What is the period of these functions?

Question

I have two functions as follows:

$x = (a-b) \cdot \cos(t) + b \cdot \cos(t\cdot(k-1))$

$y = (a-b) \cdot \sin(t) - b \cdot \sin(t\cdot(k-1))$

What are the periods of functions $x$ and $y$?

I found the similar questions (for example Principal period of $\sin\frac{3x}{4}+\cos\frac{2x}{5}$) but it does not help me to solve this issue.

The $k$ is rational number (for example - 0.35 or 1.7).

Ah well that really depends on $k$, if $k$ is $\pi+1$ for example I don't think it has any period. — Gregory Grant, Feb 12 '16 at 12:17
To be periodic I'm pretty sure you'd need $k$ to be rational. — Gregory Grant, Feb 12 '16 at 12:18
If $k$ is rational then $\cos t$ has period $2\pi$ and $\cos(kt)$ has period $2\pi/k$. The period of the sum would be the least common multiple of the period of $\cos t$ and the period of $\cos (kt)$. — Gregory Grant, Feb 12 '16 at 12:23

score 3 · Accepted Answer · answered Feb 12 '16 at 12:24

3

If $k$ is rational then $\cos t$ has period $2\pi$ and $\cos((k-1)t)$ has period $2\pi/(k-1)$. The period of the sum would be the least common multiple of the period of $\cos t$ and the period of $\cos ((k-1)t)$.

answered Feb 12 '16 at 12:24

Gregory Grant

14,874

score 2 · Answer 2 · answered Feb 12 '16 at 12:23

2

Your period for both cases will be the LCM of $2\pi\over k-1$ and $2\pi$

answered Feb 12 '16 at 12:23

Win Vineeth

3,504

score 1 · Answer 3 · answered Feb 12 '16 at 15:00

Let $$f(x)= (a-b) \cdot \cos(t) + b \cdot \cos(t\cdot(k-1))$$ with $k-1=\frac pq$; $(p,q)=1$.

We have $f(0)=a$ and the period of $f$ is the minimum positive $t$ such that $f(t)=a$.

The period of $\cos \frac {pt}{q}$ is $\frac {2q\pi}{p}$; hence $$f(2q\pi)=(a-b)\cdot\cos 2q\pi+b\cdot\cos(2q\pi)=(a-b)+b\cdot\cos( p\cdot\frac{2q\pi}{p})=a-b+b=a$$

Thus the period of $f$ is equal to $\color{red}{2q\pi}$

Note.-Concerning the expression "LCM of two irrational" I feel is a good example of non-sense.

What is the period of these functions?

3 Answers3