Four π's to make any integer

Question

This is like the "four four's" puzzle.

The challenge is to represent any integer using $\pi$,$\pi$,$\pi$,$\pi$ (uses $\pi$ exactly $4$ times).

You can use common functions on a calculator:

• Normal arithmetic operations $+ - * /$
• Square root $\surd$
• Exponential $(X^Y)$
• Negative() or minus sign $-$
• Factorial

• $\log_{10}$ or $\ln$

• Trigonometric functions $\sin, \cos, \tan$

• Parentheses $()$

You cannot use

• Floor() or ceiling () functions
• More than four $\pi$

This is similar to $\pi$ Day puzzle one to twenty, but you are allowed to use logarithms and factorials.

I have written method in the answer section. Is there any more efficient or clever method? Is there a formula using 3 $\pi$'s?

Clarification: I'm looking for creative answers outside of the rules specified above too.

Disclosure: I run the YouTube channel MindYourDecisions and am working on a video. I will credit solutions and link to this thread.

Answer 1

If arc-functions (inverse trigonometrics) are allowed:

Like in this answer, we can manage to do this using only 1 $\pi$, using the formula $$N=\tan\underbrace{\arcsin\cos\arctan\cos}_{N^2\text{ times}}\,\sin\pi$$ (copied from that answer; we only substituted $0$ with $\sin \pi$ (we might use $\tan \pi$ or so)).

Answer 2

The case of $0$ is trivial, for example:

0 = $\pi + \pi - \pi - \pi$

For a positive integer $n$, we can write:

$n = -\ln\left(\frac{\ln {\sqrt{\cdots{\sqrt{\pi}}}}}{\ln \pi}\right)\frac{1}{{\ln(-\cos(\pi) - \cos(\pi))}}$

Note:

For the repeated square roots, there are a total of $n$ square roots.

And in case I made any typos, you can check the formula for $n = 5$ written on WolframAlpha.

For a negative integer $n$, we modify the formula by omitting the minus sign at the beginning.

$n = \ln\left(\frac{\ln {\sqrt{\cdots{\sqrt{\pi}}}}}{\ln \pi}\right)\frac{1}{{\ln(-\cos(\pi) - \cos(\pi))}}$

Further explanation: the formula is modified from Paul Dirac's version for $2$'s.

Answer 3

If $\exp$ is allowed, here's a solution using only two $\pi$s.

To represent any non-negative integer $n$, let $w=\exp(-\cos(\pi))$, $u=\sqrt{...\sqrt{w}}$ ($n$ layers of $\sqrt{...}$), $s=\log(\sqrt{\exp(-\cos(\pi))})$, $n=\log(\log(u))/\log(s)$. It's easy to see that $w=e$, $u=e^{2^{-n}}$, $s=1/2$. Negate this method for negative $n$.

Just for fun, if $floor$ and $exp$ are allowed, only one $\pi$ is enough.

We only need to represent any non-negative integer $n$. First, we can get an integer $n' \geq n$, by $\lfloor \pi \rfloor!!!...!$. Then, we can try to get $x-1$ from $x$. We have $\lfloor \log(\log(\sqrt{\exp(\exp(x))}))\rfloor=\lfloor x-\log(2) \rfloor=x-1$.

Answer 4

I have to share this creative solution outside of the rules since I haven't seen this kind of formula written on any site.

Hannes W commented on my video with a method to make any number using just a single $\pi$ as follows:

$-\log\sqrt{(-\cos \pi))\%\%\cdots\%}$

Note:

There are a total of $n$ percentage symbols, and the logarithm is base $10$.

Answer 5

I also thought of Dirac's formula. Here's a formula for an arbitrary fraction:

$$\pm {m \over n} = \pm { \ln { \ln \sqrt { \sqrt { \cdots \sqrt {\pi}}} \over \ln \pi } \over \ln { \ln \sqrt { \sqrt { \cdots \sqrt {\pi}}} \over \ln \pi } }$$

where first thingy has m nested square roots the second thingy has n nested square roots.