# Can we get just $3$ from $\pi$? [closed]

Today, a friend and I solved a question and one point came up where we were discussing whether we should write $(-1)^{n-1}$ or $(-1)^{n+1}$ and quickly we remembered that it was the same thing (for some $n \in \mathbb{Z}$). Then, because we're really cool, we started coming up with different variations of what would give the same answer, i.e you could write $(-1)^{n-3}$ or $(-1)^{n+5}$ or $(-1)^{n - (1 \mod 2)}$ etc and I tried writing it in terms of $\pi$ but I wasn't sure how to get rid of all the decimal places.

Without using $\pi$ again, is it possible to add/subtract/multiply or divide $\pi$ by something to get just $3$?

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## closed as off topic by Martin, Pedro Tamaroff, Ilya, Grigory M, Zev ChonolesMay 24 '13 at 13:24

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How accurate do you want it? Exact: $$3=\frac{3}{\pi}\left(\int_{-\infty}^{\infty}e^{-x^2}\operatorname dx\right)^2$$ Approximate: $$3\approx \frac{3\ln(68925893036109279891085639286943768)}{2\pi\sqrt{163}}$$ –  Jared May 23 '13 at 19:29
"Then, because we're really cool, we started coming up with different variations of what would give the same answer." –  Sujaan Kunalan May 23 '13 at 19:30
This is not number theory. –  Pedro Tamaroff May 24 '13 at 1:02
This is one of those cases in math where someone notices something interesting, hunts around for similar sorts of things, and picks the wrong one. I think the initial question, about what other ways there are to express the sequence $-1, 1, -1, 1, \dots$ is a much more interesting and productive one. A trivial way: $i^{2n+1}$. A deep way: the sum of a Fourier series. –  dfeuer May 24 '13 at 2:41
How about $|\{ n\in\mathbb{Z} \mid 0 < n < \pi\}|$ where $|\cdot |$ denotes the cardinality of a set? –  Jeppe Stig Nielsen May 24 '13 at 9:43

How about ... $$\lfloor{\pi}\rfloor$$

Or, perhaps ... $$-3 \cos(\pi)$$

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That's a great choice. –  hyg17 May 23 '13 at 19:17
We could always use $(-3\cos(\pi))$ –  amWhy May 23 '13 at 21:30
@Amzoti All my recent "number play" with identities evaluating to $+1$ left me nimble for the task ;-) –  amWhy May 24 '13 at 0:40
Heh, I couldn't read that first one without thinking of Homer Simpson. –  Joshua Taylor May 24 '13 at 1:54
While it certainly does equal three, cos isn't one of the functions he listed in his question. –  Daniel Ball May 24 '13 at 4:47

Pff.

$$f(\pi)$$

where $f(x) = 3$.

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Haha, nicely done! –  tahatmat May 24 '13 at 6:04

$$\sqrt{\dots \sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{6\pi-9}-9}-9}-9}-9}-9}-9}\dots }=3$$

(via iteration to find a fixed point in solving $(x-3)^2=0$.)

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This deserves an infinite number of votes. –  Soham Chowdhury May 24 '13 at 13:04
@DonAntonio: I don't know what you mean here. The task is to use $\pi$ and get 3. –  NikolajK May 24 '13 at 13:28
That's right, @NickKidman: once is fine, not more. In fact I did write the same comment to almost everybody here as I, and I think many, consider this thread as almost a goof. Anyway, if I succeed to understand why your expression equals three I'll upvote your answer. –  DonAntonio May 24 '13 at 13:30
@DonAntonio: I took $x^2-6x+9=0$ and expressed it as a fixedpoint problem: $f(x):=\sqrt{6x-9},\ f(x)==x$. We get e.g. $2\to f(2)=\sqrt{3}\to f(\sqrt{3})=\sqrt{6\sqrt{3}-9}\to f(\sqrt{6\sqrt{3}-9})=\dots$. If the function $f(x)$ fulfills some conditions (alla contraction), then it's "simple" to write a program, which solves the equation $(x-3)^2$, which I designed to have $x=3$ as a solution. In my expression $\pi$ is a good iteration starting value. –  NikolajK May 24 '13 at 13:40
Oh, my: that's just Banach's fixed point theorem. Nice! +1 –  DonAntonio May 24 '13 at 13:54
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This is fun: $$\left\lceil \frac{\pi}{\sqrt[\pi]{\pi}}\right\rceil=3$$

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I've not seen that before, what is it? Lol ceiling function? –  Kaish May 23 '13 at 22:38
–  vadim123 May 23 '13 at 22:55
Pi includes its own ceiling, and one that doesn't have a gaping hole in it, and has decent overhang. –  Kaz May 24 '13 at 2:38
This too uses $\;\pi\,$ more than once. –  DonAntonio May 24 '13 at 13:32

The 3333rd digit of $\pi$ is a 3.

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Also, the 33333rd digit of $\pi$. –  A Walker May 23 '13 at 22:10
Hi, i'm in Computer Science. So this means that pretty much the 333....333rd digit of pi is always 3, right? –  Kaz May 24 '13 at 0:10
@kaz probably not.. –  ratchet freak May 24 '13 at 1:02
Ah, a statistician! –  Kaz May 24 '13 at 2:20
Maybe it happens for infinitely many numbers of the form $\frac{10^n-1}{3}$. –  dot dot May 30 '13 at 14:11
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$$\frac{\pi}{\pi} + \frac{\pi}{\pi} + \frac{\pi}{\pi}$$

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+1 for one of the very few I understood on the page –  mplungjan May 24 '13 at 12:24
This too uses $\;\pi\,$ more than once. –  DonAntonio May 24 '13 at 13:34
@DonAntonio: How about $\mathrm{TripAddSelfDiv}(\pi)$, where $\mathrm{TripAddSelfDiv}(x)=\frac{x}x+\frac{x}x+\frac{x}x$ –  dimensio1n0 Oct 8 '13 at 13:59

Let $\pi$ be the prime-counting function.

Then $3 = \pi(5)$.

Or, to use $\pi$ twice, $3 = \pi(2\pi)$.

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Why not just let $\pi$ be $3$? –  augurar Jul 1 at 4:01
@augurar, because that would be boring. Also, the symbol $\pi$ is commonly used for the prime-counting function, but is not commonly used for the number $3$. –  dfeuer Jul 1 at 6:21

You can try $3\pi^0=3$ or $(\pi\times 0) + 3 = 3$ too :-)

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So far this seems to be the only one which only add/subtract/multiply/divided and didn't re-use pi. Go you! –  Daniel Ball May 24 '13 at 4:52

$$3 \cdot e^{2 k \pi i} ~ ~ ~ \text{for all} ~ k \in \mathbb{Z}$$

... maybe that is cheating.

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I prefer to simplify the formula by removing "magic numbers" such as 3: $e^{2k\pi i} + e^{2k\pi i} + e^{2k\pi i}$. Much better. –  Dietrich Epp May 24 '13 at 1:17
@DietrichEpp But now there's still a magic number: 2 $+$'s :p –  Thomas May 24 '13 at 4:38
Then do $e^{k\pi i + k \pi i}$ :P –  Soham Chowdhury May 24 '13 at 9:03
I can't even read that in Chrome –  mplungjan May 24 '13 at 12:23
@mplungjan: I can. Have you somehow disabled MathJaX? –  dimensio1n0 Oct 8 '13 at 14:01

$$3 \equiv \pi - 4 \times \left( \frac{1}{2 \times 3 \times 4} - \frac{1}{4\times 5\times 6} + \frac{1}{6\times 7\times 8} + ... \right) \\ = \pi + \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(n)(2n+1)(n+1)}$$

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Why not this?

$$3=\frac{\pi^2}{2\sum_{i=1}^{+\infty}\frac{1}{i^2}}$$

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Yes it follows from $\zeta(2)$ ,isn't it? –  Shivam Patel Oct 10 '13 at 10:30

$$\sqrt[5]{32}+\sin^2(3)+\cos^2(3)+\sin(\pi)$$

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you forgot $=3$. –  JMCF125 Jun 14 '13 at 15:05
No, I didn't. That's a term, not an equation. –  Asaf Karagila Jun 14 '13 at 15:29
Equation is a relation between two terms which is denoted by $=$. So writing one term (the above) and then $=3$ is exactly an equation. It's a statement about two terms, and its content is that these terms are equal. –  Asaf Karagila Jun 14 '13 at 15:35
No it is not. An equation is an equality relation between two algebraic expressions (at least one, the other may have no variables), expressions that have at least one variable in them. –  JMCF125 Jun 14 '13 at 15:38
Not all equations are in algebra precalculus level. Equation, generally, is a statement in some language which includes the symbol $=$ (which in modern logic, is part of the logical symbols anyway). –  Asaf Karagila Jun 14 '13 at 15:48

$\lfloor\lceil\lfloor\lceil\lfloor\lceil\lfloor\lceil\lfloor\lceil\lfloor \pi\rfloor + \pi\rceil - \pi\rfloor + \pi\rceil - \pi\rfloor + \pi\rceil - \pi\rfloor + \pi\rceil - \pi\rfloor + \pi\rceil - \pi\rfloor$...

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Either this one's a joke or else there're lots of irrelevant "floor' functions" ...or else I just don't understand. –  DonAntonio May 24 '13 at 13:10
@DonAntonio Yes. –  Bitrex May 24 '13 at 13:18
And besides this one too uses $\;\pi\,$ more than once. –  DonAntonio May 24 '13 at 13:34
@DonAntonio You've made that comment more than once. –  Bitrex May 24 '13 at 13:44
Yes..................... –  DonAntonio May 24 '13 at 13:55

You could have $\log_{\pi}(\pi \times \pi \times \pi)$. :)

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Also, it's not really right to say "it uses only $\pi$" when there's a logarithm in it. –  leftaroundabout May 24 '13 at 10:33
OK, OK, I get it. –  Soham Chowdhury May 24 '13 at 13:03
This too uses $\;\pi\,$ more than once. –  DonAntonio May 24 '13 at 13:35

Without using π again, is it possible to add/subtract/multiply or divide π by something to get just 3?

If you try such a thing, you would have to multiply by $3/\pi$, or divide by $\pi/3$ or add $3-\pi$ or minus $\pi-3$. These are all transcendental numbers, but there is an option.

Since $\pi=-i\ln\left(-1\right)$

So, you could multiply by $\frac{3i}{\ln\left(-1\right)}$, or divide by $-\frac{i}{3}\ln\left(-1\right)$ or add $\ln \left(\left(-1\right)^ie^3\right)$ or $-\ln \left(\left(-1\right)^ie^{-3}\right)$.

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As that equals pi, and you're using it, you're using a transcendental number that happens to equal pi, just with complex numbers in it to say it isn't transcendental. +1 nonetheless –  JMCF125 Jun 10 '13 at 21:54

It seems to me the simplest may be $3=\pi-(\pi - 3)$.

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OP asked: "Without using π again" –  dr jimbob May 23 '13 at 22:25
Sure thing. let $\epsilon = \pi$. Then, $\pi - (\epsilon - 3)$. :) –  Kaz May 24 '13 at 0:12
Well, some of the other answers aren't exactly sticking to "add/subtract/multiply or divide", nor do they all use $\pi$ only once. Just saying. –  Alex Petzke May 24 '13 at 3:03
@Kaz That's a pretty big $\epsilon$ :p –  Thomas May 24 '13 at 4:06
This too uses $\;\pi\,$ more than once. –  DonAntonio May 24 '13 at 13:33

Maybe that helps you playing aroung with the numbers some more $e^{i\pi} = - 1$.

Furthermore $-3e^{i\pi} = 3$.

EDIT

Of course $\lfloor \pi\rfloor = 3$.

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