if $f$ is periodic then $f(ax+b)$ is also periodic
$begingroup$
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?
calculus analysis
asked Jan 7 at 18:59
Dimitris DimitriadisDimitris Dimitriadis
568
$endgroup$
add a comment |
$begingroup$
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?
calculus analysis
asked Jan 7 at 18:59
Dimitris DimitriadisDimitris Dimitriadis
568
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
calculus analysis
asked Jan 7 at 18:59
Dimitris DimitriadisDimitris Dimitriadis
568
$endgroup$
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?
calculus analysis
calculus analysis
asked Jan 7 at 18:59
Dimitris DimitriadisDimitris Dimitriadis
568
asked Jan 7 at 18:59
Dimitris DimitriadisDimitris Dimitriadis
568
edited Jan 7 at 19:34
Dimitris Dimitriadis
asked Jan 7 at 18:59
Dimitris DimitriadisDimitris Dimitriadis
568
asked Jan 7 at 18:59
Dimitris DimitriadisDimitris Dimitriadis
568
asked Jan 7 at 18:59
Dimitris DimitriadisDimitris Dimitriadis
568
568
$begingroup$
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
add a comment |
$begingroup$
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
$begingroup$
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
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Look at the definition of the periodic function below
'A function $f$ is periodic if there exists $Tin mathbb{R}$ 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
$$g(x+T_1)=gleft(x+frac{T}{a}right)=fleft( aleft(x+frac{T}{a} right)+b right)=f(ax+b+T)=f(ax+b)=g(x).$$
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 $f(ax+b)$.
answered Jan 7 at 19:04
Lev BanLev Ban
1,0771317
$endgroup$
$begingroup$
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
|
show 1 more comment
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Look at the definition of the periodic function below
'A function $f$ is periodic if there exists $Tin mathbb{R}$ 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
$$g(x+T_1)=gleft(x+frac{T}{a}right)=fleft( aleft(x+frac{T}{a} right)+b right)=f(ax+b+T)=f(ax+b)=g(x).$$
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 $f(ax+b)$.
answered Jan 7 at 19:04
Lev BanLev Ban
1,0771317
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
Look at the definition of the periodic function below
'A function $f$ is periodic if there exists $Tin mathbb{R}$ 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
$$g(x+T_1)=gleft(x+frac{T}{a}right)=fleft( aleft(x+frac{T}{a} right)+b right)=f(ax+b+T)=f(ax+b)=g(x).$$
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 $f(ax+b)$.
answered Jan 7 at 19:04
Lev BanLev Ban
1,0771317
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
Look at the definition of the periodic function below
'A function $f$ is periodic if there exists $Tin mathbb{R}$ 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
$$g(x+T_1)=gleft(x+frac{T}{a}right)=fleft( aleft(x+frac{T}{a} right)+b right)=f(ax+b+T)=f(ax+b)=g(x).$$
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 $f(ax+b)$.
answered Jan 7 at 19:04
Lev BanLev Ban
1,0771317
$endgroup$
Look at the definition of the periodic function below
'A function $f$ is periodic if there exists $Tin mathbb{R}$ 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
$$g(x+T_1)=gleft(x+frac{T}{a}right)=fleft( aleft(x+frac{T}{a} right)+b right)=f(ax+b+T)=f(ax+b)=g(x).$$
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 $f(ax+b)$.
answered Jan 7 at 19:04
Lev BanLev Ban
1,0771317
edited Jan 7 at 19:26
answered Jan 7 at 19:04
Lev BanLev Ban
1,0771317
answered Jan 7 at 19:04
Lev BanLev Ban
1,0771317
answered Jan 7 at 19:04
Lev BanLev Ban
1,0771317
1,0771317
$begingroup$
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
|
show 1 more comment
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$begingroup$
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
$begingroup$
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
$begingroup$
@DimitrisDimitriadis I updated my answer so that I can answer your last question.
$endgroup$
– Lev Ban
Jan 7 at 19:26
|
show 1 more comment
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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