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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?

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  • $\begingroup$ 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 $ax+b +T$. $\endgroup$ – lulu Jan 7 at 19:01
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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 $T_1=\frac{T}{a}$.

Addition

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 $aY<T$ is the period of $f$ and it is a contradiction. Therefore, $\frac{T}{a}$ is the fundamental period of $f(ax+b)$.

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  • $\begingroup$ 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 ? $\endgroup$ – Dimitris Dimitriadis Jan 7 at 19:09
  • $\begingroup$ @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 $\endgroup$ – Lev Ban Jan 7 at 19:11
  • $\begingroup$ Actually, according to the wikipedia, any real number satisfying that property is called period. The least such number is called fundamental period. $\endgroup$ – Lev Ban Jan 7 at 19:13
  • $\begingroup$ 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? $\endgroup$ – Dimitris Dimitriadis Jan 7 at 19:16
  • $\begingroup$ @DimitrisDimitriadis I updated my answer so that I can answer your last question. $\endgroup$ – Lev Ban Jan 7 at 19:26

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