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That's interesting, thanks for sharing. The main question is not answered though: " How many terms should I grab to go safe for every case? Why doesn't it suffice to take just the 1st non-zero term? " Then they work with the limit: \[\lim_{x \to 0} \frac{tan(x) - sin(x)}{x^3} \] The answer is 1/2, using the limit calculator: https://www.symbolab.com/solver/limit-calculator/\lim_{x\to0}\frac{\left(tan\left(x\right) - sin\left(x\right)\right)}{x^{^3}} The limit calculator uses L'Hôpital three times and then plugs in the value 0. What I was asing in my first post, is it possible to plug in an infinitesimal value? Is it possible to calculate limits using infinitesimals?

Normally limits are used instead of infinitesimals, but is it possible to calculate limits using infinitesimals? For example: \[ \lim_{x \to 0} \frac{sin(x) - x}{x^3} \] this is usually solved by applying L'Hopital's rule 3 times and the answer is -1/6: https://www.symbolab.com/solver/limit-calculator/\lim_{x\to0}\frac{\left(sin\left(x\right) - x\right)}{x^{^3}}

Every point of a number line is assumed to correspond to a real number. https://en.wikipedia.org/wiki/Number_line Is it possible to find points corresponding to infinitesimals on a number line? I mean finding an infinitesimal between two neighbouring points (between two real numbers). I am assuming that every point is surrounded by neighbourhood. I got this idea of neighbouring points from John L . Bells' book A Primer of Infinitesimal Analyis (2008). On page 6, he mentions the concept of ‘infinitesimal neighbourhood of 0’. But I think he would not consider his infinitesimals as points because on page 3 he writes that "Since an infinitesimal in the sense just described is a part of the continuum from which it has been extracted, it follows that it cannot be a point: to emphasize this we shall call such infinitesimals nonpunctiform."

The limit defintion of derivative in my previous post contains only the symbols h (corresponding to Δx) and dx. There is no δx. It seems to me that the introduction of "differential calculus" gives rise to symbol δx. Then there seems to appear two representations: f'(x) = dy/dx f'(x) = δy/δx I think it is possible the usage of δy/δx was chosen to escape the problem arising in real number calculus, the problem of 0/0.

I am beginning to suspect that calculus is not based on real numbers. Look at the definition of the derivative: \[\frac{dy}{dx} = \lim_{h\to\ 0}\frac{f(x+h) - f(x)}{h}\] where h is finite. What is dy/dx? An infinitesimal ratio? A ratio of two infinitesimals dy and dx ? It seems to me that we are not dealing with real numbers anymore if dy and dx are not real numbers.

I don't know. Maybe the answer can be found in the book I am studying. John L. Bell is defining the ‘derivative’ of an arbitrary given function f : R → R. For fixed x in R, define the function g: Δ → R by g(ε) = f(x + ε) so that f(x + ε) = f(x) + εf'(x) is the fundamental equation of the differential calculus in S for arbitrary x in R and ε in Δ.( Δ may be considered an infinitesimal neighbourhood or microneigbourhood of 0). Also he is stating Microaffiness Axiom: For any map f:Δ → R there exist unique a, b ϵ R such that f(ε) = a + bε for all ε ϵ Δ He draws a conclusion: Our single most important underlying assumption will be: in S, all curves determined by functions from R to R satisfy the Principle of Microstraightness. The Principle of Microaffineness may be construed as asserting that, in S, the microneighbourhood Δ can be subjected only to translations and rotations, i.e. behaves as if it were an infinitesimal ‘rigid rod’. Δ may also be thought of as a generic tangent vector because Microaffineness entails that it can be ‘brought into coincidence’ with the tangent to any curve at any point on it. Since we will shortly show that Δ does not reduce to a single point, it will be, so to speak, ‘large enough’ to have a slope but ‘too small’ to bend. I don't have all the answers to your questions. I am only studying the subject. Don't expect me to have all the answers if no-one else has been able to find them. I am looking for them in the books. I am not developing my own system of algebra.

Do you think the sum (x + 2dx) is using some different system of algebra than, for example, the sum (x + dx) ? I did not invent my own system of arithmetic. I am currently learning what John L. Bell has written in his book. I think you should ask the same question about what axioms of arithmetic are used in an infinitesimal approach: dx is nilsquare infinitesimal, meaning (dx)² = 0 is true, but dx=0 need not be true at the same time. So how is multiplication defined here? What axioms of arithmetic are being used? Maybe they are to be found is John L.Bell's book, he writes: "As we show in this book, within smooth infinitesimal analysis the basic calculus and differential geometry can be developed along traditional ‘infinitesimal’ lines – with full rigour – using straightforward calculations with infinitesimals in place of the limit concept. And in the 1970s startling new developments in the mathematical discipline of category theory led to the creation of smooth infinitesimal analysis, a rigorous axiomatic theory of nilsquare and nonpunctiform infinitesimals."

\[ f'(x) = \frac{f(x+dx) - f(x)}{dx}\] \[ f'(x + dx) = \frac{f(x + 2dx) - f(x + dx)}{dx}\] \[ f''(x) = \frac{df'(x)}{dx}\ = \frac{f'(x+dx) - f'(x)}{dx}\] from which, after a calculation (I skip writing this lengthy LaTeX code now, you may try it yourself), it is possible to get the result in my first post, the definition of second derivative: \[ f''(x) = \frac{f(x+2dx) - 2f(x + dx) + f(x)}{(dx)^2}\]

The question is: why can't John L. Bell's nilpotent infinitesimals possess inverses?

There is a book available, even for free download, A Primer of Infinitesimal Analysis by John L.Bell. It is possibly what I am looking for. The book says that: "A remarkable recent development in mathematics is the refounding, on a rigorous basis, of the idea of infinitesimal quantity, a notion which, before being supplanted in the nineteenth century by the limit concept, played a seminal role within the calculus and mathematical analysis."-direct quote Also an interesting note from the book: "A final remark: The theory of infinitesimals presented here should not be confused with that known as nonstandard analysis, invented by Abraham Robinson in the 1960s. The infinitesimals figuring in his formulation are ‘invertible’ (arising, in fact, as the ‘reciprocals’ of infinitely large quantities), while those with which we shall be concerned, being nilpotent, cannot possess inverses." -direct quote

I am not here to talk about those subjects. There are already enough books about them available. In the beginning, in my second post, I told that I am dealing with an infinitesimal approach: dx is nilsquare infinitesimal, meaning (dx)² = 0 is true, but dx=0 need not be true at the same time.

In my first post dx is an infinitesimal yes, I have obtained f''(x) = 2 using the definition in my first post

just to see how useful an infinitesimal approach is

Let's choose an example \[ f(x) = x^2 \] using the definition in my first post, obtain the second derivative of f(x)

it is possible to use division by zero: https://en.wikipedia.org/wiki/Signed_zero

What's wrong with division by zero ?

dx is nilsquare infinitesimal, meaning (dx)² = 0 is true, but dx=0 need not be true at the same time. https://en.wikipedia.org/wiki/Smooth_infinitesimal_analysis A problem seems to arise because there appears to be a division by zero in that case.

Is it possible to define the second derivative of f(x) in this way: \[ f''(x) = \frac{f(x+2dx) -2(f+dx) + f(x)}{(dx)^2} \] I am using a finite difference approximation called "Second order forward" from the link, I use dx instead of h: https://en.wikipedia.org/wiki/Finite_difference#Higher-order_differences

I have explained from the very beginning what I mean by dx: An introduction of infinity brings a duality into the definition of an infinitesimal, meaning that we have to deal with objects that both zero and non-zero at the same time. From the very beginning you evaded answering my question " is 1/∞ zero or non-zero ?" That's why you are stuck at asking what I mean by dx. I have explained everything exactly. It seems to me that you are ignoring what I have written. Also those sources you mention don't deal with the duality I mentioned, they simply ignore it. That's why it may look as though they don't agree with what I have written. But the truth is they don't even deal with it. It may look as though I am dividing by 0. But again the result depends on which order the calculation is done: take (dx)2 = 0 and divide by dx on both sides \(\frac{(dx)^2}{dx} = \frac{0}{dx}\) which is the same as \(\frac{0}{dx} = \frac{0}{dx}\) looks valid because it is 0 = 0 On the other hand if simplification is done first \(\frac{(dx)^2}{dx} = \frac{0}{dx}\) \(dx = 0 \) There is confusion only if you ignore the duality that I mentioned, I don't simply just insist that dx > 0. The confusion arises if you ignore that dx is both zero and non-zero at the same time. You are trying to force, or define, dx to be either 0 or non-zero. no, he did not apply limits, he just stated that he can ignore (dx)2 if dx is small enough, exactly what I have done. no, I am not making up my own stuff. I have seen identities like 1/∞ = 0 in many texts which seriously teach calculus. Also I have seen that many of them ignore the duality I mentioned, but not all of them. So I did not make up the duality myself, I am dealing with it. no, I don't work with such concepts. The reason is that you can't define, or force, dx to be either 0 or non-zero, because it is both. You are ignoring the problem.

No, I am not ignoring terms in order to get to correct result. I am ignoring terms that are equal to 0. The result of \(\frac{(dx)^2}{(dx)^2} \) depend on the order: if the simplification is done first then the result is different than if (dx)2 = 0 is used : \(\frac{(dx)^2}{(dx)^2} = \frac{dx}{dx} \frac{dx}{dx}\ = \frac{dx}{dx}\) <------ simplification done first \(\frac{(dx)^2}{(dx)^2} = \frac{0}{0}\) <------ (dx)2 = 0 is used So how can you be certain that \(\frac{dx}{dx} = \frac{0}{0}\) ? It could be that \(\frac{dx}{dx} = 1 \) and it does not need to be equal to \(\frac{0}{0}\). Just because dx can be non-zero, so why should \(\frac{dx}{dx} = \frac{0}{0}\) ? No, he does not use a standard limit argument, he does not use limits at all. I don't see why this very fact is so hard for you to admit. Is it because what he has shown supports what I have written? And you decided that I am doing calculations in "my style", and it would look impossible for you how studiot is also doing calculations in "my style"? You just can't admit that I am not developing my own mathematics from scratch. All that I am trying to do is understanding calculus. I use calculus, I don't invent my own calculus. I am not sure of what you want me to do.

I am not sure if I can divide (dx)2. But I am not sure about you. Maybe you can ignore (dx)2 before you continue, but not me. He did not take the limit. but you said that I am doing calculations in "my style" so that I should not end up with an extra term dx because I am ignoring (dx)2 . So why you are now telling that I am ending up with the extra term dx? Am I doing calculations in "my style" or am I not? He did not take the limit. Perhaps if he had taken the limit. That more or less looks like doing calculations in your style, not mine.

you are telling that dx and (dx)2 are the same. In that case we would have \(\frac{dx}{(dx)^2} = 1 \) Instead what we have is \(\frac{dx}{(dx)^2} = \frac{1}{dx} = \infty \) He did not calculate the limit, he just stated that " If dx is a very small quantity then (dx)2 is insignificant and may be ignored ". When I asked him to tell exactly how small dx must be so that (dx)2 can be ignored, he told me to look for an answer in a book. No, the extra term dx will not remain part of the answer just because of what studiot said " If dx is a very small quantity then (dx)2 is insignificant and may be ignored ". It would remain part of the answer only if it did not satisfy what studiot said, if it would not be small enough. All that I am asking for is for him to explain how small exactly it must be. I have told that is dx is exactly 1/∞ , and it means that it is infinitely small, only in that case (dx)2 can be ignored.

Using \( dx = 1/\infty \) \( sin(x) = x - x^3/3! + x^5/5! - ... \) \( sin(dx) = dx - {(dx)}^3/3! + {(dx)}^5/5! - ... = dx \) \( e^x = 1 + x + x^2/2! + x^3/3! + ...\) \( e^{dx} = 1 + dx + {(dx)}^2/2! + {(dx)}^3/3! + ... = 1 + dx\) because \( {(dx)}^2 \) and higher powers of dx are equal to 0 studiot already showed how to do it. Since he said that I don't use not terminology exactly, I asked how small dx is so that (dx)2 can be ignored but he said that I must look for an answer in some book. I have told that dx = 1/∞ exactly so that (dx)2 can be ignored. Am I more exact than his book?

1/∞ is not a real number. Does it make sense to have 1/∞ with the property 0 ≤ 1/∞ < x for all real numbers x ? I am not the one who is evading answering questions. I already told what does it mean: it is neither zero nor non-zero. Because it is both. You evaded answering my question. no, I don't use finite to mean non-zero. 1/∞ is not finite. You are ignoring an infinitely small distance dx, and it is not necessarily 0. Otherwise you end up with f´(x) = dy/dx = 0/0 That would not be a very good idea. are you trying to tell that I can't perform the additon x + dx ? I don't know why it should be only a notational shorthand without meaning. I sounds like forgetting or ignoring the problem. no, I can't think of any advantages in doing so.

Alright: How exact is your " If dx is a very small quantity then (dx)2 is insignificant and may be ignored so we have" ? I thought that you would understand that my sentence continued below. so does it tell exactly how small dx is?