# Thread: Sum of periodic functions

1. ## Sum of periodic functions

This theorem was inspired to me by this thread. I have found a proof of it but I'll post it later; I want to see if somebody can find another one!

Let $\displaystyle m(x), n(x)$ be continuous real-valued functions having periods $\displaystyle p,q$. Show that $\displaystyle m(x)+n(x)$ is periodic if and only if $\displaystyle p,q$ are linearly dependent over $\displaystyle \mathbb{Q}$.

2. Originally Posted by Bruno J.
This theorem was inspired to me by this thread. I have found a proof of it but I'll post it later; I want to see if somebody can find another one!

Let $\displaystyle m(x), n(x)$ be continuous real-valued functions having periods $\displaystyle p,q$. Show that $\displaystyle m(x)+n(x)$ is periodic if and only if $\displaystyle p,q$ are linearly dependent over $\displaystyle \mathbb{Q}$.
If $\displaystyle m(x) + n(x)$ has period T, then
$\displaystyle T=ap$
$\displaystyle T=bq$
where a,b are natural numbers.

hence, $\displaystyle ap=bq$
or,
$\displaystyle p=\frac{b}{a}q$

now, what is meant by linear independence here?

3. How can you be sure that the period of the sum is a multiple of both $\displaystyle p,q$? You need to show that this is true; it's the hard part of the problem.

If $\displaystyle x_1,...,x_n$ are linearly independent over $\displaystyle K$, then whenever $\displaystyle k_1x_1+...+k_nx_n=0$ with $\displaystyle k_1,...,k_n \in K$ we must have $\displaystyle k_1=...=k_n=0$. If they are linearly dependent then they are not linearly independent. In your post, $\displaystyle p-\frac{b}{a}q=0$ is a linear dependence relation over the rationals.

But you have to show that we must have $\displaystyle T=ap=bq$ for some integers $\displaystyle a,b$.

4. For a,b integers:

m(x + ap) + n(x + bq) = m(x) + n(x) (1)

Therefore ap=bq defines all periods of the sum function.

If the sum is periodic, p = b/a * q, where b/a obviously belongs to Q.

Alas c1*p + c2*q = (c1 * b/a + c2) * q, for c1=-a/b and c2=1 (both € Q) it is 0, that is if the sum is periodic p,q are linearly dependent over Q (2)

If p,q are linearly dependent over Q, p = -c2/c1 * q for c1,c2 rational numbers, then expressing c1,c2 as c1N/c1D and c2N/c2D, where c1N,c1D,c2N,c2D are integers, we get c1N*c2D*p = -c2N*c1D*q.

Taking (1) and making a=c1N*c2D and b=-c2N*c1D, ab=pq, so If p,q are linearly independent over Q, the sum function is periodic (3).

m(x) + n(x) is periodic If(3) and only if(2) p,q are linearly independent over Q.

(Sorry for the ugly proof)

5. For a,b integers:

m(x + ap) + n(x + bq) = m(x) + n(x) (1)
Ok so far.

Therefore ap=bq defines all periods of the sum function.
How so? If $\displaystyle T$ is the period of $\displaystyle m(x)+n(x)$, then $\displaystyle m(x)+n(x)=m(x+T)+n(x+T)$; but that does NOT imply $\displaystyle m(x)=m(x+T)$ and $\displaystyle n(x)=n(x+T)$. You are making a huge leap here.

If you do not use the continuity of $\displaystyle m,n$ your proof is certainly flawed because we can construct non-continuous $\displaystyle m,n$ whose sum is periodic, but whose periods are linearly independent over $\displaystyle \mathbb{Q}$.

6. Originally Posted by Bruno J.
Ok so far.

How so? If $\displaystyle T$ is the period of $\displaystyle m(x)+n(x)$, then $\displaystyle m(x)+n(x)=m(x+T)+n(x+T)$; but that does NOT imply $\displaystyle m(x)=m(x+T)$ and $\displaystyle n(x)=n(x+T)$. You are making a huge leap here.

If you do not use the continuity of $\displaystyle m,n$ your proof is certainly flawed because we can construct non-continuous $\displaystyle m,n$ whose sum is periodic, but whose periods are linearly independent over $\displaystyle \mathbb{Q}$.
Ups I'll try it later...

7. Bump. Nobody has solved this one yet... last call! I'll give the solution soon.