# Sequence of e^{inx}

• Nov 17th 2010, 02:58 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 $\displaystyle x \in \mathbb{R}$. Show that:
(i) The sequence $\displaystyle (e^{inx})$ converges if and only if $\displaystyle x \in 2 \pi \mathbb{Z}$
• Nov 17th 2010, 05:50 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 $\displaystyle x \in \mathbb{R}$. Show that:
(i) The sequence $\displaystyle (e^{inx})$ converges if and only if $\displaystyle x \in 2 \pi \mathbb{Z}$

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

Tonio
• Nov 18th 2010, 05:21 AM
chisigma
Quote:

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

Tonio

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

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

... so that the sequence converges...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Nov 18th 2010, 05:37 AM
tonio
Quote:

Originally Posted by chisigma
It is evident that if $\displaystyle x= 2\ k\ \pi$ and $\displaystyle k \in \mathbb{Z}$ is $\displaystyle \forall n$ ...

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

... so that the sequence converges...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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

Tonio