1,687 reputation
926
bio website fuz.su/~fuz
location Berlin, Germany
age 20
visits member for 3 years, 11 months
seen 21 hours ago

I am a student of computer science and mathematics at the Humboldt University of Berlin.


Sep
22
comment Proof that $a\equiv 1\,(\textrm{mod }8)$ implies $a$ is a square modulo $2^n$ for all $n$
Ah! I missunderstood. Thank you.
Sep
22
comment Proof that $a\equiv 1\,(\textrm{mod }8)$ implies $a$ is a square modulo $2^n$ for all $n$
$a=17, n=5.\ a\equiv17\mod2^5,$ but 17 is not a square... or didn't I understand your statement?
Sep
22
comment Easy proof, that $\rm e\notin \mathbb Q$
Thanks for the explanation.
Sep
22
comment Easy proof, that $\rm e\notin \mathbb Q$
The last step in the second line is not completely clear to me. Could you please elaborate why this holds?
Sep
21
comment Easy proof, that $\rm e\notin \mathbb Q$
@lhf Thank you for the interesting link.
Sep
21
comment Easy proof, that $\rm e\notin \mathbb Q$
@francis-jamet We showed that the limit exists just as Srivatsan described.
Sep
21
comment Easy proof, that $\rm e\notin \mathbb Q$
@SrivatsanNarayanan Thank you! By the way, is it clear, which proof I refer to?
Sep
21
comment Easy proof, that $\rm e\notin \mathbb Q$
@anon Fixed the first one. The second proof is not difficult, but takes too much time. That's all.
Sep
21
comment Easy proof, that $\rm e\notin \mathbb Q$
@Yuval That is actually how we did this. The problem is, that the students in my class don't like proofs containing too many “It is easy to see that...”s. The proof using this strategie filled one A4 page in the script and took 20 minutes for me to explain. (Although it is quite easy if you think about it).
Sep
13
comment Database for mathematical syntax
@3Sphere Such a reference is quite useful if you see a formula in the Internet or in a paper where the author sees no reason to include a syntax definition. So even a lookup with many definitions is great, as one can often tell the right one from the context.
Sep
5
comment Solving $8=x(2(1-\sqrt{5}))+(1-x)(2(1+\sqrt{5}))$
There is a simple way: Just use Wolfram Alpha
Aug
29
comment Reducing the time to calculate Collatz sequences
@Thijs Good point. Thank you!
Aug
29
comment Reducing the time to calculate Collatz sequences
@Qiachu But then, I would need to save all that information. Assume, that I want to test all $n$ in the intervall $[1,1\,000\,000\,000]$ - I would need about 1 GiB just for caching! But otherwise, a good idea.
Aug
8
comment How many real roots are there to $2^x=x^2$?
@Arturo Err, that would make it clearer, but the time for editing is up.
Aug
8
comment How many real roots are there to $2^x=x^2$?
@Arturo This was meant because the answerer stated, that $x = 2, x = 4$ are the only solutions.
Aug
8
comment How many real roots are there to $2^x=x^2$?
Why is $x\approx-0.76666469596212309311$ a solution?
Jul
7
comment Mathematical Career Advice to a young 16 year wannabe mathematician
@sigma.z.1980 It's good to know, that other people have the same opinion as me.
Jun
15
comment All functions $\frac{1}{f\left(y^2f(x)\right)} = \big(f(x)\big)^2\left(\frac{1}{f\left(x^2-y^2\right)} + \frac{2x^2}{f(y)}\right)$
@J. J. Ah, thanks. I didn't read the ${}^+$ in $\mathbb R^+$.
Jun
15
comment All functions $\frac{1}{f\left(y^2f(x)\right)} = \big(f(x)\big)^2\left(\frac{1}{f\left(x^2-y^2\right)} + \frac{2x^2}{f(y)}\right)$
I just spotted a possible mistake: $1/f(0)=f(0)$ does not implies $f(0)=1$. It only implies, that $f(0)\in\{-1,1\}$.
Jun
14
comment Is such a cryptographic system possible?
That's a great article!