$\Delta x\approx dx$

Question

How do you state $($small $\Delta x)\approx dx$ in terms of differential forms? $dx$ is a one-form, but I don't see why $\Delta x$ approximates a one-form if gets smaller.

Answer 1

For a function $f$, the short answer is that the differential of $f$ is set up to work that way. That is, the differential $\operatorname{d}f$ is defined to be the best linear approximation to the change in the function’s value near a given point: $$\Delta_Pf(\mathbf h) = f(P+\mathbf h) - f(P) = \operatorname{d}f_P[\mathbf h]+\text{error},$$ where the error is $o(\mathbf h)$, i.e., it goes to zero “faster” than $\mathbf h$. As $\mathbf h$ gets small, $\operatorname{d}f$ gets closer and closer to $\Delta f$.

If we have a coordinate system, then $\operatorname{d}f=\sum_i{\partial f\over\partial x_i}\operatorname{d}x_i$. In this sum, the $\operatorname{d}x_i$ are linear maps that assign to each vector the value of its $i$th coordinate, i.e., they are where you “plug in” the $i$th coordinate of the displacement $\mathbf h$. With this interpretation, $\operatorname{d}x_i=\Delta x_i$.