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#1 2007-06-26 05:32:43
- shocamefromebay
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demoive
can someone explain to me how to do demoive's theorem to find the five fifthe roots of one?
are they irrational or can they be displayed in radical form?
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#2 2007-06-26 05:41:00
- Ricky
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Re: demoive
http://en.wikipedia.org/wiki/De_Moivre's_formula
Look at the very bottom of that. Remember that 1 is in fact a complex number, 1 + 0i.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#3 2007-06-26 11:07:10
- shocamefromebay
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Re: demoive
yes
i understand taht
but when u work it out
i cant figure out what the real parts or the imaginary parts are of the answers
i cant tell if they are just irrational or if there is a way to put it in radical form
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#4 2007-06-26 13:54:42
- Ricky
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Re: demoive
I meant to correct you on that before. Something like the square root of three is irrational. In fact, the square root of any integer that isn't a perfect square is irrational. Irrational simply means that a number can not be expressed as a/b, where a and b are integers, b nonzero.
On the other hand, there are transcendental and algebraic numbers. A number is called algebraic if and only if it is the root of (solution to) a finite polynomial:
A number is transcendental if it is not a solution to any polynomial. Examples are pi and e, as well as Euler's constant.
But that's not a bad question. Is sin(2*pi*k/n) always expressible as radicals? I have a feeling yes, but I'm not certain.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#5 2007-06-26 16:25:17
- shocamefromebay
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Re: demoive
yea i get what u r sayin and all but the reall problem taht i am having is how to get
cos((4pi)/5)
as a radical form
is there a way to use the unit circle to find taht ???
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#6 2007-06-26 22:35:06
- HallsofIvy
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Re: demoive
MOST trig values cannot be written in terms of radials. pi/5 is one of them.
#7 2007-06-27 04:35:40
- Ricky
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Re: demoive
That's not true. Check out the 5th roots of unity:
http://en.wikipedia.org/wiki/Roots_of_unity
Also, here is the exact value of sin and cos of pi/5:
http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/simpleTrig.html
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#8 2007-06-27 05:07:35
- shocamefromebay
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Re: demoive
wow thanks a lot
those links halped a lot
and who said taht wikipedia was un reliable
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#9 2007-06-27 16:33:52
- shocamefromebay
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Re: demoive
is there a site that has the cos and sin of pi/7 ?????
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#10 2007-07-03 18:27:06
- Ricky
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Re: demoive
I don't know of any. But for any rational r, sin(pi*r) and cos(pi*r) are both algebraic. Note that this does not mean expressible by roots.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#11 2007-07-07 07:58:24
- shocamefromebay
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Re: demoive
what do u mean by algebraic
Last edited by shocamefromebay (2007-07-07 07:58:40)
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#12 2007-07-07 08:01:50
- Ricky
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Re: demoive
A number is algebraic if it is the root (solution) to a polynomial:
Note that such polynomials are all of finite, but arbitrary length.
Pi, e, log(11) are examples of non-algebraic (called transcendental) numbers.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#13 2007-07-07 11:40:28
- shocamefromebay
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Re: demoive
why isnt e or pi algebraic if it is the solution to
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#14 2007-07-07 12:51:11
- Ricky
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Re: demoive
The a_k coefficients must be integers.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#15 2007-07-07 14:01:14
- shocamefromebay
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Re: demoive
hmmm
but what about
isnt that algebraic
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#16 2007-07-07 18:05:46
- Ricky
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Re: demoive
Multiply both sides by 2, what do you get? Note that this can be done for any rational coefficients and it leads to the rational roots theorem.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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