Is Standard Calculus Notation Wrong?

_{Jonathan Bartlett

April 8, 2019

Intelligent Design, Mathematics}

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We usually think of basic mathematics such as introductory calculus to be fairly solid. However, recent research by UD authors shows that calculus notation needs a revision.

Many people complain about ID by saying, essentially, “ID can’t be true because all of biology depends on evolution.” This is obviously a gross overstatement (biology as a field was just fine even before Darwin), but I understand the sentiment. Evolution is a basic part of biology (taught in intro biology), and therefore it would be surprising to biologists to find that fundamental pieces of it were wrong.

However, the fact is that oftentimes fundamental aspects of various fields are wrong. Surprisingly, this sometimes has little impact on the field itself. If premise A is faulty and leads to faulty conclusions, oftentimes workaround B can be invoked to get around A’s problems. Thus, A can work as long as B is there to get around its problems.

Anyway, I wanted to share my own experience of this with calculus. Some of you know that I published a Calculus book last year. My goal in this was mostly to counter-act the dry, boring, and difficult-to-understand textbooks that dominate the field. However, when it came to the second derivative, I realized that not only is the notation unintuitive, there is literally no explanation for it in any textbook I could find.

For those who don’t know, the notation for the first derivative is . The first derivative is the ratio of the change in y (dy) compared to the change in x (dx). The notation for the second derivative is . However, there is not a cogent explanation for this notation. I looked through 20 (no kidding!) textbooks to find an explanation for why the notation was the way that it was.

Additionally, I found out that the notation itself is problematic. Although it is written as a fraction, the numerator and denominator cannot be separated without causing math errors. This problem is somewhat more widely known, and has a workaround for it, known as Faa di Bruno’s formula.

My goal was to present a reason for the notation to my readers/students, so that they could more intuitively grasp the purpose of the notation. So, I decided that since no one else was providing an explanation, I would try to derive the notation myself.

Well, when I tried to derive it directly, it turns out that the notation is simply wrong (footnote – many mathematicians don’t like me using the terminology of “wrong”, but, I would argue that a fraction that can’t be treated like a fraction *is* wrong, especially when there is an alternative that does work like a fraction). Most people forget that is, in fact, a quotient. Therefore, the proper rule to apply to this is the quotient rule (a first-year calculus rule). When you do this to the actual first derivative notation, the notation for the second derivative (the derivative of the derivative) is actually . This notation can be fully treated as a fraction, and requires no secondary formulas to work with.

What does this have to do with Intelligent Design? Not much directly. However, it does show that, in any discipline, there is the possibility that asking good questions about basic fundamentals may lead to the revising of some of even the most basic aspects of the field. This is precisely what philosophy does, and I recommend the broader application of philosophy to science. Second, it shows that even newbies can make a contribution. In fact, I found this out precisely because I *was* a newbie. Finally, in a more esoteric fashion (but more directly applicable to ID), the forcing of everything into materialistic boxes limits the progress of all fields. The reason why this was not noticed before, I believe, is because, since the 1800s, mathematicians have not wanted to believe that infinitesimals are valid entities. Therefore, they were not concerned when the second derivative did not operate as a fraction – it didn’t need to, because it indeed wasn’t a fraction. Infinities and infinitesimals are the non-materialistic aspects of mathematics, just as teleology, purpose, and desire are the non-materialistic aspects of biology.

Anyway, for those who want to read the paper, it is available here:

Bartlett, Jonathan and Asatur Khurshudyan. 2019. Extending the Algebraic Manipulability of Differentials. Dynamics of Continuous, Discrete and Impulsive Systems, Series A: Mathematical Analysis 26(3):217-230.

I would love any comments, questions, or feedback.

Comments

DS, I'm here -- just noted on the Notre Dame fire. KFkairosfocus_{April 15, 2019
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I may be talking to myself now, but FTR, I haven't come up with anything useful from the equation: d^2x/dx^2 = (2 dt^2 + 2t d^2t)/(4t^2 dt^2) = (6s ds^2 + 3s^2 d^2s)/(9s^4 ds^2) How d^2t could sometimes be 0 and sometimes not remains a mystery to me.daveS_{April 15, 2019
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PS to my #118: Regarding my question:
1) In order to evaluate d^2x/dx^2, is it necessary that we have expressed x as a function of another parameter t? (That is, is there any way other than the method you have demonstrated?)
I guess you could take your equation: d^2x = 2 dt^2 + 2t d^2t and solve it for d^2t/dt^2, obtaining: d^2t/dt^2 = (d^2x - 2)/(2t dt^2) which shows that if x is a function of t, then we can solve for d^2t/dt^2 in terms of differentials involving the dependent variable; thus working "up the chain" of dependence. I don't know if this gives us any new information though.daveS_{April 14, 2019
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johnnyb,
Perhaps a terminology would suffice. “Conditionally Independent” vs. “Unconditionally Independent”. If y = f(x), x is a “conditionally independent variable”. If there is literally no other parameter upon which x depends, then we can call it an “unconditionally independent variable.” My notation is needed on “conditionally independent” variables, but not necessarily on “unconditionally independent” variables, for which the weird part of the term drops to zero.
I believe I understand what you're saying here, and it does make sense, although it also probably exposes philosophical differences once again. I would say that the mathematician (or quasi-mathematician, in my case) makes the decision about what the independent variable is, and that's that. Maybe more to the point, she decides that she is going to differentiate y with respect to x, e.g., so she treats x as the independent variable. Further, if y is a function of x and x is a function of t, one can still differentiate y with respect to x, and the "answer" will not have anything to do with how x varies as a function of t. So the apparent dependence of d^2x/dx^2 on yet another variable (say t, as you illustrated in post #100) is a little puzzling, although presumably not a problem in the end---I guess the t's must cancel when you evaluate your entire formula for d^2y/dx^2. That's where I'm at now. I still have some questions about d^2x/dx^2. 1) In order to evaluate d^2x/dx^2, is it necessary that we have expressed x as a function of another parameter t? (That is, is there any way other than the method you have demonstrated?) 2) Is the value of d^2x/dx^2 dependent at all on the way x varies as a function of its parameter? Above I tried to calculate d^2x/dx^2 twice, once where x = t^2 and once where x = s^3, but I don't know whether the two expressions I obtained are equal. So my question is, can we prove that this equality holds? d^2x/dx^2 = (2 dt^2 + 2t d^2t)/(4t^2 dt^2) = (6s ds^2 + 3s^2 d^2s)/(9s^4 ds^2) (After typing that equation out, I realize that s = t^(3/2), which we could use to relate the s's, t's, ds's, and dt's above. I suspect one can then prove the above equation holds, but I'll have to wait until this evening to check. Perhaps its a simple consequence of the chain rule? )daveS_{April 14, 2019
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Thanks, Johnny. I know 115 some time to type up, but I followed what you were doing. Hopefully having an example will help if others ask the same questions that I did.hazel_{April 14, 2019
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JB, an important concept, the admission of ignorance in the process. x has an identity with its own core characteristics, some of which we may not know. In that light, we should be aware of the possibility that we are not exploring such or may be ignorant of such. Where, too, it is not an adequate response that we may always insert a definition that x is a variable dependent on some imaginary parameter, we are dealing with identity, cause and characteristics here. KFkairosfocus_{April 13, 2019
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Hazel - y = x^3 dy/dx = 3x^2 (d^2y/dx^2) - (dy/dx)(d^2x/dx^2) = 6x x = t^2 dx/dt = 2t (d^2x/dt^2) - (dx/dt)(d^2t/dt^2) = 2 Let's start from "(d^2y/dx^2) - (dy/dx)(d^2x/dx^2) = 6x". Now, we want dt's on the bottom. Let's begin by multiplying the whole thing by (dx/dt)^2. Since dx/dt = 2t, this yields: d^2y/dt^2 - dy/dx (d^2x)/dt^2 = 6x * (2t)^2 Since x = t^2, this becomes d^2y/dt^2 - dy/dx (d^2x)/dt^2 = 6t^2 * 2t = 24t^4 Now, note that our right side is the "wrong" derivative, but the left side isn't yet in the form we need for the derivative. How do we close the gap? Interestingly, if we multiply y' and x'', we will get the following term: "(dy/dx)((d^2x/dt^2) - (dx/dt)(d^2t/dt^2))" which, if you distribute, becomes "(dy/dx)(d^2x/dt^2) - (dy/dt)(d^2t/dt^2)" If we add this to both sides, the left-hand side will become the second derivative! (d^2y/dt^2) - (dy/dx) (d^2x/dt^2) + (dy/dx)(d^2x/dt^2) - (dy/dt)(d^2t/dt^2) = (d^2y/dt^2) - (dy/dt)(d^2t/dt^2) This is perfectly valid, as long as we do it to both sides. So, if we multiply y' and x'' on the right-hand side, that is "3x^2 * 2". Since x = t^2, this becomes "3t^4 * 2t = 6t^4". If we add that to the right side (because we added its equivalent to the left side), the right side becomes "24t^4 + 6t^4 = 30t^4", which is the correct answer. All of these are pure algebraic manipulations, involving cancelling differentials, and doing things to both sides of the equation. Knowing Faa di Bruno's formula helps you to *know* what to do (it would be harder to guess what to do without it), but it doesn't affect the valid operations on the object. Hopefully that clears it up.johnnyb_{April 13, 2019
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DaveS - Perhaps a terminology would suffice. "Conditionally Independent" vs. "Unconditionally Independent". If y = f(x), x is a "conditionally independent variable". If there is literally no other parameter upon which x depends, then we can call it an "unconditionally independent variable." My notation is needed on "conditionally independent" variables, but not necessarily on "unconditionally independent" variables, for which the weird part of the term drops to zero.johnnyb_{April 13, 2019
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johnnyb & hazel, This sentence from post #106 illustrates perhaps a philosophical difference which is getting in our way:
If you have y = f(x), but also x = g(t), then x *is not* an independent variable.
Hazel pointed out that given a function y of x, you can always reparameterize by introducing a new variable t such that x is a function of t. For example, if y = x^2 (for all real x, say), then for each x you can set t = arctan(x) and now x(t) = tan(t). If x was an independent variable before, that shouldn't change just because I artificially introduced a "dependence" of x on the new parameter. Similarly, even if we think of y as a function of x, we can often differentiate x with respect to y. We can even differentiate sin(x) with respect to sqrt(y), and so on. I still don't have a handle on this, but I don't think dependent vs independent variables are the key issue.daveS_{April 13, 2019
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Oops, at 111, I meant post 106, not post 16.hazel_{April 13, 2019
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I'll quit soon, Johnny, as I don't want you to waste your time. Also, it is really hard to follow paragraphs 3-7 of post 16, partially because of having type in plain text. I've tried writing out hings in the normal way, but can't follow some of your skipped steps. I especially don't know where the part after the sentence "So, let’s now do it with the other side of the equation:" comes from. But you conclude the the "missing"6t^4 was "hiding" in the "-(dy/dx)(d^2x/dx^2)" term. How? What would I substitute where to show that that term was equal to 6t^4?hazel_{April 13, 2019
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johnnyb, Yes, I think the algebraic proof of the formula for the second derivative of f^-1 is the most compelling demonstration so far. IIRC, the "usual" proof does use the chain rule, and does get a bit messy. Your proof looks slicker.daveS_{April 13, 2019
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Dave - I use the chain rule a lot because it is the easiest place to discuss the benefits of this method and how it allows for an algebraic manipulation. However, the practical benefits so far have actually been developing the inverse function theorem, which doesn’t involve the chain rule. Even more so, the bigger benefits I think will come from the exploratory potentials. For instance, is there some way we might be able to put d^2y/d^2x to use (pay close attention to the formula because it is not the one you normally see). Does it have a meaning that we can use? Additionally, I think that knowing the formula will help people recognize when something is displaying “dependency” behavior. As I mentioned earlier, George Montanez has an interesting idea about this which I haven’t had time to follow up on yet. It might also help in numerical applications, though I’m not sure. In any case, I think the biggest benefit will come from the exploratory abilities of the new notation to give us real information about the nature of differentials.johnnyb_{April 13, 2019
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johnnyb, Is it fair to say that the places where you see the benefits of this approach (so far) involve composition of functions and/or the chain rule?daveS_{April 13, 2019
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I'll try using johnnyb's example from the paper to see how this works. If y = x^3, and x is the independent variable, then d^2x/dx^2 = 0, so the second derivative of y wrt x is just d^2y/dx^2, and is 6x. Now if we know x = t^2, then d^2x/dx^2 = (2 dt^2 + 2t d^2t)/(4t^2 dt^2) = 1/2t^2 + d^2t/2tdt^2, so this time the second derivative does not collapse to the "usual" form. We have a d^2t/dt^2 in this term, and if t depends on another variable, we have to go through another round of this process (and stop only when we hit the "ultimate" independent variable). If t is _the_ independent variable, then d^2t/dt^2 = 0 and we're done. In both cases, the second derivative of y wrt x is 6x. It's just that when we have to use johnnyb's formula (the one on the whiteboard), the terms in that expression change in a coordinated way so as to still equal 6x.daveS_{April 13, 2019
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Hazel - We seem to be going around and around and saying the same thing. I'm not really sure what you are trying to get at. I'll go over it this time, but after this it seems we are just wasting time talking past each other. If you have y = f(x), but also x = g(t), then x *is not* an independent variable. That's why you need to keep the rest of that in - since you don't know everything about the world, it may wind up that what you thought was the independent variable wasn't, so you have to hang on to the extra. Think about integration. If I integrate dy = dx, I will get y = x + C. The "C" is there because there are facts that I don't know about the original equation that I would need to know to fill out C. It *might* be that C is zero, and therefore discardable. But I would be wrong if I just left it out because, perchance, it might wind up being zero. Likewise, it might turn out that d^2x = 0, and I can throw out the whole right-hand side. But I need to keep it in there, because I don't really have independent knowledge about the entire state of the universe. So, with @100, if I wanted the second derivative of "y" with respect to "t", that would be "(d^2y/dt^2) - (dy/dt)(d^2t/dt^2)". Using the original notation (where d^2x/dx^2 = 0), then d^2y/dx^2 = 6x. dx/dt = 2t. So, if we tried to combine these facts, we could square dx/dt and get (dx^2/dt^2), which can then be multiplied by our original second derivative: (d^2y/dx^2)(dx^2/dt^2) = d^2y / dt^2 So, let's now do it with the other side of the equation: 6x * (2t)^2 = 24x * t^2 Since x = t^2, this becomes 24t^4. However, that is *NOT* the second derivative of y with respect to t. If we substitute it in at the beginning, we would have y = t^6, and the first derivative would be 6t^5 and the second derivative would be 30t^4. We've *lost* 6t^4 somewhere along the way. Where did it go? Well, it was hiding in the -(dy/dx)(d^2x/dx^2) term. Since our original second derivative didn't take into account the fact that x might not actually be independent, when we tried to combine it with another derivative it failed. Thus, it isn't algebraically manipulable, because algebraic manipulations can affect which variable is a function of which other variable. For instance, the new notation allows you to discover an inverse function theorem for the second derivative by purely algebraic means (note - we originally thought we discovered this new, but, while I haven't found it in either any textbooks or papers, I have seen it floating on the web on occasion, and most reviewers thought it was new as well). You can determine that: x'' = -y''(1/y')^3 You can do this by pure algebraic manipulation of terms. y = f(x) y' = dy/dx y'' = d^2y/dx^2 - (dy/dx)(d^2x/dx^2) So, if we're trying to establish a formula for x'', it would look like "d^2x/dy^2 - (dx/dy)(d^2y/dy^2)". So, notice that for y'' I have three dx's on the bottom, and for x'' I have three dy's on the bottom. Therefore, to cover for this, I simply multiply by (dx/dy)^3. That gives me: y'' (dx/dy)^3 = (dx/dy)(d^2y/dy^2) - (d^2x/dy^2) This is close, in fact, it is only off by a negative sign. So, multiply both sides by -1: -y'' (dx/dy)^3 = (d^2x/dy^2) - (dx/dy)(d^2y/dy^2) Now, on the left-hand side, note that (dx/dy) can mean either x' or (1/y'). So, we can just choose (1/y'), and this becomes our formula: -y'' (1/y')^3 = (d^2x/dy^2) - (dx/dy)(d^2y/dy^2) But you can only do this if you don't assume ahead-of-time that x is an independent variable. Note that it doesn't make sense to find the second derivative of x with respect to y if x is an independent variable. Algebraic manipulation assumes that you are going to put a hold on what is dependent on whatever else.johnnyb_{April 13, 2019
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Johnny, you write,
As a friend of mine who was an early reviewer put it, the “-(dy/dx)(d^2x/dx^2)” is kind of a “hold in reserve” type of a thing. It springs into action when we know something new about x. I’m hedging my bets, saying that, if I find out something new about “x”, this is how it is going to be combined with my existing knowledge.
It seems we have agreed that if we simply have y= f(x), with x as the independent variable, then d^2x/dx^2 = 0, and so y’’ = d^2y/dx^2, as in standard notation. The quote above then says that if you find out “something new about x”, that would be expressed in the “-(dy/dx)(d^2x/dx^2)” part of your notation. Can you give an example? If you knew something new about x, such as some function x = f(t), how would that be incorporated into the second derivation of the original function involving y and x, using your quotient notation? Could you illustrate using the two functions at 100, and using the quotient form of your notation?hazel_{April 13, 2019
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Hazel - That component becomes important *only* when other equations are involved. So, your question about d^2t/dt^2 can only be answered if I know additional things about t. The "appendage", if you will, tells you how to incorporate new information into the existing formula. As a friend of mine who was an early reviewer put it, the "-(dy/dx)(d^2x/dx^2)" is kind of a "hold in reserve" type of a thing. It springs into action when we know something new about x. I'm hedging my bets, saying that, if I find out something new about "x", this is how it is going to be combined with my existing knowledge. DaveS - That's the basic idea, but I don't have a pen/paper to verify your exact calculation. Yes, different relationships between x and other variables will imply different things about how x changes, which will lead to different formulas.johnnyb_{April 13, 2019
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I don't understand. In the example at 100, in your derivation of d^2x/dx^2, you don't even use the equation about y. Also, the independent variable of the second equation is t, not x, so it would be d^2t/dt^2 that we would be interested in.hazel_{April 13, 2019
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PS: Perhaps if x = s^n, then: d^2x/dx^2 = (n - 1)/ns^n + d^2s/ns^(n - 1)ds^2daveS_{April 13, 2019
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johnnyb, Thanks, that helps. If we assume x = s^3, say, don't we get yet another expression for d^2x/dx^2? Is it comparable to what you just obtained? (6sds^2 + 3s^2d^2s)/9s^4ds^2 is what I got, which could be reduced to (2ds^2 + sd^2s)/3s^3ds^2, or even: 2/3s^3 + d^2s/3s^2ds^2 (?!)daveS_{April 13, 2019
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As for asking for a concrete example of d^2x/dx^2, let me just give the example from the paper, but write it in a different way. y = x^3 dy = 3x^2 dx d^2y = 6x dx^2 + 3x^2 d^2x x = t^2 dx = 2t dt d^2x = 2 dt^2 + 2t d^2t So, d^2x/dx^2 = (2 dt^2 + 2t d^2t)/(4t^2 dt^2) FYI - my brain is not firing on all cylinders this morning, and I've already had to edit this twice, so apologies if there are any math errors here.johnnyb_{April 13, 2019
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kf writes, "However, JB is actually doing something more direct, treating dy/dx as itself a function f ‘ (x) where this is u[x]/v[x], with substitution instances dy and dx for u and v." Steve_h pointed this out, and I seconded the issue: I think this is the heart of the flaw in Johnny's thinking. As I quoted Wikipedia, the quotient rule is about the quotient of two differentiable functions, here labeled by kf as u and v. But in what way is it legitimate to consider dy and dx as functions us and v? And if x is considered the independent variable, as Johnny has acknowledged, d^2x/dx^2 is 0, so his process just produces the standard notation for the second derivative.hazel_{April 13, 2019
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H & DS, The differences between d^2x and dx^2 are interesting. Especially if behaving like a value is part of it. d[d(x)] is obviously a second order operation giving an image, repeat of the d operation. By contrast, dx times dx seems to be some sort of multiplication of a one stage go to differential operation, if they are to operate algebraically -- for the sake of argument. That suggests that ontological difference is relevant, though obviously they may hold the same resultant value, maybe for all x. What something is as to essence and in logically accurate description is not the same as what quantitative value it holds. If there is an underlying implicit process, that would possibly make a difference: how do we creep h = [(x+h) - x] to not-quite-zero? Operationally, not merely as an idea. What is the product h^2, once it is at its not quite zero target; is that target properly dx? Likewise, on d^2 [y] what is the differential on dy = lim h-> 0 of [f(x+h) - f(x)], i.e. d^2[y]? Then, substitute x in this. What, then is d^2[x]? Thence, the ratio d^[x]:[dx]^2? Strange indeed, if valid. However, JB is actually doing something more direct, treating dy/dx as itself a function f ' (x) where this is u[x]/v[x], with substitution instances dy and dx for u and v. From this he puts forward on that calculation: d/dx[ dy/dx] = d^2y/dx^2 - [dy/dx]*[d^2x]/dx^2] Obviously, where x is indeed the final, independent variable the second term on RHS is dy/dx * 0 --> 0, yielding the result that is well known. This, on grounds that dx/dx = 1 and by substitution d/dx [1] = 0. But this is a reduction, not an equivalence. By comparison, in accounting, there is a classic equation that obtains for the balance sheet at all times. But the structure within the sides can make all the difference in the world. And, changing that internal structure ill-advisedly can be ruinous. Where the intra-side structures stand can make a significant difference, so numerical equality LHS = RHS is not the whole story. Now, what happens when y is now x itself? d/dx[ dx/dx] = d^2x/dx^2 - [dx/dx]*[d^2x]/dx^2] If dx/dx is 1, it reduces to 0 on RHS, but that is on substitution. On LHS, the slope of a constant is an operation that yields 0. What is so quantitatively and what is going on ontologically are it seems not quite identical. Curious, it seems. And, suggestive of why going to the epsilon-delta form would seem clearer conceptually. KFkairosfocus_{April 13, 2019
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KF,
He suggests, that the nature of the second order differential is something he has not got a direct or expert answer to.
I'll let johnnyb address this if he chooses, but I would *assume* this has all been hammered out if we're issuing corrections to the notation of elementary calculus. I know what d^2x/dx^2 means in the "old" system. It should be utterly routine to explain it in the new, improved system. A minimal example where d^2x/dx^2 is nonzero might be a good starting point.daveS_{April 13, 2019
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DS, In 72, JB makes an ontological point:
what d^2x/dx^2 actually *is* is a different story. It actually depends on what x is itself dependent on . . . . I can see the possibility that a truly independent variable’s second differential should always be zero, but I don’t know if I’m fully convinced of that yet. Still trying to decide.
That's being not quantity or value in the first place, leading to a discussion on what value it should take, if it is always and everywhere nil. He is not sure on genuinely independent variables. He suggests, that the nature of the second order differential is something he has not got a direct or expert answer to. Now, too, he brings to bear a relevant issue, the nature of the absolute differential dy and how we got there. Context counts and dy from x as independent is not the same in meaning as dy from t as independent variable. Or from a set of variables that leads to the partial differential contribution sum we know from further work on partial differentials. We know y is a dependent, so on what does it depend is relevant. All of this then leads to infinitesimals, and thence to constants, i.e. specific infinitesimals of form m = 1/K, K beyond any finite k in R. m is closer to 0 than any finite k in R similarly taken under the reciprocal can give, k GRT 1. Something like dx or dy is in that range but is also the result of an operation of limits as chords go to tangents. KFkairosfocus_{April 13, 2019
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hazel, I'm back at my computer, so I can view the paper again. I think johnnyb addresses my questions to some extent there, but I still am mystified by exactly what this d^2x/dx^2 is in the simplest of cases. Perhaps johnnyb can show us what it is in my example from post #91 or in his example(s) from the paper.daveS_{April 13, 2019
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But what if it's turtles all the way down? :-)hazel_{April 13, 2019
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hazel, I'm just guessing here, but perhaps in johnnyb's process, you calculate all differentials with respect to the "ultimate" independent variable at the start. And then somehow d^2x/dx^2 turns out to be nonzero in some of those cases where x depends on another variable.daveS_{April 13, 2019
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Dave, you write, "I am thinking of this situation strictly mathematically, where the independent variable is what it is, period. There is no philosophical musing about whether time is the “primary” independent variable. So if a problem is specified completely, then it should be crystal clear what d^2x/dx^2 is." I agree completely.hazel_{April 13, 2019
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