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taeto last won the day on January 19

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  1. He seems to sense that a vertical line does not have a finite slope. Give him credit for that. You have an education degree right? What is your answer if a student asks "which two points in question"?
  2. It is still a line though. If you draw it in the xy-plane it even looks suspiciously like a horizontal line, not a vertical line.
  3. If that were the case, then \(y = 1\) would be a derivative of the function that maps \(x\) to \(\sqrt{x}\) for \(x \geq 0. \) Since the line with equation \(y=1\) intersects the graph of this function in exactly one point \((1,1)\).
  4. If it is really necessary, it should be so much easier for you to present a single example of such a calculation, even without having to go all the way back to the 17'th century. Can you do this? Remember that I am asking about dividing by zero. Dividing by an infinitesimal is no problem, each of us who have been following this thread already knows how to do that.
  5. Fair enough. This is the first time I heard it referred to as just Principia Mathematica. Thanks for pointing it out Studiot, fist bumps and shake hands all around, I hope. My bad. Then the claim stands, I suppose: Newton did know how to make sense of division by zero? Somewhere in one of the three volumes and on some page?
  6. Why do you think that you would have to provide the definition of a derivative, if that is what you mean by "proof of a derivative"? Nobody asked you about that. Instead you were asked to provide some kind of reference or evidence of your claim that Newton could make sense of division by zero. Maybe it would help if you just admit that you are simply making up nonsensical stuff all the time.
  7. That cannot be entirely accurate. When Principia was written, Newton had long since passed on. Anyway, if that is where you got it from, then surely you can provide the relevant volume and page number?
  8. About that point, I think that the "standard part" of a hyperreal is only defined to be the nearest real number if the hyperreal is finite. In the context of derivatives that just means that if the hyperreal differential quotient is infinite, then you conclude that the derivative does not exist.
  9. I suspect that according to the rules of the forum, since you state here that something has been proven, then you have to provide some kind of reference or other solid evidence. Is this alleged fact something which has been proven by you personally?
  10. I think that if you look carefully enough, you will find that the derivative is calculated using hyperreals by dividing by an infinitesimal \(\Delta x\) which is nonzero. Only after dividing do you extract the standard part.
  11. Your version of something being random is very different from everyone else's. If I could write a computer program which predicts with absolute 100% certainty the winning lottery numbers in next weeks lottery, and it would be correct every time, would you still persist in a belief that the lottery numbers are drawn every week completely at random? Most people would say that they are completely predictable.
  12. Actually it does not continue "randomly", since you can write a computer program which will tell you the \(n\)'th digit after the decimal point if you input \(n.\) And the only real number that \(1/\pi\) "becomes infinitesimally close to" is the number \(1/\pi\) itself. All other real numbers have a positive real number distance. For the simple reason that the decimal expansion that you are talking about is precisely the decimal expansion of \(1/\pi\) and nothing else. Also, it is only the real numbers that have usual decimal expansions, other hyperreals do not.
  13. That makes sense. The meaning is that \(A\) is a set of four functions. The role of \(x\) remains a little ambiguous. If we tread lightly, we can surmise that \(x\) is a coordinate, and its range is the reals. Do you see what is the span(\(A\))\(=V\)? Is it true that if a function \(f\) is any one of the four functions in \(A,\) then \(f'\) belongs to \(V?\) And what if \(f\) is any function in \(V?\)
  14. To "continue forever past the decimal point" is not quite the same as being irrational. The decimal expansion of the rational number \(1/3\) is \(.333\ldots \). You seem a little too hung up on the representation of numbers. The numbers themselves do not "have decimal points" nor do they "continue forever". These kinds of descriptions only apply to whatever representation we choose, which starting from elementary school happens to be mostly the decimal point representation. But the actual properties of numbers do not depend on the way in which we happen to choose a representation of them. In the decimal notation we represent \(1/3\) as \(.333\ldots \) whereas in ternary notation we would represent the exact same number as \(.1\).
  15. Something is wrong, I may not have pinpointed it exactly. The notation is suspect; \(A\) is a set of what kind of elements, and then what is span(\(A\))? It makes sense if we read the OP as \(A = \{\exp,\exp \cos, \exp \sin, \cos, \sin \}\), and taking span(\(A\)) to be the space of all linear combinations with real coefficients of those five functions. Then the only thing to prove is that the image of the span under \(T\) is contained in the span. Hence my original comment.
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