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.

19 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.