- L's electors, if they cannot vote for L they will vote for R, because they hate N.
- R's electors, if they cannot vote for R they will vote for N, because they hate L.

## Sunday, April 20, 2008

### Election time

In Italy, there are just three parties: L (the left), R (the right), and N (north league). Elections are done, L and R get each 49\% of votes, N just 2\%. Since an agreement cannot be found, two tie-break rounds are done, where the two biggest parties are matched, and at the second round the winner is matched against the other party. Prove that there is a strategy for N's electors to make their party win the tie-break rounds, knowing that:

## Wednesday, April 16, 2008

### Minimal values

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function (any function). Let's call

$$ M := \{ x \in \mathbb{R} : \exists\epsilon_x>0: f(x) \leq f(y)\;whenever\;|y-x| < \epsilon_x\} $$

the set of points where $f$ has a (non-strict) local minimum. Let $V=f(M)$ be the set of the local minimal values. Can $V$ have positive Lebesgue measure? (note that $M$ surely can, just take as $f$ a constant function, and every point is a local minimum).

$$ M := \{ x \in \mathbb{R} : \exists\epsilon_x>0: f(x) \leq f(y)\;whenever\;|y-x| < \epsilon_x\} $$

the set of points where $f$ has a (non-strict) local minimum. Let $V=f(M)$ be the set of the local minimal values. Can $V$ have positive Lebesgue measure? (note that $M$ surely can, just take as $f$ a constant function, and every point is a local minimum).

## Friday, April 4, 2008

### An elettric circuit

There exists an electric circuit that has an interruption. The -proper- length of the interruption is $L_0$. A tract of electric wire runs along the circuit with speed $v$. The -proper- length of the wire is $L_0$ too. In circuit reference system wire length is $L<L_0$, so the circuit is never closed. In wire reference system interruption length is $L<L_0$, so at some instant the circuit is closed. How can you explain the apparent contradiction?

Update(22/04/2008): Solution posted! (show solution)

In wire reference system, the circuit is just formally closed. Indeed the time in which it remains closed is not sufficient to permit the electric current flows. In this reference system, wire length is $ L_0 / \gamma $, where $\gamma$ is $ {(1-v^2/c^2)}^{-\frac{1}{2} $. So the time during which circuit is closed is $ \Delta t = L_0 \frac{1-1/ \gamma}{v} $ and in that time a signal that has the light speed run distance $ d = \frac{c(\gamma -1)}{v \gamma} L_0 < L_0 $. In conclusion if an electrical instrument is connected to circuit, then in both reference systems it is going to do not work.

Update(22/04/2008): Solution posted! (show solution)

In wire reference system, the circuit is just formally closed. Indeed the time in which it remains closed is not sufficient to permit the electric current flows. In this reference system, wire length is $ L_0 / \gamma $, where $\gamma$ is $ {(1-v^2/c^2)}^{-\frac{1}{2} $. So the time during which circuit is closed is $ \Delta t = L_0 \frac{1-1/ \gamma}{v} $ and in that time a signal that has the light speed run distance $ d = \frac{c(\gamma -1)}{v \gamma} L_0 < L_0 $. In conclusion if an electrical instrument is connected to circuit, then in both reference systems it is going to do not work.

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