; allow me to demonstrate how to "differentiate a velocity into an acceleration". Assume a ball has been dropped off the top of a building. The ball will fall with a constant acceleration of 9.81 m/s^2, as described by the following equations:

v(t) = (9.81 m/s2) t
a(t) = v’(t) = 9.81 m/s2

Incidentally, if you perform the above experiment the equations will accurately describe the velocity and acceleration. So once again, Mathis is just spouting more of his nonsense.

[Chromium6]

Acceleration is said to be the derivative of the velocity, but it isn't. The derivative is the rate of change of the curve, and a velocity isn't a curve. The derivative of any straight line is a constant, since the rate of change of any straight line is a constant. You can't really differentiate a velocity, since there isn't any variation.

http://milesmathis.com/expon.html

Just a few comments back I gave an example where calculus was used to find the derivative of a straight line; not a curve, a straight line. So Mathis is flat out wrong. The velocity, without any doubt whatsoever, has a derivative; even when the velocity is constant (unchanging), it still has a derivative.

Wikipedia tells us that “the derivative of velocity with respect to time is the acceleration,” but that is simply false. In fact, it is upside down. You can't differentiate a velocity into an acceleration, since a velocity has a constant rate of change.

Oh yes I can; allow me to demonstrate how to "differentiate a velocity into an acceleration". Assume a ball has been dropped off the top of a building. The ball will fall with a constant acceleration of 9.81 m/s^2, as described by the following equations:

v(t) = (9.81 m/s2) t
a(t) = v’(t) = 9.81 m/s2

Incidentally, if you perform the above experiment the equations will accurately describe the velocity and acceleration. So once again, Mathis is just spouting more of his nonsense.

Well David, Milo Wolff also skates on the edge of the theoretical ice:

http://www.spaceandmotion.com/Wolff-Wav ... Matter.htm

Hope you can try and "deep charge" him like Mathis. But surprisingly, and refreshingly, Mathis is still afloat IMHO.

Machiavelli understood this human behaviour 500 years ago (1513):

There is nothing more difficult to plan, more doubtful of success, more dangerous to manage than the creation of a new system. The innovator has the enmity of all who profit by the preservation of the old system and only lukewarm defenders by those who would gain by the new system.

The Calculation by Wheeler and Feynman (W&F) wished to verify the empirical formula for the force of radiation used by Dirac Force = [da/dt]2e2/3c3
where e is the electron charge, c is the velocity of light and a is the acceleration. The mechanism of the force was unknown. They discussed this problem with Einstein who suggested a proposal by Tetrode that light (energy) transmission was not a one-way process, but two-way communication between a source molecule or atom and a receiver molecule utilizing inward and outward waves. This proposal was not popular since it appeared to violate the causality concept: Actions should appear before their causes, and because the inward waves appear to be travelling backward in time.

[LongtimeAirman]

David wrote:
That equation is the “power rule”. It was discovered by Cavalieri in the 1630’s, before the advent of calculus. It is not a “generalized derivative equation”, as Mathis claims; it is just a rule that applies *only* to polynomials.

David, We’ve been ‘round this mulberry bush before.

See: http://www.thunderbolts.info/wp/forum/phpB ... =90#p93555
Re: Lloyd Blog by David » Wed Mar 19, 2014 10:08 pm
And http://www.thunderbolts.info/wp/forum/phpB ... =90#p93558
Re: Lloyd Blog by David » Thu Mar 20, 2014 2:33 am
We go on for quite a few days.

In http://www.thunderbolts.info/wp/forum/phpB ... 105#p93612
Re: Lloyd Blog by LongtimeAirman » Fri Mar 21, 2014 2:28 pm

I answered David: The power rule is the most important derivative formula in calculus. It fell directly out of Miles' table, Miles didn't just borrow it. Cavalieri. Newton and Leibniz used it but they couldn't explain it because they were thinking of infinitesimals. Miles has done something new here and you refuse to accept it.

This table is a great simplification. I used it and found what you said wasn't there. Miles is using it as a basis to find the derivatives of all the functions you say it cannot be used for. He is putting a solid new base on the calculus. Is it your complaint that Miles hasn't reproduced 2000 years of work in the spare time he has devoted to it in the last 10 years.

(Here’s Miles’s table, for convenience)
Miles' original calculus paper: http://milesmathis.com/are.html
Miles' simplified calculus paper: http://milesmathis.com/calcsimp.html

Miles wrote:
Do you see it?
2ΔΔx = ΔΔΔx²
3ΔΔΔx2 = ΔΔΔΔx3
4ΔΔΔΔx3 = ΔΔΔΔΔx4
5ΔΔΔΔΔx4 = ΔΔΔΔΔΔx5
6ΔΔΔΔΔΔx5 = ΔΔΔΔΔΔΔx6
and so on.

...I will continue to simplify...we can cancel a lot of those deltas and get down to this:
2x = Δx2
3x2 = Δx3
4x3 = Δx4
5x4 = Δx5
6x5 = Δx6

Voila. We have the current derivative equation, just from a table

This equation is y’ = nxn-1

I’m delighted that Cavalieri discovered the “power series” in 1630. The fact that it appears in a table of constant differences by Miles doesn’t diminish its significance or prevent Miles from using it as a “generalized derivative equation”.

In answer to the oft repeated “it is just a rule that applies *only* to polynomials”:

In " Trig Derivatives foundwithout the old Calculus", http://milesmathis.com/trig.html, Miles starts with,
y = sin x = ±√(1 - cos2 x) , then substitutes
y = a = ± (1- b^2)^(1/2)
And then uses the table already described. He differentiates both a and b to get:
y' = b = cos x

Forgive my quick review.

We go on to: “In "My Calculus Applied to Exponential Functions", http://milesmathis.com/expon.html,

And, In, " The Derivatives of the Natural Log and 1/x are wrong", http://milesmathis.com/ln.html,

And just above, Re: Why Miles Mathis is wrong! by Chromium6 » Wed Aug 20, 2014 8:12 pm above, Cr6 reminds us :

Miles wrote: All of this is very ugly, as I think most people can see. We are told that we cannot find the tangent directly, even 300 years after Newton. This must be ridiculous after the publication of my long paper (http://milesmathis.com/are.html), because there I am able to find the derivative directly and precisely. I am able to derive the basic derivative equation straight from a table of differentials (and he's not the only one to do so AFAIK--CSR), and my generalized equation is exact and complete. It is not an estimate or an approximation, since we never go to zero or to an infinitesimal. Readers complain that I don't prove the equation for all numbers, only integers, but I don't need to prove it for all numbers. All numbers are defined by integers, so any extension of the basic equation is true by definition. Yes, I prove my basic and general equation from a table of integers, but the solution is not limited to integers, and this should be clear to anyone awake. Any such proof that is proved for integers is proved for all numbers, since the number line is defined by integers. Exponential notation is defined by integers. Likewise, logarithms are defined by integers and by exponential notation, so that anything proved for integers must be proved for exponents and logarithms. There is no such thing as an exponent or integer or log that is not defined by the number line, and since the number line is defined by the integers, my proof is generalized automatically. All we have to do is make a simple table of differentials, using the same method I used for integers (as I will do again below).

But you don’t accept any of that because it isn’t in a table which you have defined as the “Miles' Method”.
You even start making up function tables, which of course do not work, insisting that you are following the “Miles' method”.

David continued:
No! Mathis is wrong because his tables are based entirely on the “power rule”, which means they will only work for a polynomial. When Mathis attempts to find the derivative of a non-polynomial (for instance, “Trig Derivatives without Calculus”, http://milesmathis.com/trig.html), his method fails.

So it’s not a matter of agreeing or disagreeing, the Mathis method of differentiation simply doesn't work. Mathis can bitch and moan about limits and infinitesimals, but his proposed alternative is flawed.

You were shown that with the substitution, 1 = sin2X + cos2x the table can be used for trig functions, but you still insist, “his table will only work for a polynomial”.

David Continues:
LongtimeAirman wrote:
You said, “Once you enter the realm of non-polynomials, you have no choice but to use infinitesimals and limits; it’s just the nature of the beast”. I cannot agree. That is probably the thinking that let standard calculus depart from reality in the first place.


Fair enough. If you disagree, then give us a demonstration. Show us how to find derivatives of non-polynomials (for instance, the derivative of sin(x)), without invoking limits or infinitesimals. So far, no one has been able to do it; but if you can I'd love to see it.

Miles' actual method – throwing out infinitesimals and limits – does work. This “Once you enter the realm of non-polynomials, you have no choice but to use infinitesimals and limits”, is a very curious statement.

David wrote: “Rational trigonometry avoids direct use of transcendental functions like sine and cosine by substituting their squared equivalents. Rational trigonometry requires fewer steps to solve typical problems and avoids logical inconsistencies associated with classical trigonometry.”

You’ve been given plenty of demonstrations. Based on your past comments, it doesn’t matter if you get the answers you asked for. You will not acknowledge it. You will weasel out by repeating, yet again, the next item on your list.

REMCB

[David]

Miles' actual method – throwing out infinitesimals and limits – does work.

Alright, this calls for a demonstration. If the Mathis method actually works, as you claim it does, then show us.

This is a table of the actual differentials of log(x):

log(x) 0. .301, .477, .602, .699, .778, .845, .903, .954
Δlog(x) .301, .176, .125, .097, .079, .067, .051
ΔΔlog(x) .125, .051, .028, .018, .012, .016
ΔΔΔlog(x) .074, .023, .01, .006, .004
ΔΔΔΔlog(x) .051, .013, .004, .002

http://milesmathis.com/log.html

Find the derivative of log(x) using the above table of differentials. I claim that it can't be done. So here’s your chance to prove me wrong.

Show us exactly (step-by-step) how to find the derivative of log(x) using the above table.

[LongtimeAirman]

David wrote:

LongtimeAirman wrote:
Miles' actual method – throwing out infinitesimals and limits – does work.

Alright, this calls for a demonstration. If the Mathis method actually works, as you claim it does, then show us.

Miles Mathis wrote:
This is a table of the actual differentials of log(x):

log(x) 0. .301, .477, .602, .699, .778, .845, .903, .954
Δlog(x) .301, .176, .125, .097, .079, .067, .051
ΔΔlog(x) .125, .051, .028, .018, .012, .016
ΔΔΔlog(x) .074, .023, .01, .006, .004
ΔΔΔΔlog(x) .051, .013, .004, .002

http://milesmathis.com/log.html

Find the derivative of log(x) using the above table of differentials. I claim that it can't be done. So here’s your chance to prove me wrong.

Show us exactly (step-by-step) how to find the derivative of log(x) using the above table.

55. The Derivative of log(x) is also Wrong. http://milesmathis.com/log.html I show more problems with the Modern Calculus using my simple differential tables. I find a new slope for log(x) which is fantastically close to the current slope, but which comes from a completely different analysis. 5pp.

Not only have I not backed off, I have found a way to advance even further. In my next paper3 http://milesmathis.com/power.html, I will show that even the derivative equation for powers is proved in a faulty way. Yes, although I have confirmed the equation y' = nxn-1 in my long paper, it turns out even that proof will fall. It gets the correct slope and velocity and so on, but it is not absolutely correct as a matter of math or physics.

This is a table of the actual differentials of log(x):

log(x) 0. .301, .477, .602, .699, .778, .845, .903, .954
Δlog(x) .301, .176, .125, .097, .079, .067, .051
ΔΔlog(x) .125, .051, .028, .018, .012, .016
ΔΔΔlog(x) .074, .023, .01, .006, .004
ΔΔΔΔlog(x) .051, .013, .004, .002

The following are the real differential equations for log(x), finding line 3 from line 1. According the current definition of the derivative, line 3 is the derivative of line 1. Line 2 is the general rate of change of the curve log(x), and line 3 is the rate of change at a given interval x.

ΔΔlog(x) = Δlog(x ) – Δlog(x + 1)
Δlog(x) = log(x + 1) – log(x)
Δlog(x + 1)) = log(x + 2) – log(x + 1)
ΔΔlog(x) = log(x + 1) – log(x) – log(x + 2) + log(x + 1)
log(x)/dx = 2log(x + 1) – log(x) – log(x + 2)

In the differential table, each line is a factor of 2 separated from the previous line. What I mean is, the first real differential in line 1 is the log of 2. So we are starting with 2. If we want a slope, we have to shift the entire table by a factor of 2. Since we are finding line 3 from line 1, we must shift or multiply by 22=4.

rate of change = 4[2log(x + 1) – log(x) – log(x + 2)]

But, as before, that isn't the slope. The slope is found most easily and perfectly by this equation, as I discovered today while chewing for the third day in a row on the derivative of ax.

slope @ (x,y) = [y@(x + 1) - y@(x - 1)]/2

slope x=3 is .1505, not .145
slope x=4 is .111, not .109
slope x=5 is .088, not .087
slope x=6 is .073, not .0725
slope x=7 is .0625, not .0621
slope x=8 is .0545, not .0543
slope x=10 is .0437, not .0435

Once again, my critics have to be sweating. Those numbers are astonishing, and there is no way around it. To see a full explanation of why averaging forward slope and backward slope is actually better than going to zero, you will have to read the extended analysis and question answering in my earlier paper on the derivative for exponents2.

REMCB

[LongtimeAirman]

From: Re: Why Miles Mathis is wrong! Post by David » Thu Aug 21, 2014 3:40 am http://www.thunderbolts.info/wp/forum/phpB ... 94#pr98803

David wrote:

Miles Mathis wrote: Acceleration is said to be the derivative of the velocity, but it isn't. The derivative is the rate of change of the curve, and a velocity isn't a curve. The derivative of any straight line is a constant, since the rate of change of any straight line is a constant. You can't really differentiate a velocity, since there isn't any variation.

http://milesmathis.com/expon.html

Just a few comments back I gave an example where calculus was used to find the derivative of a straight line; not a curve, a straight line. So Mathis is flat out wrong. The velocity, without any doubt whatsoever, has a derivative; even when the velocity is constant (unchanging), it still has a derivative.

David,
Your comments are amazing. You quote Miles saying “The derivative of any straight line is a constant, since the rate of change of any straight line is a constant”, then turn around and say that “Mathis is flat out wrong. The velocity, without any doubt whatsoever, has a derivative; even when the velocity is constant (unchanging), it still has a derivative”.

Yes a constant. You and Miles agree about that.

David continues:

Miles Mathis wrote:
Wikipedia tells us that “the derivative of velocity with respect to time is the acceleration,” but that is simply false. In fact, it is upside down. You can't differentiate a velocity into an acceleration, since a velocity has a constant rate of change.

Oh yes I can; allow me to demonstrate how to "differentiate a velocity into an acceleration". Assume a ball has been dropped off the top of a building. The ball will fall with a constant acceleration of 9.81 m/s^2, as described by the following equations:

v(t) = (9.81 m/s2) t
a(t) = v’(t) = 9.81 m/s2

Incidentally, if you perform the above experiment the equations will accurately describe the velocity and acceleration. So once again, Mathis is just spouting more of his nonsense.

David, I see the nonsense, you’ve described velocity v(t) (whose units should be m/s) as an acceleration (units: m/s2).

REMCB

[David]

David, I see the nonsense, you’ve described velocity v(t) (whose units should be m/s) as an acceleration (units: m/s^2).

When Mathis has mismatched units (a common occurrence), the remedy is really quite simple; just go ahead and alter the units:

We need to dump some of those extra meter dimensions. Every time we multiply a velocity and a velocity that are in line, we have to dump one of the meter dimensions.

But in this instance, dumping won't be necessary. The units are already correct.

v(t) = (9.81 m/s2) t

Velocity equals acceleration multiplied by time. And in case you need reminding, time has units of seconds.

[Chromium6]

http://physics.info/kinematics-calculus/

The third equation of motion relates velocity to displacement. By logical extension, it should come from a derivative that looks like this …
dv = ??
dx

But what does this equal? Well nothing by definition, but like all quantities it does equal itself. It also equals itself multiplied by 1. We'll use a special version of 1, dt/dt, and then do a bit of special algebra — algebra with infinitesimals. Look what happens when we do this. We get a derivative equal to acceleration and another equal to the inverse of velocity.

-------------

THE EQUATION   V = V0 + AT IS FALSE as a field equation

by Miles Mathis

In a recent paper on the muon1, I showed that this equation doesn't work when v0 is equal to c. It doesn't work when we have a particle with a very high velocity being accelerated by a gravity field. In that case the equation is vf = v0 + 2v02t. Why?

Then we add the initial velocity. That's it. The equation is basically a definition of acceleration. But in a recent paper on the calculus and variable acceleration2, I showed that the equation can actually be calculated from the time derivative.

v = a[d(t2)]/2 = at

The 2 in the denominator comes from the halved first interval, which we must take into account in any acceleration.

So we have confirmed that part of the equation. The problem must be with the way the initial velocity and the acceleration stack up in a field. Apparently we can't just add the final velocity from the acceleration to the original velocity. Again, why?

Put simply, the reason is because if we have an initial velocity meeting an acceleration field, we have three velocities. Let us start with just the acceleration field, and no initial velocity. Let us say you have a gravity field, or any other acceleration field that creates a normal or squared acceleration. If you place an object in the field, the object will be given two simultaneous velocities. That is what an acceleration is, after all. If the object has one velocity, it has no acceleration. If it has an acceleration, it must have more than one velocity over each interval or differential. A squared acceleration, or an acceleration to the power of 2, is composed of two velocities. Therefore an object placed in any acceleration field will be given two velocities.

OK, but now we take an object that already has a velocity of its own, and we fly it into that acceleration field. It must then have three velocities over each interval. Or, to say it another way, it now has a cubed acceleration. It has three t's in the denominator.

For this not to be the case, we would have to break Newton's first law. If we don't include the initial velocity in each differential of acceleration, that means the acceleration field has somehow negated that velocity during the time of acceleration. The field would have to stop the object, then accelerate it, then give it its original velocity back at the end. Of course all that is impossible and absurd. No, the field would have to keep the initial velocity in every differential of acceleration, meaning that it would be accelerating the velocity, not the object.

Now maybe you see the problem with the equation in the title. It does not include a cubed acceleration. The “a” variable stands for a squared acceleration. The equation includes three velocities, but the first velocity has not been integrated into the acceleration. It is only added, as a separate term. But that is not the way the real field would work. The real field would accelerate the initial velocity, not just the object. As it is, the initial velocity is not summed in every differential, as it should be; it is only summed outside the acceleration. That cannot work, either as a matter of calculus or as a matter of physics. The equation is wrong, in any and all cases where a field is involved.


I will show that the reason is because the equation in the title is false as a field equation. It only works when the acceleration is not a field. It works only when the acceleration is an internal acceleration, as with a car.

In textbooks, the equation in my title is derived like this:

a= v/t
v= at

Then we add the initial velocity. That's it. The equation is basically a definition of acceleration. But in a recent paper on the calculus and variable acceleration2, I showed that the equation can actually be calculated from the time derivative.

...

A reader has pointed out to me that the dimensions appear to be off in that final equation. What causes it is that I am multiplying a distance by a distance, which appears to give us a distance squared. Go back two steps and you will see what I mean. But a distance times a distance is not a distance squared, if they are in line. A distance times a distance is a distance squared only in the case that the two distances are orthogonal, so that we have a square or something. A distance times a distance in line cannot be a distance squared anymore than a distance plus a distance can be a distance squared. This is another mainstream misunderstanding.

I have used algebra here to solve, but notice that if we use integration, we get the same problem: ∫(2x + y)dx = x(x + y). We integrate a sum but get meters squared.

This problem ties into the historical problem of velocity squared versus acceleration. Physics has never been clear about the mechanical difference. One has meters squared in the numerator and one does not, but are they really different? No. A square velocity IS an acceleration, by definition, so if we are getting different dimensions it is by ignoring some mechanics. We see that in this problem very clearly: if we multiply a velocity times a velocity in a field, we should get an acceleration. Therefore the extra length in the numerator is just that: extra. [You can now read more about this in a separate paper.]


http://milesmathis.com/voat.html

[David]

This is a table of the actual differentials of log(x):

log(x) 0. .301, .477, .602, .699, .778, .845, .903, .954
Δlog(x) .301, .176, .125, .097, .079, .067, .051
ΔΔlog(x) .125, .051, .028, .018, .012, .016
ΔΔΔlog(x) .074, .023, .01, .006, .004
ΔΔΔΔlog(x) .051, .013, .004, .002

The following are the real differential equations for log(x), finding line 3 from line 1. According the current definition of the derivative, line 3 is the derivative of line 1. Line 2 is the general rate of change of the curve log(x), and line 3 is the rate of change at a given interval x.

ΔΔlog(x) = Δlog(x ) – Δlog(x + 1)
Δlog(x) = log(x + 1) – log(x)
Δlog(x + 1)) = log(x + 2) – log(x + 1)
ΔΔlog(x) = log(x + 1) – log(x) – log(x + 2) + log(x + 1)
log(x)/dx = 2log(x + 1) – log(x) – log(x + 2)

In the differential table, each line is a factor of 2 separated from the previous line. What I mean is, the first real differential in line 1 is the log of 2. So we are starting with 2. If we want a slope, we have to shift the entire table by a factor of 2. Since we are finding line 3 from line 1, we must shift or multiply by 2^2=4.

rate of change = 4[2log(x + 1) – log(x) – log(x + 2)]

But, as before, that isn't the slope. The slope is found most easily and perfectly by this equation, as I discovered today while chewing for the third day in a row on the derivative of a^x.

slope @ (x,y) = [y@(x + 1) - y@(x - 1)]/2

To see a full explanation of why averaging forward slope and backward slope is actually better than going to zero, you will have to read the extended analysis and question answering in my earlier paper on the derivative for exponents.

http://milesmathis.com/log.html

Okay, so this is how Mathis calculates the derivative for a non-polynomial.

Step 1: Construct a table of differentials for the function.

Step 2: Ignore or discard the table; it’s never used.

Step 3: Calculate an average value for the slope.

Step 4: Grossly exaggerate that the average value “is vastly superior to the current method in both operation and answer”.

To call the Mathis method of differentiation mathematically thin, would be an understatement. It doesn’t even qualify as a solution; it’s just a rough approximation.

Leibniz devoted more than a decade to analyzing the problem. Whereas Mathis spent three whole days “chewing” on it; and as a result, his solution looks like something he spit up.

[David]

The third equation of motion relates velocity to displacement. By logical extension, it should come from a derivative that looks like this

dv/dx = ??

But what does this equal? Well nothing by definition…

Wow! It’s hard to wrap your mind around exactly what that is; the change in velocity with respect to displacement. It has units of 1/s, which is the same as frequency. But knowing that doesn’t help explain it.

And it gets even more bizarre when you consider a change in acceleration with respect to displacement:

da/dx = ??

That has units of 1/s^2. But again, a rational explanation is nowhere in sight. There could a useful application that has just been overlooked. Give it a catchy name and start using it; it’s guaranteed to confuse and bewilder virtually everyone.

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Oh by the way, I am surprised that you fellows are still hanging out here. With your new Mathis Forum (http://milesmathis.the-talk.net/) up and running, why stray from heaven? You have eliminated all dissention; wasn’t that what you wanted most? Let’s ask Spock, he may know the answer.

"After a time, you may find that having is not so pleasing a thing, after all, as wanting. It is not logical, but it is often true." - Spock