Solution to Fascinating Frequencies 1

We wanted to find the frequency of a gamma radiation particle. We know that if it had undergone beta decay, it would have released the same amount of energy, but by releasing 2 beta radiation particles, both traveling at half of c. The equation to find the energy of these particles is shown below, where m0 is the particle’s rest mass energy, ≈0.511 meV. It gives ≈8.5 microjoules of energy per particle, or ≈17 microjoules in all. We then use the second equation to find the frequency of a photon with ≈17 microjoules of energy. Our answer is thus that the frequency of the photon is ≈2.566×1024 rotations per second.

E\approx\sqrt{\frac{(m_0\times0.5c)^2}{1-\frac{v^2}{c^2}}c^2+(m_0c^2)^2}\approx 8.5\times10^{-6}J\\[32pt] r=\frac Eh\approx\frac{17\times10^{-6}J}{6.626\times10^{-34}Js}\approx\frac{2.566\times10^{28}}s

Solution to Peculiar Products

We are trying to find the value of the expression below. Note than the tan stuff approaches infinity as k increases. This is because 360/k approaches 0, and so 90- 360/k approaches 90. And as x approaches 90, tan(x) approaches infinity. So the reciprocal of the tan stuff gets smaller and smaller. So the side lengths approach zero while the number of sides increases. The shapes with more sides approximate circles. Their circumference equals the number of sides times the length of each side. K approaches 10 times more than the tan stuff. And so this will reach infinity like Exquisite Equations did.

\prod_{k=1}^\infin\Bigg(A\bigg(k,\,\frac1{tan (90\degree-\frac{360\degree}k)}\bigg)\Bigg)\\[16pt]