Is the reciprocal of a periodic function periodic too? [closed]
up vote
-2
down vote
favorite
Let us assume we have a periodic function f(r)=f(r+L), can we conclude if its reciprocal function 1/f(x) is periodic or not?
periodic-functions
asked Nov 18 at 21:38
Jared Lo
1116
closed as off-topic by Jyrki Lahtonen, amWhy, Zvi, Rebellos, user10354138 Nov 20 at 1:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, amWhy, Zvi, Rebellos, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-2
down vote
favorite
Let us assume we have a periodic function f(r)=f(r+L), can we conclude if its reciprocal function 1/f(x) is periodic or not?
periodic-functions
asked Nov 18 at 21:38
Jared Lo
1116
closed as off-topic by Jyrki Lahtonen, amWhy, Zvi, Rebellos, user10354138 Nov 20 at 1:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, amWhy, Zvi, Rebellos, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
2
Yup. Let $f(x)$ be periodic with period $p$, and let $g(x) = 1/f(x)$. Then notice, as $f(x+p)=f(x)$, $$g(x+p) = frac{1}{f(x+p)} = frac{1}{f(x)}=g(x)$$ So unless there's some nuance I'm overlooking, not having much experience in this area, the reciprocal of periodic functions should be periodic.
– Eevee Trainer
Nov 18 at 21:41
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Let us assume we have a periodic function f(r)=f(r+L), can we conclude if its reciprocal function 1/f(x) is periodic or not?
periodic-functions
asked Nov 18 at 21:38
Jared Lo
1116
Let us assume we have a periodic function f(r)=f(r+L), can we conclude if its reciprocal function 1/f(x) is periodic or not?
periodic-functions
periodic-functions
asked Nov 18 at 21:38
Jared Lo
1116
asked Nov 18 at 21:38
Jared Lo
1116
asked Nov 18 at 21:38
Jared Lo
1116
asked Nov 18 at 21:38
Jared Lo
1116
asked Nov 18 at 21:38
Jared Lo
1116
1116
closed as off-topic by Jyrki Lahtonen, amWhy, Zvi, Rebellos, user10354138 Nov 20 at 1:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, amWhy, Zvi, Rebellos, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Jyrki Lahtonen, amWhy, Zvi, Rebellos, user10354138 Nov 20 at 1:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, amWhy, Zvi, Rebellos, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
2
Yup. Let $f(x)$ be periodic with period $p$, and let $g(x) = 1/f(x)$. Then notice, as $f(x+p)=f(x)$, $$g(x+p) = frac{1}{f(x+p)} = frac{1}{f(x)}=g(x)$$ So unless there's some nuance I'm overlooking, not having much experience in this area, the reciprocal of periodic functions should be periodic.
– Eevee Trainer
Nov 18 at 21:41
add a comment |
2
Yup. Let $f(x)$ be periodic with period $p$, and let $g(x) = 1/f(x)$. Then notice, as $f(x+p)=f(x)$, $$g(x+p) = frac{1}{f(x+p)} = frac{1}{f(x)}=g(x)$$ So unless there's some nuance I'm overlooking, not having much experience in this area, the reciprocal of periodic functions should be periodic.
– Eevee Trainer
Nov 18 at 21:41
2
2
Yup. Let $f(x)$ be periodic with period $p$, and let $g(x) = 1/f(x)$. Then notice, as $f(x+p)=f(x)$, $$g(x+p) = frac{1}{f(x+p)} = frac{1}{f(x)}=g(x)$$ So unless there's some nuance I'm overlooking, not having much experience in this area, the reciprocal of periodic functions should be periodic.
– Eevee Trainer
Nov 18 at 21:41
Yup. Let $f(x)$ be periodic with period $p$, and let $g(x) = 1/f(x)$. Then notice, as $f(x+p)=f(x)$, $$g(x+p) = frac{1}{f(x+p)} = frac{1}{f(x)}=g(x)$$ So unless there's some nuance I'm overlooking, not having much experience in this area, the reciprocal of periodic functions should be periodic.
– Eevee Trainer
Nov 18 at 21:41
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
We have that
$$f(r)=f(r+L) implies frac1{f(r)}=frac1{f(r+L)}$$
and therefore $1/f(x)$ is periodic providing that $f(x) neq 0$ .
answered Nov 18 at 21:42
gimusi
87.4k74393
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We have that
$$f(r)=f(r+L) implies frac1{f(r)}=frac1{f(r+L)}$$
and therefore $1/f(x)$ is periodic providing that $f(x) neq 0$ .
answered Nov 18 at 21:42
gimusi
87.4k74393
add a comment |
up vote
0
down vote
We have that
$$f(r)=f(r+L) implies frac1{f(r)}=frac1{f(r+L)}$$
and therefore $1/f(x)$ is periodic providing that $f(x) neq 0$ .
answered Nov 18 at 21:42
gimusi
87.4k74393
add a comment |
up vote
0
down vote
up vote
0
down vote
We have that
$$f(r)=f(r+L) implies frac1{f(r)}=frac1{f(r+L)}$$
and therefore $1/f(x)$ is periodic providing that $f(x) neq 0$ .
answered Nov 18 at 21:42
gimusi
87.4k74393
We have that
$$f(r)=f(r+L) implies frac1{f(r)}=frac1{f(r+L)}$$
and therefore $1/f(x)$ is periodic providing that $f(x) neq 0$ .
answered Nov 18 at 21:42
gimusi
87.4k74393
answered Nov 18 at 21:42
gimusi
87.4k74393
answered Nov 18 at 21:42
gimusi
87.4k74393
answered Nov 18 at 21:42
gimusi
87.4k74393
87.4k74393
add a comment |
add a comment |
2
Yup. Let $f(x)$ be periodic with period $p$, and let $g(x) = 1/f(x)$. Then notice, as $f(x+p)=f(x)$, $$g(x+p) = frac{1}{f(x+p)} = frac{1}{f(x)}=g(x)$$ So unless there's some nuance I'm overlooking, not having much experience in this area, the reciprocal of periodic functions should be periodic.
– Eevee Trainer
Nov 18 at 21:41