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taeto last won the day on August 1

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  1. definition of derivative

    It seemed that you were basically explaining the standard approach to the OP, and doing it quite well at that. But seeing that he seems to have a starting point of objecting to the classical approach, it is maybe not an effective strategy to bring it in like that. Abbott is friendly and careful. It is clear that the \(\delta x\) and \(\delta y\) can be any numbers, however large, as opposed to the setting imagined by the OP. I have not come across Ferrar or Hobson. I have learned some analysis from Apostol and Rudin, and I have lectured analysis from Bartle. Could be that I am simply too old school.
  2. definition of derivative

    Again: to say "\(dx=1/\infty\)" does not make sense unless you explain it. Why actually do you think that \(1/\infty^2\) can be "ignored", and \(1/\infty\) cannot be ignored? Look pretty much similar to me. What studiot did was this. He replaced the \(h\) in the usual definition by your \(dx\), and then did what one does when using the classical definition, namely calculate the expression and calculate the limit as \(h \to 0\) (respectively \(dx\to 0\) ). The \((dx)^2\) bit means that when you divide \(h^2\) by \(h\), then you get \(h\), and \(h \to 0\), i.e. "can be ignored". However, it seems that you cannot ignore the extra term \(dx\), and it will remain a part of the answer. And you tell us that it is not zero. According to your thinking, the derivative is not zero. Another thing is that studiot correctly assumed that the \(h\) (which he called \(dx\) to humour you) is a real number, so you can calculate with it using normal arithmetic in the field \( (\mathbb{R},+,\cdot) \). If you want to do arithmetic with things that are not real, which is what you say, then you have to build a lot more theory before you can perform the similar sequence of calculations.
  3. definition of derivative

    It does make sense. You explicitly allow \(1/\infty = 0\), but that would make it a real number. Alternatively it seems fine to have a non-real \(\epsilon > 0\) that is less than every positive real. That would mean an extension to the usual linear ordering of \(\mathbb{R}\). But if you want to do calculations using such a thing, you have to explain what the rules are: what is going to be the meaning of \(x+\epsilon\) and \(\epsilon x\), and things like that. Great. So we will get to see how you use your formula to calculate the derivative at \(x=0\) of the function \(f\) given by \(f(x)=x^2\) for every \(x\in \mathbb{R}\) like I asked before? Why not show us how it actually works? Show how it is meaningful.
  4. definition of derivative

    I would be curious to see a reference for that. What does a "resulting field" look like? Sure you could join \(i\) to \(\mathbb{R}\) and get \(\mathbb{C}\), and just give the name "\(\infty\)" to \(i\). But here he wants to preserve the order relation from the real numbers, and that is one thing that will be a little harder to do.
  5. definition of derivative

    If you want to do that, you will have to assign some meaning to such a sum, because there is no a priori meaning attached to it. Maybe a better analogy would be to think of two functions f and g that have disjoint domains.
  6. definition of derivative

    It would be interesting to observe how he reacts to your own nice exposition re differentiating \( x \mapsto x^2\) at the point \(x=0\). But you are presenting a very classical approach, and I think that he is not aligned with that. I am more concerned with how he will explain the ways in which he diverges from the common understanding. Your explanation he will probably just blankly contradict. I would like to see his own explanation.
  7. definition of derivative

    If anything I said so far can be construed to implicate that I think that 113 uses terminology which is in any way consistent or meaningful, then I apologize. What are other counts by which you differ?
  8. definition of derivative

    That is the physicist's approach. From a mathematical point of view, the quantity \((dx)^2\) does depend on \(dx\), and `ignorable' is not a valid predicate. So from that point, the derivative becomes a function of the variable \(dx\). I suspect that the OP wants to consider the possibility that \(dx\) is not a real number. That leaves them with the unthankful burden of explaining the meaning of evaluating \(f(x+dx)\) for a function \(f\), which is only defined on real numbers, in a point \(x+dx\) which is not a real number.
  9. definition of derivative

    Could you demonstrate how that works if \(f(x)\) is equal to \(x^2\) for every real number \(x\): how would you calculate \(f'(0)\) using your formula?
  10. What does this mean ?

    Yes, the truth is that the limit does not exist.
  11. What does this mean ?

    It is a meme, see en.wikipedia.org/wiki/Meme The fact is that [math]\sum_{n=1}^\infty n^{-s}[/math] is convergent for every complex number [math]s[/math] with [math]\mbox{Re}\ s > 1[/math]. But the function that maps [math]s[/math] to [math]\sum_{n=1}^\infty n^{-s}[/math] is only defined for [math]\mbox{Re}\ s > 1[/math]. So there is another function, called the Riemann zeta function [math]\zeta[/math] which is defined for all complex numbers except [math]1[/math], which is a pole. The Riemann zeta function satisfies [math]\zeta(s) = \sum_{n=1}^\infty n^{-s}[/math] for all complex values [math]s[/math] with [math]\mbox{Re}\ s > 1[/math], but not elsewhere. The value of [math]\zeta(-1)[/math] happens to be [math]-1/12[/math]. For [math]s=-1[/math] the expression [math]\sum_{n=1}^\infty n^{-s}[/math] makes no sense.
  12. surviving free fall into the water from high altitude

    I saw a story about a lady whose parachute did not open and she fell free fall. She still survived, and it was said, because she was lucky to fall on her chest, to make the rib cage take most of the brunt (I suppose it was the ribs, the story did go like that). That was on land. But if it works on land it might also work on water, no?
  13. Do you masturbate?

    I believe the forum is completely public. So maybe add an "I don't know" option.
  14. Disproof of Riemann Hypothesis

    Three questions: 1. Who is wrong? 2. What are they wrong about? 3. What would be right?
  15. definition of derivative

    If your math book suggests to compute the derivative of \(y\) as the limit of the difference quotient of \(f\), without saying first that \(y=f(x)\), then it makes sense for you to throw it in the recycling bin. Maybe it can. But this is analysis, a particular branch of math that deals with functions of real and complex numbers. If you ask whether it makes sense to have a real number \(d\) with the property \(0 < d \leq x\) for all real numbers \(x\), then the answer is that such a \(d\) cannot exist, because, e.g., \(d/2\) has \(0 < d/2 < d\) in \(\mathbb{R}\) contradicts that \(d\) has the required property. If this is not the property that you would want an "infinitely small" number to have, then what is the property that you are thinking of? Will you evade to answer that question? Just writing up some string of symbols like \(1/\infty\) doesn't always point to something that makes any sense. I ask again, what does it mean to you? Apparently you can make sense of it, since you keep going on about it. Why don't you answer the question: "what does it mean?"? Now it looks like you use "finite" to mean "non-zero". In the context of analysis, distance is given by Euclidean metric, and then the only distance that is not "finite" is identically 0. You end up with \(f(x+dx)-f(x) = 0\) always, independently of \(f\) and \(x\). The most positive I can say would be something like just forget about the actual meaning of \(dx\). In ordinary usage, the functions \(x\) and \(dx\) belong to different species, they do not allow to be composed together by the binary operation of addition. It is something like trying to add a scalar to a 2-dimensional vector. The most you could do is to use your expression as a notational shorthand, which substitutes \(dx\) for \(h\) and removes the needs to write the \(\lim\) symbol. Maybe you can think of any advantages in doing so. Something like how the Leibniz notation allows to write the chain rule in the form \(\frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx}\).