Recent content by Take_it_Easy

I think that my previous reply has been deleted, or I never was able to post it and thought I did. By the way, I actually did not conceive the number 0.9999999999... aritmetically. I mean... what are these dots? The only way I saw this was through a limit of a serie. With this point of...

That's very easy to show, once you can use the theorem "f continuous on a compact => f uniformly continuous", even though I had to stress some tecnichal detail. Take eps > 0. You have to find a delta > 0 such that for every couple of elements x1, x2 in [0, +infty) such that | x1 - x2 | <...

I can't get why exp(in^2) converges. Can you explain, please?

The proof I wrote actually shows it DOES NOT converge. I can't proof it diverges to +infinity yet. My proof is very tecnical and LONG and actually there are some points a bit difficoult that I have to clear out before to call an EXACT proof. I need help!

I wish we could! :) think of \sum_{n=1}^{\infty} \left( 1 - {1 \over n} \right)^n

You are right! sqrt(2) does not exist. And in fact also 1 does not exist. 1 is the multiplication of 13/7 and 7/13 now 13/7 and 7/13 are just symbols for their decimal representations which are 13/7 = 1,85714285... 7/13 = 0,53846153... and the decimal places continue on infinitely. So...

Since 2sin(x)cos(x) = sin(2x) you can write the integrand function x/2 \cdot \sin (2x) you can use first the substitution y=2x and then use integration by part formula to integrate y/4 \cdot \sin (y) it is EASY if you choose to derive y/4 and integrate \sin(y).

This is not an homework, but just a couriosity that I have I never found a easy way to solve the question of convergence of this serie \sum_n (\sin(n))^n any help is appreciated!

well, in my modest opinion, since 1 = (-1)(-1) = (-1)(-1)1 = (-1)(-1)(-1)(-1) = ... = ((-1)^(2n))(1^m) for every n,m natural, you should put your question under some rule of irredundance to try to find a interesting answer.

The (powerful) Vitali theorem states that a bounded function f: D \subset \mathbb R \longrightarrow \mathbb R defined on a bounded domain is Riemann integrable IF AND ONLY IF it has a set of point of discontinuity of measure zero. Now in your function you have that [a,b] is the set of the...

Totally agree :)

well, there is uniform convergence of the succession of functions of circles to the function constant = 0 in the interval (say) [0,1] now the elements of the succession have a graph that has constant length \pi, but the limit has a graph of length 1. Well that's not a paradox, also more...

A classroom has 2 rows of 8 seats. There are 14 students, 5 of whom always sit in the front row and 4 whom always sit in the back. In the first row there are always 5 students that will seat. In how many ways they can sit? It is the number of injective functions from a set of 5 elements in a...

To place the two red you have to select a subset of two elements in a set of 36. So you have a number of choices equal to the combinations of 36 on 2 that is 36*35/2 = 18*35 choices. Then you have to place the four blue and you have a number of choices equal to the number of the subsets of...

You are WELCOME! Well I also noticed I made a 'print' mistake... in the last row I wrote [F:E] instead of [F: F \cap E] but I guess you noticed the mistake and you got the right meaning. See you next time!