markgvanzalk
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I went through it and the calculations seemed fine in terms of real numbers but I just don't understand how this process works logically
Fermat's method of calculus
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I went through it and the calculations seemed fine in terms of real numbers but I just don't understand how this process works logically
D
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I have a book somewhere in my library called "The History of Real Analysis." It discusses Fermat's approach.
Fermat was not noted for following the standards of rigor that became prevalent in the latter half of the 19th century.
But here is why it works.
[math] \text {Let } T(u,\ v) \text { be the equation of a line going through } (u, u^2) \text { and } (v, v^2). \\ \text {So, the slope of a line going through those points is }\\ \dfrac{v^2 - u^2}{v - u} = \dfrac{(v - u)(v + u)}{v - u} = v + u.\\ \text {That is true provided that } v \ne u.\\ \text {Let } u = v + e \implies u + v = 2u + e.\\ \text {So, the slope of a line going through those points is } 2u + e.\\ v = u \implies e = 0 \implies 2u + e = 2u,\\ \text {which is indeed the derivative of } x^2 \text { at } x = u. [/math]
Effectively, Fermat was taking a limit of the slope as u and v approached each other, but he had no way to express the idea of a limit. It worked, but no one could justify calculus according to modern ideas of rigor. So what he was doing was taking the idea of the Newton quotient (before Newton published anything), expressing it a way that gave a function in x alone and a second function for which e was a factor. When e is set equal to 0, only the function in x alone is left. The logic is flawed, but the results are fine.
Fermat was not noted for following the standards of rigor that became prevalent in the latter half of the 19th century.
But here is why it works.
[math] \text {Let } T(u,\ v) \text { be the equation of a line going through } (u, u^2) \text { and } (v, v^2). \\ \text {So, the slope of a line going through those points is }\\ \dfrac{v^2 - u^2}{v - u} = \dfrac{(v - u)(v + u)}{v - u} = v + u.\\ \text {That is true provided that } v \ne u.\\ \text {Let } u = v + e \implies u + v = 2u + e.\\ \text {So, the slope of a line going through those points is } 2u + e.\\ v = u \implies e = 0 \implies 2u + e = 2u,\\ \text {which is indeed the derivative of } x^2 \text { at } x = u. [/math]
Effectively, Fermat was taking a limit of the slope as u and v approached each other, but he had no way to express the idea of a limit. It worked, but no one could justify calculus according to modern ideas of rigor. So what he was doing was taking the idea of the Newton quotient (before Newton published anything), expressing it a way that gave a function in x alone and a second function for which e was a factor. When e is set equal to 0, only the function in x alone is left. The logic is flawed, but the results are fine.