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Thread: trig function - transcendental?

  1. #1 trig function - transcendental? 
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    Trig functions of angles which are rational multiples of π are algebraic numbers. Are all trig functions of irrational mutiples of π transcendental numbers?


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    If pi is a transendental number and a transendental number must not be the root of a non-zero polynomial with rational co-efficients while pi is the root of a non-zero polynomial with rational co-efficients. Is pi really transcendental?


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    Genius Duck Dywyddyr's Avatar
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    Quote Originally Posted by fiveworlds View Post
    while pi is the root of a non-zero polynomial with rational co-efficients
    Source please.
    Pi is transcendental.
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    No what it also says is pi is not the root of a non zero polynomial
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    Fiveworlds, you were warned before to stop posting unhelpful nonsense in the math subforum...this time you get a week of pondering the meaning of that warning.
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    Quote Originally Posted by fiveworlds View Post
    No what it also says is pi is not the root of a non zero polynomial
    So why did you give that link to support your claim that
    Quote Originally Posted by fiveworlds View Post
    pi is the root of a non-zero polynomial with rational co-efficients
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    I wish people would not go off on (nonsensical) tangents unrelated to the original question.
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    Quote Originally Posted by mathman View Post
    Trig functions of angles which are rational multiples of π are algebraic numbers. Are all trig functions of irrational mutiples of π transcendental numbers?
    Mathman, can you link to a good proof of this statement?

    Trig functions of angles which are rational multiples of π are algebraic numbers

    I believe it, but I would like to see the proof for some clues of how to solve this one. Its a good question though, its been in my head since last night.
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    Quote Originally Posted by TridentBlue View Post
    Quote Originally Posted by mathman View Post
    Trig functions of angles which are rational multiples of π are algebraic numbers. Are all trig functions of irrational mutiples of π transcendental numbers?
    Mathman, can you link to a good proof of this statement?

    Trig functions of angles which are rational multiples of π are algebraic numbers

    I believe it, but I would like to see the proof for some clues of how to solve this one. Its a good question though, its been in my head since last night.
    I have an idea. I was going to try to get it to go through but I think I'll just throw it out there and perhaps someone can take it forward. Collaborative math :-)

    Suppose x = (n/m)pi, meaning that x is a rational multiple of pi. I don't care if it's in lowest terms or not.

    Let exp(x) be the complex exponential function exp(x) = e^ix that maps the unit interval to the unit circle.

    Then (using brackets for readability only -- no other meaning intended) ...

    [exp((n/m(pi)]^m = = exp(i*n* pi) = cos(n*pi) + i * sin(n*pi)

    When n is even, cos(n*pi) is 1 and sin(n*pi) is zero.

    When n is odd, cos(n*pi) is -1 and sin(n*pi) is zero.

    So the complex number exp(((n/m)pi) is algebraic. It satisfies z^m = 1 or z^m - 1 depending on whether m is even or odd, respectively.

    A more sophisticated shorthand for all this is to note that the set of all rotations of the plane about the origin form a group, where the group operation is defined as doing the one rotation after another. In this group, an element (rotation) has finite order if and only if it is a rotation of a rational multiple of pi. And in that case, the complex number on the unit circle corresponding to that angle, is a root of unity. You can raise it to some integer power and get 1. So it's a complex algebraic number.

    What we are asked to show involves going the other way. We want to get from some trig function of a number and get back to the complex number that satisfies the polynomial z^n = 1. (If n is odd we can just replace it with 2n, it doesn't matter.)

    In other words if x = (n/m)pi then we want to show that cos(x) satisfies a polynomial. It would be nice to demonstrate the specific polynomial.

    I must be missing something obvious.
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    Quote Originally Posted by someguy1 View Post
    I have an idea. I was going to try to get it to go through but I think I'll just throw it out there and perhaps someone can take it forward. Collaborative math :-)

    Suppose x = (n/m)pi, meaning that x is a rational multiple of pi. I don't care if it's in lowest terms or not.

    Let exp(x) be the complex exponential function exp(x) = e^ix that maps the unit interval to the unit circle.

    Then (using brackets for readability only -- no other meaning intended) ...

    [exp((n/m(pi)]^m = = exp(i*n* pi) = cos(n*pi) + i * sin(n*pi)

    When n is even, cos(n*pi) is 1 and sin(n*pi) is zero.

    When n is odd, cos(n*pi) is -1 and sin(n*pi) is zero.

    So the complex number exp(((n/m)pi) is algebraic. It satisfies z^m = 1 or z^m - 1 depending on whether m is even or odd, respectively.
    Yeah, i get that. What's not clear to me from the first statement is why if z is algebraic, the real and imaginary part of must be too. I'm sure its true, I just can't see it. Working on it...
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    Quote Originally Posted by TridentBlue View Post

    Yeah, i get that. What's not clear to me from the first statement is why if z is algebraic, the real and imaginary part of must be too. I'm sure its true, I just can't see it. Working on it...
    Oh yeah this is obvious ... after I cheated and looked it up. I's so simple I wish I'd worked harder to figure it out! I googled "real and imaginary part of complex algebraic numbers" and Google turned up this mathoverlow link ...

    On the real and imaginary parts of algebraic numbers - MathOverflow

    Suppose z = x + iy is algebraic. The algebraic numbers are a field and also closed under conjugation. Using z' for conjugation,

    (z + z')/2 = x and (z - z')/2i = y are both algebraic.

    Particularizing this to the matter at hand: If n, m are integers then z = cos((n/m)*pi) + i sin((n/m)*pi) satisfies z^m = +/- 1 so z is algebraic; and therefore so are the sin and cosine of (n/m)*pi. Then the other trig functions are derived from sin and cos by division or inversion, which also preserve algebraic numbers.

    Quote Originally Posted by mathman View Post
    I wish people would not go off on (nonsensical) tangents unrelated to the original question.
    That's part of being on the Internet. But another part of being on the internet is being able to collaborate on problems. The Internet giveth, and the Internet taketh away :-)
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    [QUOTE=someguy1;426914]
    Quote Originally Posted by TridentBlue View Post

    Particularizing this to the matter at hand: If n, m are integers then z = cos((n/m)*pi) + i sin((n/m)*pi) satisfies z^m = +/- 1 so z is algebraic; and therefore so are the sin and cosine of (n/m)*pi. Then the other trig functions are derived from sin and cos by division or inversion, which also preserve algebraic numbers.
    Yeah, I was unable to see that yesterday, I just got a minute to think about it today and that came to me, I really slapped my forehead. Of course algebraic numbers are closed under conjugation.

    So, any real output of sine or cosine with (m/n)*pi as an input corresponds with a root of unity, and is itself an algebraic number. Now just some intuition on the matter.
    1) Does a natural number n times an irrational number ever make a natural? No.
    2) Therefore a number w = cos(irrational) + i*sin(irrational) can only be real (imag part 0) when raised to power of 0. So clearly, no polynomial of one term fit the bill.
    3) So the only way it could be true, is if some sum of the powers of w are equal to the conjugate of some other sum of the powers of w.
    4) I think the polynomial can't have two terms either. Suppose w^n = conjugate(w^(n-c)). If c is even, w must be a root of unity. If its odd, its square root must be a root of unity. Either way, its algebraic. Could something a proof be done by induction? We can show the base case (no one term) could we show the induction step?

    Anyway, thanks for chatting with me on this. I'm no math guru by any means, but I do enjoy thinking about it.

    PEace!
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  15. #14  
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    Since we all seem to agree that rational multiples of π have algebraic trig functions, does anyone have any thoughts about the irrational multiples of π?
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    Quote Originally Posted by mathman View Post
    Since we all seem to agree that rational multiples of π have algebraic trig functions, does anyone have any thoughts about the irrational multiples of π?
    Seems like it should be true but I haven't figure out how to prove it.

    It amounts to saying that if cos(x) is transcendental, then x is an irrational multiple of pi. And then showing that if cos(x) is transcendental then none of the other trig functions can be algebraic.

    A wrinkle is that if cos(x) and sin(x) are both transcendental, how do you know that tan(x) = sin(x)/cos(x) isn't algebraic? In which case there would be an irrational multiple of pi with an algebraic trig function even though sin and cos are both transcendental.

    I haven't made any progress, but there's at least a couple of us thinking about it, for what it's worth.

    If you need an answer in a hurry I'd try math.stackexchange or mathoverflow; but that would spoil the fun over here :-)
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  17. #16  
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    When I first posed the question I was thinking only of sin and cos. One approach that someone in another forum raised is - what are the angles in right triangles with integer sides? If they could be proven not multiples of π, that would answer the question.
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