Paul P. Mealing

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Wednesday, 10 October 2012

The genius of differential calculus


Newton and Leibniz are both credited as independent ‘inventors’ of calculus but I would argue that it was at least as much discovery as invention, because, at its heart, differential calculus delivers the seemingly impossible.

Calculus was arguably the greatest impetus to physics in the scientific world. Newton’s employment of calculus to give mathematical definition and precision to motion was arguably as significant to the future of physics as his formulation of the General Theory of Gravity. Without calculus, we wouldn’t have Einstein’s Theory of Relativity and we wouldn’t have Schrodinger’s equation that lies at the heart of quantum mechanics. Engineers, the world over, routinely use calculus in the form of differential equations to design most of the technological tools and infrastructure we take for granted.

Differential calculus is best understood in its application to motion in physics and to tangents in Cartesian analytic geometry. In both cases, we have mathematics describing a vanishing entity, and this is what gives calculus its power, and also makes it difficult for people to grasp, conceptually.

Calculus can freeze motion, so that at any particular point in time, knowing an object’s acceleration (like a free-falling object under gravity, for example) we can determine its instantaneous velocity, and knowing its velocity we can determine its instantaneous position. It’s the word ‘instantaneous’ that gives the game away.

In reality, there is no ‘instantaneous’ moment of time. If you increase the shutter speed of a camera, you can ‘freeze’ virtually any motion, from a cricket ball in mid-flight (baseball for you American readers) to a bullet travelling faster than the speed of sound. But the point is that, no matter how fast the shutter speed, there is still a ‘duration’ that the shutter remains open. It’s only when one looks at the photographic record, that one is led to believe that the object has been captured at an instantaneous point in time.

Calculus does something very similar in that it takes a shorter and shorter sliver of time to give an instantaneous velocity or position.

I will take the example out of Keith Devlin’s excellent book, The Language of Mathematics; Making the invisible visible, of a car accelerating along a road:

x = 5t2 + 3t

The above numbers are made up, but the formulation is correct for a vehicle under constant acceleration. If we want to know the velocity at a specific point in time we differentiate it with respect to time (t).

The differentiated equation becomes dx/dt, which means that we differentiate the distance (x) with respect to time (wrt t).

To get an ‘instantaneous’ velocity, we take smaller and smaller distances over smaller and smaller durations. So dx/dt is an incrementally small distance divided by an incrementally small time, so mathematically we are doing exactly the same as what the camera does.

But dx occurs between 2 positions, x1 and x2, where dx = x2 – x1

This means:  x2 is at dt duration later than x1.

Therefore  x2 = 5(t + dt)2 + 3(t + dt)

And x1 = 5t2 + 3t

Therefore  dx = x2 – x1 = 5(t + dt)2 + 3(t + dt) - (5t2 + 3t)

If we expand this we get:  5t2 + 10tdt + 5dt2 + 3t + 3dt – 5t23t

{Remember: (t + dt)2 = t2 + 2tdt + dt2}

Therefore dx/dt = 10t dt/dt + 5dt2/dt + 3dt/dt

Therefore dx/dt = 10t + 3 + 5dt

The sleight-of-hand that allows calculus to work is that the dt term on the RHS disappears so that dx/dt gives the instantaneous velocity at any specified time t. In other words, by making the duration virtually zero, we achieve the same result as the recorded photo, even though zero duration is physically impossible.

This example can be generalised for any polynomial: to differentiate an equation of the form, 
y = axb

dy/dx = bax(b-1)  which is exactly what I did above:

If y = 5x2 + 3x

Then dy/dx = 10x + 3

The most common example given in text books (and even Devlin’s book) is the tangent of a curve, partly because one can demonstrate it graphically.

If I was to use an equation of the form y = ax2 + bx + c, and differentiate it, the outcome would be exactly the same as above, mathematically. But, in this case, one takes a smaller and smaller x, which corresponds to a smaller and smaller y or f(x). (Note that f(x) = y, or f(x) and y are synonymous in this context). The slope of the tangent is dy/dx for smaller and smaller increments of dx. But at the point where the tangent’s slope is calculated, dx becomes infinitesimal. In other words, dx ultimately disappears, just like dt disappeared in the above worked example.

Devlin also demonstrates how integration (integral calculus), which in Cartesian analytic geometry calculates the area under a curve f(x), is the inverse function of differential calculus. In other words, for a polynomial, one just does the reverse procedure. If one differentiates an equation and then integrates it one simply gets the original equation back, and, obviously, vice versa.

47 comments:

Anonymous said...

Calculus is a cheat. Unless time is discrete there is no instantaneous. When one uses calculus to describe reality it is doing so with a +/- confidence interval. For most observations the approximation suffices. However, at some level of observation, the calculus is negated.

By the bye, let it be known to all and sundry that Newton discovered the calculus. Newton had a need. Leibniz was only playing parlour games. It is much easier to do something once you know it can be done.

Paul P. Mealing said...

Well, most scholars give equal credit to both Newton and Leibniz.

According to Wikipedia:

A careful examination of the papers of Leibniz and Newton shows that they arrived at their results independently, with Leibniz starting first with integration and Newton with differentiation. Today, both Newton and Leibniz are given credit for developing calculus independently.

I think calculus 'cheats' infinity, especially if one looks at integration. By making increments infinitesimal, we get an infinite sum, which makes it mathematically exact. In differentiation, if the increment is made infinitesimal then it also becomes mathematically exact and allows integration and differentiation to be inverse functions, known as the Fundamental Theorem of Calculus.

Regards, Paul.

Paul P. Mealing said...

Remember that calculus only works for a continuous function, at least over the range one is integrating or differentiating. Implicit in that is that infinitesimals are only valid for a continuous function. Which means that calculus is not an approximation.

Regards, Paul.

Anonymous said...

Indeed, calculus requires not only continuous functions, but also smooth functions.

However, calculus decidedly has nothing to do with vanishing quantities, limits or infinitesimals.

For the first and only rigorous formulation of calculus in human history, visit:

http://johngabrie1.wix.com/newcalculus

There are many errors in these comments that appear on your web page.

Paul P. Mealing said...

Hi Anonymous,

However, calculus decidedly has nothing to do with vanishing quantities, limits or infinitesimals.

What then are 'm' and 'n', if not vanishing quantities in your 'new calculus'? If they are not required, why introduce them? A point has zero dimensions, so how can you find the gradient of a tangent that has no dimensions?

I admit your system works, but I don't see anything new in it except the use of two infinitesimals instead of one.

Regards, Paul.

Paul P. Mealing said...

Just to elaborate on my previous comment.

Someone called John Gabriel wants to be nominated for the Abel Prize in mathematics for supposedly inventing a revolutionary method of doing calculus. However, careful analysis of his method reveals that there is nothing revolutionary about it at all.

The fundamental equation at the heart of his method is the following:

f'(c) = [f(c+m) - f(c-n)]/(m+n)

However, later in the calculation he makes m = n = 0. Therefore his original equation, if taken on face value, is a division by zero.

This is the enigma facing all methods of derivative calculus, including the one I described in my post above, and it is overcome by making 'm' and 'n' infinitesimal or using the method of 'limit'.

So either John Gabriel is ignorant or delusional. Either way, his dissertation is evidence that anyone can post something on the internet and claim they are some sort of genius.

Regards, Paul.

Anonymous said...

You are ignorant Mr. Mealing. Please try to understand what you read or ask nicely and I shall explain.

Mealing: What then are 'm' and 'n', if not vanishing quantities in your 'new calculus'?

Gabriel: They are horizontal distances from the endpoints of secant lines to the point of tangency.

Mealiing: If they are not required, why introduce them?

Gabriel: They are required. Without them, the secant method does not work.

Mealing: A point has zero dimensions, so how can you find the gradient of a tangent that has no dimensions?

Gabriel: I agree a point has no dimensions. Who said a tangent line has no dimensions?

Mealing: However, later in the calculation he makes m = n = 0.

Gabriel: After the difference quotient is reduced, it is perfectly legal to do this. Consider that (m+n) always divides f(c+n)-f(c-m). Therefore, if the gradient is k, then after the quotient is reduced, there will be exactly one term in k. The only (m,n) pair that belongs to the tangent is (0,0). In the New Calculus, m and n have a special relationship - something that can't be done with Newton's flawed calculus. Observe also that before the difference quotient is simplified, both m and n can NEVER be zero, because no secant line possesses a (0,0) pair.

Mealing: Therefore his original equation, if taken on face value, is a division by zero.

Gabriel: Wrong. I suggest you go back and study my New Calculus. If you have questions, I will be glad to answer.

BTW: The New Calculus is truly worthy of 10 Abel prizes. No one before me was able to formulate a rigorous formulation. Please don't spout your ignorance without careful study!

Paul P. Mealing said...

This is so much dribble. In principle, there is nothing new in your calculus. If m and n 'can NEVER be zero' then they are infinitesimals.

You have done nothing that has not been done before in differential calculus. You take small increments (infinitesimals) and make them disappear so they give you the right answer. Sorry, it's been done before.

Regards, Paul.

Anonymous said...

Mealing: This is so much dribble. In principle, there is nothing new in your calculus.

It's completely different to Newton's flawed calculus.

Mealing: If m and n 'can NEVER be zero' then they are infinitesimals.

Gabriel: Nonsense. There is no such thing as an infinitesimal. I did NOT say m and n cannot be zero. I said that before the difference quotient is simplified, m and n can NEVER be zero. After the difference quotient has been simplified, m and n are ZERO. You should try to pay attention to detail.

Mealing: You have done nothing that has not been done before in differential calculus.

Gabriel: That's just your ignorant opinion. My New Calculus is the first rigorous calculus in human history.

Mealing: You take small increments (infinitesimals) and make them disappear so they give you the right answer.

Gabriel: I do no such thing. You don't understand the definition of the New Calculus derivative:

f'(x)= { f(x+n)-f(c-m) } / (m+n)

This difference quotient is the general expression of slope for a secant line that is parallel to the tangent line. No secant line has an (m,n) pair that is (0,0). Only the tangent line has this pair, because after the quotient has been simplified, m and n no longer matter. They don't vanish. They are ZERO because the horizontal distances on each side are ZERO.
To understand this, suppose that the slope of a parallel tangent is k. Then k = { f(x+n)-f(x-m) / (m+n) }. This means that the rise is equal to k(m+n) and the run is (m+n), therefore the slope = k(m+n) / (m+n) = k.
The New Calculus is based on well-defined concepts. There is much that you don’t understand because you have not bothered studying it. Rather than spout your ignorance, ask questions.

Anonymous said...

Unless your next comment shows me that you have tried to understand, I shall probably ignore you.

If on the other hand you ask sensible questions, I might answer.

Paul P. Mealing said...

Well, you may be able to convince others, but you haven't convinced me.

The best exposition on calculus I've read is in What is Mathematics? by Richard Courant and Herbert Robbins. In particular, more recent editions (1996) contain an addendum by Ian Stewart on the 'accepted' role of infinitesimals.

If you think you're cleverer than Ian Stewart regarding anything mathematical, then good luck to you. Stewart is an esteemed mathematician.

Like all versions of calculus, you have selected some point in your calculation to make your infinitesimals (that are not infinitesimals) equal to zero. It's just that your argument for doing so is different to other arguments that effectively use the same device, as you can't do calculus without it.

Regards, Paul.

Anonymous said...

Mealing: Well, you may be able to convince others, but you haven't convinced me.

Gabriel: Others are convinced because they have studied it carefully.

The best exposition on calculus I've read is in What is Mathematics? by Richard Courant and Herbert Robbins.

Mealing: In particular, more recent editions (1996) contain an addendum by Ian Stewart on the 'accepted' role of infinitesimals.

Gabriel: There is not a single mathematician in history who has ever understood calculus the way I have, and I doubt there will ever be one in the future again, because I have formulated a rigorous calculus.


Mealing: If you think you're cleverer than Ian Stewart regarding anything mathematical, then good luck to you. Stewart is an esteemed mathematician.

Gabriel: Stewart is a monkey next to me. Newton, Leibniz and Cauchy would have given anything to see my New Calculus.

Mealing: Like all versions of calculus, you have selected some point in your calculation to make your infinitesimals (that are not infinitesimals) equal to
zero.

Gabriel: The previous statement is such nonsense, that it's hard for anyone to answer. You are stating your opinion, which by the way is not fact. So far you have not been able to show me where you think my New Calculus conforms to your false claims.

Mealing: It's just that your argument for doing so is different to other arguments that effectively use the same device, as you can't do calculus without it.

Gabriel: Again, just your opinion. You obviously haven't understood my New Calculus. It happens to those who don't pay attention to detail.

I am not interested in your opinion, only mathematics. State what you think is wrong and I will address it. State how you think my calculus is in any way the same as Newton's, and I shall provide a refutation. Other than that, I have nothing to add.

Paul P. Mealing said...

You’re right: I’m ignorant about a lot of things, but unlike you I don’t claim to be a mathematical genius of any calibre. And, yes, you could be a genius and I’m too stupid to know it. But where you see differences with traditional methods of doing calculus, I only see similarities, and those similarities don’t surprise me.

You have a different equation to the traditional method of doing calculus but you still require a means to avoid a division by zero. Therefore, in principle, the same mechanism of ‘vanishing quantities’ applies. You simply provide a different explanation.

Regards, Paul.

Anonymous said...

Mealing: You have a different equation to the traditional method of doing calculus but you still require a means to avoid a division by zero.

Gabriel: That is incorrect. Division by zero never occurs in my definition. There is no division by zero - ever.

Mealing: Therefore, in principle, the same mechanism of ‘vanishing quantities’ applies.

Gabriel: False. Nothing 'vanishes'. Every secant line has an (m,n) pair where m and n are never 0. The tangent line also has an (m,n) pair - it's always (0,0), and although it need not ever be used, it is perfectly legal to discard m and n after the quotient is simplified.

Using my definition, there are no approximations, limits or vanishing quantities. The slope of the tangent line is always equal to the slope of a parallel secant line. It's actually quite simple.

Mealing: You simply provide a different explanation.

Gabriel: I provide the first rigorous formulation of calculus in human history. Had Newton, Leibniz or any other mathematician before me, been able to accomplish what I have, calculus would be very different today.

Anonymous said...

{f(x+n)-f(x-m)} / (m+n) is the slope of a parallel secant line. It is also always the slope of the tangent line.

Since no secant line has an (m,n) pair that is (0,0), division by zero never occurs.

Realize that (m+n) ALWAYS divides f(x+n)-f(x-m) 'exactly' to produce some gradient k.

rise = f(x+n)-f(x-m)
run = m+n

There is no such thing as an instantaneous rate or slope at a point. A tangent line always has the same slope. Nothing changes. The modern definition of slope comes from Newton who was trying to find a way to determine tangent line slopes. The Ancient Greeks used angles to determine slope. Newton began to use tan(angle) to determine slope.

By so doing, lines vertical to the horizontal no longer have defined slopes using tan(angle).

Anonymous said...

It's absurd that 'mathematicians' ever subscribed to the nonsense of vanishing quantities.

Of course rise/run is meaningless when rise=0 and run=0. To wit, this is what Cauchy tried to remedy with his introduction of the flawed limit concept.

In the New Calculus, one does not even attempt to do stupid and illogical things like this. There is nothing left to hand-waving explanations or chance.

The New Calculus is what Newton was trying to accomplish. It is the reason he did not rush immediately to publish his ideas, because he knew that he was not quite there yet.

If Newton were alive today, he would be laughing at modern academics and proclaiming me as the greatest mathematician ever.

That I have corrected Newton, Leibniz and Cauchy is irrefutable. However, the part of the New Calculus I have not shared with the world will dazzle the best and brightest minds. It is truly a brand new mathematics based on tangent objects.

While the New Calculus is easily extended to multi-variable calculus in a similar way to Newton's calculus, the machinery I have discovered entails a completely different approach. As I said, it is guaranteed to dazzle.

Aey Lamauo said...

what the heck is Gabriel roleplaying as himself by posting as Anonymous and referring to himself as "Gabriel"

John Gabriel said...

It dawned on me that the language and terminology I was using with the BIG STUPID (mainstream academia) may have been too advanced for them, so I produced the following proofs using only 8th grade arithmetic:

The 8th grade proof of the new calculus derivative:

We can prove that if f(x) is a function with tangent line equation t(x)=kx+b and a parallel secant line equation
s(x)=[{f(x+n)-f(x-m)}/(m+n)] x + p, then f'(x)={f(x+n)-f(x-m)}/(m+n).

Proof:

Let t(x)=kx+b be the equation of the tangent line to the function f(x).

Then a parallel secant line is given by s(x)=[{f(x+n)-f(x-m)}/(m+n)] x + p.

So, k={f(x+n)-f(x-m)}/(m+n) because the secant lines are all parallel to the tangent line.

But the required derivative f'(x) of f(x), is given by the slope of the tangent line t(x).

Therefore f'(x)={f(x+n)-f(x-m)}/(m+n).

Q.E.D.

Also, m+n is a factor of the expression f(x+n)-f(x-m).

Proof:

From k(m+n)=f(x+n)-f(x-m), it follows that m+n divides the LHS exactly. But since m+n divides the left hand side exactly, it follows that m+n must also divide the RHS exactly. Hence, m+n is a factor of the expression f(x+n)-f(x-m). This means that if we divide f(x+n)-f(x-m) by m+n, the expression so obtained must be equal to k. This is only possible if the sum of all the terms in m and n are 0.

Q.E.D.

The previous two proofs hold for any function f.

Visit my New Calculus YT Channel for the best math videos on the web:

https://www.youtube.com/channel/UClBbBVLs3M-d3dNgU4Vop_A

Paul P. Mealing said...

Sorry John,

I couldn't post this when I received it, and then I forgot about it, so it's a few days late, and I apologise for the delay.

The key to your 'Proof' is to determine k (the slope of the tangent and the derivative f'(x)), which is to divide a function by 'm + n', where 'the sum of all the terms in m and n are 0.'

Therefore you are dividing by 0.

Regards, Paul.

John Gabriel said...

Paul,
Sorry, I got no notification that you responded.

At no time do I divide by 0 - ever. See, none of the parallel secant lines have a pair of distances that is (0,0). That the sum of the terms in m and n is zero, is proved in that proof I gave you.

Watch the following videos to understand:

https://www.youtube.com/watch?v=0MTyidv82V0

https://www.youtube.com/watch?v=QjaAMNnMLqM

John Gabriel said...

I produced this video especially for you!

https://www.youtube.com/watch?v=07xr7Wn58q4

Paul P. Mealing said...

Hi John,

I watched all your videos, including the last one. There are a couple of things I noticed.

In your last video, when you are playing with your graph the dividend does become 0 at the 1.31 and 1.35 min mark, where the slope becomes 4.8 and 3.8621 respectively, instead of the constant 3.4322. So even your so-called 'proof' doesn't quite work as you claim in your own demonstration.

The other thing is that you reverse engineer your secant line from the slope of the tangent which you derive by using the standard formula for calculus as I described in my post (above). How can you get your secant parallel to the tangent without knowing the slope of the tangent beforehand?

Calling your opponents names like 'moron', 'baboon', 'idiot' and 'numbskull' doesn't endear you to anyone watching your videos. There are many claimants to the title, 'greatest mathematician ever', all of whom tower over you, so that you have the perspective of an ant when you look up to them.

Regards, Paul.

Paul P. Mealing said...

Correction: I think 'dividend' should be 'divisor', so my mistake.

Btw, in case you think that 'perspective of an ant' is a personal insult, it applies to me too.

Regards, Paul.

John Gabriel said...

"In your last video, when you are playing with your graph the dividend does become 0 at the 1.31 and 1.35 min mark,"

It's correct. But at that point, division doesn't actually take place, because one can't divide by 0. Just because it shows 0 at that point, does not mean it's actually happening. Division by 0 is simply not possible.

"So even your so-called 'proof' doesn't quite work as you claim in your own demonstration."

It works EXACTLY as I claim.

"The other thing is that you reverse engineer your secant line from the slope of the tangent which you derive by using the standard formula for calculus as I described in my post (above)."

No, I don't reverse engineer anything. The slope of any line is found from two given points on the line.

"How can you get your secant parallel to the tangent without knowing the slope of the tangent beforehand?"

You DO NOT have to get your secant parallel to the tangent line. You can assume there are infinitely many parallel secant lines to the tangent. That was the topic of one of the videos you claimed to have watched. Since the slope of the tangent line is some k, then it's correct to assume the parallel secant line must also have a slope k. It turns out this is the case given the derivative: f'(x)=k + Q(x,m,n) where Q(x,m,n) must be 0 in order for this to be true. Finally, whilst m+n is never 0, in the case of the derivative expressed as f'(x)=k + Q(x,m,n), one concludes that setting m=n=0 is equivalent. However, not necessary because Q(x,m,n) is equal to 0.

"Calling your opponents names"

My opponents are absolute imbeciles when compared to my intelligence. I am far smarter than anyone who ever came before me. The New Calculus is no doubt the greatest feat of human intellect.

John Gabriel said...

Perspective of an ant applies to you Paul, but it does not apply to me because I am far more intelligent than you or anyone else could ever dream of being.

Not an insult, but a fact. :-)))

Paul P. Mealing said...

Hi John,

Yes, I know that there are an infinite number of secant lines parallel to the tangent but they are dependent on your values of m and n. Because there's also an infinite number of secants that aren't parallel to the tangent.

When you move your tangent your secant moves with it and as you say: "it's always parallel to the tangent line". But it's only parallel for specific values of m and n, so how do you work out your values of m and n without reverse engineering?

Regards, Paul.

John Gabriel said...

Of course the parallel secant lines are dependent on the values of m and n, but the tangent line is NOT!

A secant line is parallel to the tangent line for all appropriate (m,n). You don't know m and n! You assume that the secant line is parallel to the tangent line. It's a valid assumption as has been proved over 2000 years ago. Based on that assumption, I am able to find m and n through the auxiliary equation Q(x,m,n)=0. There is NO reverse engineering happening anywhere.

Here are the steps:

1. Assume that the secant line is parallel to the tangent line. It is a valid assumption.
2. Show that the slope of a parallel secant line will be f'(x) = [ f(x+n)-f(x-m)] / (m+n) for appropriate m and n. A high school student can do this.
3. Since the secant line is parallel to the tangent line, f'(x) must be equal to k, the slope of the tangent line.
4. f'(x)=k+Q(x,m,n) which is only possible if Q(x,m,n)=0. See, at this step we verify the assumption.
5. From Q(x,m,n)=0 you can find m and n. However, you don't need to, because the sum of all the terms in m and n is equal to 0. It turns out that in this form, you can also use the pair (0,0) for m and n, but all the infinitely many other pairs are still valid.

John Gabriel said...

In my previous comment I mentioned that one can assume there are parallel secant lines. Actually, there is no need to *assume* anything, we know there are parallel secant lines. The only case when there is not a parallel secant line, is if the point is an inflection point, which is fine because there is no tangent line at points of inflection, only half tangent lines as you find in your bogus calculus.

John Gabriel said...

Also, we don't care about the infinitely many secant lines that are not parallel because those that are, must have the same slope as the tangent line.

Paul P. Mealing said...

Okay, but if you choose m and n at random the secant line will most likely not be parallel to the tangent line. It only becomes parallel at the 'limit' when m and n become 0.

But your graph in your video shows that the secant line is parallel to the tangent which is one of your major points. So your graph is the exception rather than the rule, if you choose m and n at random.

To prove my point, I did choose random values of m and n (using the function in your video) and the slope of the secant was NOT parallel to the tangent.

Regards, Paul.

John Gabriel said...

But Paul, I have already explained to you that you CANNOT choose m and n randomly. In fact, it makes no sense. Think about it. m and n and the point of tangency have a relationship. You can always find one given the other two. So, you will always know the point of tangency. So choose m or n and then use the relationship (auxiliary equation) to find the other.

You can't choose BOTH randomly. You can choose one and then find the other (auxiliary equation which is not possible in your bogus calculus).

Of course I show the secant lines parallel. Those are the only secant lines I am interested in because they have the same slope as the tangent line.

Paul P. Mealing said...

Sorry, I've been away all day - it's Easter Sunday here, which is probably just starting for you. I assume you're somewhere in America.

Okay, so you've confirmed something I already knew: you can't 'assume that the secant is parallel to the tangent'.

'You can always find one given the other two'. I've watched your videos again, including your 'proof' of the 'auxiliary equation' being equal to k (the slope of the tangent), which is not so much a proof as a definition. The equation only equals k for certain values of m and n. If you know m and k then you can find n.

Nothing I've seen explains how you get the values of m and n, and I can't see how you can derive them without knowing the slope, k. You obviously have an algorithm in your graph that determines them as you move certain points about. But it depends completely on the secant being parallel to the tangent, so the slope of the tangent must be known to create a parallel secant.

Regards, Paul.

John Gabriel said...

You can assume there are parallel secant lines and that is sufficient.

But of course you know there is some slope k. Hey, that slope k is what you find when you set h=0 in your bogus calculus, isn't it?

The algorithm you talk about is the auxiliary equation. It is described in detail here:

https://drive.google.com/file/d/1A3ZXEZGRv6bqCAOo9ymJ4tZXJRZ17sUt/view

The slope of the tangent line is found by the finite difference which defines the slope of a parallel secant line. It is NOT known beforehand. What part of the proof do you not understand?

Proof:

1. We can assume that there are innumerably many secant lines that are parallel to the tangent line and also innumerably many secant lines that are not parallel to the tangent line.
2. We form our difference quotient [ f(x+n)-f(x-m) ] / (m+n) denoting the slope of any secant line..
3. To find any pair (m,n) such that the difference quotient [ f(x+n)-f(x-m) ] / (m+n) is equal to the tangent line slope, whose equation is given by t(x)=kx+b, we need [ f(x+n)-f(x-m) ] / (m+n) = k.
4. Simplifying [ f(x+n)-f(x-m) ] / (m+n), we get an expression without m or n and an expression Q(x,m,n) whose terms may contain x, but must contain either of m or n or both. If there is no m or n, then Q(x,m,n) is 0.
5. Since the tangent line slope does not depend on m or n, the sum of all the terms in m and/or n must be zero. This is only possible if Q(x,m,n) = 0.
6. Therefore the expression without m or n, must be the same as k, that is, k = [ f(x+n)-f(x-m) ] / (m+n).

Since k is the tangent line slope which is called the derivative and assigned the symbol f'(x), it follows that f'(x)= [ f(x+n)-f(x-m) ] / (m+n).

Objections:
"But Mr. Gabriel, you are dividing by zero!"

No, I am not. There is no secant line whose (m,n) pair is (0,0). The difference quotient is valid only if m+n > 0.

"But Mr. Gabriel, you are reverse engineering!"

No, I am not. All my assumptions are valid and I do not assume anything about the derivative, except that it must be equal to the tangent line slope.

"But Mr. Gabriel, how do you know there is a term without m or n in it?"

Two reasons: (i) k does not depend on m or n. (ii) I can prove that m+n is a factor of every term in the numerator:

From k(m+n)=f(x+n)-f(x-m), it follows that m+n is a factor of the LHS which is a product of two factors k and m+n. k(m+n) is measured by m+n exactly k times. Since m+n is a factor of the left hand side, it follows that m+n must also be a factor of the RHS, that is, the RHS is also a product of two factors. Hence, m+n is a factor of the expression f(x+n)-f(x-m), that is, m+n measures f(x+n)-f(x-m) exactly k times also. This means that if we divide f(x+n)-f(x-m) by m+n, the expression so obtained must be equal to k. This is only possible if the sum of all the terms in m and n are 0. The equation formed by setting these terms to 0 in the New Calculus is called the auxiliary equation - the first powerful feature new students learn about which is not possible in any way in the flawed mainstream calculus.

Paul P. Mealing said...

Okay, thanks for that. The link you gave is what I was looking for.

So you've worked out a clever algorithm that gives you a parallel secant to a tangent, without having to find the derivative. Maybe that's your 'great' contribution to mathematics, though you claim Euclid did it 2300 years before you. To be fair, I doubt that Euclid came up with the algorithm. I acknowledge your 'modest' contribution, if it hasn't been done before (I don't really know if it has or not).

However, as you say yourself: "...you don’t need to do any of this to find the derivative." In fact, it only gives the derivative if m and n are both 0. So it doesn't give us anything we didn't already know, and the conventional method of finding the derivative (as per my post) does it without a complex algorithm, which actually has to disappear to give the derivative.

So I would contend that you haven't contributed to calculus per se, if 'you don't need to do any of this to find the derivative.'

Both your method and the so-called 'bogus' method require 'vanishing entities', as I describe in my post, so you've contributed nothing except a clever method to find a secant parallel to a tangent without the derivative. That bit I'll acknowledge.

Regards, Paul.

John Gabriel said...

Well, actually it is a rather big deal, because no one before me realised it. You're getting very confused - Euclid knew that one could construct parallel lines to any given line. He did not realise the New Calculus.

"However, as you say yourself: "...you don’t need to do any of this to find the derivative." In fact, it only gives the derivative if m and n are both 0. "

Rubbish. The derivative is f'(x)= k + Q(x,m,n). That Q(x,m,n) is PART of the derivative. If you had bothered to read the proof and understand it, you would know that the sum of all the terms in m and n is zero. That this is equivalent to m=n=0, does not mean that m and n are ever equal to 0 or even need to be, for otherwise the secant method would not work!

Tell me, which secant line ever has the pair (0,0)? Not one!

"So it doesn't give us anything we didn't already know,"

Not entirely true. It gives you a rigorous method without any vanishing quantities, limits, infinity or any other nonsense in the bogus mainstream formulation.

"and the conventional method of finding the derivative (as per my post) does it without a complex algorithm, which actually has to disappear to give the derivative."

Only problem is that it is flawed.

"Both your method and the so-called 'bogus' method require 'vanishing entities',"

Gosh, I'd say you are really dim. My method is 100% rigorous. There are no vanishing quantities. Those secant lines have been there since eternity. Nothing approaches anything as happens in your bogus calculus. The m and n that you see in k+Q(x,m,n) have infinitely many values that will make Q(x,m,n) equal to 0. Nothing vanishes anywhere.

"so you've contributed nothing except a clever method to find a secant parallel to a tangent without the derivative."

Um, no. I did what no human before me was able to do: formulate a rigorous calculus without any bullshit (excuse my language) and that is a very big deal.

Also, the new calculus integral is different to your mainstream definition, but given that you're struggling so hard to understand the derivative, I don't hold out much hope you will understand the integral any time soon either.

I gave you the proof and also a chance to show me where you think it is wrong. You haven't done so and it seems I am wasting my time with you.

Paul, the new calculus is established fact. I am not debating it with you. I am trying to teach it to you.

Paul P. Mealing said...

Yes, I see that the Q part of your equation always equals 0, even for non-zero values of m and n. But graphically the secant line actually becomes the tangent line only when m=n=0.

All non-zero values of m and n represent a secant line, not the tangent line, as demonstrated by your own manipulation of your graphical representation.

When you say: 'you don't need to do any of this to find the derivative'; you mean you don't need to solve the auxiliary equation. But you do need to solve it if you want to know the non-zero values of m and n for a specific secant. Of course, when the secant is the tangent, then m=n=0 as above, and the formula disappears, and therefore becomes irrelevant at the tangent - it no longer exists.

I understand that your 'method' finds the derivative equation f'(x), by using a secant rather than the tangent itself. But unless m=n=0, it doesn't represent the tangent. You've found an alternative method to find f'(x) from any secant, not just the tangent.

In the example I use in my post I'm using a physical phenomenon, not a graph where to obtain an 'instantaneous' velocity, dt must go to 0. Now, I know you could graph it and find the answer using your Q, but physically it requires an infinitesimal duration.

You haven't changed calculus. Schrodinger's equation still works using the same formula for derivatives we've always used, which, in the end, is the same as yours.

I admit what you've done is clever, but it hasn't changed the fundamental formula we use for differential calculus. It's finding a derivative from a secant, but the Q part is zero (as you point out) so it makes no difference to the formula we've always used.

Regards, Paul.

John Gabriel said...

"But graphically the secant line actually becomes the tangent line only when m=n=0."

No. The secant line never becomes a tangent line. What you're saying is impossible nonsense. Secant lines are not secant lines if they become tangent lines. That is brainwashing you have experienced from your mainstream education. Geometric object do NOT change.

"All non-zero values of m and n represent a secant line, not the tangent line, as demonstrated by your own manipulation of your graphical representation.'

FALSE! They represent both the slope of the parallel secant lines and the tangent line.

"Of course, when the secant is the tangent, then m=n=0 as above, and the formula disappears, and therefore becomes irrelevant at the tangent - it no longer exists."

The formula DOES NOT disappear! If I wrote k + Q(t) where Q(t) = 5t-5t, then Q(t) DOES NOT disappear. It has a value of ZERO regardless of the value of t! Nothing disappears.

"I understand that your 'method' finds the derivative equation f'(x), by using a secant rather than the tangent itself. But unless m=n=0, it doesn't represent the tangent."

Really?! Did you graduate from high school? Didn't you learn in high school that parallel lines have equal slopes???!!! The derivative equation f'(x) ALWAYS represents the tangent line. It does not care about the value of m or n. In fact, m and n are irrelevant to the tangent line.

"You've found an alternative method to find f'(x) from any secant, not just the tangent."

CORRECT! Now contrast this statement with the idiocy you wrote earlier and ask if you haven't contradicted yourself!

"In the example I use in my post I'm using a physical phenomenon, not a graph where to obtain an 'instantaneous' velocity, dt must go to 0."

Your example is total and complete anti-mathematical NONSENSE.. There is no such thing as an instantaneous rate. It is not possible. There are MANY reasons, but consider that the minute you calculate a rate, it is already in the past. Moreover, time is not composed of infinitesimals (which are also a MYTH) so that you can divide by something infinitely small.

"Now, I know you could graph it and find the answer using your Q, but physically it requires an infinitesimal duration."

Rubbish. You CANNOT even produce ONE infinitesimal. Sorry Paul, I do not rely on FAITH. My intelligent brain guides me. Nothing "appears" to me. I know exactly what is happening.

"You haven't changed calculus."

I HAVE!! My New calculus is the FIRST AND ONLY rigorous formulation of calculus in human history. More intelligent aliens probably know of it already.


"Schrodinger's equation still works using the same formula for derivatives we've always used, which, in the end, is the same as yours."

The results of mainstream calculus are generally correct (definition of tangent line is WRONG in mainstream calculus), BUT it's formulation is FICTION.

"I admit what you've done is clever, but it hasn't changed the fundamental formula we use for differential calculus."

It has changed everything! It is RIGOROUS. That is a very big deal.

"It's finding a derivative from a secant, but the Q part is zero (as you point out) so it makes no difference to the formula we've always used."

Correct! However, it is RIGOROUS. That's a HUGE difference. Also, the new features such as the auxiliary equation and numerous features you still haven't gotten to know yet because you are stumped on the derivative, are NOT possible in your bogus calculus.

To say I haven't made a difference is FALSE. Not a single human before me was able to realise the method. Gee, I'd say that is not only clever, it's a historic accomplishment!

John Gabriel said...

"In the example I use in my post I'm using a physical phenomenon, not a graph where to obtain an 'instantaneous' velocity, dt must go to 0. Now, I know you could graph it and find the answer using your Q, but physically it requires an infinitesimal duration."


There is NO such thing as Instantaneous Rate. No time unit is indivisible! 

To understand what is time, you need to read my article:

https://www.linkedin.com/pulse/what-time-john-gabriel

I can't help becoming irate when any academic who talks about "instantaneous" anything. 

One of my readers commented on my YT video at 
https://www.youtube.com/watch?v=WrfywOcx7K4&lc=UgzB2oIYfvhBoUmMlYp4AaABAg   
which is part of the Academic Ignorance and stupidity Series: 

@tthirupathy 

Very useful and fundamental video to understand both differentiation and integration. If you explain this using f(x)=x squared and  f '(x) =2x , it will be easier to grasp the meaning. I always used to understand mathematics definitions using physics. For example the motion of a car ( assuming the movement is uninterruptible) , if the average speed of a car between any two locations was V km/hr then there was at least one instant where the speed indicator displayed V km/hr. If you are able to think and understand this action , you might have accepted instantaneous rate of change there by the mainstream calculus. When we are in child age , this may not be possible. For beginners we should emphasis to understand mathematics using real-time physics examples. Practice of mathematics without understanding proofs is a useless study. It doesn't make any sense.  This video should be kept as first introduction of NC . Ok let's see?. 

My response: 

It is correct to say that there is at least one instant where the speed is V km/hr, but that is the speed at time t . It is NOT an instantaneous speed because t is measured in a given time unit which is  NOT indivisible , therefore to talk about an instant is plainly stupid!? 

Explanation: 

One of the dumbest concepts is "instantaneous rate" which is absolute rot.  Let me educate the lot of you who think otherwise! 

Velocity = {  distance covered   }   /  {   elapsed time }   

or in symbols: 

Velocity = {  s(t_final) - s(t_initial)  }  /  {  t_final - t_initial  } 

where  s(t) is the input from an analog device or in theoretical examples just the position function. Did you get this?  In airplanes, rockets, cars, etc, NO calculus is used to determine the velocity!  Most of these systems are *causal systems*. Read up on that! 

There is NO such thing as an instantaneous rate!!!!! 

Try thinking for yourselves!  It can be very refreshing as you become better at it! 

https://www.youtube.com/watch?v=MgUB0pILNj8 

Paul P. Mealing said...

I am not an academic, but I actually agree with what you say about 'instantaneous' - I make the same point in my post (if you've read it) that time is indivisible, but mathematically you can have an ideal that doesn't exist in the physical world.

The point is that you can calculate, using the calculus formula (the exact same as yours), to any degree of 'instantaneous' you want, as long as the x in f(x) is 'Real' and the function is continuous. Note: I put instantaneous in inverted commas because I know it's a fudge, and I specifically say: '…even though zero duration is physically impossible'; in my post. I say x is Real, but you can do calculus with complex numbers as well, as I'm sure you know.

Regards, Paul.

John Gabriel said...

You cannot have an ideal in mathematics. Concepts have to be well formed otherwise they are nonsense. Mathematics came from Platonism which supports the view that well-formed concepts or noumena are independent of the human mind or any other mind.

There is no such thing as "real number". Neither Dedekind Cuts nor Cauchy sequences are valid constructions.

Rather than use the incorrect word "instantaneous", you can just say "at time t".

I don't consider complex numbers to be actual numbers, but quasi-number objects. Of course I am aware of calculus with these objects - most of it is theoretical nonsense with no practical applications.

Zero duration is a myth. Duration by definition is greater than 0.

BTW: YOu did not publish my previous comment which refutes what you wrote about the secant and tangent lines. It refutes the claims you made in your comment time stamped 02 April, 2018 15:43

John Gabriel said...

I see you did publish it. Sorry then.

Paul P. Mealing said...

I've posted all your comments but one went to spam and I didn't see it, but it's now posted.

I actually agree with this statement: "Mathematics came from Platonism which supports the view that well-formed concepts or noumena are independent of the human mind or any other mind." But those 'noumena' include complex numbers and transcendental numbers like pi and e, all of which are made famous in Euler's identity.

I think you are making a big deal out of very little. You've found another way to derive a well known formulation. You derive the slope of a secant instead of the tangent and get the same answer because parallel lines have the same slope by definition. That's not a big deal.

John Gabriel said...

I am going to be directing a lot of traffic to this page. Hope you don't mind.

Many people ask the same questions as you do and I am tired of repeating myself. So this page can serve as a record. In fact, I may just create a PDF of it if you don't mind?

John Gabriel said...

"But those 'noumena' include complex numbers and transcendental numbers like pi and e, all of which are made famous in Euler's identity."

pi and e are noumena, but complex numbers are not. If one thinks of a complex number as a vector, it makes sense, but to call it a number is wrong, because a number is the measure of a magnitude:

https://www.linkedin.com/pulse/how-we-got-numbers-john-gabriel-1

So complex number theory can be useful in proving theorems, for example the fundamental theorem of algebra. The idea of complex number is not well formed, unless you realise the concept as a vector.

"I think you are making a big deal out of very little. You've found another way to derive a well known formulation. You derive the slope of a secant instead of the tangent and get the same answer because parallel lines have the same slope by definition. That's not a big deal."

It's not a big deal?!!! Ha, ha. Think about what you wrote there. Since Newton (and before), no one has been able to realise this *rigorous* method which I realised. It's also a noumenon by the way! Whilst I am the first to realise it, does not mean I own it.

It's ingenious!!! Never mind a big deal. I think you are either being shortsighted and ignorant or dishonest if you think it is not a big deal, because it is a big deal. In all human history, no one before me realised it. And don't forget what came out of it: auxiliary equation (had many powerful uses), new definition for integral, first constructive proof of the mean value theorem, etc...

Sorry Paul, it's a big deal alright! Chuckle.

Paul P. Mealing said...

Yes, i is not a number per se - I've made that point myself (on this blog) - it can't quantify anything. You call it a vector and I've called it a dimension, because that's how it's expressed graphically and algebraically. But it's a very important concept because we wouldn't have quantum mechanics without it.

We do agree on some things.

Maybe I am shortsighted or ignorant but not dishonest. History will decide in the end and I can accept that.

John Gabriel said...

Okay. I'll say one more thing about this and leave it at that:

So mainstream calculus has built up all this machinery: real numbers, limits, infinity, etc. and here I come along with a sound analytic geometry formulation, and you think it is no big deal. Well, history will decide.

"Devlin also demonstrates how integration (integral calculus), which in Cartesian analytic geometry calculates the area under a curve f(x), is the inverse function of differential calculus. In other words, for a polynomial, one just does the reverse procedure. If one differentiates an equation and then integrates it one simply gets the original equation back, and, obviously, vice versa."

Whilst every academic before me knew about the mean value theorem, not a single one was able to prove it constructively. Yes, Newton and Leibniz knew that the reverse procedure yields a derivative. However, none understood why.

Well, the astounding truth is that the fundamental theorem of calculus is derived in ONE step from the mean value theorem. Had I not discovered the New Calculus, I would not even have been able to prove it using your mainstream formulation, which I did here:

https://drive.google.com/open?id=0B-mOEooW03iLZG1pNlVIX2RTR0E

But not without creating a patch (the poditional derivative) so that it can be done because your mainstream calculus is broken beyond repair:

https://drive.google.com/open?id=0B-mOEooW03iLVVg3QWtOdkxUbVk

I would never have realised the Gabriel polynomial which is a closed form of Taylor's series. Too much to discuss here. But you can get an idea here:

https://drive.google.com/open?id=0B-mOEooW03iLc0JhR00xbnY4dms

John Gabriel said...

"I think you are making a big deal out of very little. You've found another way to derive a well known formulation. You derive the slope of a secant instead of the tangent and get the same answer because parallel lines have the same slope by definition. That's not a big deal.

02 April, 2018 22:59"

Just to set things straight:

"You've found another way to derive a well known formulation."

i. I have found the ONLY sound way of determining a derivative. What you learn in your bogus calculus is unsound nonsense and a FLAWED formulation.

ii. "That's not a big deal."

It's a very big deal because no one before me was able to do this. You saying this is dishonest.

iii. "You derive the slope of a secant instead of the tangent and get the same answer because parallel lines have the same slope by definition. That's not a big deal."

Indeed, I derive the slope of the tangent line using a parallel secant line, not just any secant line. It's a big deal when you didn't know it! As for this matter, no one before me knew it also.