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5h
comment Counterexample to Maximum modulus principle
There are examples that are better in that they don't require us to take a stand on how singularities relate to the statement. Consider $f(z) = 1/(1+|z|^2)$ on any disk containing the origin, for example.
5h
comment Simpson's Rule approximation
The OP and a followup comment both say $f^{(4)}(1)$, so I think I do mean $\theta=1$. (Not that I'm convinced that's correct.)
18h
comment Simpson's Rule approximation
You say $f^{(4)}(1)=42$, but is $\theta=1$ where the largest absolute value of $f^{(4)}(\theta)$ occurs?
2d
comment Probability - die - The number of throws until a $5$ and a $6$ have been obtained.
When I read the title, I amplified it to "Probability - die die die!"
2d
comment Verifying multiplicative inverses of modulo n are the elements that are relatively prime to n
You have proved: $a$ is a unit implies $\gcd(a,n)=1$. There's also a converse to be proved.
2d
comment Verifying multiplicative inverses of modulo n are the elements that are relatively prime to n
@JohnMartin: indeed, the deeper (and more important for this application) result is the converse, that if $\gcd(a,b)=1$ then there are integers $x,y$ such that $ax+by=1$.
2d
comment Find $n$ for which $\frac{(n!)^2-(n+2)!}{(n+1)!}$ is an integer
It is an integer for $n=5,7,8,9,11,13,14,15,17,19,20,\dots$. Indeed it is an integer for all $n\ge5$ such that $n+1$ is not prime (which are the integers for which $n+1$ divides $n!$.
Apr
26
comment Calculate Euler inverse function
Hint: if $13\mid\phi(n)$, then either $13^2\mid n$ or some prime $p\mid n$ with $p\equiv1\pmod{13}$.
Apr
26
revised $\bigcup_{n=1}^{\infty}G_n $ is a multiplicative subgroup of the group $\mathbb C\setminus\{0\}$
edited title
Apr
26
comment Question about the solutions to quadratic congruence $x^2\equiv -1(\mbox{mod}\;p)$
I don't believe this has been proved yet, although it's probably within the power of current technology. Conjecturally, for example we believe that there are infinitely many primes $q$ for which $q^2+1=2p$ for some prime $p$; for these primes $p$, one of the solutions to $x^2\equiv-1\pmod p$ is the prime $q$.
Apr
26
comment Question about the solutions to quadratic congruence $x^2\equiv -1(\mbox{mod}\;p)$
This isn't related to Artin's conjecture.
Apr
26
comment If $f$ has a pole of order $m$ at $z_0$ find the order of the pole of $g(z) = \frac{f'(z)}{f(z)}$ at $z_0$.
"$f$ has a pole of order $m$ at $z_0$" means that $(z-z_0)f(z)$ is analytic and nonzero in a neighborhood of $z_0$; thus its Laurent series there looks like $f(z) = \frac{c_{-m}}{(z-z_0)^m} + \frac{c_{-(m-1)}}{(z-z_0)^{m-1}} + \cdots$. (So, for example, $f'$ will have a pole of order $m+1$.) Hint: use the fact that $g(z) = \frac d{dz}(\log f(z))$.
Apr
25
comment Is $\frac{1}{11}+\frac{1}{111}+\frac{1}{1111}+\cdots$ an irrational number?
One could search MathSciNet or Google Scholar for "Lambert series transcendental" to see if anyone's made progress on that question.
Apr
25
awarded  Enlightened
Apr
25
awarded  Nice Answer
Apr
25
answered Is $\frac{1}{11}+\frac{1}{111}+\frac{1}{1111}+\cdots$ an irrational number?
Apr
25
comment Function in $L^2(\mathbb R^n)$
... give it more than one minute....
Apr
25
comment Suppose that $a, b ∈ N$ are relatively prime. Prove that, for any $k ∈ N$, $a^k$ and $b$ are relatively prime.
Yes, this is correct! Here's the same proof, without induction: if $1=\alpha a+\beta b$, then $$1=(\alpha a+\beta b)^k = (\alpha^k) a^k + \bigg( \sum_{j=1}^k \binom kj \alpha^{k-j} a^{k-j} \beta^j b^{j-1} \bigg) b.$$
Apr
24
comment Function in $L^2(\mathbb R^n)$
If that's the question you want answered, then you should put that question in the main post (or make a new post), not in a comment. (Your proof in the post looks essentially correct; what you're proving is that any function $f(X)$ such that $f(X) \ge 1$ when $|X|\ge R$ cannot possibly be in $L^2(\Bbb R^n)$, or indeed in any $L^p(\Bbb R_n)$.)
Apr
24
comment Proving that ${p \choose r}$ is an integer for a prime $p$ and $0 < r < p, r \in \mathbb{Z}$
The classical proof is to show that $\binom pr = \binom{p-1}{r-1}+\binom{p-1}r$ and then (trivially) prove by induction that they're all integers.