# if $f$ is periodic then $f(ax+b)$ is also periodic

Let as define a function $$f: R \to R$$ which is periodic, with fundamental period T. Approve that $$f(ax+b)$$ is also periodic and the fundamental period is $$\frac{T}{a}$$.

My solution: if $$y = ax + b$$ then there is a fundamental period $$T_2$$ so that $$f(y+T) = f(y)$$ . Due to the fact that f is periodic then the equation is true. So the first part of the exercise has been solved (I think).

Edited: My solution: if $$y = ax + b$$ then there is a $$T_2 \in R$$ so that $$f(y+T) = f(y)$$ . Due to the fact that f is periodic then the equation is true. So the first part of the exercise has been solved (I think).

Now, we must approve that $$T_2 = \frac{T}{a}$$. Obviously this is true but how can we write it using a mathematic way?

• Your argument for periodicity appears to be circular : "Since it is periodic it has a period and since it has a period it is periodic". Just notice that replacing $x$ with $x+\frac Ta$ takes $ax+b$ to $ax+b +T$. – lulu Jan 7 at 19:01

Look at the definition of the periodic function below

'A function $$f$$ is periodic if there exists $$T\in \mathbb{R}$$ such that $$f(x)=f(x+T).'$$

So the mathematical way to write this proof is as below.

Let $$g(x)=f(ax+b)$$ then note that $$g(x+T_1)=g\left(x+\frac{T}{a}\right)=f\left( a\left(x+\frac{T}{a} \right)+b \right)=f(ax+b+T)=f(ax+b)=g(x).$$

Thus, $$g(x)=f(ax+b)$$ is periodic function with the period $$T_1=\frac{T}{a}$$.

As for finding fundamental period, we need to find the least such $$T_1$$ satisfying $$f(ax+b+T_1)=f(ax+b).$$

In this case, we need the fundamental period of $$f$$. Let's say that is $$T$$.

Assume there is $$Y<\frac{T}{a}$$ such that $$f(a(x+Y)+b)=f(ax+b)$$. Then observe that $$f(ax+b)=f(a(x+Y)+b)=f(ax+b+aY).$$

Since we can vary $$x$$ so that $$ax+b$$ is just arbitrary real number. Then $$aY is the period of $$f$$ and it is a contradiction. Therefore, $$\frac{T}{a}$$ is the fundamental period of $$f(ax+b)$$.

• So, to approve that $f(ax+b)$ is periodic, you must assume that the period is $\frac{T}{a}$. But, no one tell us that period is $\frac{T}{a}$. How could we approve that $f(ax+b)$ is periodic without knowing the period ? – Dimitris Dimitriadis Jan 7 at 19:09
• @DimitrisDimitriadis In order to prove that a function is periodic, you don't need to specify the period. The thing you should do it to show that there is a real number $T$ such that $g(x)=g(x+T)$. en.wikipedia.org/wiki/Periodic_function – Lev Ban Jan 7 at 19:11
• Actually, according to the wikipedia, any real number satisfying that property is called period. The least such number is called fundamental period. – Lev Ban Jan 7 at 19:13
• Suppose that we know only that $f$ is periodic with period $T$. How could we say that $f(ax+b)$ is periodic and how can we find the new period? – Dimitris Dimitriadis Jan 7 at 19:16