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A long time ago, I studied mathematics. I used to know three or four things about complex analysis, topological vector spaces, and point set topology. But I've forgotten at least five of them.


3h
comment Decide the smooth function $r : \mathbb R \rightarrow \mathbb R$ of the equation $r(t)^2 + r'(t)^2 = 1$.
The list is not quite exhaustive.
4h
comment Prove that $a+ib$ is prime in $\Bbb Z[i]$, of $a^2+b^2$ is prime in $\Bbb Z$.
Can you show that $\mathbb{Z}[i]$ is a PID?
15h
comment Convergence of $ \sum_{n=1} ^\infty \frac {1}{n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} )}$
Please do not use \dfrac in question titles, as that takes up too much vertical space in the questions list. See the Guidelines for use of $\LaTeX$ in titles on meta.
16h
comment Convergence of $ \sum_{n=1} ^\infty \frac {1}{n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} )}$
Do you know some better approximation for $$\sum_{k=1}^n \frac{1}{k}\,?$$ You've probably come across it, you just need to remember.
16h
answered Prove for any integer $N$ that there exists $n > N$ where $n!-1$ is not a prime
19h
comment Show that $\operatorname{rank}(A+B) \leq \operatorname{rank}(A) + \operatorname{rank}(B)$
@Pierre-YvesGaillard No, the question was erroneously merged with one asking about the composition one and a half years ago. The merge is now undone, and your answer is back where it belongs.
19h
comment Show that $\operatorname{rank}(A+B) \leq \operatorname{rank}(A) + \operatorname{rank}(B)$
Merge undone, thanks for drawing attention.
23h
revised $U(R) \equiv U(B) \equiv U(N)$, trying to find unique values that result in indifference.
No need to state it in the title that the question was answered
1d
revised Criteria for the existence of zero-divisors and idempotent elements in the integers modulo $n$
Small correction, fix of typo, and we need a < n, otherwise consider n(n-1) etc.
1d
answered How to prove the identity $\prod_{n=1}^{\infty} (1-q^{2n-1}) (1+q^{n}) =1 $ for $|q|<1$?
1d
comment How to prove the identity $\prod_{n=1}^{\infty} (1-q^{2n-1}) (1+q^{n}) =1 $ for $|q|<1$?
Is complex analysis a tool you can use?
1d
comment If $E$ is a closed set there exist a set $S$ such as $E=S'$
I thought of the one we sketched out here in the comments. But you could, if you want, also write the other [or only the other].
1d
revised Limits of sequences connected with real and complex exponential
edited title
1d
comment Seperating points in the complex plane
If $R_i = R_j$, how does that constrain the location of $q$?
1d
comment Using Intermediate value theorem and Rolle's theorem
But in your approximate evaluation of $f(0.01)$, you should not write $f(0.01) = 2\cdot (-4.6) + 0 + 7$, since that's not true, it's $f(0.01) \approx \dotsc$.
1d
comment Is there an analytic bounded function on $\omega \subset \mathbb C\setminus]-\infty;0]$ such that $|f(x)|\leq e^{-x^{1/2}}$
What about $e^{-\sqrt{z}}$ with the principal value of the square root? There are more you can construct in a similar way.
1d
comment If $E$ is a closed set there exist a set $S$ such as $E=S'$
Basically yes. The details need to be done for a complete proof of course, and you need not only $S' \subset E\setminus E'$ but also $S' \supset E\setminus E'$. You probably didn't mention that in the comment because that's the obvious part. Will you write your proof in an answer?
1d
comment Is there an analytic bounded function on $\omega \subset \mathbb C\setminus]-\infty;0]$ such that $|f(x)|\leq e^{-x^{1/2}}$
Not $\sqrt{z}$, since $\sqrt{x} > e^{-x^{1/2}}$ for $x \geqslant 1$. But using $\sqrt{z}$, you can construct nonzero analytic functions satisfying the desired inequality.
1d
comment Is there an analytic bounded function on $\omega \subset \mathbb C\setminus]-\infty;0]$ such that $|f(x)|\leq e^{-x^{1/2}}$
Trivially, there is $f\equiv 0$. Non-trivially, do you know what $\lvert e^w\rvert$ is? And can you see a connection between $\mathbb{C}\setminus ]-\infty,0]$ and $\sqrt{z}$?
1d
comment The inverse of a bijective holomorphic
Aside: We can show that $f^{-1}$ is holomorphic with the residue theorem, without first showing that $f'$ has no zeros. That is quite nice, and also a generalisation of that is often useful. Using the integral formula or another variant/consequence of the residue theorem to directly show that $f'$ has no zeros, I have no immediate idea. Let me think a little.