# Sequence of e^{inx}

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• Nov 17th 2010, 03:58 PM
shmounal
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 $x \in 2 \pi \mathbb{Z}$
• Nov 17th 2010, 06:50 PM
tonio
Quote:

Originally Posted by shmounal
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
• Nov 18th 2010, 06:21 AM
chisigma
Quote:

Originally Posted by tonio
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

$\chi$ $\sigma$
• Nov 18th 2010, 06:37 AM
tonio
Quote:

Originally Posted by chisigma
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