My Math Forum Cube roots of -1

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December 30th, 2013, 05:16 AM   #11
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Re: Cube roots of -1

Quote:
Originally Posted by mathbalarka
Quote:
 Originally Posted by agentredlum The result I'm looking for is ...
I'd represent it as

$\sqrt{3}= -ie^{i \pi/3} + ie^{-i\pi/3}$

And transform everything to $^{i}$ by the simple yet interesting identity

$e^\pi= i^{-2i}$

We can generalize this to any elementary number (i.e., an algebraic number with minimal polynomial having solvable Galois group), see Gauss sums and Kronecker-Weber theorem and the later for anybody who wants to dig deep in these things.
That's very nice, but I asked for a single term.

December 30th, 2013, 05:28 AM   #12
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Re: Cube roots of -1

Quote:
 Originally Posted by agentredlum That's very nice but i asked for a single term.
First thing I would have asked you, or any transcendental number theorist would, is what do you mean by a "single term".

But, for the sake of keeping everything fair, I'd rather not but simply say that yes, mine indeed is not representable in a single term (or apparently so). But if simplified, you get

$\sqrt{3}= i^{1/3} - i^{5/3}$

which is less "uglier" than that of the one you presented.

 December 30th, 2013, 05:43 AM #13 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 234 Re: Cube roots of -1 Terms are separated by addition and/or subtraction, so, for instance, $x^2 \ + \ x \ - \ 5$ has 3 terms as written, and $3xy^2zw^3$ is a single term as written. So how would you represent 3 in terms of your beautiful method using $\ i \$?
December 30th, 2013, 05:44 AM   #14
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Re: Cube roots of -1

Quote:
 Originally Posted by agentredlum Bravo! I actually did it a little differently , $\Large i ^{i} \= \ \large e^{ \tiny{-} \large \frac{ \pi}{2}}$ Now, on RHS we want to change e to 3, get rid of $\ \pi \$, get rid of the negative and keep the 2 in the denominator for use as square root $$$\Large i^{ \Large i}$$^{ \Large { \large \frac{ \tiny{-} \ln 3}{ \pi}} \= \ $$\Large {e} ^{ \tiny{-} \frac{ \pi}{2}}$$^{ \tiny{-} \frac{ \ln 3}{ \pi} }$ And the result follows on the LHS.
Nice method!

Could this be what you're looking for:

$\sqrt{3}= \sqrt{\lfloor\cos(i \cdot i)^{\large i \cdot i} \cdot \cos(i \cdot i)^{\large i \cdot i} \rfloor}$

December 30th, 2013, 05:52 AM   #15
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Re: Cube roots of -1

Quote:
 Originally Posted by agentredlum Terms are separated by addition and/or subtraction, so, for instance, $x^2 \ + \ x \ - \ 5$ has 3 terms as written, and $3xy^2zw^3$ is a single term as written. So how would you represent 3 in terms of your beautiful method using $\ i \$?
That is an ambiguous definition, you do understand that, no? Anyways, let's not discuss it here. I'll create a topic on NT section soon on some conjectures regarding this problem I have in my mind, so I hope to see you there.

As for representing 3, perhaps square it?

PS : My representation is nothing of much "beauty" although it is of interest, yes. Have you searched the keywords I have italicized?

December 30th, 2013, 05:54 AM   #16
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Re: Cube roots of -1

Quote:
 Originally Posted by Hoempa $\sqrt{3}= \sqrt{\lfloor\cos(i \cdot i)^{\large i \cdot i} \cdot \cos(i \cdot i)^{\large i \cdot i} \rfloor}$
Hmm, interesting, but unfortunately I don't like this much Agent's and your previous representation makes thing so much interesting (topic coming up on NT)

December 30th, 2013, 05:56 AM   #17
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Re: Cube roots of -1

Quote:
Originally Posted by Hoempa
Quote:
 Originally Posted by agentredlum Bravo! I actually did it a little differently , $\Large i ^{i} \= \ \large e^{ \tiny{-} \large \frac{ \pi}{2}}$ Now , on RHS we want to change e to 3 , get rid of $\ \pi \$ , get rid of the negative and keep the 2 in the denominator for use as square root $$$\Large i^{ \Large i}$$^{ \Large { \large \frac{ \tiny{-} ln 3}{ \pi}} \= \ $$\Large {e} ^{ \tiny{-} \frac{ \pi}{2}}$$^{ \tiny{-} \frac{ ln 3}{ \pi} }$ And the result follows on the LHS.
Nice method!

Could this be what you're looking for:

$\sqrt{3}= \sqrt{\lfloor\cos(i \cdot i)^{\large i \cdot i} \cdot \cos(i \cdot i)^{\large i \cdot i} \rfloor}$
Very nice. It's in radians , right?

December 30th, 2013, 06:12 AM   #18
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Re: Cube roots of -1

Quote:
 Originally Posted by mathbalarka As for representing 3, perhaps square it?
Then you need to perform multiple additions and several multiplications using your result for $\ \sqrt{3}$

All i have to do is put a factor of 2 in front of the $\ iln3 \$ using my single term result for $\ \sqrt{3}$

Quote:
 Originally Posted by mathbalarka PS : My representation is nothing of much "beauty" although it is of interest, yes. Have you searched the keywords I have italicized?
Yes i looked at a couple of wiki articles , interesting.

December 30th, 2013, 06:16 AM   #19
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Re: Cube roots of -1

Quote:
 Originally Posted by agentredlum Then you need to perform multiple additions and several multiplications using your result for $\ \sqrt{3}$ All i have to do is put a factor of 2 in front of the $\ iln3 \$ using my single term result for $\ \sqrt{3}$
I understand you. I am posting the NT topic now, it will take 5-10 minutes to type it in. What we need is a better definition of "single term" or "multiple term" to work our way in and avoid ambiguity.

 December 30th, 2013, 06:47 AM #20 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Cube roots of -1 I have posted it here, feel free to add anything. I have to go now, bye!

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