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} + ....}}}}.




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