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Abr001am

 In the search of a question

When you are looking for a specific question (using Approach0 ...
Abr001am
May 15, 2020 10:27
ok i appreciate.
Abr001am
May 15, 2020 10:23
Thanks very much for this valuable collect, i will bookmark them for later check.
Abr001am
May 15, 2020 10:20
no it's not mine, i feel like there was more questions before about fermat semi primes written in the form $2^n+1$
Abr001am
May 15, 2020 10:19
@MartinSleziak ok sorry, i found topics there about requesting deleted questions with sql
 

 REHAB

Help Alex write regex to autoflag questions. Seeking multiple ...
Abr001am
May 15, 2020 07:02
Hello dear folks, may i ask if there is deleted questions about fermat semi primes that can be found somehow ?
 

 Mathematics

Associated with Math.SE; for both general discussion & math qu...
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Abr001am
May 7, 2020 14:29
sorry 2^n+1
Abr001am
May 7, 2020 14:26
where q, p are primes
Abr001am
May 7, 2020 14:26
Hello, how can i find questions related to the problem $2^n=pq$ i can't find any in the main. thatks a lot.
Abr001am
Jun 4, 2018 20:11
@anon yes well, i gave you an assimilation.
Abr001am
Jun 4, 2018 20:10
@anon well it's almost like wheras a system of equations is solvable or not, but in this case there is a determinant.
Abr001am
Jun 4, 2018 20:03
how to avoid, or atleast forecast something like that ?
Abr001am
Jun 4, 2018 20:02
i'm coping with telescoping series and i often come accross a circular definition.
Abr001am
Jun 4, 2018 19:59
yes there is 3 variables, i noticed now :D
Abr001am
Jun 4, 2018 19:57
4 integrals ? lol that's enough for me to blowout of overthinking.
Abr001am
Jun 4, 2018 19:55
Waiting waiting
Abr001am
Jun 4, 2018 16:40
here i was trying to understand why the area of a circle drawn upon a sphere is bigger same as is but as a cap in an euclidean isometry. It was long time ago.
Abr001am
Jun 4, 2018 16:28
triangles in noneuclidean round planes (spherical or ovalic ... etc) have >$\pi$ sum of inner angles.
Abr001am
Jun 3, 2018 15:30
@Waiting take care , i was just making certitude of an answer of mine.
Abr001am
Jun 3, 2018 15:18
@Waiting if it's true that means an integral can be calculated in more than two ways yielding to completely different results .
Abr001am
Jun 3, 2018 15:15
now, if let $u=ln(x)$ and calculate $f(x)$ we get $\int u e^u dx $ right @waiting which is $xe^x-e^x+C$ by parts.
Abr001am
Jun 3, 2018 15:11
@Waiting oh yeah i forgot to integrate the constant. that's true
Abr001am
Jun 3, 2018 15:10
+c , let it be. so it's correct.
Abr001am
Jun 3, 2018 15:09
@waiting Ok if we have $f(x)=\int ln(x) dx$ if we integrate it by parts we get $f(x)=xln(x)-1 $ right ?
Abr001am
Jun 3, 2018 15:06
@Waiting can i ask you a question ? (about integrals i mean)
Abr001am
Jun 3, 2018 14:59
Yes, i don't have time much time as years, i hope i can grasp on the principle within few months.
Abr001am
Jun 3, 2018 14:56
@Waiting well, we are all made of stars (this is a serbian proverb iirc).
Abr001am
Jun 3, 2018 14:55
the way robjohn used telescoping series in each sum even where it's not intended to be used is really marvellous.
Abr001am
Jun 3, 2018 14:53
yes, i'm now calculating some sums
Abr001am
Jun 3, 2018 14:50
was sarcastically meaning 'keeping track of his answers'
Abr001am
Jun 3, 2018 14:48
hello waiting, i'm stalking robjohn's awsome archive of telescoping series
Abr001am
May 28, 2018 04:06
it seems impossible to give an approximate sum without theta summation.
Abr001am
May 28, 2018 04:03
well i didn't want to onebox all this stuff, i just gave you the message id.
Abr001am
May 28, 2018 04:02
24 hours ago, by Symposium
We have $\displaystyle \prod_{0 \le k \le n}\left(1-x^{2^{k}}\right) = \frac{\left(1-x\right)}{\left(1-x^{2^{n+1}}\right)}\prod_{1 \le k \le n+1}\left(1-x^{2^{k}}\right) = \frac{\left(1-x\right) }{\left(1-x^{2^{n+1}}\right)}\prod_{0 \le k \le n}\left(1-x^{2^{k+1}}\right). $

But $\displaystyle \left(1-x^{2^{k+1}}\right) = \left(1-x^{2^{k}}\right)\left(1+x^{2^{k}}\right)$, thus $\displaystyle \prod_{0 \le k \le n}\left(1-x^{2^{k+1}}\right) =\prod_{0 \le k \le n}\left(1-x^{2^{k}}\right)\prod_{0 \le k \le n}\left(1+x^{2^{k}}\right). $
Abr001am
May 28, 2018 03:58
44833160
Abr001am
May 28, 2018 03:54
@Symposium sounds a nifty idea but (sorry for my obtuseness) i'm not being able to link the euler product with my sum, the expanded expression yields to all kinds of exponents aside 2^k
Abr001am
May 27, 2018 04:06
i tried expandin terms, without any good aim, i think the best way is to go around this by theta theorem
Abr001am
May 27, 2018 04:04
oh well <=1 yes
Abr001am
May 27, 2018 04:03
@AkivaWeinberger that says x<=1, because if it's strictly positive, the sum flies to infinity
Abr001am
May 27, 2018 03:58
@GFauxPas dude lol no, just for each variank $k$, because for the special case $k=2$, the question exists.
Abr001am
May 27, 2018 03:57
but then i suspected it existed under different form.
Abr001am
May 27, 2018 03:57
I thhought of posting a question in this subject in main for every general term $k$
Abr001am
May 27, 2018 03:56
integer.
Abr001am
May 27, 2018 03:55
worked!
Abr001am
May 27, 2018 03:54
Hello, Is there a concise, not concise, even asymptotic way to find a sum $ \sum_i x^{k^i} $ ?
 

 Reverse Engineering

General discussion for reverseengineering.stackexchange.com
Abr001am
May 30, 2018 17:57
hello, is it ontopic to share a re experience and ask some help to bypass evaluation period of some software if i remove the brand appearing obviouosly but the software can be identified easily by other characteristics that i can't hide ?
 

 /dev/chat

General discussion for unix.stackexchange.com. If you have a q...
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Abr001am
May 26, 2018 08:40
He was so pretentious to install all the cloud from the beginning, in case it ends over like the prophecy said.
Abr001am
May 26, 2018 08:37
venturebeat.com/2017/11/04/the-end-of-the-cloud-is-coming
Abr001am
May 26, 2018 08:37
i'v read an article about an eventual end of the cloud. Lemme dig it out.
Abra001
May 26, 2018 08:33
if you omit all the intranets, the residue is the cloud.
Abra001
May 25, 2018 21:21
the apostrophe is clearely escaped in the first example.