Are Trigonometric Polynomials with Integer Coefficients Countable?

  • Thread starter Thread starter cragar
  • Start date Start date
  • Tags Tags
    Polynomials Trig
cragar
Messages
2,546
Reaction score
3

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.
 
Physics news on Phys.org
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top