How to Prove Accumulation Points of a Sequence: an = e^(2 * π * i * n*x)

bl77c

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Let x be an irrational real number. We define an = e^(2 * π * i * n) for all n ∈ N (where N represents the set of natural numbers). Show that every z ∈ C (the set of complex numbers) with |z| = 1 is an accumulation point of the sequence (an).
I dont know what to do exactly, what happens if x is rational and how does this sequence look.
 
Let x be an irrational real number. We define an = e^(2 * π * i * n) for all n ∈ N (where N represents the set of natural numbers). Show that every z ∈ C (the set of complex numbers) with |z| = 1 is an accumulation point of the sequence (an).
I dont know what to do exactly, what happens if x is rational and how does this sequence look.
Please check the original. There ought to be some place where x is used, shouldn't there?
 
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Sorry, i forgot to write it down , the correct sequence is n = e^(2 * π * i * n*x)
 
Let x be an irrational real number. We define \(an = e^{2\pi i n x}\) for all n ∈ N (where N represents the set of natural numbers). Show that every z ∈ C (the set of complex numbers) with |z| = 1 is an accumulation point of the sequence (an).
I don't know what to do exactly, what happens if x is rational and how does this sequence look.
Okay, now please tell what you do know, so we can help you work this out. Our goal is to help you learn how to solve problems, not to do it for you:

Can you picture what this sequence looks like in some particular case -- maybe first for a rational value of x like 1/2? This sort of experimentation is often a vital part of both solving problems and learning general concepts.

Also, please state the definition you were given for an accumulation point, and perhaps look back at an example you've been given.
 
Okay, now please tell what you do know, so we can help you work this out. Our goal is to help you learn how to solve problems, not to do it for you:

Can you picture what this sequence looks like in some particular case -- maybe first for a rational value of x like 1/2? This sort of experimentation is often a vital part of both solving problems and learning general concepts.

Also, please state the definition you were given for an accumulation point, and perhaps look back at an example you've been given.
In the given definition, for every ε > 0, there exist infinitely many n such that |z − an| < ε. I recognize that for rational x, the sequence will eventually repeat, implying that there are not infinitely many n such that |z − an| < ε. In fact, for certain ε values, there might not be a single n that satisfies the condition.
Now for the case where x is irrational. For every n ∈ N, the values of n*x are different due to the irrationality of x. Since x is not periodic, I can generate my n*x in a way that 2πnx = m2πy, where m ∈ N and y is an angle I created. Therefore, I can reach any point z, except in some specific cases, like -1, 1, i, -i but i can still reach them very closely
 
For every n ∈ N, the values of n*x are different due to the irrationality of x.
n*x would be different for rational x too. I wonder if you mean n*x modulo 1?
Since x is not periodic, I can generate my n*x in a way that 2πnx = m2πy, where m ∈ N and y is an angle I created.
I find this very cryptic. What angle did you create?
 
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