A riddle on the ‘Law’ of Excluded Middle

A medieval riddle that exposes the false axiom embedded in the ‘Law’ of Excluded Middle.

A riddle on the ‘Law’ of Excluded Middle:

On Mt Olympus, Zeus turns a lamp on. After half a minute, he switches it off. A quarter of a minute later, he switches it on…ad infinitum.

Let’s suppose we represent the on state by +1 and the off state by -1. Given that:

\[\begin{equation} \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+... = 1 \tag{1} \end{equation}\]

at the end of one minute is the lamp on or off?

Grandi’s series:

The infinite series:

\[\begin{equation} \sum_{n=0}^{\infty} (-1)^n \tag{2} \end{equation}\]

has no definite value but given that the partial sum:

\[\begin{equation} S_N = \sum_{n=0}^N (-1)^n \tag{3} \end{equation}\]

equals \(1\) when \(N\) is even and \(0\) when \(N\) is odd, we may intuit that its sum should be \(\frac{1}{2}\).

This may be made precise using the method of Césaro summation.

Césaro summation:

Let \(\big(a_n\big)_{n=1}^{\infty}\) be a sequence, and let:

\[\begin{equation} S_N = \sum_{n=1}^N a_n \tag{4} \end{equation}\]

be its Nth partial sum.

The sequence \((a_n)\) is called Césaro summable, with Césaro sum \(A \in \mathbb{R}\), if as \(n \to \infty\) the arithmetic mean of its first \(n\) partial sums \(s_1,s_2,...,s_n\) tends to \(A\):

\[\begin{equation} \lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^N s_n = A \tag{5} \end{equation}\]

The resulting value is called the Césaro sum, and may be understood from a frequentist perspective as an expectation.

Geometric series and analytic continuity:

For \(z \in \mathbb{C}\), we may consider the geometric series:

\[\begin{equation} f(z) = \sum_{n=0}^{\infty} z^n \tag{6} \end{equation}\]

which is analytic in \(D = \{z \in \mathbb{C} \big\lvert \lvert z \rvert < 1 \}\) and satisfies the functional equation:

\[\begin{equation} f - z \cdot f = 1 \tag{7} \end{equation}\]

Solving for \(f\) allows us to analytically continue the geometric series:

\[\begin{equation} \forall z \in \mathbb{C} \setminus \{1\}, f(z) = \frac{1}{1-z} \tag{8} \end{equation}\]

which is analytic on \(z \in \mathbb{C} \setminus \{1\}\).

It follows that:

\[\begin{equation} \sum_{n=0}^{\infty} (-1)^n = \frac{1}{1-(-1)} = \frac{1}{2} \tag{9} \end{equation}\]

Leibniz’ probabilistic argument:

Leibniz noted that for any \(N \geq 1\),

\[\begin{equation} P(\sum_{n=0}^N (-1)^n =1) = P(\sum_{n=0}^N (-1)^n = 0) = \frac{1}{2} \tag{10} \end{equation}\]

and therefore:

\[\begin{equation} \lim_{N \to \infty} \mathbb{E}\big[\sum_{n=0}^N (-1)^n \big] = \frac{1}{2} \tag{11} \end{equation}\]

which is equivalent to the method of Césaro summation. The reason why the two methods agree is due to the existence and uniqueness of analytic continuity.

Conclusion:

This riddle is actually an effective introduction to Quantum Probability theory and Ramanujan summation. We never observe a superposition of Quantum States, only expected values because a deterministic Universe only allows for well-defined measurements.

What you can’t measure, you can’t understand and we live in a comprehensible Universe. One where Occam’s razor is applicable.