I was playing around with numbers when I noticed that $\sqrt e$ was very somewhat close to $\phi$
And so, I took it upon myself to try to find a way to express the golden ratio in terms of the infamous values, $\large\pi$ and $\large e$
The closest that I've come so far is:
$$
\varphi \approx \sqrt e - \frac{\pi}{(e+\pi)^e - \sqrt e}
$$
My question is,
Is there a better (more precise and accurate) way of expressing $\phi$ 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:
$\phi$ 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:
$\phi = \dfrac{\pi^e}{\pi^e + e^{\,\textrm{ln}\left(\pi\right)\times e}}+\sqrt{\dfrac{\tfrac{\pi e + \pi e + \pi e + \pi e + \pi e}{\pi e}}{\tfrac{\pi e + \pi e + \pi e + \pi e + \pi e + e \pi e^{\,i\pi}}{\pi e}}}$
$\sqrt e \approx 1.64872$ is not "very close" to $\phi \approx 1.61803$.
Here is a very good approximation: $$ \phi \approx \frac{1967981\,\pi-314270\,e}{3293083} $$
The error is about $2 \times 10^{-16}$.
This relation was found using FindIntegerNullVector[N@{Pi,E,(1+Sqrt[5])/2}] with Mathematica (sadly, Wolfram Alpha does not understand this).
RIES can find solutions to approximation problems like this. Running
ries -NlLeE -s -l6 2.718281828459045235360287471
(which says "find $e$ without using logarithms, exponentials, or e itself" -- $\pi$ and $\phi$ are already in by default)
gives, among others,
$$\phi\approx e-\sqrt[3]{\pi}$$ $$\phi\approx\frac{5(1+\pi)}{e}-6 \qquad \text{(to 7 decimal places)}$$
ries -SpfnrsqSCT+-*/^v -s -l6 2.718281828459045235360287471
(which says "find $e$ using only $\pi$, $\phi$, and the operators $+-\cdot/\sqrt[n]{\ \ \ }\ \text{^}\sin\cos\tan$"... basically, banning numbers too) gives
$$\phi\approx e/\sqrt{\pi-1/\pi} \qquad \text{(2 places)}$$ $$\phi^2\approx(e-1/\pi)^2-\pi \qquad \text{(3 places)}$$
and many more. (If I was more careful I would have had it solve for $\phi$ in the first place...)
An approximation: $$\phi \approx \frac { 7\pi }{ 5e } =1.618018$$
Here is another suggestion:
$$\phi=\frac{\pi}{\pi+\pi}+\sqrt{\frac{e+e+e+e+e}{e+e+e+e}}$$
If you define the sequence $a_1 = a_2 = -e^{i\pi}$, $a_k = a_{k-1} + a_{k-2}$, then $\lim_{n \rightarrow \infty} \frac {a_{n+1}}{a_n} = \phi$.
It seems all the answers so far approaching this from a theoretical perspective are approaching this in terms of exact answers, but we can say a lot about when good approximations are possible too. Of course, some answers have already provided silly ways to do this exactly, so approximations may seem unnecessary, but it provides a nice avenue for some basic transcendental number theory.
It is an unsolved problem, which virtually everyone believes to be true, that $\frac e \pi$ is irrational. Let's assume for the moment that this is true. Then it's a trivial corollary of a well-known theorem that if $\alpha$ is an irrational number, and $\beta$ is any real number, there exist arbitrarily good approximations $p + q \alpha \approx \beta$ with $p,q$ integers. That means, taking $\alpha = \frac e \pi$ and $\beta = \frac \phi \pi$, we can find integers $p,q$ such that $p e + q \pi$ approximates $\phi$ to any tolerance you desire.
One such approximation could be $357 \pi - 412 e = 1.61646... \approx 1.61803... = \phi$, which is accurate to one part in 1000. One can do better, but this at least demonstrates the principle. If the 357 and 412 bother you, you may imagine that I've written a sum with 729 terms on the left hand side instead, 357 of which are $\pi$ and 412 of which are $-e$.
So what if, against all bets, $\frac e \pi$ is rational? Then the opposite is true. There is a single best approximation to $\phi$ of the form $p e + q \pi$, which is not exact, and there are infinitely many choices of $p$ and $q$ which yield the same approximation. This is because, in that case every number of the form $p e + q \pi$ is a rational multiple of $e$ with denominator dividing $d$ the denominator of $\frac e \pi$ when written as an integer fraction in lowest terms. Of course, none of these can be exact, since they're all either 0 or transcendental, while $\phi$ is algebraic, and since the set of all such numbers is discrete (being just $\frac{e}{d}\mathbb Z$ where $d$ is the denominator mentioned above), $\phi$ is not in its closure. That is to say, the irrationality of $\frac e \pi$ is equivalent to the existence of arbitrarily good approximations to $\phi$ of the form $p e + q \pi$ for integers $p$ and $q$. Of course, the current lower bounds on $d$ are likely to be extremely large since we know plenty of digits of both $e$ and $\pi$ and haven't yet found any such rational number with value $\frac e \pi$, so there are going to be very good approximations for all practical purposes, but eventually there has to be a single best one, in exactly the same way that there's a single best integer approximation to $\phi$ (namely 2).
Luckily, even in this case we can still construct arbitrarily good approximations to $\phi$ based on $e$ and $\pi$; just not in the same way. Of course, for some $n$, it must be true that $\sqrt[n] \frac{e}{\pi}$ is irrational (this is true for any real number other than 0 and 1, and $\frac e \pi$ is clearly neither). We can play exactly the same game as we did before to get arbitrarily good approximations of the form $p \sqrt[n] e + q \sqrt[n] \pi$ to $\phi$ with $p$ and $q$ integers. If the appearance of this $n$ bothers you, we can even take $n$ to be a power of 2 so that $\sqrt[n] {}$ can be written as a repeated composition of $\sqrt {}$, i.e. $\sqrt[8]{x}=\sqrt {\sqrt {\sqrt{x}}}$.
Note that in all cases above, it's (as far as I know) unknown whether the forms given can exactly represent $\phi$, though all bets are to the negative. Certainly there are no known cases in which it does represent $\phi$ exactly, since that would give a proof that $e$ and $\pi$ are not algebraically independent (a major unsolved problem). In principle, there could be cases where it's definitely known that the form does not represent $\phi$ exactly, but really there's just about nothing about problems like this so it would surprise me if there are any cases known.
form this $\phi = 2\cos(\frac{\pi }{5})$ and euler formula $e^{ix} = \cos(x) + i\sin(x)$ you can conclude this one $\phi= 2e^{i\frac{\pi}{5}}-2i\sin(\frac{\pi}{5})$. [check]
Using the 5 constants of Euler's identity $ e^{i\pi} + 1 = 0 $ it is possible to include $ \varphi $ into an equation to give an identity containing six constants as follows: $$ e^{\frac{i\pi}{1+\varphi}} + e^{-\frac{i\pi}{1+\varphi}} + e^{\frac{i\pi}{\varphi}} + e^{-\frac{i\pi}{\varphi}} = 0 $$ See article and OEIS Sequence A193537
If we are allowing non-closed-form expressions then we have the following infinte series representations:
From Cloitre, Borwein and Chamberland a BPP formula in a $\verb/non-integer base/$
$$\pi^2=50\sum_{k=0}^\infty{1 \above 1.5pt \phi^{5k}}\Bigg({\phi^{-2}\above 1.5pt (5k+1)^2 } -{\phi^{-1}\above 1.5pt (5k+2)^2 }- {\phi^{-2}\above 1.5pt (5k+3)^2 }+ {\phi^{-5}\above 1.5pt (5k+4)^2 } +{2\phi^{-5}\above 1.5pt (5k+5)^2 } \Bigg)$$
There is also the following:
$$\phi=2\sum_{n=0}^\infty(-1)^{n}{\left(\frac{\pi}{5}\right)^{2n} \above 1.5pt (2n)!}$$
; which is play on a well known infinite series for the square root of $2.$ In particular just consider numbers written as $2\cos(\frac{\pi}{k})$ and take $k=5.$
I have found this very close to $\phi$:
$\phi \sim \dfrac{e^{e^2}}{\left(\pi^2-\dfrac{10e^2-16e-4}{6e^2-116e+69}\right)^3}$ and the error is close to $3.13\times 10^{-10}$
An approximation with an accuracy similar to that of $\pi\approx3$ (error<5%) is given by the sixth root of Gelfond's constant,
$$e^{\frac{\pi}{6}}\approx \phi$$
with rational term series
$$e^{\frac{\pi}{6}}=\sum_{k=0}^{\infty}\frac{\left(e^{\frac{\pi}{2}} - (-1)^k e^{-\frac{\pi}{2}}\right)\Gamma\left(\frac{k+i}{2}\right)\Gamma\left(\frac{k-i}{2}\right)}{4\pi k!}$$
Here's one I've found: $$\phi\approx\frac e\pi\left(\frac1{\sqrt3}+\frac1{\sqrt{\pi^{\sqrt{163}-\sqrt\pi}}}\right)+\sqrt{\frac{e^{\sqrt[3]{\pi}+1}}{3\pi}}=1.618131648\cdots$$
And here's some more interesting near-integer identities...
This is an approximation for $\pi$ by Ramanujan $$\pi\approx\frac95+\sqrt\frac95 = \frac{9+3\sqrt5}5 = \frac65(\varphi+1) = \frac65\varphi^2$$ The expression give us $$\varphi\approx \sqrt{\frac56\pi} = 1.6180\color{red}{2159}\dots\quad(E\sim 1.24\cdot10^{-5})$$
I've found this xkcd on random approximations. It gives, among others $$\sqrt5\approx\frac2e+\frac32\quad\text{and}\quad\sqrt5\approx\frac{13+4\pi}{24-4\pi}$$ These expression give us $$\begin{aligned} \varphi&\approx\frac1e+\frac54 ~=~ 1.61\color{red}{7879}\dots&&(E\sim1.55\cdot 10^{-4})\\ &\approx\frac{37}{8(6-\pi)} = 1.6180339\color{red}{0472}\dots&&(E\sim8.40\cdot10^{-8}) \end{aligned}$$
This one was found with the help of Wolfram Mathematica. $$\begin{aligned} \frac{24499\pi-17844e}{5599\pi} &= 1.618033988749894\color{red}{9426}\dots\quad(E\sim 9.44\cdot10^{-17}) \end{aligned}$$
This one is not an identity for the Golden Ratio, but it does link it to $\pi$ and $e$, as well as Pascal's triangle:
$\frac{e-\frac{1}{e}}{2}=\frac{\pi^2}{3!}-\frac{\pi^4-3\pi^2}{5!}+\frac{\pi^6-5\pi^4+6\pi^2}{7!}-\frac{\pi^8-7\pi^6+15\pi^4-10\pi^2}{9!}+\cdots$
The connection to the Golden Ratio can be seen from the fact that, ignoring the negative signs, the coefficients are those of the Fibonacci polynomials. They can thus be read from Pascal's triangle:
It follows from combining Euler's identity with the following:
$e^x=\dfrac{\sum\limits_{k=1}^\infty\frac{F_k(x)}{k!}}{\sum\limits_{k=1}^\infty{\left(-1\right)^{k-1}\cdot \frac{F_k(x)}{k!}}}$
