- #1

epr1990

- 26

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[tex]\sum_{k=1}^\infty\frac{\mu (k)-\varphi (k)}{k}\log \left( 1-\frac{1}{\phi^k} \right) = \prod_{k=1}^\infty \left( 1-\frac{1}{\phi^k} \right)^{2\pi i\frac{\mu (k)-\varphi (k)}{k}}[/tex]

where [itex]\mu , \varphi , \phi [/itex] are the Möbius function, Euler totient function, and golden ratio respectively.

Now, at first, this looks almost nonsensical because it demonstrates equality between a product and its logarithm. I.e. exponentiating gives product=e^product^2*pi*i (or you can take the logarithm if you can see it better this way, personally, I'm better with products than sums, and definitely the logarithm of an infinite sum which you would most likely need to refactor and deal with possible rearrangement issues... yikes). It follows that, in a sense, this form is redundant, for we must have

[tex]\left(\prod_{k=1}^\infty \left( 1-\frac{1}{\phi^k} \right)^{\frac{\mu (k)-\varphi (k)}{k}}\right)^{2\pi i} = 1 [/tex]

Furthermore, this implies

[tex]\sum_{k=1}^\infty\frac{\mu (k)-\varphi (k)}{k}\log \left( 1-\frac{1}{\phi^k} \right) = 1[/tex]

and

[tex]\prod_{k=1}^\infty \left( 1-\frac{1}{\phi^k} \right)^{\frac{\mu (k)-\varphi (k)}{k}} = e[/tex]

Finally, using Dirichlet convolution and inversion and the basic properties of the logarithm and geometric series, we can show that

for 0 < x < 1

[tex]\sum_{k=1}^\infty\frac{\mu (k)}{k}\log \left( \frac{1}{1-\frac{1}{x^k}} \right) = x[/tex]

and

[tex]\sum_{k=1}^\infty\frac{\varphi (k)}{k}\log \left( \frac{1}{1-\frac{1}{x^k}} \right) = \frac{x}{1-x}[/tex]

Thus,

[tex]\sum_{k=1}^\infty\frac{\mu (k)}{k}\log \left( \frac{1}{1-\frac{1}{\phi^k}} \right) = \frac{1}{\phi}[/tex]

and

[tex]\sum_{k=1}^\infty\frac{\varphi (k)}{k}\log \left( \frac{1}{1-\frac{1}{\phi^k}} \right) = \frac{\frac{1}{\phi}}{1-\frac{1}{\phi}} = \phi[/tex]

The point of this result is that it suggests a deep relation between the prime numbers and chaos,

I was hoping that anyone who reads this would share their thoughts and insights on this relationship, or possibly similar results.