Are Trigonometric Polynomials with Integer Coefficients Countable?

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SUMMARY

The set of all trigonometric polynomials with integer coefficients is countable, as established through the mapping of coefficients to prime numbers. The polynomial is represented as t(x) = a + Σ a_n cos(nx) + Σ b_n sin(nx), where the sums are finite and n is a natural number. The proof relies on demonstrating that finite subsets of a countable set are also countable, utilizing mathematical induction to show that subsets with no more than n elements are countable. This foundational concept confirms the countability of trigonometric polynomials with integer coefficients.

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cragar
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Homework Statement


Prove that the set of all trigonometric polynomials with integer coefficients is countable.

Homework Equations


t(x)= a+\sum a_ncos(nx)+ \sum b_n sin(nx)
the sum is over n and is from 1 to some natural number.

The Attempt at a Solution


So basically we have to look at all the possible trig polynomials of all finite lengths.
with some natural a out front.
let's first look at the ones where n=1
so we have a+cos(1x)+bsin(x)
How about we map these to the first prime number 2.
since a can be anything, if a is 1 then this t(x) will go to 2 and if a is 2 then t(x) will go to 2^2
if a is three then t(x) goes to 2^3.
Now for the sum from n=1 to 2 will map these to the next prime and do the same process as above with the constant out front. Is this the right idea.
I am mapping these to prime numbers so we can get a unique mapping and we don't have to worry about sending 2 things to one thing.
 
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cragar said:

Homework Statement


Prove that the set of all trigonometric polynomials with integer coefficients is countable.

Homework Equations


t(x)= a+\sum a_ncos(nx)+ \sum b_n sin(nx)
the sum is over n and is from 1 to some natural number.

The Attempt at a Solution


So basically we have to look at all the possible trig polynomials of all finite lengths.
with some natural a out front.
let's first look at the ones where n=1
so we have a+cos(1x)+bsin(x)
How about we map these to the first prime number 2.
since a can be anything, if a is 1 then this t(x) will go to 2 and if a is 2 then t(x) will go to 2^2
if a is three then t(x) goes to 2^3.
Now for the sum from n=1 to 2 will map these to the next prime and do the same process as above with the constant out front. Is this the right idea.
I am mapping these to prime numbers so we can get a unique mapping and we don't have to worry about sending 2 things to one thing.

In general the set of all finite subsets of a countable set is countable. That's the proof you want. There's nothing about trig polynomials that makes it any easier. Try to prove that first.
 
Dick said:
In general the set of all finite subsets of a countable set is countable. That's the proof you want. There's nothing about trig polynomials that makes it any easier. Try to prove that first.

Going from the hint Dick has given, start by showing that the set of subsets with no more than ##n## elements is countable. Since you're working finite here, induction should be good I believe.
 

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