Recent content by libero.ego

Does that claim holds though? Because (0,1) and N have not the same cardinality..

I see, i misunderstood the point then. Maybe this: The function sequence f_n (x)=sin(x)^{n} converges to the function: f(x)= 1, if x=\frac{n\pi}{2} f(x)= -1, if x= \frac{3n\pi}{2} f(x)= 0, elsewhere Then, if n\in N, f(n)=0, \forall n, so the limit of f(n), should be zero(not sure...

By the way, for the same reason also sin(x)^x^2 doesn't exist, numerically it gives a result probably because the algorythm stops iteracting and approssimates.

Using the fact that, a succession a_n has limit if and only if any sub-succession has, then you can see that if you pick the subsuccession n=Pi/2 r, r natural, then you have non convergence.

Hi Spreco, do this: u1=u2, 50 sin(314t + ∏/3)=100 sin(314t - ∏/6) => sin(Pi t + Pi/3)=2 sin(Pi t - Pi/6), sin(a+b) = sin a cos b + cos a sin b, then: sin(Pi t) cos(Pi/3) + cos(Pi t) sin(Pi/3) = 2 sin(Pi t) cos(Pi/6) - 2 cos(Pi t) sin(Pi/6) sin(Pi t)= a, cos(Pi t)= b, sqrt(3) / 2 a + 1/2 b =...