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 $\pi$ ?

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    $\begingroup$ $\sqrt e$ is not "very close" to $\phi$ $\endgroup$ – lhf Jul 29 '13 at 3:09
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    $\begingroup$ @lhf: I've corrected it. Funfact: It's more closer to $\phi$ than David Feinberg's lonely number. $\endgroup$ – Nick Jul 29 '13 at 10:56
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    $\begingroup$ @Dan: My mistake, sorry. I accidentally swapped a plus for a minus. It's a bit closer now. $\endgroup$ – Nick Jul 29 '13 at 10:58
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    $\begingroup$ @Nick I've had a go at retagging. There might be some more conceptual things appropriate, but this will do for now. $\endgroup$ – Sharkos Jul 29 '13 at 11:16
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    $\begingroup$ It is notable that $ \pi /2 \approx \phi $ $\endgroup$ – Mrigank Shekhar Pathak Mar 6 '17 at 11:11

16 Answers 16


$e$ 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 $\phi$.

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    $\begingroup$ The other answers are cute, but this adds a lot more depth. Thanks for the contribution! $\endgroup$ – Beska Jul 29 '13 at 1:31
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    $\begingroup$ Bravo! I think the OP's underlying question is whether $\phi$ is fundamentally connected to $\pi$ and/or $e$ much in the same way as $\pi$ and $e$ are interconnected. The answer is - of course - "no, it's not", for the reasons outlined by this answer. $\endgroup$ – Euro Micelli Jul 29 '13 at 1:33
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    $\begingroup$ Would you consider $\varphi = e^{i\pi/5}+e^{-i\pi/5}$ "a nontrivial way"? I suppose it's a member of "$e^{i\pi}$ and its family".. $\endgroup$ – A. Rex Jul 29 '13 at 9:01
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    $\begingroup$ $+1$. I don't know why the above joke is upvoted more than this answer. $\endgroup$ – Mo_ Jul 29 '13 at 11:03
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    $\begingroup$ And I don't get, why is this answer upvoted at all. It is't even the answer. $\endgroup$ – gukoff Jul 29 '13 at 19:43

The following is exact. :-)


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    $\begingroup$ +1 So cute! ${}{}$ $\endgroup$ – Jyrki Lahtonen Jul 28 '13 at 21:14
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    $\begingroup$ when I saw this, I knew that moment, the course of my life is changing $\endgroup$ – user85461 Jul 28 '13 at 21:24
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    $\begingroup$ This might be the most upvoted goofy answer...nice!+1 $\endgroup$ – DonAntonio Jul 28 '13 at 21:27
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    $\begingroup$ This makes a great math joke! $\endgroup$ – zerosofthezeta Jul 29 '13 at 1:35
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    $\begingroup$ I should have expected an answer like this the moment I posed my question. $\endgroup$ – Nick Jul 30 '13 at 13:40

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:

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

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    $\begingroup$ Even though it comes from the fact that $\varphi = \frac{1+\sqrt{5}}{2}$, it's neat seeing it in this way. $\endgroup$ – Cameron Williams Jul 29 '13 at 3:17
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    $\begingroup$ This answer is related to the geometric fact if you form a star inside a regular pentagon, the ratio of the side of the star to that of the pentagon is $2\cos\frac{\pi}{5} = \varphi$. $\endgroup$ – achille hui Jul 29 '13 at 7:05
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    $\begingroup$ @CameronWilliams: I'm not sure what you mean, though of course that is the golden ratio. If you change the $5$s in my expression to $6$s, say, you don't get $(1+\sqrt{6})/2$. (In fact, you get $\sqrt{3}$.) $\endgroup$ – A. Rex Jul 29 '13 at 8:55
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    $\begingroup$ This is a great answer to a not particularly great question. 'Your idle curiosity is the gateway to deep and beautiful mathematics which you are capable of understanding' - the best thing a teacher can tell you. Sad that the 'funny' answer has more than three times as many upvotes $\endgroup$ – jwg Jul 29 '13 at 11:25
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    $\begingroup$ (+1) I didn't see your answer until after I posted, so I deleted mine. However, note that $$ \begin{align} \color{#C00000}{\left(e^{i\pi/5}+e^{-i\pi/5}\right)}^2 &=e^{i2\pi/5}+2+e^{-i2\pi/5}\\ &=\left(e^{i2\pi/5}+1+e^{-i2\pi/5}\right)+1\\ &=-\left(e^{i4\pi/5}+e^{-i4\pi/5}\right)+1\\ &=\color{#C00000}{\left(e^{i\pi/5}+e^{-i\pi/5}\right)}+1 \end{align} $$ $\endgroup$ – robjohn Jul 31 '13 at 0:42

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.

  • $\begingroup$ Good one but I think the OP was searching for a finite expression. $\endgroup$ – Paracosmiste Jul 29 '13 at 11:31
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    $\begingroup$ Any hints for why this works? $\endgroup$ – user7530 Jul 30 '13 at 8:49
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    $\begingroup$ @metacompactness: Don't go presuming such things. This is a beautiful answer. $\endgroup$ – Nick Jul 30 '13 at 13:06
  • $\begingroup$ @Nick you didn't specify if you wanted a finite expression or an infinite one but your two approximations suggest a finite expression. $\endgroup$ – Paracosmiste Jul 30 '13 at 15:11
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    $\begingroup$ If you want an approximate finite expression, then you can just cut off the continued fraction. $\endgroup$ – A. Rex Jul 31 '13 at 21:47

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:

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

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    $\begingroup$ Care to explain the downvote? $\endgroup$ – Mark K Cowan Jul 28 '13 at 22:43
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    $\begingroup$ This is a joke just as vadim's answer. $\endgroup$ – Paracosmiste Jul 29 '13 at 11:30
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    $\begingroup$ These "joke" solutions are perfectly correct. Easy math isn't wrong math. If you look at the original question, these answers are following all the rules. $\endgroup$ – Matt Jul 29 '13 at 17:10
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    $\begingroup$ @Matt: Sure, they are correct and are very much amusing but they can be simplified down to $(1+\sqrt 5)/2$ by even those who aren't that good in math. The best answer should be one that is both correct and isn't so intuitively understandable to the common man. $\endgroup$ – Nick Jul 30 '13 at 13:15
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    $\begingroup$ @Nick, why should it be good if something is not understandable by the common man? Is maths about creating secrets that only mathematicians understand at the end? No. It's about giving precise solutions to problems that we find interesting. The simpler the solution, the better. If the solution looks like a joke, then you phrased the question unprecise. $\endgroup$ – Turion Aug 2 '13 at 11:50

$\sqrt e \approx 1.64872$ 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...)

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    $\begingroup$ Nice to know about ries! $\endgroup$ – lhf Jul 31 '13 at 3:05
  • $\begingroup$ Neat!$\left . \right .$ $\endgroup$ – Thomas Aug 2 '13 at 11:57
  • $\begingroup$ On my machine, $e - \sqrt[3]\pi$ is about 1.254, which doesn't seem like a very good approximation. $\endgroup$ – Tanner Swett Aug 2 '13 at 21:40
  • $\begingroup$ @TannerSwett: It's possible I did the conversion wrong as I typed it up. As you can see from the commands I actually asked RIES to find $e$ not $\varphi$ and so I converted before posting. $\endgroup$ – Charles Aug 3 '13 at 0:54

An approximation: $$\phi \approx \frac { 7\pi }{ 5e } =1.618018$$

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    $\begingroup$ Good attempt. Surprisingly, this was one of the approximations I did initially. $\endgroup$ – Nick Jul 30 '13 at 13:32
  • $\begingroup$ J. DePompeo (2004) wrote the equivalent $$\frac{5\phi e}{7\pi}\approx 1$$ (formula 34 in mathworld.wolfram.com/AlmostInteger.html) $\endgroup$ – Jaume Oliver Lafont Jan 26 '16 at 10:43

Here is another suggestion:


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    $\begingroup$ yet another joke!! $\endgroup$ – Paracosmiste Jul 29 '13 at 11:30
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    $\begingroup$ The original are always better ;-) $\endgroup$ – Luc M Jul 29 '13 at 15:42

If you define the sequence $a_1 = a_2 = -e^{i\pi}$, $a_k = a_{k-1} + a_{k-2}$, then $\lim_{n \rightarrow \infty} \frac {a_{n+1}}{a_n} = \phi$.

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    $\begingroup$ You can even let $a_1 = e$ and $a_2 = \pi$ if you'd like. The initial values don't much matter as long as they're positive. $\endgroup$ – A. Rex Jul 29 '13 at 8:57
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    $\begingroup$ ...which boils down to the Fibonacci sequence, for anyone who is curious as to how it works! $\endgroup$ – Mark K Cowan Jul 31 '13 at 15:26

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]

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    $\begingroup$ This is a good approach! Check out my answer for how to turn this into an especially clean expression. $\endgroup$ – A. Rex Jul 30 '13 at 0:16

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


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

$$e^{\frac{\pi}{6}}=\sum_{k=0}^{\infty}\frac{\left(e^{\frac{\pi}{2}} - (-1)^k e^{-\frac{\pi}{2}}\right)\Gamma\left(\frac{k+i}{2}\right)\Gamma\left(\frac{k-i}{2}\right)}{4\pi k!}$$

Rational series representation of $e^\pi$


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...

  • $\begingroup$ This is similar to my initial aproach and those coincidences in the link are nice. What makes 163 special? $\endgroup$ – Nick Aug 1 '18 at 19:32
  • $\begingroup$ It's the largest Heegner number, and $\exp(\pi\sqrt{163})$ is very close to an integer. This has more 'coincidences'. $\endgroup$ – TheSimpliFire Aug 2 '18 at 7:22

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 $k=5.$


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