ConformalInfinity comments on When circular reasoning is logical evidence

[ConformalInfinity]

Very interesting, I am in a similar position with respect to learning the relevant mathematics as you know from my first comment. One thing that your sequence resembles to me is the divergent infinite sum 1-$1+1-1+1-1+1-\mathrm{1...}$ This sequence does not get closer and closer to any particular value, so from the most standard perspective, its sum is undefined. However, the partial sums alternate from 1, to 0, back to 1 again and continue to do so ad infinitum, which means that their average is $1/2$. A different way of looking at this sum is through the formula $1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}+...=\frac{1}{1-x}$ . If x = −1, then the series $1-1+1-1+...$ becomes $\frac{1}{1-x}=\frac{1}{2}$ . This suggests that, to the extent that this sum has a value, it is $\frac{1}{2}$. Although this series is not the same as your sequence, if we take 1 to represent true, 0 to represent false and subtraction from 1 to represent logical negation, then the equivalent sequence is the sequence of partial sums:$1,0,1,0,1,0,...$ and $\frac{1}{2}$ now represents the concept of ‘tralse’ . In the case of your sequence, it would be the set of partial products of an infinite product$(-1)(-1)(-1)(-1)(-1)(-1)(-1)...$ . How to obtain a generalization of the value of an infinite product when it does not converge, I am not sure. One possibility is to operate on a logarithmic scale, on which multiplication is equivalent to addition. Multiplication by −1 is equivalent to raising $e$ to the power of $i\pi$ , which suggests that the infinite product is given by $e^{i\pi (1+1+1+1+1+1+...)}=e^{i\pi +i\pi +i\pi +i\pi +i\pi +i\pi +i\pi ...}=e^{\frac{i\pi }{1-\frac{1}{1}}}$

. In complex analysis, there is a well defined infinity which is the reciprocal of 0, therefore $e^{i\pi +i\pi +i\pi +i\pi +i\pi +i\pi +i\pi ...}=e^{\frac{i\pi }{1-\frac{1}{1}}}=e^{\infty }=\infty$. Though $\infty$ is not $0$, it is the single other number on the Riemann sphere which is its own negative, and is therefore also a solution to the equation $x=-x$ !

I have found another way to shoehorn into your interpretation the notion of ‘tralse’ . I’m not sure if this meaningful or not, but I didn’t know that calculation would produce a result compatible with the idea of ‘tralsity’ until I carried it out, and it did produce such a result in an unexpected way .

I will mirror your disclaimer about the idea being newly encountered and not clearly explained.