mathwonk said:
However modern differential geometry and differential calculus has succeeded in giving a precise, although somewhat complicated, meaning to these separate symbols, in which it is entirely correct to say that the derivative is an actual quotient of the two differentials dy and dx.
I hope no one minds if I butt in with some questions. This is something I've been puzzling about for a long time. Anyone who's just learning this stuff should not pay too much attention to this post, as I'm sure it will contain misunderstandings on my part and unnecessary complications as I try to connect various ideas that I've read and struggled with.
mathwonk said:
If you picture the family of tangent lines to the graph of y = f(x), they represent the graphs of a family of linear functions. For each point x=a on the x axis, the linear function represented by the tangent line at x=a, is the function of h defined by dfa(h) = f'(a).h, i.e. multiplication by f'(a).
For the tangent line, I get: f(a) + f'(a).(h-a). Translating the origin to (a,f(a)), this becomes df(a,h) = f'(a).h. This is going to be a bit of a mouthfulm but... should these coordinate systems, copies of
R2, located at each point (a,f(a)) be thought of as one set of coordinate spaces representing the components, in coordinate bases, of the cotangent spaces (i.e. vector spaces of continuous linear functions of tangent vectors) associated with each point of Euclidean 1-space? Or something like that...
mathwonk said:
For the function y=x, this assigns the identity function taking h to 1.h at every point.
This reminds me of something I read about Hamilton making a distinction between numbers in their role as functions acting on other numbers, and numbers in their passive role as the arguments of others, so 1(h) = 1.h. In one dimension, is the idea that one copy of the real numbers plays the role of a chart for Euclidean 1-space, others play the role of components of tangent vectors in the coordinate tangent bases associated with each point, and others the role of components of cotangent vectors in the coordinate cotangent bases associated with each point?
The differential seemed to begin as a function of two variables, a and h, then a was held fixed and it starts being treated as a function only of h. Is df(a,h) a cotangent vector field, and and df(h), at some fixed a, the actual cotangent vector which is the value of this field at the point a? (AFTERTHOUGHT: Oh, on rereading, I see you already said this in slightly different language.)
mathwonk said:
The quotient of two linear functions b.h/c.h, is the constant function b/c, whose value is the quoptient of the multipliers in the two linear functions.
b and c are cotangent vectors located at the same point, right?
\frac{\mathrm{d}f(h)}{\mathrm{d}g(h)} \bigg|_a = \frac{f'(a) \, \mathrm{d}x(h)}{g'(a) \, \mathrm{d}x(h)} \equiv \frac{f'(a)}{g'(a)} \frac{1}{1}=\frac{f'(a)}{g'(a)}
Or maybe b.h/c.h means diving the values at each point of these cotangent vector fields to give a scalar field. How do we know that the variable h on the bottom has the same value as the variable h on the top: is the principle that we hold it as fixed at some arbitrary non-zero value for the purpose of this definition?
mathwonk said:
In this sense the quotient dy/dx = df/dx is a function whose value at x=a is the constant function of h whose value is the quotient of the multipliers f'(a)/1. I.e. the value of dy/dx at x=a is indeed f'(a).
In other words, a quotient of cotangent vectors is defined by dividing their components in some basis, and treating dx/dx as 1? (So this is a function Q : T* x T* --> R?)
mathwonk said:
In a more general setting, we introduce a "vector field" as a family of tangent vectors to our surface or other space. Vectors are variables on which linear functions act. Then a covector field is a family of linear functions, and can be paired at each point with the vectors in our vector field to give a number at each point, i.e. a function.
Should the reciprocal 1/c be regarded as a tangent vector, as suggested by the notation, and by the fact that the linear function b could be said to act on the variable c to give a scalar.
\mathrm{d}x\left ( \frac{\partial }{\partial x} \right ) = \frac{\mathrm{d} x}{\mathrm{d} x} = 1
In this view, is differentiation in single-variable calculus a special case of the scalar product of a cotangent vector and a tangent vector?
mathwonk said:
A differential is just a covector field, i.e. something that pairs with a vector field to give a function.
Q : T* x T --> R, rather than Q : T* x T* --> R? (This function being a scalar field, the derivative, f'(x)?)