# Tag Info

2

By "almost equal", you mean "almost everywhere equal". Which means that the two functions agree except possibly on a set of measure zero. You can kind of visualize the measure of a subset of the real line as being its "length". And in fact, this is exactly what the measure is for intervals: that is, the measure of $(a,b)$ is $b-a.$ Now, what we call ...

6

We give a proof of the strict inequality $$\frac{1 + p(1) + p(2) + \cdots + p(n-1)}{p(n)} < \sqrt{2n}.$$ The square root arises by first proving an upper bound $(k-1)/2 + (n/k)$ for any integer $k \geq 1$, and then minimizing over $k$; this kind of thing is seen often in analysis, but isn't a common tactic in the inequalities that appear in competition ...

0

In the "Update" (the day after the Question itself was posted), the OP mentions a recursion for counting the partitions of $n$ into $k$ distinct parts, each part at most $M$: $$p_k(\leq M, \mathcal{D},n) = p_{k-1}(\leq M-1, \mathcal{D},n-k) + p_k(\leq M-1, \mathcal{D},n-k)$$ and asks "How can I prove this?". To see this, separate the ...

1

Let $x$, $y$ $\in \mathcal{X}$ be such that $P(x)\cap P(y) \neq \emptyset$. Then there is some $c \in P(x)$, $P(y)$. Assume for a contradiction that $P(x) \neq P(y)$ . Then there is $A \in F$ such that $x,c \in A$ but $y \notin A$ (or there is $B \in F$ such that $y,c \in B$ but $x \notin B$. This case follows similarly) Since $F$ is closed under taking ...

1

2 examples I like: Fix any natural number $n\geqslant 2$ and define $\sim$ on $\mathbb Z$ by $x\sim y$ if and only if $x$ and $y$ differ by a multiple of $n$ (this means that $x-y$ is a multiple of $n$). Then the equivalence class of $x$ is $[x]=\{\ x+kn\ |\ k\in\mathbb Z\ \}$, so $\mathbb Z$ is partitioned into the subsets of integers that have the same ...

1

Consider the set $A = \{a,b,c,d,e,f,g,h,i \}$, and look at the array: $$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}$$Define $\sim$ on $A$ by saying that $x \sim y \iff x$ and $y$ are in the same column. This way, we have $a \sim d$, $b \sim h$, $f \sim i$, $a \not\sim e$, $h \not\sim c$, for example. I'll leave ...

0

$xRy$ if and only if $x=y$ you can see $R$ is equivalence relation

1

Consider an elementary school. Define an equivalence relation on the children in the school based on $\text{thing}_1 \sim \text{thing}_2$ if and only if they are in the same room. Then every child is assigned to a room and so every child lands in one and only one of the equivalence classes. Thus the set of children is the union of all of the equivalence ...

1

(Too long for a comment.) Just to give the exact value of $(2)$, we have the identity, $$A = \frac{1}{e^{5\pi/6}\,\eta\big(\tfrac{2\,i}{5}\big)} = \frac{2^{19/8}(-1+5^{1/4})\,\pi^{3/4}}{e^{5\pi/6}\,(-1+\sqrt{5})^{3/2}\,\Gamma\big(\tfrac{1}{4}\big)} = 0.0887758\dots$$ which is approximated by, $$B ... 1 HINT: Note that$$\binom{\lambda_i'}2=\sum_{k=0}^{\lambda_i'-1}k=\sum_{k=1}^{\lambda_i'}(k-1)\;.\tag{1} Now $\lambda_i'\ge k$ if and only if there are at least $k$ parts $\lambda_j$ of $\lambda$ such that $\lambda_j\ge i$. In terms of the Ferrers diagram of $\lambda$, the $i$-th column has $\ge k$ elements if and only if there are at least $k$ rows with ...

Top 50 recent answers are included