No equations...need one line answer only.

No tricks here.. $e$, $\pi$, $\phi$ are regularly used universal constants.

Here is the one line answer:

Starting with $e$ and $\pi$ as Fibonacci seeds, do hundred iterations and take the ratio of last two terms to get to $\phi$.

If you want, you can do more to improve accuracy to multiple decimal places.


closed as off-topic by Omega Krypton, El-Guest, Glorfindel, Brandon_J, Jeff Zeitlin May 31 at 12:29

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  • $\begingroup$ No equations please..just one sentence answer..can be elegantly explained in about 15 words or less.. $\endgroup$ – Uvc May 31 at 11:06
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    $\begingroup$ I am really puzzled by downvotes..puzzle is explicitly stated..very concise..no equations needed..simple understanding of mathematical process is all you need to give the answer!! $\endgroup$ – Uvc May 31 at 11:38
  • $\begingroup$ Is this right and can it be expressed in words? $\phi = e^{i \pi / 5} + e^{-i \pi / 5} $ $\endgroup$ – Weather Vane May 31 at 12:54
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    $\begingroup$ Yeah: that's not exact. The ratio of terms from a Fibonacci series oscillates around $phi$ with alternate terms above and below, only being $phi$ at the infinite term. $\endgroup$ – Weather Vane May 31 at 14:33
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    $\begingroup$ So, the intended answer is equivalent to "Starting with $e$ and $\pi$, chuck both in the dumpster. Do Fibonacci to get the answer." How very elegant. $\endgroup$ – Bass May 31 at 23:20


The product of e square and pi cube is approximately equal to six hundred divided by phi cube.

In formula

$$\pi \approx \bigg(\frac{600}{(e\phi)^2}\bigg)^{1/3}$$
EDIT: $$e^2\pi^3 \approx \frac{600}{\phi^2}$$

Found @


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    $\begingroup$ Not the one I am looking for..you start with e and pi..end up with phi $\endgroup$ – Uvc May 31 at 11:22
  • $\begingroup$ @UVC Is it fine now? $\endgroup$ – Ak19 May 31 at 11:28
  • $\begingroup$ No..you start with e and Pi and end up with Phi..that process can be stated in one sentence $\endgroup$ – Uvc May 31 at 11:30
  • $\begingroup$ This is accurate only to 3 decimal places. $\endgroup$ – Gareth McCaughan May 31 at 11:35
  • $\begingroup$ @Gareth..if we are thinking on the same lines, it should be lot more accurate after 20 steps or so...let me do some quick calculations and will get back to you $\endgroup$ – Uvc May 31 at 14:23

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