# Tag Info

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No, there is no known closed form for the partition function. Here is the sequence on OEIS, and also the Wolfram MathWorld page on this subject is very thorough. Although, looking at this MathSE post, I might be wrong ...

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Given any small positive real $\varepsilon > 0$, for any partition $P = \{x_0, x_1, \ldots, x_n\}$ of $[0, 1]$ such that $0 = x_0 < x_1 < \cdots < x_n = 1$, since $\mathbb{Q}$ is dense in $\mathbb{R}$, for each $i$, there exists $q_i \in \mathbb{Q}$ such that $x_i - \varepsilon < q_i < x_i$, it then follows that $$M_i = \sup_{I_i} f(x) - ... 1 Another way: Let P = \{x_0,x_1,\ldots,x_n\} be a partition of [0,1], and let M_i denote the supremum of f on [x_{i-1},x_i]. There must exist a rational number q in the interval [(x_{i-1} + x_i)/2,x_i], and so it follows that M_i \geq (x_{i - 1} + x_{i})/2. This gives:$$U(f,P) = \sum_{i = 1}^{n}M_i(x_i - x_{i - 1}) \geq \sum_{i = ...

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Suppose otherwise. Write $A = \{a_1, \dotsc, a_n\}$, $B = \{b_1, \dotsc, b_n\}$, and $C = \{c_1, \dotsc, c_n\}$, each indexed increasingly, such that $a_1 < b_1 < c_1$. Thus $a_1 = 1$. Set $k = b_1 - 1$, so that $\{1, \dotsc, k\} \subseteq A$. Now, for any $b \in B$, if any of $b+1, \dotsc, b+k$ are in $C$, then we have a forbidden triple. Similarly ...

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Let $S\subset\mathbb{N}$ be the subset of valid block sizes (in the example you gave, $S$ is the set of odd numbers.) Let $g(z)$ be the generating function for $S$; that is, $g(z)=\sum_{k\in S}z^k$. Let $a_n$ be the number of partitions of a queue of length $n$; note that $a_0=1$ since the empty queue can be partitioned in exactly one way (the empty ...

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HINT: Number the elements of $S$ from $1$ through $2n$. In order to form a partition of $S$ into $n$ pairs, we can begin by pairing element $1$ with one of the other $2n-1$ elements. That leaves $2n-2$ elements still to be paired.

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Let $\lambda=\langle\lambda_1,\ldots,\lambda_r\rangle$. Let $F$ be the Ferrers diagram for $\mu$. For $k\in[r]$ let $F_k$ be what remains of $F$ when each row longer than $\lambda_k$ has been shortened to length $\lambda_k$. Verify that $\sum_{i,j}\min\{\lambda_i,\mu_j\}=\sum_{k=1}^r|F_k|$. We now determine how many times each column of $F$ is counted in ...

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The idea here is that all but finitely many of the points in $\{ 1/n : n \in \mathbb{N} \}$, where the discontinuities in $f$ are, are very close to $0$. So you can work this way. Let $\varepsilon >0$. Choose $N \in \mathbb{N}$ such that $1/(N+1)<\varepsilon/2$. Then make the first two points of the partition be $0$ and $1/(N+1)$. Now you just have ...

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$$\sum_{i=0}^{n-1} f(x_i) \, \Delta x_i = \sum_{i=0}^{n-1} \sqrt{\frac{4i^2}{n^2}} \left( \frac{4(i+1)^2}{n^2} - \frac{4i^2}{n^2} \right).$$ Etc. Why don't you know where to begin? Are you unaware of what Riemann sums are? If so, maybe you could ask about that.

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