# Profound Similarities

I was messing around with some ideas the other day and noticed some similarities between some very beautiful identities. $\displaystyle e^{\theta\imath} = \cos\theta + \imath\sin\theta$, $\displaystyle \varphi^{n} = \frac{L_{n} + F_{n}\sqrt{5}}{2}$

Here are identities for each of the functions above: $\displaystyle \cos\theta = \frac{e^{\theta\imath} + e^{-\theta\imath}}{2}$, $\displaystyle \sin\theta = \frac{e^{\theta\imath} - e^{-\theta\imath}}{2i}$ $\displaystyle L_n = \varphi^{n} + (1-\varphi)^{n}$, $\displaystyle F_n = \frac{\varphi^{n} - (1-\varphi)^{n}}{\sqrt{5}}$

Cosine is an even function, and the nth Lucas number is also given by an even function. Sine is and odd function, and the nth Fibonacci number is given by one also. But what should be remarkable is the similarity of the definitions.

Here’s something else that is really awesome. One should notice $e^{\theta\imath} = \cos\theta + \imath\sin\theta$ and $e^{-\theta\imath} = \cos\theta - \imath\sin\theta$ are conjugates. One should also notice that $\varphi = \frac{1 + \sqrt{5}}{2}$ and $1 - \varphi = \frac{2}{2} - \frac{1+\sqrt{5}}{2} = \frac{1-\sqrt{5}}{2}$ are conjugates as well.

Now, I can define an operator and it’s conjugate or dual that can be used to simply write the relationship between all the above functions and more. I’ll use the symbol $\heartsuit$, because I don’t know what it’s real use is and I <3 this relation. $\displaystyle \heartsuit F = F + \overline{F}$, $\displaystyle \heartsuit^{\ast} F = F - \overline{F}$,

where $\overline{F}$ represents the conjugate of a function $F$.

Now it’s possible to write: $\displaystyle \heartsuit\varphi^{n} = L_n$, $\displaystyle \frac{1}{\sqrt{5}}\heartsuit^{\ast}\varphi^{n} = F_n$ $\displaystyle \frac{1}{2}\heartsuit e^{\theta\imath} = \cos\theta$, $\displaystyle \frac{1}{2\imath}\heartsuit^{\ast}e^{\theta\imath} = \sin\theta$

Now if I let $\overline{e^{\theta\imath}} = e^{-\theta\imath}$ and then generalize (perhaps wrongly, but wait and see what happens!) $\overline{e^{\theta}}=e^{-\theta}$, then I can write: $\displaystyle \frac{1}{2}\heartsuit e^{\theta} = \cosh\theta$, $\displaystyle \frac{1}{2}\heartsuit^{\ast}e^{\theta} = \sinh\theta$

This is in perfect accordance with the (lesser?) known identities $\cos(\theta\imath) = \cosh\theta$ and $\sin(\theta\imath) = - \sinh\theta$ and the identities: $\displaystyle \cosh\theta = \frac{e^{\theta} + e^{-\theta}}{2}$, $\displaystyle \sinh\theta = \frac{e^{\theta} - e^{-\theta}}{2}$ $\blacksquare$

Just so you can stare at them all together, here they are: $\displaystyle \heartsuit\varphi^{n} = L_n$, $\displaystyle \frac{1}{\sqrt{5}}\heartsuit^{\ast}\varphi^{n} = F_n$ $\displaystyle \frac{1}{2}\heartsuit e^{\theta\imath} = \cos\theta$, $\displaystyle \frac{1}{2\imath}\heartsuit^{\ast}e^{\theta\imath} = \sin\theta$ $\displaystyle \frac{1}{2}\heartsuit e^{\theta} = \cosh\theta$, $\displaystyle \frac{1}{2}\heartsuit^{\ast}e^{\theta} = \sinh\theta$