## Archive for January, 2011

### Willems on theoretical engineering

January 15, 2011

Jan Willems on theoretical engineering research:

One of the differences in working in a mathematics versus a theoretical engineering environment is psychological. Both fields feel insufficiently appreciated. Mathematics reacts to this by blaming the ‘others’, for misunderstanding them. Engineering reacts by blaming ‘themselves’, for not doing ‘the right thing’. Discussions about theoretical engineering research often feels like visiting a graveyard in the company of Nietzsche. From the beginning of my career until now, I have always been hearing that ‘the field is dead’, ‘circuit theory is dead’, ‘information theory is dead’, ‘coding theory is dead’, ‘control theory is dead’, ‘system theory is dead’, ‘linear system theory is dead’,$\mathcal{H}_{\infty}$ is dead’. Good science, however, is always alive. The community may not appreciate the vibrancy of good ideas, but it is there. The absence of this impatience is one of the things that makes working in a mathematics department simply more pleasant.

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Source:

Jan C. Willems, In Control, Almost from the Beginning Until the Day After Tomorrow [pdf], European Journal of Control, vol. 13, p. 75, 2007.

### How many votes did Alice get?

January 5, 2011

Last month’s Ponder This challenge was as follows:

In an election, Charles came in last and Bob received 24.8% of the votes. After counting two additional votes, he overtook Bob with 25.1% of the votes. Assuming there were no ties and all the results are rounded to the nearest promille (one-tenth of a percent), how many votes did Alice get?

And here‘s the solution. It was the easiest Ponder This ever.

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My solution:

Let $a, b, c \in \mathbb{N}$ be the number of votes gotten by Alice, Bob, and Charles, respectively. If Charles came in last, then $a > c$ and $b > c$. Since Bob received 24.8% of the votes and he was not the last one, then we deduce that Alice was the first with more than 50% of the votes. Therefore, we have $a > b > c$.

In the first counting, Bob got 24.8% of the votes. Since the results are rounded off, we have

$\displaystyle \frac{b}{a + b + c} = 0.248 + \delta_1$

where $|\delta_1| < 0.001$ is the round-off error. Hence, we obtain

$(0.248 + \delta_1) a + (\delta_1 - 0.752) b + (0.248 + \delta_1) c = 0$.

Multiplying both sides by $10^3$, and letting $s_1 := 10^3 \delta_1$, we get

$(248 + s_1) a + (s_1 - 752) b + (248 + s_1) c = 0$.

Note that, since $|\delta_1| < 0.001$, we have that $|s_1| < 1$. After counting two additional votes, Charles overtook Bob with 25.1% of the votes. Let $c' = c+2$ be the new number of votes that Charles got. Hence,

$\displaystyle \frac{c+2}{a + b + c + 2} = 0.251 + \delta_2$

where $|\delta_2| < 0.001$ is the round-off error. Thus, we have

$(0.251 + \delta_2) a + (0.251 + \delta_2) b + (\delta_2 - 0.749) c = 1.498 - 2 \delta_2$

and, multiplying both sides by $10^3$, and letting $s_2 := 10^3 \delta_2$, we get

$(251 + s_2) a + (251 + s_2) b + (s_2 - 749) c = 1498 - 2 s_2$

where $|s_2| < 1$. Since there were no ties and these two additional votes were enough for Charles to overtake Bob, then we have $c+2 > b > c$. As $b,c$ are natural numbers, we conclude that $b = c +1$.

Hence, we have a linear system of equations in $(a,b,c) \in \mathbb{N}^3$

$\left[\begin{array}{ccc} (248 + s_1) & (s_1 - 752) & (248 + s_1) \\ (251 + s_2) & (251 + s_2) & (s_2 - 749) \\ 0 & 1 & -1 \end{array}\right] \left[\begin{array}{c} a \\ b \\ c \end{array}\right] = \left[\begin{array}{c} 0 \\ 1498 - 2 s_2 \\ 1 \end{array}\right]$

where $|s_1|, |s_2| < 1$. Note that $a,b,c$ must be natural numbers, which means that there should exist $(s_1, s_2) \in (-1,1)^2$ such that the linear system of equations has a solution in $\mathbb{N}^3$. Let us relax the system of equations by making $s_1, s_2 = 0$ and allowing the unknowns to take values in $\mathbb{R}$ rather than $\mathbb{N}$. Then, we obtain the new linear system of equations

$\left[\begin{array}{ccc} 248 & -752 & 248 \\ 251 & 251 & -749 \\ 0 & 1 & -1 \end{array}\right] \left[\begin{array}{c} a \\ b \\ c \end{array}\right] = \left[\begin{array}{c} 0 \\ 1498 \\ 1 \end{array}\right]$

whose solution is $(a^*, b^*, c^*) \approx (84.664, 41.168, 40.168)$. Taking the floor of each component, we obtain a solution in $\mathbb{N}^3$

$(a,b,c) = (84, 41, 40)$.

Note that $41 / 165 \approx 0.248485$, and $42 / 167 \approx 0.251497$, meaning that $s_1 = 0.485 < 1$ and $s_2 = 0.497 < 1$, which are admissible values. Note also that $c+2 > b > c$. Hence, all the conditions are satisfied. We thus conclude that Alice got 84 votes.