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Re: The New Calculus: Never division by ZERO. No bullshit limit theory. Sound analytic geometry.
Posted:
Nov 10, 2017 11:29 PM


Den fredag 10 november 2017 kl. 14:59:54 UTC+1 skrev John Gabriel: > On Friday, 10 November 2017 08:45:22 UTC5, Zelos Malum wrote: > > Den fredag 10 november 2017 kl. 14:37:59 UTC+1 skrev John Gabriel: > > > On Friday, 10 November 2017 08:01:38 UTC5, John Gabriel wrote: > > > > "It confuses me that you can set m+n equal to zero after you cancel out m+n in the ratio. Surely this is wrong?" > > > > > > > > Well, consider first that I proved the sum of all the terms in m and n must be zero using 8th grade algebra (see http://thenewcalculus.weebly.com) but also note that the slope of the tangent line does NOT depend on m or n so given the derivative as an expression in x and Q(x,m,n), it follows Q(x,m,n) must be zero, that is, m=n=0. > > > > > > > > So to summarise: > > > > > > > > i. There is NEVER any division by zero in the New Calculus, only in the bogus calculus. > > > > > > > > f'(x)= [ f(x+n)f(xm) ]/ (m+n) > > > > > > > > m+n is NEVER zero because no parallel secant line has the (0,0) distance pair for (m,n). Even after the ratio is reduced, that is, > > > > > > > > f'(x) = f'(x) + Q(x,m,n) > > > > > > > > any of the (m,n) pairs used before the factor is cancelled are still valid. Those parallel secant lines NEVER disappear!!! That shit only happens in your bogus limit calculus. It is easily proved that Q(x,m,n) = 0 as demonstrated in the 8th grade proof. > > > > > > > > Now when you deal with wellformed concepts, you are able to reach sound conclusions. For example, after the difference quotient is simplified in the New Calculus (not bogus calculus!), you can if you like, set m=n=0 because the tangent line does not depend on m or n. > > > > > > > > ********************** > > > > Moreover, because straight lines DO NOT have a derivative (only a slope), you will notice that for any straight line, there will NOT be any expression in m or n, that is, Q(x,m,n) is ALWAYS 0. Example: y=kx => y' = k(x+n)k(xm) / (m+n) > > > > > > > > => y' = k(m+n)/(m+n) + Q(x,m,n) = k + Q(x,m,n) = k + 0 = k. > > > > > > > > You see? Perfection is hard to rubbish. Does the slope of a straight line ever cease to be? Does it require m or n? NO. > > > > ********************** > > > > > > > > ii. Unlike your bogus calculus in which the meaning of h changes before and after the factor h is cancelled, the meaning of m and n remains unchanged: They are DISTANCES from the point of tangency to the endpoints of the parallel secant line. Those (m,n) pairs have been there since past perpetuity and continue to be there forever. They will never "vanish" or shrink or any shit like that. They are sound geometric objects: line segments! Chuckle. No infinitesimal differentials or any other kakka. > > > > > > > > iii. Your bogus calculus is such diarrhea that it never makes any sense. It was attacked by Berkeley because he was a smart guy. He knew Newton was onto something but he wanted solid mathematics, not mythology and ghosts. > > > > > > > > iv. Because the New Calculus is well formed and the FIRST and ONLY rigorous formulation in human history, new theorems and results are realised that are NOT possible in your diarrhea mainstream calculus. For example, even if you try to copy the New Calculus in your bogus calculus, you will fail! > > > > > > > > Consider the Auxiliary Equation Q(x,m,n)=0. The "analog" in bogus calculus is: > > > > > > > > f'(x) = lim { h > 0 } f'(x) + h > > > > > > > > But what the hell does f'(x) + h mean? It's nonsense. Why? Because Newton who was inferior to me intellectually was using NONPARALLEL secant lines. In other words, he was experimenting. He did not know how to find the derivative in a sound coherent way as I do in my New Calculus. Newt was the master of approximation and experimentation  that's what made him such a physicist genius. But I am the true mathematical genius  Newt would have loved or hated me. > > > > > > See entire explanation here: > > > > > > https://drive.google.com/open?id=1DuqrOjc5EIXoHpj6SnzUhFy4RPAAKVcE > > > > Any function f? > > f(x)=sin x > > > > as others have shown, that does not have the factor m+n in it when you do f(x+n)f(xm) > > > > So instantly false. > > Liar. You are instantly disproved: > > https://docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnx0aGVuZXdjYWxjdWx1c3xneDoxYmRlMTZjMDFlYjZlMmQ
You merely assert it is k there, you haven't shown it is a factor of it. Show me how you factor it out exactly for that.



