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Sameer
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 Complex roots   « on: Sep 17th, 2007, 11:21pm » Quote Modify

Another one from my book:

Find all the roots of the equation: (x - 1)n = xn where n is a positive integer.
 « Last Edit: Sep 17th, 2007, 11:22pm by Sameer » IP Logged

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Aryabhatta
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 Re: Complex roots   « Reply #1 on: Sep 18th, 2007, 1:33am » Quote Modify

Clearly x = 0 is not a root.

Divide by xn and we are done, right?
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towr
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 Re: Complex roots   « Reply #2 on: Sep 18th, 2007, 9:44am » Quote Modify

So x = 1/(1- r), where r is any of the nth roots of 1.
i.e., if memory serves me right, r= ei 2 k/n with k=0..n-1
 « Last Edit: Sep 18th, 2007, 9:46am by towr » IP Logged

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pex
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 Re: Complex roots   « Reply #3 on: Sep 18th, 2007, 2:02pm » Quote Modify

on Sep 18th, 2007, 9:44am, towr wrote:
 So x = 1/(1- r), where r is any of the nth roots of 1. i.e., if memory serves me right, r= ei 2 k/n with k=0..n-1

Yes, except that k=0 won't work, obviously.
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Sameer
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 Re: Complex roots   « Reply #4 on: Sep 18th, 2007, 9:46pm » Quote Modify

on Sep 18th, 2007, 1:33am, Aryabhatta wrote:
 Clearly x = 0 is not a root.   Divide by xn and we are done, right?

Yep.

on Sep 18th, 2007, 2:02pm, pex wrote:
 Yes, except that k=0 won't work, obviously.

Well in this case k = 0 does work!! (I think!!!)
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Aryabhatta
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 Re: Complex roots   « Reply #5 on: Sep 18th, 2007, 11:40pm » Quote Modify

on Sep 18th, 2007, 9:46pm, Sameer wrote:
 Yep.     Well in this case k = 0 does work!! (I think!!!)

The original equation has exactly n-1 roots (counting multiplicity) as it is a polynomial of degree n-1.

So we have to reject one root.
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pex
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 Re: Complex roots   « Reply #6 on: Sep 19th, 2007, 5:40am » Quote Modify

on Sep 18th, 2007, 11:40pm, Aryabhatta wrote:
 The original equation has exactly n-1 roots (counting multiplicity) as it is a polynomial of degree n-1.   So we have to reject one root.

Yes, and k = 0 leads to r = 1, leaving x = 1/(1 - r) undefined.
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Grimbal
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 Re: Complex roots   « Reply #7 on: Sep 19th, 2007, 5:52am » Quote Modify

Now compute Re(x)
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pex
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 Re: Complex roots   « Reply #8 on: Sep 19th, 2007, 6:12am » Quote Modify

on Sep 19th, 2007, 5:52am, Grimbal wrote:
 Now compute Re(x)
That's interesting! I get that Re(x) = 1/2 for all of the roots. Derivation:
 hidden: xk = 1 / (1 - exp(2k pi i / n))  = (1 - exp(-2k pi i / n)) / (2 - exp(2k pi i / n) - exp(-2k pi i / n))  = (1 - cos(2k pi / n) + i sin(2k pi / n)) / (2 - 2cos(2k pi / n))  = 1/2 + 1/2 i [ sin(2k pi / n) / (1 - cos(2k pi / n)) ],   and clearly, the real part is 1/2, independent of k.
 « Last Edit: Sep 19th, 2007, 6:13am by pex » IP Logged
Sameer
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 Re: Complex roots   « Reply #9 on: Sep 19th, 2007, 9:18am » Quote Modify

Yep, I just thought infinity could be one of the roots!! Maybe not.. Yep I get the real part to be 1/2 too!! My original answer was in a reduced form 1/2(1+icotk/2) k=1..n-1 which easily showed what the real part was.
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 Re: Complex roots   « Reply #10 on: Sep 19th, 2007, 6:38pm » Quote Modify

If n is even, x can be 1/2.
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Sameer
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 Re: Complex roots   « Reply #11 on: Sep 19th, 2007, 6:55pm » Quote Modify

on Sep 19th, 2007, 6:38pm, srn347 wrote:
 If n is even, x can be 1/2.

The question asks for general solution over all n.
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 Re: Complex roots   « Reply #12 on: Sep 23rd, 2007, 3:29pm » Quote Modify

It wuold have to be one where x=x-1, so x=infinity.
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towr
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 Re: Complex roots   « Reply #13 on: Sep 23rd, 2007, 11:16pm » Quote Modify

on Sep 23rd, 2007, 3:29pm, srn347 wrote:
 It wuold have to be one where x=x-1, so x=infinity.
And yet again you fail.
There are n-1 solutions that don't tend to infinity. And since infinity isn't a complex number it can't be a solution, so only those other n-1 remain.
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 Re: Complex roots   « Reply #14 on: Sep 24th, 2007, 6:25pm » Quote Modify

It has to be complex? Take the n root and x=x-1, therefore it must be infinity(positive or negative).
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Sameer
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 Re: Complex roots   « Reply #15 on: Sep 24th, 2007, 8:24pm » Quote Modify

on Sep 24th, 2007, 6:25pm, srn347 wrote:
 It has to be complex? Take the n root and x=x-1, therefore it must be infinity(positive or negative).

ok to make it easier to understand, if you have complex set available, what are all the roots of x3 = 1?
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towr
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 Re: Complex roots   « Reply #16 on: Sep 25th, 2007, 1:26am » Quote Modify

on Sep 24th, 2007, 6:25pm, srn347 wrote:
 It has to be complex?
Complex numbers include the real numbers, FYI. Infinity is neither a real number, nor an imaginary number nor a combination. Hence it can't be a solution.

Quote:
 Take the n root and x=x-1, therefore it must be infinity(positive or negative).
Wrong. By simply taking the nth root at both sides, you lose n-1 solutions (and there are only n-1 solutions in the first place)
If e.g. x2=(x-1)2, then you don't get the solution by taking x=(x-1); so why would you think this is allowed for other n?

For x2=(x-1)2, just simplify the equation:
(x-1)2 = x2 -2x + 1, therefore x2=x2 -2x + 1, or x=1/2.
You even suggested x=1/2 yourself for even n!
 « Last Edit: Sep 25th, 2007, 1:26am by towr » IP Logged

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 Re: Complex roots   « Reply #17 on: Sep 25th, 2007, 5:07pm » Quote Modify

The roots of x3=1 are 1, -1/2 + sqrt(3)(i)/2, and -1/2 -sqrt(3)(i)/2
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Sameer
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 Re: Complex roots   « Reply #18 on: Sep 25th, 2007, 6:12pm » Quote Modify

on Sep 25th, 2007, 5:07pm, srn347 wrote:
 The roots of x3=1 are 1, -1/2 + sqrt(3)(i)/2, and -1/2 -sqrt(3)(i)/2

Good, now look at towr's example and what do you learn?
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srn437
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 Re: Complex roots   « Reply #19 on: Sep 25th, 2007, 8:04pm » Quote Modify

That there are no real solutions for all natural number exponents(which I already new).
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Obob
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 Re: Complex roots   « Reply #20 on: Sep 25th, 2007, 8:30pm » Quote Modify

Except whenever n is even, x=1/2 is a solution...
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Sameer
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 Re: Complex roots   « Reply #21 on: Sep 25th, 2007, 8:44pm » Quote Modify

on Sep 25th, 2007, 8:04pm, srn347 wrote:
 That there are no real solutions for all natural number exponents(which I already new).

What part of the title of this problem confuses you? Are you genuinely trying to learn or just wasting my time?
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"Obvious" is the most dangerous word in mathematics.
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Proof is an idol before which the mathematician tortures himself.
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srn437
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 Re: Complex roots   « Reply #22 on: Sep 26th, 2007, 7:23pm » Quote Modify

As already stated, you need to have all the n's answered. x would have to equal n-1.
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Sameer
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 Re: Complex roots   « Reply #23 on: Sep 26th, 2007, 9:39pm » Quote Modify

Note to Mods: Can you please delete posts that are irrelevant to this thread? You get my drift!!
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple

Proof is an idol before which the mathematician tortures himself.
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srn437
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 Re: Complex roots   « Reply #24 on: Sep 26th, 2007, 9:48pm » Quote Modify