Rebellion #1: The Natural Logarithm of Zero

This blog will become more serious progressing to the actual math of the proof. Until now, it’s been about introducing people of varying backgrounds to the idea of the Riemann Hypothesis. While there are other forms of the Riemann Zeta Function that will be talked about later, what will be fundamental to the proof is the ability to calculate the \ln{0} as a limit. Now that seems obvious, but I’m sure what I’m about to state here will start some discussion. That’s a good thing.


I’ve seen many people calculate the \lim_{m \to 0} \ln{m} as -\infty with no imaginary term since it is discontinuous depending on whether approaching from the left or right side. Inspecting limits of those forms, typically look something like:

\lim_{m \to 0^+} \ln{m} = -\infty

When calculating the left-side limit imaginary terms become introduced to handle the negative.

\lim_{m \to 0^-} \ln{m} = \ln{-1*|m|} = \ln{e^{\pi{i}}*|m|} = \ln{e^{\pi{i}}} + \ln{|m|} = \pi{i} -\infty

Now these aren’t exactly incorrect, but they ignore the fact that both sides of the limit have multiple values. Multiplying anything be -1^{2k} where k is a positive or negative integer doesn’t change value. but in this case it offers us a lot of insight. Reworking both sides, the limits develop as follows:

\lim_{m \to 0^+} \ln{-1^{2k}*m} = \lim_{m \to 0^+} \ln{e^{2k\pi{i}}*m} = \lim_{m \to 0^+} \ln{e^{2k\pi{i}}} + \ln{m} = 2k\pi{i} -\infty

\lim_{m \to 0^-} \ln{-1^{2k}*-1*|m|} =\lim_{m \to 0^-} \ln{e^{\pi{i}*(2k + 1)}*|m|} =\lim_{m \to 0^-} \ln{e^{\pi{i}*(2k + 1)}} + \ln{|m|} = (2k+1)\pi{i} -\infty

To look for continuity, the left-side limit is set equal to the right.

(2k+1)\pi{i} - \infty = 2k\pi{i} - \infty

While infinities don’t typically cancel, in this case they both are a representation of Re\{\ln{0}\} and thus are the same value. Another way of thinking about infinity in this case, which also fits well with this proof, is that of the largest possible integer. Since zeta functions and alternate representations of it are the sum of a series of values, the number of terms in the series shares this same representation. Anyway, removing the real infinities and looking only at the imaginary terms of the equation, it continues

(2k +1)\pi{i} = 2k\pi{i}

2k+1 = 2k

At this point it looks as if there is no continuity as this statement is not true. However, k is defined as any integer. Rearranging allows another limit to be taken.

\lim_{k \to \infty} \frac{2k + 1}{2k} = 1

Taking the \lim_{k \to \infty} or \lim_{k \to -\infty} the above statement holds true.

\lim_{m \to 0} \ln{m} = -\infty -\infty\pi{i} = -\infty + \infty\pi{i}

This reveals the fact that the ln{0}, or at least the limit of it, is actually dual-valued. The \pi term is left in for syntax purposes. It should be noted that again, this is a representation of infinity representing the largest possible integer.

This representation of \ln{0} will be vital to the upcoming proof and it cannot be completed without it.

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