Anonymous
10/31/2025, 11:04:07 PM
No.16832825
>>16832816
Are you serious? The set of approximated lengths does have an upper bound, it's 4. Limit and arc-length do not commute due to the previously mentioned issue with derivatives.
To demonstrate, imagine that the curves were parametrized f_n(t) = (x_n(t), y_n(t)) for t in [0, 2pi), such that f_n(0) = (1, 0), and f_n'(t) -> (-sin(t), cos(t)) uniformly. Denote by L_n the arc-length of f_n, then we have
lim_{n->inf} L_n = int_0^2pi sqrt((lim_{n->inf} x_n'(y))^2 + sqrt(lim_{n->inf} y_n'(t))^2) dt = int_0^2pi sqrt((-sin(t))^2 + (cos(t))^2)dt = 2pi
i.e., it works perfectly. The only step that fails in OP's version is commuting the limit into x_n' and y_n'.
Are you serious? The set of approximated lengths does have an upper bound, it's 4. Limit and arc-length do not commute due to the previously mentioned issue with derivatives.
To demonstrate, imagine that the curves were parametrized f_n(t) = (x_n(t), y_n(t)) for t in [0, 2pi), such that f_n(0) = (1, 0), and f_n'(t) -> (-sin(t), cos(t)) uniformly. Denote by L_n the arc-length of f_n, then we have
lim_{n->inf} L_n = int_0^2pi sqrt((lim_{n->inf} x_n'(y))^2 + sqrt(lim_{n->inf} y_n'(t))^2) dt = int_0^2pi sqrt((-sin(t))^2 + (cos(t))^2)dt = 2pi
i.e., it works perfectly. The only step that fails in OP's version is commuting the limit into x_n' and y_n'.