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The Proof for the Current Derivative for Powers is False
by Miles Mathis
Yes, I will show that the proof of y' = nx^{n1} is false. Not only unnecessary, but false. I will reprove it by a simpler and more transparent method.
Many readers don't understand why I would attack the calculus, so I try to begin all these papers by reminding them that physics has hit several major walls in the past century. Pure mathematicians may not be aware of that, so they may not be aware that we have empirical evidence that their maths are failing. I get emails telling me that the calculus is the greatest thing ever invented and that I am either an ingrate or a monster for looking closely at it. But I am looking closely at it for a reason. The two pillars of 20th century physics, Quantum Mechanics and General Relativity, both hit similar walls. Many people know that they have failed to be unified, but most people don't know that both have to be renormalized. Renormalization is a big part of QED, as is admitted, and GR also requires a sort of renormalization, a push that is hidden in the tensor calculus and in the field definitions. Renormalization was perfected by a famous physicist named Richard Feynman, and he is notorious for calling his own creation “hocuspocus” that was “not mathematically legitimate.” He also called it a shellgame. What does renormalization do? It removes zeroes and infinities from equations that are imploding or exploding. Why are these equations imploding and exploding? No one knows. Richard Feynman was the top mathematical physicist of his time, and Edward Witten is the top mathematical physicist now. Witten has posed just this question at Claymath, as one of the Millennium prizes. He wants to know why the point maths of QED and QCD are failing. I have no intention of submitting for this prize, since I know they will not like my answer. But my answer is that the maths are failing because they are based on the calculus, and that the calculus is failing in QED because it is based on the point and on a move to zero. This has also affected the search for unification, since the mainstream is trying to unify by quantizing gravity. But since they have misdefined the photon and other particles as point particles, based on a misunderstanding of the calculus, this effort is wasted.
Therefore, my work on the calculus is neither capricious nor insolent. It may seem to overreach at times, but it is always focused. It is focused on redefining the derivative and on jettisoning the point from all equations.
The next complaint I hear is that I seem to have an aversion for limits and infinities. In fact, I don't^{1}. I believe some problems are best solved with limits: I just don't think the calculus is one of them. The calculus can be solved by simple number relations, because that is what creates the equalities. As it turns out, proving the calculus with limits is not only unnecessary and inefficient, it is false. It breaks rules and finds fake numbers. It also warps fields and allows for particles and motions that cannot exist. The problems embedded in the calculus are what have caused many of the physical problems in the past century.
Currently, modern mathematicians use the calculus to find a derivative and a slope of the tangent by taking Δx to zero. I have shown^{2} many reasons they can't do that (and don't need to do that), but the main reason is the one I will concentrate on in this paper: it changes the given curve. If you go below 1 for the change of your independent variable, you will have changed the curve. This is important, because unless you also monitor that change, you will get the wrong answer for your curve at x. I will show you what I mean straight from the tables for x^{2} and x^{3}.
Let Δx=1 (x= 1, 2, 3, 4...)
x^{2} = 1, 4, 9, 16, 25, 36, 49, 64, 81
x^{3} = 1, 8, 27, 64, 125, 216, 343
Δx^{2} = 3, 5, 7, 9, 11, 13, 15, 17, 19
Δx^{3} = 7, 19, 37, 61, 91, 127
ΔΔx^{2} = 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
ΔΔx^{3} = 6, 12, 18, 24, 30, 36, 42
ΔΔΔx^{3} = 6, 6, 6, 6, 6, 6, 6, 6
Let Δx=.5 (x=.5, 1, 1.5, 2...)
x^{2} = .25, 1, 2.25, 4, 6.25, 9
x^{3} = .125, 1, 3.375, 8, 15.63
Let Δx=.25 (x=.25, .5, .75, 1...)
x^{2} = .0625, .25, .5625, 1, 1.5625
x^{3} = .01563, .125, .4219, 1, 1.95
If Δx=.5, then y = x^{2} no longer has its original rate of change or curvature, as you see. It has exactly ¼ the curvature it originally had. The curve y = x^{3} loses much of its original curvature, too: it retains only 1/8 of its curvature. If we continue taking Δx toward zero, by making Δx=.25, this outcome is magnified. y = x^{2} has 1/16 of its curvature, and y = x^{3} has 1/64 of its curvature.
This shouldn't be happening, and is not usually known to happen. You will not see the curves analyzed in this way.
A critic will say, “Of course the curve is straightening out. That is the whole point. We are going to zero to magnify the curve. When you magnify a curve, its loses its curvature at a given rate, depending upon the magnification. Your curve x^{2} at Δx=.5 IS the same curve, it is just four times smaller. ”
True, but the curve should lose its curve at the same rate you magnify it. If all the calculus were doing is magnifying the curve, then if you magnified 2 times, the curve would lose half its curve. If you are approaching zero in a defined and rigorous manner, your magnification and curvature should change together. But here, you magnify by 2 by halving your Δx, but your curvature has shrunk to ¼ with x^{2} and to 1/8 with x^{3}. That is not a quibble, that is a major problem. If you change your curve, you change your tangent.
My will critic will answer, “It doesn't matter how much the curve changes as we go in. We are going into a point, and the tangent only hits at a point. Therefore the curvature won't change at that point.”
Wow, that sounds like pettifogging to me. By that argument you can make the slope anything you want to at any point on any curve. If changing the curvature doesn't really change the curvature, then curvature has no meaning.
Currently, the calculus just ignores this problem, or dodges it with oily answers like that last one. To approach a limit in this way while your given curve is changing would require a very tight proof to convince me it is legal, and I have never seen one. If you dig, you find that it requires an infinite line of proofs to “prove” the legality of the first move to zero. For example, if you go to Wikipedia, you will see the first in this line of proofs. Wiki starts by telling us that the difference quotient
has the intuitive interpretation that the tangent line to ƒ at a gives the best linear approximation to ƒ near a (i.e., for small h). This interpretation is the easiest to generalize to other settings.
But to tighten this up a bit, they next let the slope of the secant Q(h) go to zero, and tell us
If the limit exists, meaning that there is a way of choosing a value for Q(0) which makes the graph of Q a continuous function, then the function ƒ is differentiable at the point a, and its derivative at a equals Q(0).
They still have not proved anything there, they have just juggled some terms. Notice they say, “IF the limit exists.” In fact, they admit in the next sentence that the quotient is undefined at h=0, which means the limit they have just created does not exist. You cannot choose the value h=0, so their function is nullified.
Some will say that is an unnecessarily harsh judgment, but it is no more than the truth. Every point on every curve becomes a limit with the modern calculus, since whenever you approach a value of x, you are approaching a limit to find the derivative at that point. Q(0) exists not at the limit of some given curve, it exists at every point on that curve. Any point you desire to find a derivative for becomes your limit of zero. So a curve is just a compendium of limits. A curve becomes a sum of zeroes. Zeno knew^{1} that was a paradox 2500 years ago, but the modern calculus still boldly embraces it.
Wiki admits that taking Δx (their h) to zero is a problem:
The last expression shows that the difference quotient equals 6 + h when h is not zero and is undefined when h is zero. (Remember that because of the definition of the difference quotient, the difference quotient is never defined when h is zero.) However, there is a natural way of filling in a value for the difference quotient at zero, namely 6. Hence the slope of the graph of the squaring function at the point (3, 9) is 6, and so its derivative at x = 3 is ƒ '(3) = 6.
More generally, a similar computation shows that the derivative of the squaring function at x = a is ƒ '(a) = 2a.
Do you see what they just said? After 300 hundred years, this is the rigor we get. Wiki tells us there is “a natural way of filling in a value for the difference quotient at zero.” That just means that we already know what the derivative is by looking at differentials. We know the answer, so we push the difference quotient to match it. That is the “natural way” of solving this.
True, there are other more complex methods for proving the move to zero. In fact, there are three centuries worth of proofs, in hundreds of thousands of pages, from Newton and Leibniz and Euler and Lagrange and Cauchy and Riemann so on, all different and all in different notations. But if the answer were clear, don't you think it could have been presented a bit more quickly and easily than that? One would think that if the move to zero were legal, it could have been shown immediately. In my experience, only things that aren't true require proofs of a million pages over many centuries.
I think that just from what I have said, it is clear that the move to zero is illegal. You cannot go to a limit to analyze a curve when your curve is changing at a different rate than your approach to the limit.
To solve, modern mathematicians simply shrink Δx to suit themselves, never noticing or caring that it must change the curve of the given curve. In other words, they take a graph like the one below, draw the forward and backward slopes (or secants, as the case may be), then begin making them smaller and closer to their chosen point. Because it all looks perfectly legal on the graph, no one ever questions the legality of it. But I have just shown it is strictly illegal. If you go below Δx=1, you will change your curve. If you have made your Δx twice as small and at the same time your curve is 4 times smaller, then your absolute curvature has changed. There is no way around it.
But even if one or all of the millions of pages of proofs are correct, it doesn't matter. Why should we choose to solve this problem with a million pages of difficult proofs, when we can solve it by looking at a few tables of simple differentials? Why do teachers and textbooks and Wiki reference all these complex proofs and never show us the simple tables?
Regardless of the status of all these proofs, going to zero wasn't necessary to begin with. We can find specific slopes as well as general slope equations by several other methods, and none of them use limits. We don't need to go below Δx=1, because the forward slopes and backward slopes will give us the slope at x by a simple average. Since x is changing at a constant rate on the graph, the forward slopes and backward slopes are the same size differentials, by definition. The constancy of change in x assures us that our given value of x is at the midpoint between forward and backward slopes. Just look at the graphs: the change in x is always the same.
My critic will say, “What you say is true of squared acceleration, but you clearly don't understand cubed acceleration. You can't find distances from cubed accelerations by averaging, since the distance in the second period is much greater than the distance in the first.” Well, that is also true of squared acceleration. With a squared acceleration, the distance in the second period is much greater than the distance in the first. So that isn't the reason we can't (at first) seem to average. The reason we can't seem to average with powers above 2 is that the power 2 changes at a constant rate of 2, but higher powers don't.
Let me show you exactly what I mean. We can find a slope for x^{2} very simply and accurately by averaging forward and backward slopes, as you see from this graph. However, another similar graph tells us we cannot get the current value of the slope that way for x^{3}. Why? It is because the curve x^{2} is changing 3, 5, 7, 9. You can get that either from the table or the graph. It is changing 2 each time. The curve x^{2} has a fundamental acceleration of 2. Therefore we can average in one step. The average of 5 and 7 is 6, which is the slope at x=3. But the curve x^{3} is changing 7, 19, 37. It appears we can't average.
The modern calculus tells us this is why we have to go to zero. We can't average forward and backward slopes with most functions, therefore we have to solve by going to zero. But that is false. With x^{3} we don't have to go to zero any more than we did with x^{2}. We can find a derivative with a simple average. Like this.
Since x^{3} is changing 7, 19, 37, it has a fundamental acceleration of 6n (where n=1, 2, 3). You can see that in the last two lines in the table above. That being the case, our acceleration could be written as this series:
1, 1 + 6, 1 + 12, 1 + 18, 1 + 36...
That is where the numbers 1, 7, 19, 37 come from. So, if we want to find the slope at 3, say, that will be between the numbers 19 and 37. Just consult the graph. I have shown that we cannot average 19 and 37 directly, because that would give us the number 28, which is not the current slope. But since the curve is is achieved by a 1+ series, we can subtract the one away from each term. If we do that, then our forward and backward slopes at x=3 will be 18 and 36, in which case we can find the current slope by averaging. (18 + 36)/2 = 27. That is the current slope at 3. So we could find a slope just by averaging, even with an acceleration of 6n.
You will say, “Wait, you just changed your curve by doing that. You just proved that changing the curve was forbidden, then you did it. You subtracted 1 away from your series, and you now have this series:
Δx^{3} = 6, 18, 36, 60, 90
Those are the rates of change for 0, 6, 24, 60, 120, 210, not x^{3} = 1, 8, 27, 64, 125, 216.”
True, but the curve 0, 6, 24, 60, 120, 210 is still an acceleration of 6n, therefore it is an acceleration above x^{2}, therefore you CAN find an average acceleration for powers above 2. You can't find it just by adding two numbers and dividing by 2, but you can find it. In this case, it is the forward slope minus 1 plus the backward slope minus 1, over 2. It is still an average, it is still very simple, and it doesn't require using a limit.
m@(x,y) = {[Δy@(x+1)] – 1} + {[Δy@(x)] – 1}
2
The same analysis applies to x^{4}:
m@(x,y) = {[Δy@(x+1)] – 12} + {[Δy@(x)] – 12}
2
Because we can average forward and backward slopes like this with a general equation, it means the process is not an accident or push.
Δx^{5} = 1, 31, 211, 781, 2101, 4651, 9031
m@(x,y) = {[Δy@(x+1)] – (10x2 + 1)} + {[Δy@(x)] – (10x2 + 1)}
2
We can average powers above 2 because they are constant. They are constant not as the power 2 is constant: the power 2 is constant at the first rate of change. But all simple powers are constant in that they increase in a consistent manner, by a process that can be broken down. We can see that right from the tables. If we take enough changes of any power, we see that it is constant at a fundamental level. That is what 6, 6, 6, 6 is telling us about x^{3}. Two rates down, it is constant. Therefore it is constant. That was my point in a recent paper on “variable” acceleration. Cubed acceleration is not really variable. It is constant. It can be averaged, if you do it in the right way. It is a consistent increase, therefore it can be analyzed in a straightforward manner, as we are doing here. We don't need limits, we can just use simple number relations.
Although I have shown we can average forward and backward slopes with all powers, the slope equations get very complicated as we advance into the higher powers. We also encounter a problem with finding slopes for values of x near 1, since we are subtracting large numbers from our Δy's. This means we need a better way to generalize our slope equation. I have already shown how to do that in my long paper on the derivative. I will gloss it again here.
We will pull the general equation straight from the tables. We will start with the smaller powers. Since x^{3} is changing 7, 19, 37, it has a fundamental acceleration of 6n (where n=1, 2, 3). You can see that in the last two lines in the table above. Since x^{2} has a fundamental acceleration of 2, the fundamental acceleration of x^{3} is 3 times that of x^{2} over each interval. Six is three times two. We can write that as f x^{3} = 3x^{2}, where f means fundamental acceleration.
If we are physicists, or logical people of any stripe, that proof of the derivative of x^{3} is much preferable to the current one. We don't go to zero, we don't talk of limits or functions or infinitesimals or any of that. We pull the general derivative equation straight from a table of differentials, and in doing so we see right where all the numbers are coming from. Now we just need to generalize that equation. We can do that by analyzing other powers. By studying the simple tables^{2}, we find that all other powers obey the same relationship we just found between x^{2} and x^{3}.
f x^{n} = nx^{n1}
The differentials themselves give us the derivative equation for powers. This means we don't need any other proof of it. A table of differentials is all the proof we need. It is a proof by “show me.” You want me to prove that a dog is white, so I show you the white dog. You want me to prove that the derivative equation for powers is f x^{n} = nx^{n1}, so I show you the tables, with the numbers sitting right next to each other. If you require a proof beyond that, we must call you a confused and meddlesome person, and we recommend you go into set theory, where you can write thousandpage books proving tautologies (while ignoring much greater real problems sitting on your desk).
I will answer one more question here before I move on to the other more important questions on my desk. A close reader will ask, “We can write the series 0, 6, 24, 60, 120, 210... as x^{3} – x, and you have shown that both the curve x^{3} – x and the curve x^{3} can be written as accelerations of 6n. By your abbreviated and direct proof, both curves should have a derivative of 3x^{2}. But they don't. The derivative of x^{3} – x is 3x^{2} – 1. How do you explain that?”
Once again, I am not here to show that the current derivatives for powers are wrong. I am here to show that the proofs are wrong. I admit the derivatives are different for x^{3} – x and x^{3}, but that difference can be shown and generalized without using limits. In this case, the difference is caused by the first term in the series. The first term in one series is 1 different from the other, and so is the derivative. So the difference in equations can be shown by simple demonstration, or by pointing to a table. It doesn't require limits or difficult proofs. I have not exhausted all the demonstrations, or answered all questions. I am only here to suggest that every question has a simpler answer than the one we have so far been shown, one that can be achieved without limits. All calculus questions can be answered by studying the tables, since the tables supply the actual number relations that generate the calculus. Fundamentally, calculus is about these number relations, not about limits or approaches to zero.
Because the calculus is not about limits and can be proved without limits, it cannot find solutions at points or instants. My method differs from the modern calculus not only in its simplified proofs, but in its definitions. Because Δx is always 1 and cannot go below one, our derivatives and solutions are always found over a defined interval of 1. Instantaneous velocities and accelerations are impossible, as are point particles and all other solutions at points. This solves many of the problems of QED and General Relativity. It solves renormalization directly, since the equations are never allowed to become abnormal to begin with. And it disallows "mass points" in the field equations. If you cannot have math at a point, you cannot have mass at a point. Modern physicists have been fooled by the calculus into thinking they can or should be able to do things they simply cannot do. My correction to the calculus disabuses them of this mistaken notion. They have had problems with points in their math and their fields because points do not exist, in either math or fields. Only intervals exist. Only intervals can be studied mathematically. This is why they call it the differential calculus. It is a calculus of differentials, and differentials are always intervals. Just check the epsilon/delta proof. It is defined by differentials, not points. Mathematicians at all levels and in all centuries always seem to forget that whenever it is convenient.
^{1}http://milesmathis.com/zeno.html
^{2}http://milesmathis.com/are.html
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