go to homepage Trig Derivatives found Since I first published my paper on the calculus several years ago, I have gotten many angry emails like this one: If this paper was useful to you in any way, please consider donating a dollar (or more) to the SAVE THE ARTISTS FOUNDATION. This will allow me to continue writing these "unpublishable" things. Don't be confused by paying Melisa Smith--that is just one of my many noms de plume. If you are a Paypal user, there is no fee; so it might be worth your while to become one. Otherwise they will rob us 33 cents for each transaction.
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go to short calculus paper
without the old Calculus
by Miles Mathis
You are wrong! Mathematics is a science about numbers. Graphs, plots are for illustration—you can prove nothing from them.
y = sin x
y' = cos x
How do you prove that by your method?
Yuri
Even my mother, who is a professional mathematician, has failed to see how I can incorporate my table into an analysis of all functions. She never got angry, but she has used my silence as proof that my method has limited use.
Now, I said in a footnote to that paper that my method applied to all of calculus and all functions, not just differentials or polynomials. It applies to trig functions, logarithms, integrals, and so on. I assumed that anyone who understood my argument would see that immediately. I didn't even bother to write a follow-up paper on integration, it seemed so clear to me that anyone could just read up the tables instead of down. I was busy with other important problems and decided to let that paper hang, along with any paper that specifically addressed trig functions. Frankly I had hoped that someone might come along and see my point, and that they would do the dirty work of advancing my theory into these other alleys. Once I have solved a problem, I tend to get bored, and stating the obvious does not really inspire me to write.
However, I now see, years later, that I was mistaken in assuming that my initial paper would penetrate into the mathematical community. It has been turned down for publication in all the top forums, for what I think are political reasons. So I have recently gone back and simplified my argument, self-publishing a shorter and simpler paper, argued in what I consider to be an extremely transparent manner and language. I hope that this paper may eventually make some headway in the mainstream, even if I continue to be blocked by the higher-ups.
Beyond that, I have decided to publicly solve Yuri's trig problem for him, knowing full well that it won't be the further miracle anyone needs. No matter what I do or how I do it, I now expect most of the status quo to find a way to dismiss it out of hand. They weren't bothered by the fact that the current equation has been hanging from skyhooks for 350 years, and so they won't be impressed to see the equation finally grounded. Anyone who studies my table and doesn't undergo an epiphany is someone who is pretty much unreachable, and solving this trig problem with the table won't reach them either. But here goes.
So, Yuri, watch closely, my friend. I will do it so quickly and so easily, you will no doubt think it is nothing. I will show you how to do it without limits, without going to zero, without infinite series, and without the current derivation of the calculus. I will do it using only my table of exponents and the constant differential.
y = sin x = ±√(1 - cos2 x)
Notice that we are still dealing here with exponents. The cosine is squared and that is the important fact here, not the fact that we are dealing with trig functions. From a rate of change perspective, the trig function is meaningless. A sine or cosine is just a number, like any other. It is written as function of an angle, but that does not affect the rate-of-change math at all. The cosine of x is a single variable, and we could rewrite it as b if we wanted to, to simplify the variable for the rate of change math. Likewise, we could rewrite sin x as a, if we desire. All we have to do is make sure we don't confuse sine and cosine, since they vary in different ways, but we can mark them anyway we want.
Let a = sin x
b = cos x
Therefore, we could rewrite the equation as
y = a = ±√(1 - b2)
Square both sides
y2 = 1 - b2
Since sine and cosine are co-dependent, we can differentiate either side, or both sides, starting with either side we like.
Let z = 1 - b2
z = y2
Δz = z' = 2y (from my table of integer exponents)
Δ(1 - b2) = 2y
Now switch sides and differentiate again
2y = Δ(1 - b2)
2Δy = 2y' = 2b (once again, straight from the table of exponents)
y' = b = cos x
You will say that I just followed normal procedure, but I didn't, since whenever I use the equation nzn-1 = Δzn I pull it from my table of exponents and constant differentials, not from current sources, which I have shown are all faulty. I prove this equation using a constant differential, not a diminishing differential or a method using limits. My table shows that with the exponent 2, you only have to go to a third sub-change in the rate of change chart in order to find a straight line, or a constant rate of change. This means that you aren't anywhere near zero, and aren't anywhere near an infinite series of any kind. You are two steps below the given rate of change for this problem (which is an acceleration or its pure math equivalent) and two steps is two steps, not an infinite number of steps. In any rate of change problem, we simply aren't dealing with infinite series, points, or limits. We are dealing with subchanges, and we are seeking a line of constant differentials. Not a point, a line. This is why my method is so important.
It does not matter in this problem that the curve was created by sine or cosine. The way the curve was created does not concern us in calculus. All we need is at least one dependence. If we have that dependence then we can use the definition of exponent and integer to create the table, and that table will straighten our curve out in a definite and finite number of steps—the number of steps being absolutely determined by the exponent itself.
An infinite series is only created by an infinite exponent. But an exponent signifies a change, and a change requires time, so that an infinite exponent would imply infinite time. We do not need to solve equations concerning infinite time, not in physics and not in mathematics. Therefore we have no need of infinite series in rate of change problems.
My analysis of the Cartesian graph (in my long paper) was necessary to discover the problem with calculus, and therefore this analysis certainly transcends "illustration", especially since that word has been thrown at me in a pejorative sense. I never claimed that calculus was all about graphs, or implied that the graph was the central feature of either calculus or of my argument. But I would never have discovered what I did without an in-depth analysis of the graph and the way the curve is created there, and I could never fully explain my method without using the graph to make it clear.