Can the golden ratio accurately be expressed in terms of $e$ and $\pi$

The following is exact. :-)

$$\phi=\frac{\frac{\pi}{\pi}+\sqrt{\frac{e+e+e+e+e}{e}}}{\frac{e}{e}+\frac{\pi}{\pi}}$$

$e$ and $\pi$ are transcendental numbers, that is to say they are not the solution of any polynomial with rational coefficients. It's not hard to see that if $x$ is transcendental, then the following are also transcendental:

• $x \pm c$ for any rational number $c$,
• $kx$ for any nonzero rational number $k$ (so $x/k$ too),
• $x^n$ for any whole number $n > 1$,
• $\sqrt{x}$ and indeed $\sqrt[n]{x}$ for any whole number $n > 1$.

$\phi$ is not transcendental: it is the solution to a simple quadratic polynomial, so it won't be the result of any operations like the above applied to a transcendental number. The only way you're going to get an expression exactly equal to $\phi$ using only one of $e$ and $\pi$ and the above operations is by not really using them at all, e.g. cancelling them out completely, as other answers do.

What if you use both $e$ and $\pi$? Well, somewhat absurdly, it's not even known if $e + \pi$ is irrational, let alone transcendental! So with our current level of knowledge we can't say whether you can make $\phi$ exactly in a nontrivial way. However, I think most mathematicians would be extremely surprised if it turned out that $e + \pi$ or anything like it (apart from $e^{i\pi}$ and its family, of course!) were not transcendental, and hence mostly useless for constructing something like $\phi$.

At the time of writing, three of the other answers simply express the golden ratio by using expressions like $e/e$ and $\pi/\pi$ to get small integers. The fourth and final one discusses why a good solution is unlikely.

I believe using the imaginary unit $i=\sqrt{-1}$ results in the following very elegant solution: $$ \varphi = e^{i\pi/5} + e^{-i\pi/5}. $$

Edit: robjohn notes that one can directly derive the fundamental identity for the golden ratio $\varphi^2 = \varphi + 1$ from this expression:

$$ \begin{align} \color{#C00000}{\left(e^{i\pi/5}+e^{-i\pi/5}\right)}^2 &=e^{i2\pi/5}+2+e^{-i2\pi/5}\\ &=\left(e^{i2\pi/5}+1+e^{-i2\pi/5}\right)+1\\ &=-\left(e^{i4\pi/5}+e^{-i4\pi/5}\right)+1\\ &=\color{#C00000}{\left(e^{i\pi/5}+e^{-i\pi/5}\right)}+1 \end{align}. $$

If the discussion is not limited to closed-form expressions, it's worth adding that Ramanujan's first letter to Hardy contains an identity that, with a slight rearrangement, allows one to precisely express $\phi$ in terms of $\pi$ and $e$:

$\phi =\sqrt{\frac{1}{2} \left(5+\sqrt{5}\right)}-\cfrac{e^{-\frac{2 \pi}{5}}}{1+\cfrac{e^{-2\pi}} {1+\cfrac{e^{-4\pi}} {1+\cfrac{e^{-6\pi}} {1+\ddots} } } }$

Although $\phi$ can be found in the radicand (and is thus not isolated on the LHS), Ramanujan's insight is certainly beautiful and is noteworthy for uniting three of the pillars of number theory.

Rather than just give you a fish, I'll teach you how to fish:

$(\phi - 1)\phi = 1$

$\phi^2 - \phi - 1 = 0$

$\phi = \dfrac{1 + \sqrt{5}}{2}$

Now replace the integers there with a load of self-cancelling $\pi$/$e$ terms which ultimately give you the values 1, 5, (2 or 4) to taste (-:

Throw in some complex numbers too if you're feeling brave

For example:

$\phi = \dfrac{\pi^e}{\pi^e + e^{\,\textrm{ln}\left(\pi\right)\times e}}+\sqrt{\dfrac{\tfrac{\pi e + \pi e + \pi e + \pi e + \pi e}{\pi e}}{\tfrac{\pi e + \pi e + \pi e + \pi e + \pi e + e \pi e^{\,i\pi}}{\pi e}}}$