taeto

Not really with anything. If you say "Assume a set S exists such that 2+2=5", then you reach a contradiction. In which case you just resolve that no such set exists, and you deduce that 2+2 is not 5. If you say "Assume a set S exists such that P. Then Q", where P is an undecidable proposition, and you prove your statement, then you have proved an implication: if S exists, then Q is true. This happens very often in algebra, which has lots of undecidable statements, such as P="G is an abelian group" and the likes. Then by showing "P => Q: x and y are in G implies x+y=y+x" you have shown that Q is true given a particular circumstance P.

And interestingly, this is actually not completely correct. There are some cases, even in ordinary arithmetic, when it can happen that there is no conflict at all with rules or axioms to prevent the existence of the humperdink. Examples arise when a humperdink means a solution to certain "diophantine equations"; they are built from addition and multiplication of integer numbers and some variables that can be given integer number values. A humperdink then means an assignment of values to the variables for which the equation is satisfied. An example of a diophantine equation would be something like \(x^3 + 2x^2y^2 - 4y^3 +5 =0.\) For this particular one it is probably either easy to solve it, or to prove that no solution exists, when integer numbers are substituted in for \(x\) and \(y.\) But it is not too hard to imagine that it might take some amount of work to decide either way. Some diophantine equations have been constructed for which it is impossible to prove that they have solutions, and impossible as well to prove that they have no solutions. Therefore in such a case, you cannot simply assert that one exists, nor assert that none exists. Whenever you make such an assertion, you have to prove it. It is not enough that it does not conflict with rules or axioms. Of course Gödel was the pioneer of such discoveries.

I still have no idea what you mean by your \(dx\). It is used in various roles in Calculus, Analysis and Differential Geometry, and none of them agree with what you say about it. Now you state \((dx)^2=0.\) When you use an equality \(=\) sign, do you actually mean that the things on either side are identical? And your original post assumes that it makes sense to divide by \(dx\), right? So then if you take \((dx)^2=0\) and divide by \(dx\) on both sides, does dividing the same quantity by the same quantity produce different results depending on whether the quantity is on the LHS or the RHS of the equality sign? If you carry out this division step, do you see why it is confusing at least to some people when you insist that \(dx>0?\) When you try to understand calculus, you should become familiar with the standard limit argument and be able to recognize that studiot applied it. It looks like you are making up your own stuff. Have you ever seen identities like \(dx = 1/\infty\) or expressions like \(x+dx\) in any text which seriously teaches calculus? Anyway, you are obviously not developing any "mathematics", seeing that you only work with undefined concepts. Show how to calculate the derivative \(g'(0)\) of the function \(g\) defined by \(g(x)=x\) if \(x\) is a real number, and \(g(x)=0\) if \(x\) is not a real number.

You incorrectly attribute it to wtf. But wtf took it as a quote from a quora post by the OP.

You do not have to show that one number is greater than the other in the ordering. You can compare them by just saying that they are different, by the construction.

I had thought that you were attached to the idea that you may "ignore" \(dx^2\) at any time before you finish the calculation. Happy to see that is not the case. You are "ignoring" terms only when doing so will lead you to the correct result? You have not volunteered much motivation to show that you "definition" of a derivative is somehow more practical than the standard definition. You refer to studiot's calculation, in which he uses a standard limit argument. How about another different case to show how your suggestion may be the superior one. Now let g be the function for which g(x) = x if x is real, and g(x) = 0 if x is not real. Of course in real analysis you only consider domains that are subsets of the set of real numbers, accordingly you would compute a derivative of g'(0) = 1 at x = 0, if x = 0 is contained in an open interval of the domain of g. But that would be by the classical definition of derivative. What would your answer be, and how would you arrive at it?

Are you sure that you can divide by \((dx)^2\)? Do we not have to ignore \((dx)^2\) before we continue? This is precisely the reason why we can see that studiot was doing the classical computation. When he divides his version of \((dx)^2\) by \(dx\) he will end up with a term \(dx\) the limit of which is \(0\), meaning that as an additive term it can be" ignored", that is, already considered to add \(0\) to the result. However, when you divide \((dx)^2\) by \(dx\) you get \(dx\), a term which you insist on being nonzero, hence it cannot be ignored in the same sense. Apparently studiot used "ignoring" to mean that we can replace additive terms which have limit equal to \(0\) by \(0\) itself when we take a limit. This is a very precise notion. But you seem to have a different idea of what it means to "ignore". I.e. if we do calculations your style, do we get \(1 = \frac{dx}{dx} = \frac{(dx)^2}{(dx)^2} = \frac{0}{0}\) because we can "ignore" \((dx)^2\)?

I am glad if it was funny. You are american though. And it must be different for the brits. Myself I had to go to Canada for my phd studies. I probably have very little idea of your respective jargons. I am from Denmark, as it says in my location tag. I wanted to write "on the ball". You and studiot are on the ball. Just keep staying on top of it, please.

For somebody who is not sure about the precise definitions it may seem plausible that well-ordering implies ability to count all the elements by going up the order from first element to next, and so on. That however is not true about well-orderings. You can well-order the natural numbers as 1,3,5,7,...,2,4,6,8,..., i.e. the odd numbers are in their usual order, and the even numbers as well, but every even number is larger than every odd number in this ordering. You can check that it is a well-order. And if you start from the left and go up infinitely long, you only get to count the odd numbers. Any well-ordering of the reals has similar properties, though much more complicated, well, actually impossible, to describe.

Sorry about that to both of you. I meant to write something else but slipped. Apologies.

Fair enough. I forgot about the exact title. The thread is divided among contributors who either do not know what is in the thread or do not remember what it is about. Plus Strange and studiot, they are always on the game.

Strange: there is no proof. It is just an illustration of how the Collatz sequences work, using the standard tree representation. I suggest that it is allowed to discuss a problem in the forum without making a claim to have solved it, no?

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.

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.

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.

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.

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.

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.

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?

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.

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?

Yes, the truth is that the limit does not exist.

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.

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?

I believe the forum is completely public. So maybe add an "I don't know" option.