Disguised forms of 2
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Disguised forms of 2
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.
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.
 NathanielJ
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Re: Disguised forms of 2
Try playing with WolframAlpha's closedform 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.
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.
Last edited by NathanielJ on Wed Oct 13, 2010 12:38 am UTC, edited 1 time in total.
 phlip
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Re: Disguised forms of 2
[math]\int_0^{\infty}{e^{x}dx}e^{i\pi}[/math]
[math]\frac{333}{3\cdot3\cdot33\cdot33}[/math]
[math]\frac{333}{3\cdot3\cdot33\cdot33}[/math]
Last edited by phlip on Wed Oct 13, 2010 12:37 am UTC, edited 1 time in total.
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Re: Disguised forms of 2
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.
[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
([1;2,2,2,...])^{2}
Hiding in plain sight!
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she/they
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Re: Disguised forms of 2
NathanielJ wrote:Try playing with WolframAlpha's closedform 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.
Re: Disguised forms of 2
$$π(prime_2)$$
All I've got.
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 Eebster the Great
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Re: Disguised forms of 2
sec(arctan(sec(arctan(sec(arctan(sec(arctan(0)))))))) = 2
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Re: Disguised forms of 2
phlip wrote:[math]\frac{333}{3\cdot3\cdot33\cdot33}[/math]
I had no idea two had so many threes in it!
This and Eebster's are my favorite.
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 BlackSails
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Re: Disguised forms of 2
~ 137*(electron charge)^2/((4*pi*permittivity of free space)*planck's constant*speed of light)
 phlip
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Re: Disguised forms of 2
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{(555)(5\cdot55)}{5\cdot5\cdot5}}[/math]
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Re: Disguised forms of 2
[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.
... the notation to get the sup/subscripts above/below the sigma no longer eludes me.
Last edited by WarDaft on Thu Oct 14, 2010 4:25 am UTC, edited 2 times in total.
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 phlip
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Re: Disguised forms of 2
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
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{(555)(5\cdot55)}{5\cdot5\cdot5}}[/math]
Does that only work with 5s?
Re: Disguised forms of 2
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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 phlip
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Re: Disguised forms of 2
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{nnn}{n}[/imath] to get 10 a fair bit, though... but only as part of a larger thing.Mike_Bson wrote:Does that only work with 5s?
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 BlackSails
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Re: Disguised forms of 2
Sorry, its the reduced planck's constant, and should be 2*pi in the denominator.
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Re: Disguised forms of 2
Let me try some of those, phlip.
$$\sqrt[6]{\frac{6666}{66}\frac{6\cdot 6\cdot 6+6}{6}}$$
$$\sqrt[6]{\frac{6666}{66}\frac{6\cdot 6\cdot 6+6}{6}}$$
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Re: Disguised forms of 2
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]
Re: Disguised forms of 2
[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?
Re: Disguised forms of 2
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
Do the usual change of base rules work with log base i?
That is, does log_{i}(a)=ln(a)/ln(i)?
If so I know how to actually deal with log base i, despite it being a strange base.
e^{a}e^{b}=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.
That is, does log_{i}(a)=ln(a)/ln(i)?
If so I know how to actually deal with log base i, despite it being a strange base.
e^{a}e^{b}=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.
Re: Disguised forms of 2
mikel 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 welldefined but couldn't really follow it.
Re: Disguised forms of 2
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
Wow, I feel stupid. So how would you evalute that expression? I saw something on wikipedia about it not being welldefined 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
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]
[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
[imath]\sqrt[8]{7*7*77*77133_4}[/imath]
I fail at trying 7...
EDIT: YAY! [imath]\sqrt[3]{\frac{\frac{\frac{7!}{7*77}}{7}}{2\frac{1}{7}}}[/imath]
EDIT2: primes in sequence [imath](2*3^5/(7*11)*131719)/23 = 2\frac{4}{1771}[/imath]
EDIT3: Also, 10
I fail at trying 7...
EDIT: YAY! [imath]\sqrt[3]{\frac{\frac{\frac{7!}{7*77}}{7}}{2\frac{1}{7}}}[/imath]
EDIT2: primes in sequence [imath](2*3^5/(7*11)*131719)/23 = 2\frac{4}{1771}[/imath]
EDIT3: Also, 10
Re: Disguised forms of 2
[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]
Re: Disguised forms of 2
[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]
(The real root, anyway)
[math]\sqrt[5]{5\cdot5+5+\sqrt[5]{5\cdot5+5+\sqrt[5]{5\cdot5+5+\sqrt[5]{\dots}}}}[/math]
(The real root, anyway)
Re: Disguised forms of 2
[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.
Somewhat simple, but I like it.
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Re: Disguised forms of 2
[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.
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
[math]\frac{e^\pi\pi}{10}[/math]
Exactly 2; the difference is due to rounding errors.
Exactly 2; the difference is due to rounding errors.
Re: Disguised forms of 2
[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}}}}[/math]
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Re: Disguised forms of 2
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.
Re: Disguised forms of 2
[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.
Non est salvatori salvator,
neque defensori dominus,
nec pater nec pater,
nihil supernum.
neque defensori dominus,
nec pater nec pater,
nihil supernum.
Re: Disguised forms of 2
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^3e^%283%29%29^%281%2F3%29+x^2%2F%28x^3%2Be^%286%29%29]WA[/url]
Re: Disguised forms of 2
Fixed. That 6 was supposed to be a 3.
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nec pater nec pater,
nihil supernum.
neque defensori dominus,
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Re: Disguised forms of 2
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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