A medieval riddle that exposes the false axiom embedded in 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?

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.

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.

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 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.

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.

For attribution, please cite this work as

Rocke (2022, Sept. 30). Kepler Lounge: A riddle on the 'Law' of Excluded Middle. Retrieved from keplerlounge.com

BibTeX citation

@misc{rocke2022a, author = {Rocke, Aidan}, title = {Kepler Lounge: A riddle on the 'Law' of Excluded Middle}, url = {keplerlounge.com}, year = {2022} }