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        riddles >> putnam exam (pure math) >> Complex roots
        
(Message started by: Sameer on Sep 17th, 2007, 11:21pm)
Title: Complex roots
Post by Sameer on Sep 17th, 2007, 11:21pm
Another one from my book:

Find all the roots of the equation: (x - 1)n = xn where n is a positive integer.
Title: Re: Complex roots
Post by Aryabhatta on Sep 18th, 2007, 1:33am
Clearly x = 0 is not a root.

Divide by xn and we are done, right?
Title: Re: Complex roots
Post by towr on Sep 18th, 2007, 9:44am
So [hide]x = 1/(1- r), where r is any of the nth roots of 1.
i.e., if memory serves me right, r= ei 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/suppi.gif k/n with k=0..n-1
[/hide]
Title: Re: Complex roots
Post by pex on Sep 18th, 2007, 2:02pm

on 09/18/07 at 09:44:05, towr wrote:
So [hide]x = 1/(1- r), where r is any of the nth roots of 1.
i.e., if memory serves me right, r= ei 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/suppi.gif k/n with k=0..n-1
[/hide]

Yes, except that [hide]k=0 won't work, obviously.[/hide]
Title: Re: Complex roots
Post by Sameer on Sep 18th, 2007, 9:46pm

on 09/18/07 at 01:33:41, Aryabhatta wrote:
Clearly x = 0 is not a root.

Divide by xn and we are done, right?


Yep.


on 09/18/07 at 14:02:24, pex wrote:
Yes, except that [hide]k=0 won't work, obviously.[/hide]


Well in this case k = 0 does work!! (I think!!!)
Title: Re: Complex roots
Post by Aryabhatta on Sep 18th, 2007, 11:40pm

on 09/18/07 at 21:46:53, 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.
Title: Re: Complex roots
Post by pex on Sep 19th, 2007, 5:40am

on 09/18/07 at 23:40:17, 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.
Title: Re: Complex roots
Post by Grimbal on Sep 19th, 2007, 5:52am
Now compute Re(x)
Title: Re: Complex roots
Post by pex on Sep 19th, 2007, 6:12am

on 09/19/07 at 05:52:29, Grimbal wrote:
Now compute Re(x)
That's interesting! I get that [hide]Re(x) = 1/2 for all of the roots.[/hide] Derivation:[hideb]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.[/hideb]
Title: Re: Complex roots
Post by Sameer on Sep 19th, 2007, 9:18am
Yep, I just thought infinity could be one of the roots!! Maybe not.. Yep I get the real part to be [hide] 1/2 [/hide] too!! My original answer was in a reduced form [hide] 1/2(1+icotkhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2) k=1..n-1 [/hide] which easily showed what the real part was.
Title: Re: Complex roots
Post by srn347 on Sep 19th, 2007, 6:38pm
If n is even, x can be 1/2.
Title: Re: Complex roots
Post by Sameer on Sep 19th, 2007, 6:55pm

on 09/19/07 at 18:38:38, srn347 wrote:
If n is even, x can be 1/2.


The question asks for general solution over all n.
Title: Re: Complex roots
Post by srn347 on Sep 23rd, 2007, 3:29pm
It wuold have to be one where x=x-1, so x=infinity.
Title: Re: Complex roots
Post by towr on Sep 23rd, 2007, 11:16pm

on 09/23/07 at 15:29:59, 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.
Title: Re: Complex roots
Post by srn347 on Sep 24th, 2007, 6:25pm
It has to be complex? Take the n root and x=x-1, therefore it must be infinity(positive or negative).
Title: Re: Complex roots
Post by Sameer on Sep 24th, 2007, 8:24pm

on 09/24/07 at 18:25:41, 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?
Title: Re: Complex roots
Post by towr on Sep 25th, 2007, 1:26am

on 09/24/07 at 18:25:41, 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!
Title: Re: Complex roots
Post by srn347 on Sep 25th, 2007, 5:07pm
The roots of x3=1 are 1, -1/2 + sqrt(3)(i)/2, and -1/2 -sqrt(3)(i)/2
Title: Re: Complex roots
Post by Sameer on Sep 25th, 2007, 6:12pm

on 09/25/07 at 17:07:44, 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?
Title: Re: Complex roots
Post by srn347 on Sep 25th, 2007, 8:04pm
That there are no real solutions for all natural number exponents(which I already new).
Title: Re: Complex roots
Post by Obob on Sep 25th, 2007, 8:30pm
Except whenever n is even, x=1/2 is a solution...
Title: Re: Complex roots
Post by Sameer on Sep 25th, 2007, 8:44pm

on 09/25/07 at 20:04:03, 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?
Title: Re: Complex roots
Post by srn347 on Sep 26th, 2007, 7:23pm
As already stated, you need to have all the n's answered. x would have to equal n-1.
Title: Re: Complex roots
Post by Sameer on Sep 26th, 2007, 9:39pm
Note to Mods: Can you please delete posts that are irrelevant to this thread? You get my drift!!
Title: Re: Complex roots
Post by srn347 on Sep 26th, 2007, 9:48pm
Try asking a question with an answer!
Title: Re: Complex roots
Post by Grimbal on Sep 27th, 2007, 12:34am

on 09/26/07 at 19:23:11, srn347 wrote:
As already stated, you need to have all the n's answered. x would have to equal n-1.

I guess you mean x-1.

Sorry to tell you that, but your understanding of mathematics is a complete mess.
Title: Re: Complex roots
Post by towr on Sep 27th, 2007, 2:08am

on 09/26/07 at 19:23:11, srn347 wrote:
As already stated, you need to have all the n's answered. x would have to equal n-1.
Perhaps you misunderstand the question. What is asked for is not a single value for x that works for all n; what is asked for is an expression which given any n, provides values for x that satisfy the equation.


on 09/26/07 at 21:39:48, Sameer wrote:
Note to Mods: Can you please delete posts that are irrelevant to this thread? You get my drift!!
If this final attempt at explanation fails, I'll be glad to. So that should be in under a day.
Title: Re: Complex roots
Post by srn347 on Sep 27th, 2007, 4:46pm
Oh individual values. Obviously the evens are already solved. Some exponents may have multiple solutions.
Title: Re: Complex roots
Post by pex on Sep 28th, 2007, 12:10am

on 09/27/07 at 16:46:01, srn347 wrote:
Oh individual values. Obviously the evens are already solved. Some exponents may have multiple solutions.

srn347 - A complete solution was outlined in the first three replies to this thread. Indeed, there are multiple solutions: n-1 for every n.


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