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by Miles Mathis

Abstract: I will show that the current derivative of the natural log and the current derivative of 1/x are both wrong. In doing so, I will show the magnificent cheat in the current derivation of dln(x)/dx, embarrassing every living mathematician.

I have been prodded by recent events to add an introductory paragraph here, explaining why I would want to attack the calculus. These papers have only been up about a week, but I can already see the firestorm ahead.

I come to this problem from physics. In my theoretical research over the last decade, I have come to the conclusion that many of the problems in QED and GR are caused by fundamental and long-standing disclarities in the calculus. These problems include the cause of the need for renormalization, the cause of the failure of unification, and the cause of all the point problems in both QED and GR. All the modern maths are based on the calculus, and a problem with the calculus would infect every manipulation that depends upon it. For this reason, these latest papers are not proof that I am contrary and crazed, they are just one more indication that I have the courage to go where my nose leads me, the consequences be damned. My corrections to the calculus are incomplete and will remain incomplete for many months or years, no doubt. But since the calculus has been incomplete for centuries, perhaps millennia, I do not feel especially pressed to apologize. I will continue to put up what I have as I discover it, and if some of turns out to be wrong, well, so what. No one gets everything right.

The derivative for the natural log is currently found by this method:

dln(x)/dx = limd→0[ln(x+d) - ln(x)]/d =
lim ln[(x+d)/x]/d = lim(1/d) ln(1 + d/x) = lim[ln(1 + d/x)1/d]
Set u=d/x and substitute:
limu→0 [ ln (1 + u)1/ux ]
= 1/x ln [limu→0(1 + u)1/u]
= 1/x ln(e) = 1/x

That derivation is false. It is false because it contains huge errors, errors I can point out very easily. I will lead with the biggest error.

Look closely at this manipulation:

limu→0 [ ln (1 + u)1/ux ] = 1/x ln [limu→0(1 + u)1/u]

As they pull the 1/x down from the exponent and put it in front of the ln (which is legal) they also shift the limu→0 forward, so that it is now in front of the ln (which is not legal). It is gloriously illegal. You cannot separate the ln from its number. This is important, because as confirmation of the last step, we are sent to the “definition” of e:

e = limn→0(1 + n)1/n

Notice that does not read

e = limn→0ln(1 + n)1/n

Nor are the two equations equivalent. Because they are not equivalent, we can see that shifting the limn→0 is not just a matter of preference. With the term limu→0 [ ln (1 + u)1/ux ], we are monitoring the natural log as we approach the limit. With the term limu→0(1 + u)1/u we are monitoring only the sum and exponent as we approach the limit. Not at all the same. In fact, this is what the brackets mean in the term

limu→0 [ ln (1 + u)1/ux ]

The mathematicians write down the brackets and then ignore them, transferring the limit inside them to suit themselves.

So the final substitution of e for limu→0(1 + u)1/u is illegal. The limu→0 has no reason or right to be there. The correct derivation should go like this:

limu→0 [ ln (1 + u)1/ux ] = limu→0 [1/x ln (1 + u)1/u]

If they can shift the lim inside the brackets, why not shift it anywhere? Why not

[ ln (1 + limu→0u)1/ux ]

They don't do that, obviously, because they don't want to destroy their e substitution. By the same token, you can't shift 1/x outside the limit. 1/x is a curve, which will change the way the limit is approached once again. Remember, this is a curve equation, which means that it is not only u that is changing: x is also changing. Go back a step and you will see that more clearly. Before we brought 1/x down, it was with u in the exponent like this: 1/ux. That means that the rate of change of u and the rate of change of x are affecting one another. They are tied together. Which means that the change of u cannot be considered separately. In other words, the exponent 1/ux isn't approaching infinity like 1/u is in the definition of e.

Again, that is why the limit sign must stay outside the brackets.

That is not really an error. It is a fudge. A big cheat. In other words, it was not done by accident. This proof has very heavy prints of finessing on it, and it is amazing that no one catches these mathematicians in their pawing and gnawing.

The gods of math grimace and gnash their teeth.

But there are other problems: even the first equation

dln(x)/dx = limd→0[ln(x+d) – ln(x)]/d

is false. As I will prove again below, the variable d must be equal to 1; so we have no ratio. The rate of change of the independent variable must be defined as one, to get the differentiation tables to work. It must also be defined as one so that we can apply these differentials to the number line, where the rate of change of the integers is one, by definition. You can't push the differential of your independent variable to zero without pushing the dependent variable to zero as well, and this just makes the graph smaller: it doesn't change the relationships of the numbers. If it doesn't change the relationships, it can't give you a solution.

You will say that d isn't the independent variable, but it is. It was originally dx or Δx that was going to zero, and the moderns have just relabeled it as d or h or whatever in order to hide this fact.

How do we begin to correct such deep-seated problems? If the calculus were healthy, we would not see such brazen and long-lived cheats in full view, so we must assume it is a total mess. In my long paper on the derivative1, I started over from the beginning, but here I will gloss and simplify, limiting myself to the solutions of these two specific derivatives: ln(x) and 1/x.

The derivative is currently defined in several ways. At Wikipedia, in the second sentence on the page, the derivative is defined as

Loosely speaking, a derivative can be thought of as how much a quantity is changing at a given point.

I think we have had enough of loose speech, and a quantity cannot change at a point, by definition. What this author means is, “how much a drawn curve is changing at a point on a graph.” But since a point on a graph is an ordered pair like (x,y), a “point” on a graph is not really a point. It is not a point in space and it is not a dimensionless entity. An ordered pair has two dimensions, by definition of the word “pair.”

Somewhat less loosely, the derivative is defined as the rate of change of the given curve, and it is defined as the slope of the tangent at x. To begin our analysis, let us look at the table of the actual differentials of ln(x):

ln(x)   0, .6931, 1.099, 1.386, 1.609, 1.792, 1.946, 2.079, 2.197
Δln(x)    .6931, .4055, .2877, .2231, .1823, .1542, .1335, .1178
ΔΔln(x)   .2876, .1178, .0646, .0408, .0281, .0207, .0157
ΔΔΔln(x)   .1698, .0532, .0238, .0127, .0074, .0050
ΔΔΔΔln(x)   .1166, .0294, .0111, .0053, .0024
ΔΔΔΔΔln(x)   .0872, .0183, .0058, .0029

The derivative, if defined as the general rate of change of the function, is the second line. The problem, as you can clearly see, is that we aren't straightening out our curve, as we were with the power functions. We can find a derivative of ln(x), of a sort, since we can certainly find a rate of change of the curve. But that rate of change is increasing as we look further and further down the chart. Therefore, no derivative of ln(x) will ever give us a straight line or a velocity. Every derivative of ln(x) is a greater curve than ln(x) itself.

There is another problem. The current derivative of ln(x) is said to be 1/x. But you can see immediately that isn't true. The rate of change of line two is not 1/x. Nor is there any constant we can use to make the rate of change 1/x, since there is no value for z that will make the rate of change of line two z/x.

But let us be more rigorous. The calculus was invented for the power functions, and with power functions, the derivative is not really the general rate of change of the curve. The derivative, as currently used, is the rate of change at x. To say it again, it is the rate of change at a specific interval, not the rate of change of the whole curve.

To be even more rigorous, the derivative equation is an equality between two rates of change: the rate on the left side of the equality and the second rate on the right side. For example, if we go to the differential tables for powers to analyze the function y' = x3, we find that it is really an equality like this

3Δx2 = ΔΔx3

I have shown that can be read as “the rate of change of the curve x2 times 3 is equal to the second rate of change of the curve x3”. The derivative is not simply the rate of change of the curve on the right side. The derivative is a relationship of two curves. The derivative is telling us a number equality between two different curves. That being so, as we study this problem, we have to compare the tables for ln(x) to the tables for 1/x, to see if the rates of change ever equal eachother.

1/x   1, .5, .3333, .25, .2, .1667, .1429, .125, .1111, .1, .0909
Δ1/x    .5, .1667, .0833, .05, .0333, .02381, .0179
ΔΔ1/x    .3333, .0833, .0333, .01667, .00952, .00591
ΔΔΔ1/x   .25, .05, .01667, .00714, .00361
ΔΔΔΔ1/x   .2, .0333, .00953, .00353

To find the right lines to analyze, we can follow the method we used in the tables for powers and analyze the current derivative of ln(x) from its table. We are told:

dln(x)/dx = 1/x

Which means the rate of change of the curve 1/x is equal to the second rate of change of the curve ln(x).

Δ1/x = ΔΔln(x)

Is that true? Let us take a look:

Δ1/x   .5, .1667, .0833, .05, .0333, .02381, .0179
ΔΔln(x)   .2876, .1178, .0646, .0408, .0281, .0207, .0157

Well, the two curves are very close, which explains the relative success of the current derivative; but the rates are not a match. But we can check in a few other places, to be sure. The equality may not be at the first rate of change. In the power tables, the equalities are sometimes further down the chart, as with higher powers.

ΔΔ1/x    .3333, .0833, .0333, .01667, .00952, .00591
ΔΔΔln(x)    .1698, .0532, .0238, .0127, .0074, .0050

Again, close, but we appear to be diverging. Let's try another:

ΔΔΔ1/x    .25, .05, .01667, .00714, .00361
ΔΔΔΔln(x)   .1166, .0294, .0111, .0053, .0024

Definitely diverging. The best match was with Δ1/x.

I have shown that the derivative 1/x is a good approximation, but it is not a match. This tends to confirm my critique of the proof, since if the e substitution were allowed, this derivative should be perfect or close to perfect. The definition of e using the limit is not an approximation, since it allows you to be as exact as you want to be. It would not cause the kind of error we see here in the differentials.

The current derivative was found by looking at the curve on the graph and actually drawing some tangents. After long years of this, it was decided that the slopes seemed to follow 1/x. Then the math was simply pushed to create a proof of that educated guess. But instead of guessing and then pushing equations, let us calculate the equation right from the numbers in the table. We want an equation that gives us those numbers above. It looks kind of difficult at first, but we just remember the definition of a differential. We remember how the table was made, and what it means. We remember that we can always write ΔΔln(x) as Δln(x) – Δln(x + 1), since that is the definition of a rate of change, and it comes right out of the table, as you see. The numbers in that line are found by that method.

By the same method, we can write Δln(x) as ln(x + 1) – ln(x). Again, that comes right out of the table, and the order is reversed since ln(x) is getting larger while Δln(x) is getting smaller. And we can rewrite Δln(x + 1) as ln(x + 2) – ln(x + 1). Combining all that gives us

ΔΔln(x) = [ln(x + 1) – ln(x)] – [ln(x + 2) – ln(x + 1)]
= ln[(x + 1)/x] – [ln(x + 2)/(x + 1)] =
ln[(x + 1)/x]/[(x + 2)/(x + 1)]
= ln[(x + 1)2/(x2 + 2x)]

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

rate of change = 4 ln[(x + 1)2/(x2 + 2x)]

But is that the slope? No. As with the exponential functions, to find a slope we just find an average of the forward slope and the backward slope, like this:

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

The slope at x=2 is .5495, not .5.
The slope at x=3 is .3464, not .3333.
The slope at x=4 is .255, not .25.
The slope at x=5 is .203, not .2.
The slope at x=6 is .1685, not .1667.
The slope at x=7 is .1435, not .1429.

You can see that these new slopes are very close to the current ones. To see a full argument for why averaging forward slope and backward slope is actually better than going to zero, you will have to read the full analysis in my paper on the exponential functions.

Once I made these tables, I could see that ln(x) wasn't acting like a power curve. It wasn't straightening out as we differentiated. The first solution I considered was integrating. Because our curve is increasing instead of decreasing, we can try integrating instead of differentiating. To find a velocity, you need to straighten out the curve, and differentiating is further curving this curve, you see. So reversing the process might have helped us. We can seek an integral or anti-derivative of ln(x) just by reversing our chart, like this:

Δln(x)    .6931, .4055, .2877, .2231, .1823, .1542, .1335, .1178
ln(x)   0, .6931, 1.099, 1.386, 1.609, 1.792, 1.946, 2.079, 2.197
∫ln(x)    .6931, 1.792, 3.178, 4.787, 6.579, 8.525, 10.60, 12.80
∫∫ln(x)    .6931, 2.48, 5.663, 10.45, 17.03, 25.55, 36.16, 48.95

All I have done is turn the table upside down, then try to find more lines of differentials. But since we are going the opposite direction of what has historically been called differentiation, I call these lines integrals. They are found by adding instead of subtracting, and that is what integration originally meant.

But, as you now see, we find the same problem in this direction in the table. Our curve is again increasing. It switched directions, and then immediately began increasing again. So we can't find a straight line, or what has been called the velocity, by either the old-fashioned differentiation or integration. Calculus was invented in relation to the power functions, and ln(x) is not a power function. The power functions could be related to one another, because the rate of change of a higher power was equal to the rate of change of a lower power. As we found the derivative, we took higher powers to lower powers, thereby straightening them out. Here, we can't do that. Both up and down the table, we get more curvature.

It is also very important to notice that with this function, differentiating never lowers the rate of change. With power series, finding a derivative always meant decreasing the curve of the function, which means the rate of change of the curve was decreasing. But with ln(x), neither differentiating nor integrating decrease the curvature. Both increase it. In problem solving, this must be important.

The current calculus has solved this dilemma by 1) Ignoring it. They pretend these simple tables don't exist; 2) Fudging the equations to match the slope they know exists, from studying the lines on real graphs. If the slope looks like the slope of 1/x, they assume that it is, and they push their math to find 1/x, using chain rules and finessed limits and fake substitutions and whatever else they need to hammer the solution down.

The current method has found a fair approximation with 1/x, but my method finds the right answer with no approximation. I never go to zero and don't use infinitesimals. My rates of change are absolutely correct relative to one another, and there is no margin of error. My method is vastly superior to the current method in both operation and answer. My operation is simple and transparent, with no fudgy manipulations. My notation is much simpler than the current method. And my answer is correct, whereas the current answer is simply incorrect. The current answer is nothing but a poor estimate of my answer, and I have explained why in my long paper on the derivative and its history.

Now let us look at the derivative of 1/x. The current calculus treats this function as equivalent to x-1, and uses the power equation nxn-1:

dx-1 /dx = -1x-2 = -1/x2

But we can't apply the power equation to 1/x to find a derivative. If we did, then the anti-derivative, which is supposed to be ln(x), would be x0/0. But the biggest problem is that the differentials themselves tell us the rate of change of 1/x is not 1/x2 . The only reason mathematicians chose that derivative is because it looked right on the graph. But if they had made this table, they would have seen it doesn't look right, even at a first glance.

1/x    1, .5, .3333, .25, .2, .1667, .1429, .125, .1111, .1,
Δ1/x    .5, .1667, .0833, .05, .0333, .02381, .0179
ΔΔ1/x    .3333, .0833, .0333, .01667, .00952, .00591
ΔΔΔ1/x    .25, .05, .01667, .00714, .00361
ΔΔΔΔ1/x    .2, .0333, .00953, .00353

Defined as it is, the derivative is the third line, and the third line is not changing as 1/x2. It is not changing like 1, 4, 9, 16, 25; it is changing 1, 4, 10, 20, 35. We can write the line as 1/3, 1/12, 1/30, 1/60, 1/105. Dividing each denominator by 3 gives us the series 1, 4, 10, 20, 35. So we just need to write an equation for that series. That is the famous pyramid series, and the equation is known to be x(x + 1)(x + 2)/6.

But can we derive that equation without knowing the pyramid equation? Yes, we can take it straight from the table again, using the differentials themselves.

ΔΔ1/x = [Δ1/(x + 1)] – Δ1/x
Δ1/(x + 1) = 1/x – [1/x + 1)]
Δ1/x = [1/(x + 1)] – [1/(x + 2)]
ΔΔ1/x = 1/x – [1/x + 1)] – [1/(x + 1)] + [1/(x + 2)]
= 1/x + [1/(x + 2)] – [2/(x + 1)]
= [x2 + 3x + 2 – 2x2 – 4x + x + x2]/[x(x + 1)(x + 2)]
= 2/[x(x + 1)(x + 2)]

Because the first term in the line ΔΔ1/x is .3333, we have to multiply by 3 to get our first term up to 1. So,

d(1/x)/dx = 6/(x3 + 3x2 + 2x)

But is that the slope? No, it is the rate of change. The slope is found by the equation
slope @ (x,y) = [y@(x + 1) - y@(x - 1)]/2

For x=3, the slope is -.125, not -.1111.
For x=4, the slope is -.0666, not -.0625.
For x=5, the slope is -.0416, not -.04.
For x=6, the slope is -.0285, not -.0278.
For x=7, the slope is -.0208, not -.0204.
For x=8, the slope is -.0159, not -.01563.
For x=10, the slope is -.0101, not -.01.

Those values are astonishingly close, and I think even my biggest critics must be sweating a little.

I have found a new rate of change and a new slope, both of which give us values that are very close to the current derivative. The new slope gives us values that are closer, but that doesn't mean it is the one to choose as the replacement for the derivative. I have some work left to do in showing why this averaged slope equation either can or cannot apply to 1/x. We know it can be used on x2 without any problem, but that does not mean it applies here.

As with the exponential functions, these two functions behave in strange ways. I remind you, for example, that ln(x) increases its curvature with both differentiation and integration. I said that this must affect solutions in physics, and what I meant is that in physics we find a velocity from a curve by straightening out the curve. By the modern interpretation, the velocity is always equal to the tangent, which is always equal to the derivative, but that isn't true even with power functions. It is true only with the power 2. I showed in a recent paper on cubed acceleration2 that the velocity is found from the second derivative, not the first. This is because to find the velocity, we straighten out the curve. For higher powers, it takes more derivatives to do that. But with ln(x), you can't straighten out the curve, no matter how many derivatives you take. You can find a tangent and a derivative, as I just showed, but you can't assign either one to a velocity. To find a velocity for ln(x), you have to go off the table and outside the normal equations of differentiation and integration. I will show how to do that in an upcoming paper, but for now I simply point out that if the curves increase in both directions on the table, they must have a turn-around point. This means that there must be a straight line hidden between two lines on the table, and with ln(x), that straight line must be found near ln(x). The same can be said for 1/x. The derivative cannot be the velocity, because the all the derivatives are still a curves. The velocity must have a constant rate of change, by definition, and our derivative equation does not. Of course, the slope may give us the velocity directly, but it may not. I will have to study the problem more before I come to a final decision.

Conclusion: By ignoring the differential tables themselves, the calculus has vastly overcomplicated the methods and proofs. In doing so, they have actually gotten the wrong answer for many derivatives, as with these two. Moreover, they have been forced to fudge many proofs. The fudge above with ln(x) is not what one would call subtle, and the modern calculus is full of fudges like that. For another, see my explosion of the proof of dex/dx.3

Of course, Newton and Leibniz started all this by cheating in their own proofs. Like the modern derivative proofs, the original proofs were written from the end backwards. Both men knew the answer they wanted, and they forced their proofs to show it. I have already shown this with Newton's proof1, and the same can be said for that of Leibniz.

I still haven't figured out the exact history of this mistake. Was it done because no one has been in possession of these simple tables I am making; or were the tables suppressed because they did not allow for solutions at an instant and point, and everyone desired solutions at an instant and point? I suspect the latter. The tables are so simple they could not have been overlooked or misread. My first paper on the derivative1 was probably refused for publication for the same reason this problem has been buried from the beginning. Mathematicians don't want to clarify any of the proofs or manipulations or definitions of the calculus, because that would require that they give up their instantaneous solutions. It would also require that they admit being wrong for 300 hundred years, and admit that they have been pushing proofs for the same amount of time.

Unfortunately, no one else seems to have made the connection between this historical choice to look away, and the historical problems with the point in QED and General Relativity. I have shown that renormalization is the price physics is paying for the math department's refusal to face facts. Equations are exploding and imploding in many fields, and mathematicians pretend not to know why. Edward Witten, supposedly a master of mathematics, asks in the Millennium Prize why the math of QED is such a mess. He should know that the calculus itself is the main cause of the mess.


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