@MikeMiller The author is talking about a bivariate generating function in $q,x$ that is "the generating function for partitions with distinct parts each $\leqslant n$."
Experience shows that 20% of the people reserving tables at a certain restaurant never show up. If the restaurant has 50 tables and takes 52 reservations, then the probability that it will be able to accommodate everyone is $1-14(\frac 45) ^{52}$
@AlexanderGruber I guess I got it. We use $x$ to count blocks in the partition (if need be) and $q$ to specificate the number being partitioned. Thus, let $[[k,n]]_t$ denote the number of partitions of $t$ into $k$ distinct blocks of size at most $n$. Then $$\sum_{k\geqslant 0}\sum_{t\geqslant 0}[[k,n]]_t q^tx^k$$ is the generating function of partitions into different parts of size at most $n$.
One can see that $\sum_{t\geqslant 0}[[k,n]]_t q^t=A_k^n(q)$ is the generating function of the partitions into $k$ distinct blocks of size at most $n$.
Hi! If $f$ is class $C^{1}$ and uniformly continuous, can we conclude that the derivative is bounded? I think the answer is no, but how can I construct an explicitly such function ?
@BalarkaSen What if it has the whole real numbers as domain and empty range? Then I don't think there would be such a function? Because for each $a$, there won't be a $b$ ... right?
If $K$ is a number field, and $v$ is a valuation on $K$, then $R_v$ the valuation ring of $v$ must contain $\mathbb{Z}$ - i guess this probably easy but I dont see it?
@Karl Applying the standard trick of $y = x + \delta/2$, thus $|\sqrt{x} - \sqrt{x + \delta/2}| < \epsilon$. But if we choose $x$ to be something very small in $(0, 1)$, then this is $|\sqrt{x} - \sqrt{x + \delta/2}| < \sqrt{\delta/2}$. But a proper choice of $\delta$ can satisfy the inequality, so this idea was lame. What am I missing?
@Sawarnik Not really. I really did ignore you this time.
I am looking to prove that if $f$ is a complex analytic function (holomorphic) on some domain $D$, and $|f(z)|=1 \ \forall \ z \in D$. Then $f$ must be constant
I am always confused what theorem to apply / use to prove these types of questions.
@BalarkaSen That was fun. It is actually uniformly continuous (though not uniformly convergent, which we both like to mistakenly write). Just set $\delta=\epsilon^2$
The open mapping theorem states that if U is a domain of the complex plane C and f : U → C is a non-constant holomorphic function, then f is an open map. Eg from the Rieman mapping theoren there exists a one-to-one mapping $T$ from $D$ onto $C^*$, the unit disk. But since $|z|=1$, such a mapping can not exists.
@N3buchadnezzar There is nothing requiring anything one-to-one. We have a holomorphic $f$ on the domain $D$, and we want to show it is constant. Under the assumption that it is not constant, we don't have any one-to-one guarantee. All we have is that $f$ would be an open mapping. In particular, $f(D)$ would be open.
Let $\binom nk_q$ be the Gaussian $q$-binomial. Then it is known this counts the $k$ dimensional subspaces of an $n$ dimensional vector space over $\Bbb F_q$, @Daniel.
