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I continue to get letters complaining that my correction to the calculus h→0 h But, as before, that is both unnecessary and false. We don't go to a limit, because h is neither zero nor approaching zero. Instead, we make a simple table of differentials. a = 1 1,1,1,1,1,1,1,1,1,1,1,1,1 a = 2 2, 4, 8, 16, 32, 64, 128, 256 e = 2.718 e, 7.389, 20.086, 54.6, 148.4, 403.4, 1097a = 3 3, 9, 27, 81, 243, 729, 2187, 6561 a = 4 4, 16, 64, 256, 1024, 4096 a = 5 5, 25, 125, 625, 3125, 15625 a = 6 6, 36, 216, 1296, 7776, 46656, 279936 Δa1 0 Δa2 2, 4, 8, 16, 32, 64, 128, 256 Δ e 4.67, 12.7, 34.5, 93.8, 255, 693.6Δa3 6, 18, 54, 162, 486, 1458, 4374 Δa4 12, 48, 192, 768, 3072, 12288 Δa5 20, 100, 500, 2500, 12500 Δa6 30, 180, 1080, 6480, 38880 ΔΔa2 2, 4, 8, 16, 32, 64, 128, 256 ΔΔ e 8.03, 21.8, 59.3, 161.2, 438.6ΔΔa3 12, 36, 108, 324, 972, 2916 ΔΔa4 36, 144, 576, 2304 ΔΔa5 80, 400, 2000, 10000 ΔΔa6 150, 900, 5400, 32400 ΔΔΔ e 13.77, 37.5, 101.9, 277.4ΔΔΔa3 24, 72, 216, 648 ΔΔΔa4 108, 432, 1728 ΔΔΔa5 320, 1600, 8000 ΔΔΔa6 750, 4500, 27000 What can we tell already? Well, we can tell that the current derivative for y = a ^{x} is probably wrong. The current derivative is
da ^{x}/dx = a^{x}ln(a)
But a cursory glance at the table tells us that might be wrong. We can see from the table that if a = 2, we have a rather special situation. The rate of change of the first curve y = 2 ^{x} (line 2 in the table above) is 2^{n}. The rate of that change (line 9) is 2^{n}, and the change of that change (line 15) is also 2^{n}. Therefore, the derivative of a^{x} when a = 2 appears to be a^{x}. This means that for the current derivative to be correct, the value ln(a) for a = 2 needs to be 1. But it isn't. The natural log of 2 is about .693.
What I will now do is derive the proper derivative, straight from the table. Since I showed in my power tables ^{1} and natural log tables^{7} that the derviative is actually the second rate of change of our given curve, we have to study line 15 in relation to line 2.
a = 2 2, 4, 8, 16, 32, 64, 128, 256 ΔΔa2 2, 4, 8, 16, 32, 64, 128, 256 Then we find one line directly from the other, using the basic differential equations: Δa ^{x} = a^{x+1} - a^{x}Δa ^{x+1} = a^{x+2} - a^{x+1}ΔΔa ^{x} = [a^{x+2} - a^{x+1}] - [a^{x+1} - a^{x}]ΔΔa ^{x} = a^{x+2} - 2a^{x+1} + a^{x}But we aren't finished. Let us compare line 17 to line 4. The first term in line 4 is 3, and the first term in line 17 is 12. To compare the rates of change, we have to mesh the two series of numbers, which means we have to multiply line 17 by 1/4. But that can't be our general transform, since it doesn't work on lines 5 and 18, or on lines 6 and 19. The general transform is 1/(a - 1) ^{2}. Which makes our derivativeda ^{x}/dx = [1/(a - 1)^{2}][a^{x+2} - 2a^{x+1} + a^{x}]This means that our snap analysis of a=2 was correct. The transform reduces to 1, and so ln(a) cannot apply. This new derivative equation also gives us a good number for e. If we let x=2, the derivative equals 7.39, which is the present value of e^{2}=7.39. Let us look at some other numbers
The slope at e, x=1 is 2.71828, which confirms the current number.The slope at e, x=2 is 7.393, which confirms the current number.The slope at e, x=3 is 20.086, which confirms the current number.However, if we calculate the slopes for other values of a, we find a large mismatch with current values: The slope at a=2, x=2 is 4, not 2.77. The slope at a=3, x=2 is 9, not 9.9. The slope at a=4, x=4 is 256, not 355. It appears that the derivative equation reduces to a ^{x}, which was our first guess from the table. da^{x}/dx = a^{x}But the slope is either not the derivative here, or we need an extra manipulation to get the slope from the derivative. The slopes just calculated for values of "a" other than e cannot be right. So let us seek the tangent and slope, damn the derivative and the rate of change of the curve. My critics have told me that the calculus has long since moved past graphs and tables of differentials, but in the case of the slope and the tangent, that cannot be true. The slope and tangent are defined relative to the graph. These curve equations represent accelerations, but unless x and y are orthogonal on a graph, we won't get a curve. In real life, you can accelerate in a straight line, remember. So these accelerations were put on a graph, with x and y at right angles, specifically in order to create a curve we could analyze. The slope is defined as Δy/Δx. Currently, the analysis takes Δx to zero to find a solution, but have shown that is both impossible and unnecessary (and I will show it again right now, in a novel and damning way). The current method allows the calculus to find solutions at an instant and point, which is impossible. It is unnecessary, since we can find the slope without doing that. Once again, we can pull them straight from the table, without going to zero or any limit. But we will also consult a graph as we go, to see what this means there. If we let Δx=1, then we can find a slope by the first method I have written on the graph. (4 + 2)/2 = 3. The slope at x=2 is 3. You can see that is just averaging the forward slope and the backward slope. But the historical calculus was never satisfied with that answer. Mathematicians thought, “Why not take Δx below 1, and get a more exact answer?” So they did what I have begun to do on the graph. They looked at a smaller sub-slope, where Δx=.5. Using that smaller interval, they found a slope of 2.828. Then, by going to zero, they found a limit for that slope at 2.77. Since 2.77 is 4ln(2), they thought they had found the slope. The problem there is that if a=2 is your base, your denominator in your slope cannot be less than 2. To see why, you have to go back to line 1 in our table above. If the base is a=1, then you get a constant differential of 1, as you see. But you want a smaller differential, so you think, “I will just use smaller values for x. I won't use 1, 2, 3. I will use .00001, .00002, .00003.” Try it, and see what happens. No matter how small you make your x's, you still get 1, 1, 1, 1, 1. Since a=2 is defined relative to a=1, what you cannot do with a=1, you cannot do with a=2. Or, reverse this logic. Say you demand that you be able to find smaller values for a=2 in line 2. So you ignore me and just do it. Instead of 1, 2, 3, you start with .5, 1, 1.5. This gives you smaller differentials, and this allows you to take the equation 1.66 + 1.17 = 2.828 straight from the table, confirming the first step toward zero as shown on the graph. OK, but now you will have to do the same for all the values of a on the table. You say, fine. But if you do that, you will have a very strange-looking table: a = .5 .707, .5, .354, .25, .177 a = 1 1,1,1,1,1,1,1,1,1,1,1,1,1 a = 2 1.41, 2, 2.83, 4, 5.66, 8, 11.3, 16 a = 3 1.73, 3, 5.2, 9, 15.6, 27, 46.8 Do you see the problem? You have made your Δx smaller, but it has skewed your entire solution. The rate of change of the line a=2 is not what it was before. You have changed your original curve! These two curves are not equivalent:
a = 2 2, 4, 8, 16, 32, 64, 128, 256 a = 2 1.41, 2, 2.83, 4, 5.66, 8, 11.3, 16 One curve is not double the other one, as you want it and need it to be. The first curve is the second curve squared. To say it another way: when you lowered your value for Δx, what you wanted was to put your curve under a magnifying glass. You wanted to look closer at it, moving in closer to that value of x. This is how the history of calculus is taught. This is precisely what the inventors and masters tell us they were doing. They were magnifying parts of the curve to study it. What they thought they were doing is this: when they halved their Δx, they thought they had magnified the curve by 2. In other words, in going from Δx=1 to Δx=.5, they thought they were twice as close to zero, and therefore twice as close to the limit and the answer. But I have just proved that this assumption was wrong. They were not twice as close. They were not in any proper approach to a limit. In going from Δx=1 to Δx=.5, they had not halved the curvature, they had actually gone to the square root of the curvature, so their magnification was not working like they thought it was. Going to zero historically looked like a great idea, since it seemed to promise a more exact slope. But in going below Δx=1, the calculus has actually falsified its solution. It has found what appears to be a more exact solution only by changing its original curve. You cannot legally go below Δx=1, because that differential is what defined the curve to begin with. A smaller differential will give you a different curve and a different rate of change. What this means for our solution is that the slope at x=2 on our graph is not 4 or 2.77. It is simply 3. Our differential Δx cannot be taken below 1, due to our definitions and givens. If you go below 1, you are cheating and you are getting the wrong answer. If you desire precision in your answer, you do not take Δx to xero, your make your 1 smaller. Meaning, you set up your graph where x=1 angstrom instead of 1 meter. Our derivative method above therefore does not yield a slope. To find a slope, you use differentials from the table, but you solve in this way: slope @ (x,y) = [y@(x + 1) - y@(x - 1)]/2 This new slope equation skews the solution for e^{x}. If we find a slope at x=2, the slope is 8.68, not e^{2} = 7.39. The slope of e^{x} is not e^{x}.What does all this mean? It means that the calculus has been very sloppy in its math and definitions. The calculus needs to be more rigorous in defining what it wants to find from the curve. In physical situations, what the calculus wants from a curve is a velocity, but I will show below that these curves won't give them that. Velocity is defined in a rigorous manner, and you can't get a velocity from these curves. In pure math, the calculus claims to want to find a rate of change at a point, but since there is no such thing, we won't be able to find that either. We have just found a slope, but what does that apply to, if not to a rate of change at a point or to a velocity? Well, it applies to a rate of change at (x,y), which is the rate of change at two number values, which is a rate of change at two distances from the origin. In other words, it is a rate of change at the end of two defined intervals. As such, it is not the rate of change at a point in space. It may loosely be defined as a rate of change at a "position" in space, but that position is defined relative to other positions, and is always represented by differentials, as in distances from the origin. But why did we find different values for the slope and the derivative here? Aren't they the same? Not really. Again, it is a lack of rigor that has doomed us throughout history. With a=2, x=2, we found a value of 4 for the derivative and of 3 for the slope. Which is correct? Both are correct, and either can be used in math or physics. The number 4 is the change in y between x=2 and x=3. The number 3 is the average change in y midway between x=1 and x=3, and since x is changing at a constant rate, that gives us the correct value at x=2. Remember, the curvature here comes from y accelerating, not x. We put in consistent values for x, so x by itself is acting like a velocity or the pure math equivalent. No matter how big or small you make change in x, you always insert steadily increasing values, remember, as in 1, 2, 3. We never study curve equations by putting in accelerating values for x, as in 1, 4, 9, 16. So, if we define the derivative as the rate of change after a given time, rather than the rate of change at a given time, the derivative will equal the slope. In that case we can just use my simplified slope equation. By saying "after a given time," I am not implying that we are calculating a total change from zero or the origin, I am just reminding you that we are finding a time at specific x, and that x is telling us a distance from the origin. You will say, "If we signify a time or position 'after some time,' haven't we signified an instant or a point? Isn't the endpoint of any interval a point?" No, the end"point" of any interval is a position in time or space, but not an instant in time or a point in space. The position "after 6 seconds" is not at an instant, since after 6 seconds your clock does not stop running. A second is defined as an interval between ticks, but not even ticks happen at an instant. Just as you can't measure a second with complete accuracy, you can't have an event at a instant or point. In physics and math, there are only intervals, measured with more or less accuracy. Then you will say, "But when we actually draw a tangent to a curve on a graph, we can measure a slope more accurately than you have allowed here. Are you saying we have cheated in that also?" Yes, that is what I am saying. For instance, let us study the graph I just posted. The distance 1 is about 5/8 of an inch there. That size differential therefore defines the graph and the curve on it. You say you can tell the difference between a slope of 2.773 and 3 on that graph. First of all, accurate slopes and tangents are very difficult to find by hand, especially to curves that are curving so slightly. I doubt you or anyone else can find that accurate a slope by hand. That is why these equations were developed in the first place: you can't do it by hand or eye. But even if you could, you would find a slope of 3 at x=2, not a slope of 2.773. You are sure you would find a slope of 2.773, but you simply trust the calculus too much. Now you say, "But you can't be right. You are averaging two lengths of curve that aren't even close to the same. The curve above that point is about twice as long as the curve below. Therefore your average has to be just a wild approximation. And yet you claim it is more accurate than the calculus which goes to zero to find precision. You must be mad!" No, you must be blind not to see that the averaging here will give us precisely the right answer without any approach to zero, since the interval above and the interval below are exactly the same size, by definition. In saying they aren't, you give me the length of the curve or of y, but that is not what defines the intervals above and below. What defines them is x, and x is the same size in both places. For instance, if x were t instead, then the horizontal axis would be time. In that case, the time during the interval above and the time during the interval below would be equal. Because the intervals are equal in this way, we can average without any qualms. And that average will give us the right answer, without any approximating or error. Since x has no acceleration, all the acceleration is with y, which means we have a constant acceleration, which means we can average like this without any problem. Going to zero is not only unnecessary, it gets the wrong answer. I have now joined the proofs for a h→0 h Those two equations aren't analogous in form or theory. In the first equation, h is not in an exponent; but in the second equation, it is. That can't work. And it can't work for another fundamental reason. Look again: in the second formulation, h is exponential in the numerator, and not exponential in the denominator! So how can h be approaching zero at the same rate in both places? Exponents don't change at the same rate as normal variables. The rates aren't even close, and every high school kid knows that. This is another blunder of titanic proportions. If this basic derivative is wrong, then we must assume that the bulk of the differential equations in the standard math tables are also wrong. And if the bulk of the differential equations are wrong, then the bulk of the integrals are also wrong. As you can now see, my correction to the calculus requires that we recheck every single derivative and integral known to man. How in the name of all that is holy could the entire world neglect to check the derivative against a simple list of differentials like this? I can see why mathematicians would prefer to generalize their derivative equations without making a table every time, but you would think they would make the table the first time they calculated a basic type of derivative, like this fundamental exponential derivative, just to be sure they weren't doing their chain rules wrong or something. This is just more proof that the history of math is a cesspool of false equations. Some will try to squirm out of this by telling me that the calculus can't be pulled from these tables I am making, but if this is the case, they will have to explain to everyone how and why my first table in my long paper so successfully and easily proved the equation y' = nx ^{n-1}. I think it is clear that my method of finding differentials and rates of change is both fundamental and straightforward. This method shows that the most used equation of calculus, y' = nx^{n-1} is correct. But it also shows that modern proofs are using a different method when finding that equation and when finding the equation da^{x}/dx = a^{x}ln(a). I have just shown that the two methods for deriving the equations can't be the same, since the differentials from the tables confirm the first equation and refute the second.
All those mathematicians who have, since 1820, moved into sexier fields because there was nothing left to do in calculus are looking more and more foolish. I have shown that “pool ball mechanics” is a house of cards, and now I have shown that calculus is another house of cards. There appears to be plenty of work to do, and I can't do it all myself.
Something else is strange about the calculus. Notice that the derivative of a curve is a tangent, which is a straight line. But (we are told) the derivative of a velocity with respect to time is an acceleration. In one instance, we get a straight line from a curve; in the other, we get a curve from a straight line. We are told we can differentiate a line into a curve and differentiate a curve into a line. This paradox is caused by another imprecision in language. Acceleration is said to be the derivative of the velocity, but it isn't. The derivative is the rate of change of the curve, and a velocity isn't a curve. The derivative of any straight line is a constant, since the rate of change of any straight line is a constant. You can't really differentiate a velocity, since there isn't any variation. If you differentiate a velocity in the differential tables, you only get a line of numbers like 6, 6, 6, 6, 6. That can tell you a distance, but it can't tell you an acceleration. Therefore, when we “differentiate” dv, we aren't differentiating a velocity. This is because dv is calculus shorthand for Δv, and Δv isn't a velocity; Δv is already an acceleration, by definition. An acceleration is defined as a change in velocity. When we find an acceleration “by going to zero,” what we are really finding is an “instantaneous” acceleration from an average acceleration. But, unless the acceleration is variable, the acceleration at any instant will BE the average acceleration. Instantaneous acceleration = average acceleration. This means that the equation
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