![]() Again, these properties is how it is defined, and makes no claim of the nature of f(x) in any statement. For this to be the case, I then am able to derive r. There is no mention of g(x) in the MVT statement, this is how we define g(x) by definition, which will satisfy the property irrelevant of f(x). Of course, I don't know r, and so can add an arbitrary condition on g(x) such that I can now choose an r. I define some arbitrary function, g(x), such that g(x)=f(x)-rxįor r. The idea is I have my functions f(x) which I can not make any definitions on besides the original statement, otherwise would not be general. Whether or not it is well defined is a different topic. ![]() We have not yet stated its purpose, this is how we DEFINE g(x) and r, and yet not limit f(x) in any way. Remember, we have said no property of r, there is not said one thing of r in the statement. What we are trying to prove is that there is a c in between interval such that f'(c)=(f(b)-f(a))/(b-a). Hint: you can't use c in your proof because that's what you need to prove. ![]() By choosing r in the way you do to "satisfy" Rolle's theorem, that is, f(a)-ra=f(b)-rb, you are in fact using the very c you set out to prove exists. Assuming a property in a matter that f'(c)=(f(b)-f(a))/(b-a) would be circular, since I am considering f(x) to be a general function, so this couldn't just be a definition, but would adding a specific property I never proved, of f(x). Assuming any property we want from it is not circular, as long as it is a well defined property. We define g(x) to be this function such that g(a) = g(b), and r is arbitrary, we have not made a statement yet of its purpose beforehand. No one said it has to be, g(x) is NOT f(x). f(b) is not necessarily equal to f(a) in the MVT." "You can't just choose an r so that g satisfies the conditions of Rolle's theorem, because:Ī. J-O-H-N Gabriel - discoverer of the new calculus. I will not give you the answer because you are dishonest, ignorant and incompetent. Mythmaticians the last 400 years have failed. Hint: you can't use c in your proof because that's what you need to prove.īig Fail! You have been failing since the inception of Wikipedia. f(b) is not necessarily equal to f(a) in the MVT.ī. ![]() You can't just choose an r so that g satisfies the conditions of Rolle's theorem, because:Ī. We now want to choose r so that g satisfies the conditions of Rolle's theorem. Since f is continuous on and differentiable on (a, b), the same is true for g. ĭefine g(x) = f(x) − rx, where r is a constant. That's a loss for future real mathematicians. It hasn't been published anywhere because I am generally not liked by stupid people in the mainstream. So common sense in fact, that most with a high school education can learn the new calculus without taking a lifetime of learning.Īll this aside, you cannot refute that my constructive proof of the MVT was never realised by anyone before me. My ideas are not revolutionary, they are common sense. If one is not part of the academic cabal, the chances of publishing anything are zero. As for accepted knowledge, it is delusional to think that only it should be published. That's something you can't say about mainstream calculus whose results are generally correct, but the formulation is flawed. To someone not in mainstream academia, the new calculus makes perfect sense. There are no real mathematicians in mainstream academia, only mythmaticians. Kegelkugel ( talk) 20:04, 2 February 2017 (UTC) Limits and infinitesimals are not generally accepted outside of mainstream academia. From a mathematicians point of view, your approach just looks wrong and informal. That's why your "new formulation of calculus" is very misleading for someone who is not a professional mathematician. Limits and infinitesimals are totally accepted in mathematics. Using the first and only rigorous formulation of calculus in human history:Ģ601:646:8B82:E709:EC2A:A37E:89CF:63E5 ( talk) 19:33, 2 February 2017 (UTC) ĭear John, Wikipedia should be a source of (widely) accepted knowledge, not a platform to spread revolutionary ideas. Using the flawed mainstream calculus and a patch I, John Gabriel, introduced to prove it: The first and only constructive proof of the MVT 26 First Mean Value Theorem for Definite Integrals: proof is incorrect.22 Government of Beijing celebrates the mean value theorem.9 The section on divided differences in the article.1 The first and only constructive proof of the MVT.
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