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#### All VC-dimension Tue, Apr 12, 2016 02:00 CEST

Definitions of VC-dimesion and $$\varepsilon$$-nets. $$d$$ hyperplanes intersection bounds Sun, Jan 17, 2016 00:00 CET

We bound the position of the $$0$$-cells of an arrangement of hyperplanes in $$\mathbb{R}^d$$. This allows, for example, to build an hypercube that intersects all cells of the arrangement. Such an hypercube must contain at least one point of each cell of the arrangement. When $$q > 0$$, in order to fix which point of a $$q$$-cell we want to include in the hypercube, it suffices to add the $$n$$ hyperplanes of equation $$x_i = 0$$ to the arrangement. With those additional hyperplanes, the arrangement is such that each $$q$$-cell of the arrangement with $$q > 0$$ contains a $$0$$-cell of the arrangement, hence, we only need the hypercube to intersect, for each $$0$$-cell $$\nu$$ of the arrangement, an hypersphere of center $$\nu$$ and arbitrarily small radius. The inequalities of the polyhedral set defining our hypercube will thus only depend on the position of the $$0$$-cells of our arrangement. Polyhedral sets Fri, Jan 15, 2016 01:00 CET

A polyhedral set in $$\mathbb{R}^d$$ is the intersection of a finite number of closed halfspaces, and a (convex) polytope is a bounded polyhedral set. This definition of a polytope is called a halfspace representation (H-representation or H-description). Symbols Sun, Nov 8, 2015 01:00 CET

$$\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}$$ Probabilistic primality testing Fri, Jul 31, 2015 02:00 CEST Fibonacci numbers Mon, Jun 29, 2015 02:00 CEST

The Fibonacci numbers are defined as $$f_0 = 0,\ f_1 = 1$$ and, for $$i \ge 2,\ f_i = f_{i-1} + f_{i-2}$$. Here is the beginning of the Fibonacci sequence:

$$0, 1, 1, 2, 3, 5, 8, 13, 21, \ldots$$

We generalize the definition above by changing the two initial values, for example with $$f_0 = 4,\ f_1 = 6$$ we obtain

$$4, 6, 10, 16, 26, 42, \ldots$$ Inverse sum equations Mon, Jun 29, 2015 02:00 CEST

We are given $$k > 0 \in \mathbb{R}$$ and $$a_1, a_2, \ldots, a_n \ge 1 \in \mathbb{N}$$ such that

$$a_1 < a_2 < \cdots < a_n.$$

We want to solve the following equation

$$\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} = k.$$ Binomial coefficient tricks Wed, Jun 24, 2015 00:00 CEST

$$\binom{n}{k} = \binom{n}{n-k}$$

holds because keeping $$k$$ elements from $$n$$ elements is equivalent to discarding $$n-k$$ elements from $$n$$ elements. Complex numbers division Tue, Jun 23, 2015 02:00 CEST

$$\frac{a+bi}{c+di} = \ldots$$ Gaussian elimination Fri, Jun 19, 2015 02:00 CEST