# Euler, φ and ρ

[Note: I need to end the equations with a period to persuade WordPress/Latex to put sane spacing between lines.

Euler’s famous equation $e^{iπ} + 1 = 0$ is usually considered the “most beautiful” in mathematics.

We could also write it in the so-called “ugly” format as $e^{iπ} = -1$.

Now we know that for the golden ratio φ,

${1 \over {φ}} - φ = -1$.

So let’s rewrite Euler using that:

$e^{iπ} = {1 \over {φ}} - φ$.

Now switch the -φ to the other side:

$e^{iπ} + φ = {1 \over {φ}}$.

and multiply through by φ to get rid of that ugly fraction.

$φe^{iπ} + φ^{2} = 1$.

Now we know that for the plastic ratio ρ,

$ρ + 1 = ρ^{3}$.

which means that

$1 = ρ^{3} - ρ$.

So replace 1 in Euler with its equivalent in terms of ρ:

$φe^{iπ} + φ^{2} = ρ^{3} - ρ$.

and switch the negative ρ to the other side:

$φe^{iπ} + φ^{2} + ρ = ρ^{3}$.

Now we can take the cube root on each side (swapped sides for next step)

$ρ = \sqrt[3]{φe^{iπ} + φ^{2} + ρ}$.

Now we can replace the ρ on the right hand side with its equation, ad infinitum:

$ρ = \sqrt[3]{φe^{iπ} + φ^{2} + \sqrt[3]{φe^{iπ} + φ^{2} + \sqrt[3]{φe^{iπ} + φ^{2} + \sqrt[3]{φe^{iπ} + φ^{2} + ....}}}}$.

And voila, we have an equation for ρ in terms of φ, e, i and π.

We actually could have taken a short cut at $φe^{iπ} + φ^{2} = 1$  above, since we know that

$ρ = \sqrt[3]{1 + \sqrt[3]{1 + \sqrt[3]{1 + \sqrt[3]{1 ....}}}}$.

and then just replaced the 1s with $φe^{iπ} + φ^{2}$.

If we had to use the tau τ version of Euler, which is $e^{iτ} = 1$, then we could write it as

$ρ = \sqrt[3]{e^{iτ} + \sqrt[3]{e^{iτ} + \sqrt[3]{e^{iτ} + \sqrt[3]{e^{iτ} + ....}}}}$.

And similarly for φ:

$φ = \sqrt{e^{iτ} + \sqrt{e^{iτ} + \sqrt{e^{iτ} + \sqrt{e^{iτ} + ....}}}}$.