# Periodic Function problem

• August 26th 2007, 02:58 PM
Periodic Function problem
Let $f(x)=cosx+cos \pi x$
a) Show that $f(x)=2$ has a unique solution.
b) Show that f(x) is not periodic.

My proof so far:

Now, I realize that f(x) = 2 iff x=0 in this case, but will that be enough for a proof?

And since f(x) = 2 only once, then it cannot be periodic.
• August 26th 2007, 03:10 PM
Jhevon
Quote:

Let $f(x)=cosx+cos pi x$
a) Show that $f(x)=2$ has a unique solution.
b) Show that f(x) is not periodic.

My proof so far:

Now, I realize that f(x) = 2 iff x=0 in this case, but will that be enough for a proof?

And since f(x) = 2 only once, then it cannot be periodic.

i would think that's enough

how did you show that f(x) = 2 iff x = 0 ?
• August 26th 2007, 03:16 PM
Really, I only know that cosx = 1 and cospix = 1 when x = 0, and I just can't find anything else x can equal to in order to get the same result.
• August 26th 2007, 03:28 PM
Jhevon
Quote:

Really, I only know that cosx = 1 and cospix = 1 when x = 0, and I just can't find anything else x can equal to in order to get the same result.

well, that won't work as a proof!

use the sum-to-product formula for cosine: $\cos \alpha + \cos \beta = 2 \cos \left( \frac {\alpha + \beta }{2} \right) \cos \left( \frac {\alpha - \beta}{2} \right)$

...or maybe, you can continue with your line of thought, but add that since the range of cosine is [-1,1], the only way we can get two is if both terms are 1, since neither can be greater than 1, and if one of them is less than 1, the other would have to be greater than 1 to add up to 2 etc etc etc ...
• August 26th 2007, 04:10 PM
Plato
The following is true: $a \le 1,\;b \le 1,\quad a + b = 2\quad \Rightarrow \quad a = 1\;\& \;b = 1$.
$\cos (t) = 1\quad \Rightarrow \quad t = 2k\pi$.
$\cos (\pi t) = 1\quad \Rightarrow \quad \pi t = 2k\pi \quad \Rightarrow \quad t = 2k$
$2k\pi = 2k\quad \Rightarrow \quad k = 0$

Thus the only solution is $x=0$!
• August 26th 2007, 10:08 PM
CaptainBlack
Quote:

Originally Posted by Plato
The following is true: $a \le 1,\;b \le 1,\quad a + b = 2\quad \Rightarrow \quad a = 1\;\& \;b = 1$.
$\cos (t) = 1\quad \Rightarrow \quad t = 2k\pi$.
$\cos (\pi t) = 1\quad \Rightarrow \quad \pi t = 2k\pi \quad \Rightarrow \quad t = 2k$
$2k\pi = 2k\quad \Rightarrow \quad k = 0$

Thus the only solution is $x=0$!

This is not quite right. The two k's need not be the same, so we have:

$
t=2k_1 \pi
$

and

$
t=2 k_2
$

for some $k_1, k_2 \in \bold{Z}$, then eliminating $t$ between these we have:

$k_1 \pi=k_2$,

but as $\pi$ is irrational this is imposible unless $k_1=k_2=0$.

(PS it is obvious that the proof must hinge on $\pi$ being irrational so any
proof that does not use this fact must be deficient in some manner)

RonL
• August 27th 2007, 10:36 AM
ThePerfectHacker
Plato solved this problem the best way. There is a puzzle which asks if $f(x) = \sin x + \sin x^{\circ}$ is a periodic function. The solution is to note that $f'(x)$ is also periodic and its maximum value is at $x=0$. And argue like above there is no other maximum value.
• August 27th 2007, 01:39 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Plato solved this problem the best way.

Unfortunatly not as Platos proof would also show that $f(x)=\cos(x)+\cos (7 x)$ is aperiodic, but its not.

RonL
• August 27th 2007, 02:35 PM
Plato
If the proof that I posted contained all the symbols that I had intended, then it would be correct. I do understand that we need two different possible values of K. I thought that I had used a K’ for the second. But several edits must have dropped some symbols; what was posted above is flawed.

The function is $f(x) = \cos (x) + \cos (\pi x)$ and as noted both of those terms must be 1 if the sum is two.
We get $\cos (x) = 1\quad \Rightarrow \quad x = 2K\pi$. Because it is the same x in the sum we get the following.
$\cos \left[ {\pi \left( {2K\pi } \right)} \right] = \cos \left( {2\pi ^2 K} \right) = 1\quad \Rightarrow \quad 2\pi ^2 K = 2\pi K'$

This means that $\pi K = K'$ and $\pi$ being irrational means $K = K' = 0$ or $x = 0$.