- #1
lokofer
- 106
- 0
"Definition2 of derivative...
We have the defintion (taking the limit) for hte derivative:
[tex] \frac{f(x)-f(a)}{x-a} [/tex] for an Euclidean Space...
But what keeps us from defining another metric (on an Euclidean or other abstract space) so the derivative takes de form:
[tex] \frac{(df(x),f(a))}{d(x,a} [/tex] so "d" is a distance in the form that you can define "infinitesimal2 elements on an space and that for x=a ,d=0
Then the "abstract" definition of integral is:
[tex] \sum_{i} f(X_i ) d(X_{i+1},X_{i}) [/tex]
of course in the limit that the distance [tex] || X_{i+1}-X_í}||\rightarrow 0 [/tex] and ¿what happens if we had an "Infinite" dimensional space...so
- it is "numerable" (ie: R^{n})
- it's not "numerable" (function space)
We have the defintion (taking the limit) for hte derivative:
[tex] \frac{f(x)-f(a)}{x-a} [/tex] for an Euclidean Space...
But what keeps us from defining another metric (on an Euclidean or other abstract space) so the derivative takes de form:
[tex] \frac{(df(x),f(a))}{d(x,a} [/tex] so "d" is a distance in the form that you can define "infinitesimal2 elements on an space and that for x=a ,d=0
Then the "abstract" definition of integral is:
[tex] \sum_{i} f(X_i ) d(X_{i+1},X_{i}) [/tex]
of course in the limit that the distance [tex] || X_{i+1}-X_í}||\rightarrow 0 [/tex] and ¿what happens if we had an "Infinite" dimensional space...so
- it is "numerable" (ie: R^{n})
- it's not "numerable" (function space)