Disguised forms of 2

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svensen
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Re: Disguised forms of 2

Postby svensen » Wed Oct 27, 2010 9:14 pm UTC

Dopefish wrote:I don't think n! is defined for anything except non-negative integers.

The gamma function is the extension to factorial, that works for whatever you want to feed it. Given the stuff above didn't at all look like a non-negative integer (or even real), it strictly speaking probably should have been a gamma function involved.


I really hate to get bogged down in semantics, but I disagree.

The factorial is defined for all complex numbers except for negative integers. That's why when I punch 9.5! into my calculator it returns a number, not "undefined". For numbers apart from non-negative integers, It happens to be defined in terms of the gamma function, but that doesn't make it not valid.

It's the same way that the sine function is defined in terms of the exponential function or that the gamma function itself is defined in terms of an integral.

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Re: Disguised forms of 2

Postby Talith » Wed Oct 27, 2010 9:40 pm UTC

Let's not get too distracted in this thread (or we should make a new thread) but the factorial function is defined recursively on the positive integers and is given by n!=n(n-1)!, 0!=1. The gamma function is defined in terms of an integral and happens to be equal to the factorial operator for positive integer values. So no, the factorial isn't defined for all complex numbers, the gamma function is, and when your calculator tells you that 9.5! is defined, that's probably because your calculator uses the gamma function in place of the factorial. This is much like a calculator will return i when you input sqrt{-1} if the calculator is sophisticated enough; it doesn't return i because the square root function is defined from all reals to all reals (because it isn't), it uses a different function which is the square root function from the complex numbers to the complex numbers, which can be defined.

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Re: Disguised forms of 2

Postby skeptical scientist » Wed Oct 27, 2010 10:30 pm UTC

svensen wrote:The factorial is sometimes defined for all complex numbers except for negative integers. That's why when I punch 9.5! into my calculator it returns a number, not "undefined". For numbers apart from non-negative integers, It happens to sometimes be defined in terms of the gamma function, but that doesn't make it not valid.

It's the same way that the sine function is sometimes defined in terms of the exponential function or that the gamma function itself is defined in terms of an integral.

Fixed.

Mathematicians are sometimes consistent about definitions, and sometimes not. So, the sine function might be defined in terms of the exponential function. It might also be defined by a power series, or defined by a unit circle. There is no single definition which is "the" definition, with the others merely being properties that follow from the single definition. This is just like how 0 is sometimes a natural number, and sometimes not, depending on who you ask. There are certain conventions which simply fail to be universal, so you have to accept that sometimes people may be using different conventions from your own, and not call them "wrong" to do so.
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Re: Disguised forms of 2

Postby Kurushimi » Wed Oct 27, 2010 11:13 pm UTC

svensen wrote:
Dopefish wrote:I don't think n! is defined for anything except non-negative integers.

The gamma function is the extension to factorial, that works for whatever you want to feed it. Given the stuff above didn't at all look like a non-negative integer (or even real), it strictly speaking probably should have been a gamma function involved.


I really hate to get bogged down in semantics, but I disagree.

The factorial is defined for all complex numbers except for negative integers. That's why when I punch 9.5! into my calculator it returns a number, not "undefined". For numbers apart from non-negative integers, It happens to be defined in terms of the gamma function, but that doesn't make it not valid.

It's the same way that the sine function is defined in terms of the exponential function or that the gamma function itself is defined in terms of an integral.

If you have a TI calculator, try punching 9.6! and see what happens.

I have never seen the factorial function being used with anything except integers. The definition I see used most often doesn't even allow for nonnegative numbers so I don't think that it would be correct to say the factorial of something other than an integer.

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Re: Disguised forms of 2

Postby vilidice » Sun Oct 31, 2010 4:17 pm UTC

The half-derivative of x^2 evaluated at pi is 2.

kind of an awesome relationship actually, there's a more general rule for other parabolas too.

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Re: Disguised forms of 2

Postby silverhammermba » Sun Nov 07, 2010 9:26 pm UTC

[math]2 = 1 + \cfrac{2}{1 + \cfrac{2}{1 + \cfrac{2}{\ddots}}}[/math]

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Re: Disguised forms of 2

Postby Blatm » Sat Nov 13, 2010 2:03 am UTC

[math]2 = \sum_{n=0}^{\infty}\sum_{i=n}^{2n}\sum_{k=0}^{2(i-n)} (-1)^{n+i}\frac{ (2n)!(2i)!(4n+1)}{16^n (2n-i)!(n!)^2 i! k!}\left((-1)^ke - \frac{1}{e}\right)[/math]

I'd be very impressed if anyone figured out where this is from.

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Re: Disguised forms of 2

Postby Sagekilla » Sun Nov 14, 2010 1:12 am UTC

vilidice wrote:The half-derivative of x^2 evaluated at pi is 2.

kind of an awesome relationship actually, there's a more general rule for other parabolas too.


Do you mean of x? Because the half derivative of x is [imath]2\sqrt{\frac{x}{\pi}}[/imath]

Evaluate that at pi and you'd get 2.
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Re: Disguised forms of 2

Postby Mavrisa » Sun Nov 14, 2010 4:22 am UTC

Heh, I'm surprised nobody's mentioned [imath]\frac{3}{\pi} + \frac{\pi}{3} \approx 2.0021272[/imath]
I know the pi equals three people would just say this is 1 + 1, but it looks unobvious at first glance. Your call to decide whether or not it's cheating.
Or perhaps: [imath]log(1-\sqrt{2}+e^2-3 \pi+\pi^2) \approx 2.00413444[/imath]
Then there's always slightly under: pi times one of the roots of 5 x^3-7 x^2-7 x+6 = 0 (the root in question is about 0.636287, and pi times that is about 1.9989553)
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Re: Disguised forms of 2

Postby cyanyoshi » Thu Nov 18, 2010 6:42 am UTC

[math]2 = \sqrt[3]{10+\sqrt{108}}+\sqrt[3]{10-\sqrt{108}}[/math]

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Re: Disguised forms of 2

Postby hencethus » Thu Nov 18, 2010 8:40 am UTC

Here's my contribution:
[math]{{sin([i^{-i} \cdot i^{-i}]^e + e^{\pi e})}\over{cos(e^{\pi e})\cdot sin([i^{-i} \cdot i^{-i}]^e)}}=2[/math]
Which of course just means that
[math](i^{-2i})^e = e^{\pi e}[/math]
Is that cheating?

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Re: Disguised forms of 2

Postby Kurushimi » Thu Nov 18, 2010 9:01 pm UTC

[math]x_0 = e[/math]
[math]x_{n+1} = x_n^2 - 4x_n + 6[/math]
[math]lim_{n->\infty} x_n[/math]
Last edited by Kurushimi on Wed May 04, 2011 2:19 pm UTC, edited 1 time in total.

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Re: Disguised forms of 2

Postby Mike_Bson » Fri Nov 19, 2010 7:23 pm UTC

Not sure if anyone has posted this one. . .

[math]([cos(cos(cos(cos(cos(cos(cos(5)))))))]^2)*4 = 1.9999051[/math]

That;s take the cosine of 5 seven times, square it, and multiply it by 4. I found this randomly playing on my calculator, I took the cosine of 5 seven times, and noticed it was about .707, close to sqrt(1/2).

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Re: Disguised forms of 2

Postby Blatm » Sun Nov 21, 2010 7:21 am UTC

[math]2 = \sum_{n=1}^{\infty}\sum_{j=n}^{2n-1}\sum_{k=0}^{2(j-n)+1} (-1)^{n+j} \frac{(4n^2-n)(2n)!^2(2j)!}{4^n(2(j-n)+1)n!^2(2n-j-1)!j!}\left[\frac{2\pi}{(2(j-n)+1)!} + (-1)^{\frac{k(k-1)(k-2)(k-3)}{24}}\frac{2^{\frac{k+1}{2}}(i^k + (-i)^k + (-1)^k - 3) }{k(2(j-n)-k+1)!k!\sqrt{3 + (-1)^{k+1}}}\right][/math]

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Re: Disguised forms of 2

Postby Eebster the Great » Sun Nov 21, 2010 10:06 am UTC

[math]2 =
{\frac{-a}{4} - \frac{1}{2}{\sqrt{\frac{a^2}{4} - \frac{2b}{3} +
\frac{2^{\frac{1}{3}}\left( b^2 - 3ac + 12d \right) }
{3{\left( 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd +
{\sqrt{-4{\left( b^2 - 3ac + 12d \right) }^3 +
{\left( 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd \right) }^2}} \right) }^{\frac{1}{3}}} +
\left(\frac{{ 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd + {\sqrt{-4
{\left( b^2 - 3ac + 12d \right) }^3 +
{\left( 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd \right) }^2}} }}
{54}\right)^\frac{1}{3}}} -
\frac{1}{2}{\sqrt{\frac{a^2}{2} - \frac{4b}{3} -
\frac{2^{\frac{1}{3}}\left( b^2 - 3ac + 12d \right) }
{3{\left( 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd +
{\sqrt{-4{\left( b^2 - 3ac + 12d \right) }^3 +
{\left( 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd \right) }^2}} \right) }^{\frac{1}{3}}} -
\left(\frac{{ 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd + {\sqrt{-4
{\left( b^2 - 3ac + 12d \right) }^3 +
{\left( 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd \right) }^2}} }}
{54}\right)^\frac{1}{3} -
\frac{-a^3 + 4ab - 8c}
{4{\sqrt{\frac{a^2}{4} - \frac{2b}{3} +
\frac{2^{\frac{1}{3}}
\left( b^2 - 3ac + 12d \right) }{3
{\left( 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd +
{\sqrt{-4
{\left( b^2 - 3ac + 12d \right) }^3 +
{\left( 2b^3 - 9abc + 27c^2 +
27a^2d - 72bd \right) }^2}} \right) }^
{\frac{1}{3}}} +
\left( \frac{{ 2b^3 - 9abc + 27c^2 +
27a^2d - 72bd +
{\sqrt{-4
{\left( b^2 - 3ac + 12d \right) }^3 +
{\left( 2b^3 - 9abc + 27c^2 +
27a^2d - 72bd \right) }^2}} }
}{54}\right)^\frac{1}{3}}}}}}} //[/math]

where

a = -24
b = 72
c = -96
d = 48

But that feels like cheating because it's just a solution to [imath]3(x-2)^4 = 0[/imath] and also because I can't get the whole expression to fit on this forum (is there a way to make it scroll?).

I get a bit more with imath:

[imath]2 =
{\frac{-a}{4} - \frac{1}{2}{\sqrt{\frac{a^2}{4} - \frac{2b}{3} +
\frac{2^{\frac{1}{3}}\left( b^2 - 3ac + 12d \right) }
{3{\left( 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd +
{\sqrt{-4{\left( b^2 - 3ac + 12d \right) }^3 +
{\left( 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd \right) }^2}} \right) }^{\frac{1}{3}}} +
\left(\frac{{ 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd + {\sqrt{-4
{\left( b^2 - 3ac + 12d \right) }^3 +
{\left( 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd \right) }^2}} }}
{54}\right)^\frac{1}{3}}} -
\frac{1}{2}{\sqrt{\frac{a^2}{2} - \frac{4b}{3} -
\frac{2^{\frac{1}{3}}\left( b^2 - 3ac + 12d \right) }
{3{\left( 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd +
{\sqrt{-4{\left( b^2 - 3ac + 12d \right) }^3 +
{\left( 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd \right) }^2}} \right) }^{\frac{1}{3}}} -
\left(\frac{{ 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd + {\sqrt{-4
{\left( b^2 - 3ac + 12d \right) }^3 +
{\left( 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd \right) }^2}} }}
{54}\right)^\frac{1}{3} -
\frac{-a^3 + 4ab - 8c}
{4{\sqrt{\frac{a^2}{4} - \frac{2b}{3} +
\frac{2^{\frac{1}{3}}
\left( b^2 - 3ac + 12d \right) }{3
{\left( 2b^3 - 9abc + 27c^2 + 27a^2d -
72bd +
{\sqrt{-4
{\left( b^2 - 3ac + 12d \right) }^3 +
{\left( 2b^3 - 9abc + 27c^2 +
27a^2d - 72bd \right) }^2}} \right) }^
{\frac{1}{3}}} +
\left( \frac{{ 2b^3 - 9abc + 27c^2 +
27a^2d - 72bd +
{\sqrt{-4
{\left( b^2 - 3ac + 12d \right) }^3 +
{\left( 2b^3 - 9abc + 27c^2 +
27a^2d - 72bd \right) }^2}} }
}{54}\right)^\frac{1}{3}}}}}}} //[/imath]

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Re: Disguised forms of 2

Postby Mike_Bson » Mon Nov 22, 2010 1:53 am UTC

@Ebester

Quartic formula much? Glad it only exists when the degree is less than or equal to four, any higher and you'd just be unleashing hell on us. . .

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Re: Disguised forms of 2

Postby Dadel » Mon Nov 22, 2010 8:17 am UTC

We had this as an exercise in Fourier Analysis, and I thought of this thread.

Show that the function g(x) = 1/sqrt(|x|) for 0 < |x| <= 1, 0 otherwise, is integrable.

The integral from -1 to 1 equals 4, so squirt it.

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Re: Disguised forms of 2

Postby Eebster the Great » Mon Nov 22, 2010 10:50 am UTC

Mike_Bson wrote:@Ebester

Quartic formula much? Glad it only exists when the degree is less than or equal to four, any higher and you'd just be unleashing hell on us. . .

There is a formula that can solve any solvable quintic and was apparently printed in one book in 2004 on three whole pages. I can't get my hands on that, though.

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Re: Disguised forms of 2

Postby martin878 » Wed Nov 24, 2010 2:08 pm UTC

I was keen to use a search algorithm to find something good. I wrote one that looks for signs:

[math]\left(\frac{3}{2^1} - \frac{1}{2^2} - \frac{4}{2^3} - \frac{1}{2^4}
- \frac{5}{2^5} + \frac{9}{2^6} - \frac{2}{2^7} - \frac{6}{2^8}
+ \frac{5}{2^9} - \frac{3}{2^{10}} - \frac{5}{2^{11}} + \frac{8}{2^{12}}\right)\pi = 2.00798[/math]

I know it's a bit off topic because it's not a disguised form of 2, just a silly long way of getting close. But engineering people might find it fun.

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Re: Disguised forms of 2

Postby martin878 » Wed Nov 24, 2010 2:21 pm UTC

Also (where [imath]p_i[/imath] is the [imath]i[/imath]th prime)
[math]\sum_{i=1}^{59}{\frac{1}{p_i}} = 2.00235[/math]

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Re: Disguised forms of 2

Postby Mike_Bson » Wed Nov 24, 2010 9:20 pm UTC

Eebster the Great wrote:
Mike_Bson wrote:@Ebester

Quartic formula much? Glad it only exists when the degree is less than or equal to four, any higher and you'd just be unleashing hell on us. . .

There is a formula that can solve any solvable quintic and was apparently printed in one book in 2004 on three whole pages. I can't get my hands on that, though.

Ah. I've heard it multiple ways; that there is no quintic formula, and that there is a quintic formula, but it has stuff other than radicals. I guess it's the latter?

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Re: Disguised forms of 2

Postby Eebster the Great » Wed Nov 24, 2010 9:25 pm UTC

Mike_Bson wrote:
Eebster the Great wrote:
Mike_Bson wrote:@Ebester

Quartic formula much? Glad it only exists when the degree is less than or equal to four, any higher and you'd just be unleashing hell on us. . .

There is a formula that can solve any solvable quintic and was apparently printed in one book in 2004 on three whole pages. I can't get my hands on that, though.

Ah. I've heard it multiple ways; that there is no quintic formula, and that there is a quintic formula, but it has stuff other than radicals. I guess it's the latter?

There is no general algorithm for solving quintics via root extraction, but there are some using other functions such as the inverse of x5+x+1=0. But that's not what I'm talking about. Some particular quintics are solvable by root extraction, and there are methods of determining which ones are. There is a formula which can solve any solvable quintic, and it is apparently extremely long and complicated.

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Re: Disguised forms of 2

Postby Hackfleischkannibale » Thu Nov 25, 2010 12:05 am UTC

Using Gregory Numbers:

[math]\Biggl[{\sin \left(\frac{5}{\sqrt{50}}\left(1+\sum_{i=1}^{\infty}\left(\frac{(2i-1)!!}{(2i)!!}\cdot\frac{1}{(2i-1)50^i}\right)\right)+\frac{2}{\sqrt{325}}\left(1+\sum_{i=1}^{\infty}\left(\frac{(2i-1)!!}{(2i)!!}\cdot\frac{1}{(2i-1)325^i}\right)\right)-\frac{2}{\sqrt{3250}}\left(1+\sum_{i=1}^{\infty}\left(\frac{(2i-1)!!}{(2i)!!}\cdot\frac{1}{(2i-1)3250^i}\right)\right)\right)}\cdot {}[/math][math]{\cos\left(\frac{4}{\sqrt{26}}\left(1+\sum_{i=1}^{\infty}\left(\frac{(2i-1)!!}{(2i)!!}\cdot\frac{1}{(2i-1)26^i}\right)\right)-\frac{1}{\sqrt{57122}}\left(1+\sum_{i=1}^{\infty}\left(\frac{(2i-1)!!}{(2i)!!}\cdot\frac{1}{(2i-1)57122^i}\right)\right)\right)}\Biggr]^{-1}[/math]
where n!! denotes the double faculty: 1*3*...*n, if n>0 and odd, 2*4*...*n if n>0 and even.
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Re: Disguised forms of 2

Postby PrinsValium » Wed May 04, 2011 9:57 am UTC

Eebster the Great wrote:[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^\dots}}[/math]
This actually has two real solutions (one being exactly 2 and the other about 4) and infinite complex solutions.


This got me curious... Generalizing to [imath]x = {{a^a}^a}^\dots[/imath] for which values of [imath]a[/imath] does this have a finite value? If we substitue [imath]a = \sqrt[b]{b}[/imath] to get [imath]x = \sqrt[b]{b}^{\sqrt[b]{b}^{\sqrt[b]{b}^\dots}}[/imath] and the equation [imath]x = \left(\sqrt[b]{b}\right)^x[/imath] taking a very special log of both sides [imath]\log_{\sqrt[b]{b}}{x} = x[/imath]. Then it isn't too far fetched (at least to me) to think this would have a solution [imath]x = b[/imath], and thus [imath]b = \sqrt[b]{b}^{\sqrt[b]{b}^{\sqrt[b]{b}^\dots}}[/imath]

I agree that the series [imath]c_{n+1} = \left(\sqrt[2]{2}\right)^{c_{n}}, \; c_1 = \sqrt[2]{2}[/imath] converges to [imath]2[/imath], but is this series the same thing as writing [imath]\sqrt[2]{2}^{\sqrt[2]{2}^{\sqrt[2]{2}^\dots}}[/imath] (or is the latter really a nonsensical notation?)?

The problem is of course the inverse to the url=http://www.wolframalpha.com/input/?i=x^x^-1]function[/url [imath]y = \sqrt[x]{x}[/imath]. It is well defined on the interval [imath]y \in (0,1][/imath] (?) as well as at [imath]y=e[/imath]. But what about the interval [imath]y \in (1,e)[/imath]. If it makes sense to view the power tower as something else than a series as above can it be seen how it would equal the greater of the two solutions to [imath]x = \left(\sqrt[b]{b}\right)^x, \; e < b < \infty[/imath]? i.e. the equation [imath]x = \left(\sqrt[b]{b}\right)^x, \; b = \sqrt[2]{2} = \sqrt[4]{4}[/imath] is satisfied by [imath]x=2\; or \; 4[/imath].
Last edited by PrinsValium on Wed May 04, 2011 11:14 am UTC, edited 1 time in total.

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WarDaft
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Re: Disguised forms of 2

Postby WarDaft » Wed May 04, 2011 11:12 am UTC

The solutions of x = bx are not all valid values of x for x = [imath]b^{b^{b^{...}}}[/imath]. Otherwise, the entire class of epsilon numbers are valid values!
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Re: Disguised forms of 2

Postby skeptical scientist » Wed May 04, 2011 1:43 pm UTC

Here's one I discovered yesterday on a friend's tie:[math]2=\left(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\right)^2.[/math]
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Re: Disguised forms of 2

Postby PrinsValium » Tue May 10, 2011 2:42 pm UTC

WarDaft wrote:The solutions of x = bx are not all valid values of x for x = [imath]b^{b^{b^{...}}}[/imath]. Otherwise, the entire class of epsilon numbers are valid values!


Ok, I'll take your word for it. I think it will be a bit too much for me too read up on those [imath]\epsilon[/imath]-numbers =) I thought that perhaps it could depend on how things were done, sort of like with [imath]\sum_{n=1}^{\infty}\frac{(-1)^n}{n}[/imath]. But, is it then correct to say that [imath]x = {{a^a}^a}^\dots[/imath] is defined for [imath]0<a\leq e[/imath] and [imath]a=\sqrt[x]{x}[/imath], if we keep to the realm of the real numbers?

One more thing on the same topic.
I think I'm correct that [imath]c_n = \sqrt[n]{n}^{c_{n-1}}, \; c_1 = 1, \; \lim_{n \to \infty}{c_n} = 1[/imath]

But what about [imath]\sqrt[2]{2}^{\sqrt[3]{3}^{\sqrt[4]{4}^\dots}}[/imath]

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Re: Disguised forms of 2

Postby skeptical scientist » Tue May 10, 2011 3:46 pm UTC

PrinsValium wrote:But what about [imath]\sqrt[2]{2}^{\sqrt[3]{3}^{\sqrt[4]{4}^\dots}}[/imath]

That converges to something between 1 and 2, since it's increasing, and bounded above by [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^\dots}}=2[/imath]. You can get better upper lower bounds by noting that all tails are between 1 and 2. From this, the actual value is somewhere between 1.94146112352 and 1.94146112389. links go to Mathematica and Rick Astley, respectively.
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Re: Disguised forms of 2

Postby NathanielJ » Tue May 10, 2011 11:06 pm UTC

skeptical scientist wrote:From this, the actual value is somewhere between 1.94146112352 and 1.94146112389. links go to Mathematica and Rick Astley, respectively.


More decimal places can be found here, by the way.
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Lorenz
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Re: Disguised forms of 2

Postby Lorenz » Tue May 10, 2011 11:37 pm UTC

By making a tower [imath]x_1^{x_2^{x_3^{x_4^{x_5}}}}[/imath] where each [imath]x_n[/imath] is one of the following:
[imath]\pm(1, \pi, e , \phi =GoldenRatio, \gamma = EulerGamma).[/imath]

The closest I could come up to 2 was
[imath]\phi^{\phi^{\phi^{-\gamma}}}[/imath][imath]\approx 1.99941[/imath]

The runner-up was
[imath]\phi^{\pi^{\pi^{-1}}}[/imath][imath]\approx 1.99923[/imath]

nomadiq
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Re: Disguised forms of 2

Postby nomadiq » Thu May 12, 2011 8:45 pm UTC

[imath]7\pi/11[/imath] is pretty awesome. Using [imath]7\pi/22[/imath] as a start its pretty easy to generate approximate integers.

Using e instead of [imath]\pi[/imath] we can do:
[imath]20 e / 27[/imath]

Lovely!

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Qaanol
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Re: Disguised forms of 2

Postby Qaanol » Thu May 12, 2011 11:36 pm UTC

nomadiq wrote:[imath]7\pi/11[/imath] is pretty awesome. Using [imath]7\pi/22[/imath] as a start its pretty easy to generate approximate integers.

Using e instead of [imath]\pi[/imath] we can do:
[imath]20 e / 27[/imath]

Lovely!

14e/19 is closer.
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nomadiq
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Re: Disguised forms of 2

Postby nomadiq » Fri May 13, 2011 3:45 am UTC

Qaanol wrote:
nomadiq wrote:[imath]7\pi/11[/imath] is pretty awesome. Using [imath]7\pi/22[/imath] as a start its pretty easy to generate approximate integers.

Using e instead of [imath]\pi[/imath] we can do:
[imath]20 e / 27[/imath]

Lovely!

14e/19 is closer.


Nice!

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John Citizen
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Re: Disguised forms of 2

Postby John Citizen » Fri May 13, 2011 1:38 pm UTC

[imath]2=\huge{-\frac{1+\exp{(\frac{1}{2i\pi} \cdot \int_0^{\infty} \frac{\log{\frac{-i\pi+t-\log{t}+\log{(-\log{\sqrt{2}})}}{i\pi+t-\log{t}+\log{(-\log{\sqrt{2}})}}}}{1+t}dt)(-1+\log{(-log{\sqrt{2}})})}}{\log \sqrt{2}}}[/imath]
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Quantum Sunshine
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Re: Disguised forms of 2

Postby Quantum Sunshine » Mon May 16, 2011 9:47 pm UTC

[math]2={1 \over {e^{i\pi \over 3}-{\sqrt{3} \over 2}i}}[/math]


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