Happy Tau Day! - [General Solution Revealed]

\[\Huge{Happy}\text{ }\Huge{Tau}\text{ }\Huge{Day!}\]

Yes! It's Tau Day, the day we have all been waiting for (or at least, I have). For those who don't know, Tau is $2\pi$ . People support Tau because it makes more sense to use Tau when calculating with radians. This is because a fourth of a circle is a fourth of Tau, but only a half of $\pi$ .

Anyways, I have decided to put up a friendly challenge! I'm pretty sure you have heard of puzzles like the four nines puzzle. Now, I introduce to you: the four Taus puzzle!

Let's see how many whole numbers we can come up with only using four Taus and any mathematical symbols, as long as you don't include any other numbers. This is a friendly challenge and is not timed, to account for people living in other time zones.

Good Luck! 6.28318...

#Geometry

Easy Math Editor

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`2 \times 3`	$2 \times 3$
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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

I'll start the challenge:

$T^{T}-T^{T}=0$ $\log_{T}(T^{\frac{T}{T}})=1$ $\log_{T}(T)+\log_{T}(T)=2$

The rest are up to you!

Ved Pradhan - 11 months, 3 weeks ago

$\frac{T+T+T}{T} = 3$ and I think that's about it?

Elijah L - 11 months, 3 weeks ago

No, I think there's more if we think hard enough.

Ved Pradhan - 11 months, 3 weeks ago

Contest Midway Hint:

There is actually a formula to write any whole number using four Taus and basic symbols! I will reveal this way some time tomorrow, but until then, here is the solution for $4$ :

$\cos(T)+\cos(T)+\cos(T)+\cos(T)=4$

@Elijah L

Ved Pradhan - 11 months, 3 weeks ago

Right, I forgot about trigonometric functions.

Elijah L - 11 months, 3 weeks ago

Yes, you can use trigonometric functions. Let me clarify: you can use any function that would be available on a basic scientific calculator, along with unique base logarithms (so we don't waste time in change of base).

Ved Pradhan - 11 months, 3 weeks ago

Contest Progress:

Well, there hasn't been much progress from the Western Hemisphere. Let's hope that the Eastern Hemisphere has more up their sleeves! Meanwhile, here is a solution for $5$ :

$(\cos(T)+\cos(T)+\cos(T))!-\cos(T)=5$

As said before, I will reveal the general solution for any natural number sometime tomorrow. Until then, good luck!

Ved Pradhan - 11 months, 3 weeks ago

Then we can easily make 6, 7, 24, 719, 720, and 721.

Elijah L - 11 months, 3 weeks ago

We can make 47, 48, and 49 in the following manner:

$47 = ((\cos(T) + \cos(T) + \cos(T))!)!! - \cos(T)$

$48 = ((\cos(T) + \cos(T) + \cos(T))!)!! + \sin(T)$

$49 = ((\cos(T) + \cos(T) + \cos(T))!)!! + \cos(T)$

This is by using the double factorial function. Similarly, 17, 18, and 19 can be achieved with triple factorial, and 11, 12, 13 with quadruple factorial...

Elijah L - 11 months, 3 weeks ago

@Elijah L : You are correct! Because $\cos(T)=1$ , a lot of values can be made by combining the factorial functions. However, they skip so many numbers in between. Because you've gotten plenty of values so far, maybe you can try to see how to generalize it for any natural value of $n$ ? If you get stuck, you can continue finding values that cannot simply be done by composing the factorial function.

Ved Pradhan - 11 months, 3 weeks ago

I love International $\pi$ Day, but now I also love Tau Day!

Thanks for spreading Tau Day into my heart, brain, soul and conscience, @Ved Pradhan!

Yajat Shamji - 11 months, 3 weeks ago

@Yajat Shamji : You're welcome!

Ved Pradhan - 11 months, 3 weeks ago

Hello, everyone!

Thanks for everyone who participated in this challenge! If you learned something about Tau or math in general, I would consider this a success!

Now, the suspense is over. Here is the general solution to find any natural number using four Taus!

For any natural number $n$ :

$-\log_{(\cos(T)+\cos(T))}(\log_{T}(\sqrt{\sqrt{\cdots\sqrt{T}}}))=n$

where there are $n$ square roots inside of the second logarithm. Try to prove why this is true by yourself. If you get stuck, look at Mind Your Decision's video on four $\pi$ s. I will post a link in a reply to this post.

Once again, I would like to thank @Elijah L and @Yajat Shamji for participating in this friendly challenge. I hope you had fun and learned something! Thanks!

Ved Pradhan - 11 months, 3 weeks ago

Here is the link to Mind Your Decision's video on four $\pi$ s: https://www.youtube.com/watch?v=tXunUuvx6_c. If you get stuck proving the general solution, look at this video, then try to adapt it to Tau instead of $\pi$ .

Ved Pradhan - 11 months, 3 weeks ago

Oh, it looks like the mentions didn't work. @Elijah L @Yajat Shamji

Ved Pradhan - 11 months, 2 weeks ago

Though the answer has been revealed, I will post a solution for 2020.

$\pi\left(\tau\ln\left(\Gamma\left(\text{exp}(\tau)\right)\right)-\Gamma(\tau)+\ln\left(\sqrt{\text{exp}\left(\sqrt{\text{exp}(\tau)}\right)}\right)\right)=2020$

( $\pi(x)$ is the prime-counting function)

X X - 11 months, 2 weeks ago

Wow, @X X, that's so cool! I've never heard of the prime counting function. I'm going to search that up right now. That's a very creative solution!

Ved Pradhan - 11 months, 2 weeks ago

The floor function popped up in my mind at first, since it guarantees the outcome would be an integer, but that is a bit uninteresting. So I thought of $\pi(x)$ , which is the number of primes less than or equal to $x$ , since it also guarantees an integer outcome.

The general solution is also great!

X X - 11 months, 2 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$