Disguised forms of 2

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Disguised forms of 2

Postby Dopefish » Tue Oct 12, 2010 11:43 pm UTC

Ello people,

I've recently found myself in a situation where it would amuse me to come up with as many forms of 2 (or at least approximately, say, to two decimal places) as possible, that people looking at it wouldn't realise it was 2 unless they actually calculated it, but at the same time not being obviously 'constructed' i.e. not tan(1.107148) and similar with long decimal components. Using radicals instead of those decimal parts is acceptable however. Additionally, I encourage the use of pi, e, and complex numbers, as well as whatever else you folks can come up with.

For example, I've found that cosh(3^(1/4)) is approximately 2, and it looks innocent enough. I'm fairly confident that there's bound to be a couple forms involving logs and lns, but playing around I haven't been able to find anything that produces either a number close enough, or that can be expressed simply enough.

Extra points for things that are exactly equal to 2, without looking like it.

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Re: Disguised forms of 2

Postby Qaanol » Wed Oct 13, 2010 12:18 am UTC

[math]x = \sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}\neq 0[/math]
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Re: Disguised forms of 2

Postby NathanielJ » Wed Oct 13, 2010 12:20 am UTC

Try playing with WolframAlpha's closed-form generator. Enter something like 2.0012 into it and see what comes up at the bottom. For example:

http://www.wolframalpha.com/input/?i=2.0012

Comes up with 44/(7pi) and 2/(3 log^3(2)). Change the 2.0012 to 1.9987 (for example) and you'll get a bunch more. Repeat.
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Re: Disguised forms of 2

Postby phlip » Wed Oct 13, 2010 12:30 am UTC

[math]\int_0^{\infty}{e^{-x}dx}-e^{i\pi}[/math]
[math]\frac{33-3}{3\cdot3\cdot3-3\cdot3-3}[/math]
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Re: Disguised forms of 2

Postby squareroot » Wed Oct 13, 2010 12:35 am UTC

This requires using a different branch of the logarithm than you usually would, but

[math]\frac{log_i (e^{e i \pi})}{e}[/math]

is a pretty annoying one, at least to me.
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Re: Disguised forms of 2

Postby Sizik » Wed Oct 13, 2010 12:56 am UTC

([1;2,2,2,...])2

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Re: Disguised forms of 2

Postby Dopefish » Wed Oct 13, 2010 1:02 am UTC

NathanielJ wrote:Try playing with WolframAlpha's closed-form generator. Enter something like 2.0012 into it and see what comes up at the bottom. For example:

http://www.wolframalpha.com/input/?i=2.0012

Comes up with 44/(7pi) and 2/(3 log^3(2)). Change the 2.0012 to 1.9987 (for example) and you'll get a bunch more. Repeat.


Ah, I knew I'd seen something somewhere that could 'guess' closed forms like that, that'll make things a bit easier to get approximate things.

Some good ones (that I'm assuming are exactly 2) as well. I particularly like the

[math]\frac{log_i (e^{e i \pi})}{e}[/math]

one, as it's got all sorts of goodies to it that have nothing to do with 2.

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Re: Disguised forms of 2

Postby Qaanol » Wed Oct 13, 2010 5:28 am UTC

Let x be any even prime.
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Re: Disguised forms of 2

Postby Mike_Bson » Wed Oct 13, 2010 5:35 am UTC

$$π(prime_2)$$

All I've got.

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Re: Disguised forms of 2

Postby Eebster the Great » Wed Oct 13, 2010 6:09 am UTC

sec(arctan(sec(arctan(sec(arctan(sec(arctan(0)))))))) = 2

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Re: Disguised forms of 2

Postby skeptical scientist » Thu Oct 14, 2010 2:45 am UTC

phlip wrote:[math]\frac{33-3}{3\cdot3\cdot3-3\cdot3-3}[/math]

I had no idea two had so many threes in it!

This and Eebster's are my favorite.
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Re: Disguised forms of 2

Postby BlackSails » Thu Oct 14, 2010 3:15 am UTC

~ 137*(electron charge)^2/((4*pi*permittivity of free space)*planck's constant*speed of light)

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Re: Disguised forms of 2

Postby phlip » Thu Oct 14, 2010 4:15 am UTC

BlackSails wrote:~ 137*(electron charge)^2/((4*pi*permittivity of free space)*planck's constant*speed of light)

Really?

Also:[math]\sqrt[5]{\frac{5!}{5}+\frac{(55-5)(5\cdot5-5)}{5\cdot5\cdot5}}[/math]

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Re: Disguised forms of 2

Postby WarDaft » Thu Oct 14, 2010 4:16 am UTC

[imath]max(x=\displaystyle\sum_{n=0}^{∞}{\frac{1}{x^{n}}})[/imath]


... the notation to get the sup/subscripts above/below the sigma no longer eludes me.
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Re: Disguised forms of 2

Postby phlip » Thu Oct 14, 2010 4:18 am UTC

WarDaft wrote:... the notation to get the sup/subscripts above/below the sigma eludes me for the moment.

You got the notation right, it just only does it that way if you use [math], not [imath]. The inline one is twerked to be not as tall, so it flows better inline.

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Re: Disguised forms of 2

Postby Mike_Bson » Thu Oct 14, 2010 4:19 am UTC

phlip wrote:
BlackSails wrote:~ 137*(electron charge)^2/((4*pi*permittivity of free space)*planck's constant*speed of light)

Really?

Also:[math]\sqrt[5]{\frac{5!}{5}+\frac{(55-5)(5\cdot5-5)}{5\cdot5\cdot5}}[/math]

Does that only work with 5s?

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Re: Disguised forms of 2

Postby WarDaft » Thu Oct 14, 2010 4:20 am UTC

phlip wrote:
WarDaft wrote:... the notation to get the sup/subscripts above/below the sigma eludes me for the moment.

You got the notation right, it just only does it that way if you use [math], not [imath]. The inline one is twerked to be not as tall, so it flows better inline.


No, there's some command that will do it in imath too, but I can't remember it for the life of me.

Edit: Aha! Got it.

Also..
phlip wrote:
BlackSails wrote:~ 137*(electron charge)^2/((4*pi*permittivity of free space)*planck's constant*speed of light)

Really?

I had wondered if they meant something else, but no, that's not it either.
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Re: Disguised forms of 2

Postby phlip » Thu Oct 14, 2010 4:24 am UTC

Mike_Bson wrote:Does that only work with 5s?
I think so... I try to avoid the ones that work with any number, like [imath]\frac{n+n}{n}[/imath], 'cause they're boring. I notice I do use [imath]\frac{nn-n}{n}[/imath] to get 10 a fair bit, though... but only as part of a larger thing.

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Re: Disguised forms of 2

Postby BlackSails » Thu Oct 14, 2010 4:37 am UTC

Sorry, its the reduced planck's constant, and should be 2*pi in the denominator.

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Re: Disguised forms of 2

Postby skeptical scientist » Thu Oct 14, 2010 8:03 am UTC

Let me try some of those, phlip.

$$\sqrt[6]{\frac{6666}{66}-\frac{6\cdot 6\cdot 6+6}{6}}$$
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Re: Disguised forms of 2

Postby kjsharke » Thu Oct 14, 2010 2:54 pm UTC

BlackSails wrote:Sorry, its the reduced planck's constant, and should be 2*pi in the denominator.


I like that one, but you can cancel those things, no?
137*(electron charge)^2/((permittivity of free space)*planck's constant*speed of light)

Looking at triangles and obfuscating cosine:
[math]\frac{1}{e^{i*\arctan{\sqrt{15}}}+e^{-i*\arctan{\sqrt{15}}}}[/math]

edit -- if you want pi as well, we can do sine instead:

[math]\frac{i}{e^{i(\pi/2-\arctan{\sqrt{15}})}-e^{i(\arctan{\sqrt{15}}-\pi/2)}}[/math]

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Re: Disguised forms of 2

Postby pizzazz » Thu Oct 14, 2010 8:44 pm UTC

[math]\frac{log_i (e^{e i \pi})}{e}[/math]

one, as it's got all sorts of goodies to it that have nothing to do with 2.


This doesn't make any sense to me...
[math]e^{e i \pi}[/math] becomes [math]e^e * e^{i \pi}=-e[/math]
But how can you take a base i logarithm of a number whose absolute value is greater than one?

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Re: Disguised forms of 2

Postby mike-l » Thu Oct 14, 2010 8:47 pm UTC

pizzazz wrote:This doesn't make any sense to me...
[math]e^{e i \pi}[/math] becomes [math]e^e * e^{i \pi}=-e[/math]
But how can you take a base i logarithm of a number whose absolute value is greater than one?


Since when is [imath]a^{bc} = a^ba^c[/imath]? In fact, the rule [imath]a^{bc} = (a^b)^c[/imath] only holds on the reals anyway.

Also, [imath]i^i[/imath] is real, though multivalued. So log base i makes sense (though also being multivalued)
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Re: Disguised forms of 2

Postby Dopefish » Thu Oct 14, 2010 9:29 pm UTC

Do the usual change of base rules work with log base i?

That is, does logi(a)=ln(a)/ln(i)?

If so I know how to actually deal with log base i, despite it being a strange base.

eaeb=e(a+b) for the benefit of those temperarely forgetting how to deal with their exponents.


Anyway, some cool stuff showing up. The simpler (in the sense of less writing, not necessarily easier math), the better.

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Re: Disguised forms of 2

Postby pizzazz » Fri Oct 15, 2010 4:04 am UTC

mike-l wrote:
pizzazz wrote:This doesn't make any sense to me...
[math]e^{e i \pi}[/math] becomes [math]e^e * e^{i \pi}=-e[/math]
But how can you take a base i logarithm of a number whose absolute value is greater than one?


Since when is [imath]a^{bc} = a^ba^c[/imath]? In fact, the rule [imath]a^{bc} = (a^b)^c[/imath] only holds on the reals anyway.

Also, [imath]i^i[/imath] is real, though multivalued. So log base i makes sense (though also being multivalued)


Wow, I feel stupid. So how would you evalute that expression? I saw something on wikipedia about it not being well-defined but couldn't really follow it.

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Re: Disguised forms of 2

Postby KrO2 » Fri Oct 15, 2010 4:41 am UTC

squareroot wrote:[math]\frac{log_i (e^{e i \pi})}{e}[/math]


In limerick form:

Take the logarithm base i
Of e to the i times e pi
Divided by e
And soon you will see
That the answer is quite simplified.

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Re: Disguised forms of 2

Postby mike-l » Fri Oct 15, 2010 2:05 pm UTC

Wow, I feel stupid. So how would you evalute that expression? I saw something on wikipedia about it not being well-defined but couldn't really follow it.


Well, it's equal to
[math]\frac{ei\pi}{e\ln i}=\frac{i\pi}{\ln i}[/math]
Now since [imath]e^\frac {i\pi}2=i[/imath], a possible value of ln i is [imath]\frac {i\pi}2[/imath] which makes the above expression 2!
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Re: Disguised forms of 2

Postby jestingrabbit » Fri Oct 15, 2010 2:32 pm UTC

Another way to go is

[math]\frac{\log_i (e^{e i \pi})}{e} = \log_i \left(\exp\left(\frac{e i \pi}{e} \right) \right) = \log_i (\exp(i \pi)) = \log_i(-1)=2.[/math]
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Re: Disguised forms of 2

Postby tckthomas » Tue Oct 19, 2010 6:57 am UTC

[imath]\sqrt[8]{7*7*7-7*7-7-133_4}[/imath]

I fail at trying 7...

EDIT: YAY! [imath]\sqrt[3]{\frac{\frac{\frac{7!}{7*7-7}}{7}}{2\frac{1}{7}}}[/imath]

EDIT2: primes in sequence [imath](2*3^5/(7*11)*13-17-19)/23 = 2\frac{4}{1771}[/imath]

EDIT3: Also, 10

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Re: Disguised forms of 2

Postby ++$_ » Tue Oct 19, 2010 8:20 am UTC

[math]\sum_{0 \le a_1 < a_2 < \dots < a_7 \le 104 \atop (a_j, 105) = 1\ \forall j}\exp\left(2\pi i(a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7) \over 105\right)[/math]

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Re: Disguised forms of 2

Postby Blatm » Tue Oct 19, 2010 11:52 pm UTC

[math]\sqrt{\frac{-1}{\sum_{n=1}^{\infty}(-1)^nn}}[/math]
[math]\sqrt[5]{5\cdot5+5+\sqrt[5]{5\cdot5+5+\sqrt[5]{5\cdot5+5+\sqrt[5]{\dots}}}}[/math]
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Re: Disguised forms of 2

Postby Sagekilla » Wed Oct 20, 2010 4:11 am UTC

[math]\frac{4}{p^2 + p} \sum_{n = 1}^{p} \int_{-\infty}^{\infty} \delta(\frac{x}{n}) dx[/math]

Somewhat simple, but I like it.
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Re: Disguised forms of 2

Postby SunAvatar » Wed Oct 20, 2010 4:33 am UTC

[math]1+\frac{1}{1+\frac{1}{194+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{194+\frac{1}{1+\frac{1}{2+\frac{1}{\cdots}}}}}}}}}[/math]
This is slightly under 2, as opposed to slightly over, but I think it's still admissible.
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Re: Disguised forms of 2

Postby andy11235 » Wed Oct 20, 2010 4:48 am UTC

[math]\frac{e^\pi-\pi}{10}[/math]

Exactly 2; the difference is due to rounding errors.

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Re: Disguised forms of 2

Postby chapel » Wed Oct 20, 2010 5:26 am UTC

[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}}}}[/math]

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Re: Disguised forms of 2

Postby Eebster the Great » Wed Oct 20, 2010 7:09 am UTC

chapel wrote:[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}}}}[/math]

This actually has two real solutions (one being exactly 2 and the other about 4) and infinite complex solutions.

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Re: Disguised forms of 2

Postby SunAvatar » Wed Oct 20, 2010 7:35 am UTC

[math]\int_{0}^{\sqrt[3]{e^3 - e^{-3}}}\frac{x^2}{x^3+e^{-3}}dx[/math]
Last edited by SunAvatar on Wed Oct 20, 2010 7:55 am UTC, edited 1 time in total.
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Re: Disguised forms of 2

Postby taemyr » Wed Oct 20, 2010 7:43 am UTC

SunAvatar wrote:[math]\int_{0}^{\sqrt[3]{e^3 - e^{-3}}}\frac{x^2}{x^3+e^{-6}}dx[/math]

-1?

[url=http://www.wolframalpha.com/input/?i=integrate+from+0+to+%28e^3-e^%28-3%29%29^%281%2F3%29+x^2%2F%28x^3%2Be^%28-6%29%29]WA[/url]

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Re: Disguised forms of 2

Postby SunAvatar » Wed Oct 20, 2010 7:55 am UTC

Fixed. That 6 was supposed to be a 3.
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Re: Disguised forms of 2

Postby jaap » Wed Oct 20, 2010 8:04 am UTC

Eebster the Great wrote:
chapel wrote:[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}}}}[/math]

This actually has two real solutions (one being exactly 2 and the other about 4) and infinite complex solutions.

I don't think so. The equation [imath]x = \sqrt{2}^x[/imath], which the power tower must satisfy, has the real solutions 2 and (exactly) 4. That doesn't mean that the expression itself has two real possible values. Its value is defined as a limit of partial power towers, and that can only be one value (if it exists at all). That limit happens to be 2 rather than 4.


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