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Math Help - Sequence of e^{inx}

  1. #1
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    Sequence of e^{inx}

    Hi, Ive been set this problem and am struggling to see how to solve it. Any help much appreciated, thanks

    Fix x \in \mathbb{R}. Show that:
    (i) The sequence  (e^{inx}) converges if and only if x \in 2 \pi \mathbb{Z}
    Last edited by mr fantastic; November 17th 2010 at 06:08 PM.
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  2. #2
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    Quote Originally Posted by shmounal View Post
    Hi, Ive been set this problem and am struggling to see how to solve it. Any help much appreciated, thanks

    Fix x \in \mathbb{R}. Show that:
    (i) The sequence  (e^{inx}) converges if and only if x \in 2 \pi \mathbb{Z}

    Hint: e^{inx}=\cos nx+i\sin nx , and we know (I hope...) that a complex sequence converges iff its real and imaginary parts converge.

    Tonio
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by tonio View Post
    Hint: e^{inx}=\cos nx+i\sin nx , and we know (I hope...) that a complex sequence converges iff its real and imaginary parts converge.

    Tonio
    It is evident that if x= 2\ k\ \pi and k \in \mathbb{Z} is \forall n ...

    \displasytyle e^{i n x} = \cos 2 \pi k n\ + i\ \sin 2 \pi k n = 1 + i\ 0

    ... so that the sequence converges...

    Kind regards

    \chi \sigma
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    Quote Originally Posted by chisigma View Post
    It is evident that if x= 2\ k\ \pi and k \in \mathbb{Z} is \forall n ...

    \displasytyle e^{i n x} = \cos 2 \pi k n\ + i\ \sin 2 \pi k n = 1 + i\ 0

    ... so that the sequence converges...

    Kind regards

    \chi \sigma

    Well, yes: that's the trivial part. I suppose the OP was interested in the other direction.

    Tonio
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