Exercise 1
One morning in Springfield, it started snowing at heavy but constant rate. Homer Simpson had just started his own snowplow business. His snowplot started out at 08:00 AM and at 09:00 AM it had gone 2 miles. By 10:00 AM it had gone 3 miles. Assuming that the snowplow removes a constant volume of snow per hour, determine the time at which it started snowing.
Solution
Exercise 2
\(a\) and \(b\) are randomly chosen real numbers in the interval \([0,1]\), that is both \(a\) and \(b\) are standard uniform random variables. Find the probability that the quadratic equation \(x^2+ax+b= 0\) has real solutions.
Solution
The equation \(x^2 + A x + B=0\) has real solutions if its determinant is positive or zero. The determinant of a second order equation \(ax^2 + bx + c=0\) is found as \(\Delta = b^2 - 4ac\). So in our case we have \(\Delta = A^2 - 4B\). We know that both \(A\) and \(B\) are \(\sim U[0,1]\). Follow this guide
- We first find the pdf of the variable \(Y=A^2\). We note that \(R_Y=[0,1]\). As usual, we start with the CDF. For \(y \in [0,1]\), we have:
\(F_Y(y) &= P(Y \leq y) \\\ &=P(X^2 \leq y) \\\ &=P(-\sqrt{y} \leq X \leq \sqrt{y}) \\\ &=\frac{\sqrt{y} -(-\sqrt{y}}{1-(-\sqrt{y})} \\\ &=\sqrt{y}\) since \(X \sim U[0,1]\). Thus, the CDF of \(Y\) is given by
\(F_Y(y) = \begin{cases} 0 & \text{ for } y \leq 0 \\\ \sqrt{y} & \text{ for } 0\leq y \leq 1 \\\ 1 & \textrm{ for } y \geq 1 \end{cases}\) Note that the CDF is a continuous function of \(Y\), so \(Y\) is a continuous random variable.
Exercise 3
Solve the equation:
\[\sqrt{x+\sqrt{x+{\sqrt{x+\sqrt{x+\sqrt{x...}}}}}} =x\]Solution
Take the squares we get: \({x+\sqrt{x+{\sqrt{x+\sqrt{x+\sqrt{x...}}}}}} =x^2\) and the infinite square appears in the left hand side. However we know that it evaluates to \(x\), hence we get \(x+x=x^2\), and the solutions are \(x_1=0\) and \(x_2=2\).
Exercise 4
Solve the infinite tetration \(x^{x^{x^{x^{x^{...}}}}} = 2\)
Solution
The solution is simple to obtain. Since the tetration is infinite, we have that the left exponent \((\cdot)^{x^{x^{x^{...}}}}\) is equal to 2. Hence \(x^2=2\), and the solution is \(x=\sqrt{2}\).
### N-queens problem
def n_queens(n, board=[]):
if n == len(board):
return 1
count = 0
for col in range(n):
board.append(col)
if is_valid(board):
count += n_queens(n, board)
board.pop()
return count
def is_valid(board):
current_queen_row, current_queen_col = len(board) - 1, board[-1]
# Check if any queens can attack the last queen.
for row, col in enumerate(board[:-1]):
diff = abs(current_queen_col - col)
if diff == 0 or diff == current_queen_row - row:
return False
return True
Let's talk!
I'm Carlo Nicolini — I am interested on the reliability of AI reasoning systems (interpretability, inference-time methods, probabilistic language programming) and on quantitative portfolio optimization (I am a maintainer of skfolio). If you're working on something in these areas and think we might collaborate, chat, discuss, I'm happy to talk about it!
The best way to reach me is on via DM on LinkedIn.