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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



And transform everything to by the simple yet interesting identity



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.

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

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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



which is less "uglier" than that of the one you presented.
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December 30th, 2013, 05:43 AM   #13
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Re: Cube roots of -1

Terms are separated by addition and/or subtraction, so, for instance,



has 3 terms as written, and



is a single term as written.

So how would you represent 3 in terms of your beautiful method using ?

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December 30th, 2013, 05:44 AM   #14
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Re: Cube roots of -1

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Originally Posted by agentredlum

Bravo!

I actually did it a little differently ,



Now, on RHS we want to change e to 3, get rid of , get rid of the negative and keep the 2 in the denominator for use as square root



And the result follows on the LHS.

Nice method!

Could this be what you're looking for:

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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,



has 3 terms as written, and



is a single term as written.

So how would you represent 3 in terms of your beautiful method using ?
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?
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December 30th, 2013, 05:54 AM   #16
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Re: Cube roots of -1

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Originally Posted by Hoempa
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)
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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 ,



Now , on RHS we want to change e to 3 , get rid of , get rid of the negative and keep the 2 in the denominator for use as square root



And the result follows on the LHS.

Nice method!

Could this be what you're looking for:

Very nice. It's in radians , right?
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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

All i have to do is put a factor of 2 in front of the using my single term result for

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.

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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

All i have to do is put a factor of 2 in front of the using my single term result for
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.
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December 30th, 2013, 06:47 AM   #20
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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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