User:Mikewarbz

Experimental Results
The following table shows the quantity of square-free numbers composed from the product of primes up to Nmax. For example there are 135 square-free numbers less than 1000 that are constructed from 3 primes. The entries that are in bold italics indicate the peaks of the distributions which can be seen to require more primes as Nmax increases. The shift of the peaks and the 'spreading out' of the distribution is the beginning of a trend described by the Erdős-Kac theorem.

We can also see from the table that:
 * The quantity of square-free numbers built from an even number of primes is approximately the same as those built from an odd number of primes.
 * The total quantity of square-free numbers (which includes 1) rapidly approaches the predicted quantity of $$\frac{N}{\zeta(2)}$$ which is approximately N x 0.6079271018...

If we select a random square-free number then it seems there is a 50-50 chance that it will have either an odd or an even number of primes. If you can prove that the parity of the primes in a square-free number can be modelled like the toss of a fair coin - heads for even, tails for odd - then you have proved the Riemann Hypothesis.

Illuminated switch
A illuminated light switch has an internal light source either a neon or a LED which allows the user to locate the switch in the dark. Most European illuminated switches are two pole requiring the live and neutral wires to pass into the switch which enables the neon to be powered directly from the mains via a resistor. The internal light source in a single pole illuminated switch derives its power when the switch is OFF from current passing through the external light bulb.

Just a line break or two



This definition can be extended to Real numbers


 * $$M(x) = \sum_{1\le k \le x} \mu(k)$$

As an integral
Using the Euler product one finds that


 * $$ \frac{1}{\zeta(s) }= \prod_{p} (1-p^{-s})= \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}$$

where $$\zeta(s)$$ is the Riemann zeta function and the product is taken over primes. Then, using this Dirichlet series with Perron's formula, one obtains:


 * $$ \frac{1}{2\pi i}\oint_{C} \frac{x^{s}}{s\zeta(s)} \, ds = M(x) $$

where C is a closed curve encircling all of the roots of $$\zeta(s).$$

Conversely, one has the Mellin transform


 * $$\frac{1}{\zeta(s)} = s\int_1^\infty \frac{M(x)}{x^{s+1}}\,dx$$

which holds for $$ \mathrm{Re}(s)>1$$.

A curious relation given by Mertens himself involving the second Chebyshev function is:


 * $$ \Psi (x) = -M\left( \frac{x}{2} \right) \log(2)-M \left( \frac{x}{3} \right) \log(3) - M \left( \frac{x}{4}\right )\log(4) + \cdots $$

A good evaluation, at least asymptotically, would be to obtain, by the method of steepest descent, an inequality:


 * $$ \oint_C F(s)e^{st} \, ds \sim M(e^{t}) $$

assuming that there are not multiple non-trivial roots of $$ \zeta (\rho) $$ you have the "exact formula" by residue theorem:


 * $$ \frac{1}{2 \pi i} \oint_C \frac{x^s}{s \zeta (s)} \, ds = \sum_\rho \frac{x^\rho}{\rho \zeta'(\rho)} - 2+\sum_{n=1}^\infty \frac{ (-1)^{n-1} (2\pi )^{2n}}{(2n)! n \zeta(2n+1)x^{2n}} $$

Weyl conjectured that Mertens function satisfied the approximate functional-differential equation


 * $$ \frac{y(x)}{2}-\sum_{r=1}^N \frac{B_{2r}}{(2r)!}D_t^{2r-1} y \left(\frac{x}{t+1}\right) + x\int_0^x \frac{y(u)}{u^{2}} \, du = x^{-1}H(\log x) $$

where H(x) is the Heaviside step function, B are Bernoulli numbers and all derivatives with respect to t are evaluated at t = 0.

Titchmarsh (1960) provided a Trace formula involving a sum over mobius function and zeros of Riemann Zeta in the form


 * $$ \sum_{n=1}^{\infty}\frac{\mu(n)}{\sqrt{n}} g \log n = \sum_t \frac{h(t)}{\zeta'(1/2+it)}+2\sum_{n=1}^\infty \frac{ (-1)^{n} (2\pi )^{2n}}{(2n)! \zeta(2n+1)}\int_{-\infty}^{\infty}g(x) e^{-x(2n+1/2)} \, dx, $$

where 't' sums over the imaginary parts of nontrivial zeros, and (g, h) are related by a Fourier transform so
 * $$ \pi g(x)= \int_{0}^{\infty}h(u)\cos(ux) \, du. $$

As a sum over Farey sequences
Another formula for the Mertens function is


 * $$M(n)= \sum_{a\in \mathcal{F}_n} e^{2\pi i a}$$  where   $$ \mathcal{F}_n$$   is the Farey sequence of order n.

This formula is used in the proof of the Franel–Landau theorem.

As a determinant
M(n) is the determinant of the n &times; n Redheffer matrix, a (0,1) matrix in which aij is 1 if either j is 1 or i divides j.