Approximation of $e$ using $\pi$ and $\phi$?

$$e \approx \frac{4 \phi +3 \pi-5}{4}$$

where $~\phi~$ is a Golden ratio .

Is it possible to construct better approximation of $e$ using $\pi$ , $\phi$ and integers ?

Solution 1:

Yes : $$e \approx \pi+\phi +\frac{1-\pi\cdot\phi}2 \approx 2.718$$ $$e \approx \frac{-\phi+2\pi-1+2\pi^2}3-\pi\cdot\phi \approx 2.71825$$ $$e \approx \frac{6\phi-2\pi+2\pi\cdot\phi}5 \approx 2.71828$$

Should you accept only linear expressions : $$e \approx \frac{46\phi+173\pi-183}{160} \approx 2.718281828459$$ $$e \approx \frac{45483\phi-2961\pi-18765}{16748} \approx 2.718281828459045235360$$ In fact we may find approximations as precise as you want. But note that their expression is as long as the decimal expansion obtained!

(Answering Peter) Of course it is more funny to find such relations yourself so let's use the free and excellent pari/gp software (the .exe should be fine for Windows (*)) to get some other relations using the $\textrm{lindep}$ function. This function implements the PSLQ and LLL algorithm (the integer relation algorithms suggested by lhf) that is for a given vector $[x_1,x_2,\cdots,x_n],\ x_i \in \mathbb{R}$ it returns a vector $[a_1,a_2,\cdots,a_n],\ a_i\in \mathbb{Z}$ such that $\sum_{i=1}^n a_i\cdot x_i \approx 0$ (in fact to a given precision : 22 digits in our first example) :

\p 100 (100 digits precision should be enough)
phi=(sqrt(5)+1)/2
lindep([exp(1),phi,Pi,1],22)

returns $[-16748, 45483, -2961, -18765]~$ (with $\sum_{i=1}^n a_i\cdot x_i \approx 10^{-17}$)

%/16748

returns $[-1, 45483/16748, -2961/16748, -18765/16748]$ that is : $$e \approx \frac{45483\phi-2961\pi-18765}{16748} \approx 2.718281828459045235360$$

These algorithms are also implemented in Mathematica, Maple and others but let's continue with pari/gp and search this time a multiplicative answer (to a precision of 5 digits) :

lindep([1,log(phi),log(Pi)],5) (note that $1=\log(\exp(1))$)

returns [-7, -14, 12]~ or (divided by 7)
[-1, -2, 12/7]~

so that $-1 -2\log(\phi)+\frac{12}7\log(\pi) \approx 0$
or $1 \approx -2\log(\phi)+\frac{12}7\log(\pi)$
and the exponential of this becomes : $$e\approx \frac{\pi^{\frac{12}7}}{\phi^2}\approx 2.7182$$

You may even find expressions like these proposed by Robert Israel using :

lindep([1,phi,Pi,Pi^2,Pi^3,-exp(1),-exp(1)*phi,-exp(1)*Pi,-exp(1)*Pi^2,-exp(1)*Pi^3],{precision})

These very powerful methods have many applications in diophantine equations and cryptography and for discovering exact relations using 'experimental mathematics'!

But it's time to let you play with that too so : have fun!!

(*) On Windows you may encounter the typical problems with console applications (unnatural rectangular copy/paste and windows resize). To avoid frustrations with gp I would recommend to use Console2 and launch gp using a syntax like :
C:\tools\Console2\Console.exe -r "/k C:\tools\pari\gp.exe"
(supposing that Console2 and pari are in C:\tools).
If you want to use Ctrl-C and Ctrl-V as usual you'll have to change the default settings (right click/Edit/Settings/Hotkeys). A text may be selected with 'shift' down before mouse-drag.
Hoping all this will help you to appreciate this excellent tool (great for number theory but not only!). There is even a version of pari for android paridroid but I didn't try it...

Solution 2:

Of course. $ \phi \approx 1.618033989$ and $ e \approx 2.718281828$ so $\displaystyle e \approx \frac{2718281828}{1618033989} \phi .$ Extend to whatever accuracy you desire.

Edit in response to comment: If you require using $\pi$ as well, then we can do $$\displaystyle e \approx \frac{2718281828}{2\cdot 1618033989} \phi + \frac{2718281828}{2\cdot 3141692654} \pi.$$

Solution 3:

Try: $${\frac {16+31\,\phi+6\,\pi +25\,{\pi }^{2}-14\,{\pi }^{3}}{11+25\,\phi +4\,\pi -48\,{\pi }^{2}+12\,{\pi }^{3}}} $$