Mixing
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Do real valued continuous periodic functions form a Vector Space over the Field of all functions?
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Preface - I'm very new to Abstract Algebra, so if my terminology/interpretations are off, my apologies.
So I was thinking about real valued continuous non zero functions (from here I'll just refer to them as functions) and that when they are periodic of the same period (from here Ill just say periodic) we observe that the product of two periodic functions is periodic as is the sum of two periodic functions. After some easy work, I found that under the real definition of addition and multiplication that they form a Field with the Zero Element $0(x) = 0$ and the One Element $1(x) = 1$ and additive and multiplicative inverses defined as for real values, i.e. $-a(x) = -1 times a(x)$ and $a(x)^-1 = frac1a(x)$
Anyways, it got me thinking about how structures that have periodic and non periodic functions could be defined. I may be wrong, but given that the sum of two periodic functions is periodic and the product of a periodic with a non periodic is periodic that we can define a Vector Space $(V,+,times)$ over the Field $F$ where
$V$ = set of periodic functions
$F$ = set of all non-zero functions
Vector addition is the real addition of two periodic functions. Scalar Multiplication is the real multiplication of a non-zero function with a periodic function.
I believe this may be OK, but I'm concerned that the Scalar Field contains Vectors - i.e. a periodic function is a function and thus in F. I suppose I could circumnavigate this by defining F as non periodic non-zero functions but I'm not sure if I'm allowed to have it as it currently is defined.
Are there any pointers/references people can suggest.
Thanks,
David
Ps - Another form of this would be if we took integers divisible by a given integer k (addition of two divisible by k values and scalar mult of any integer with one divisible by k).
Edit _ as pointed out if can’t occur as the function set do not form a field.
abstract-algebra vector-spaces
asked Jul 29 at 11:59
user3053801
1008
 |Â
show 2 more comments
up vote
1
down vote
favorite
Preface - I'm very new to Abstract Algebra, so if my terminology/interpretations are off, my apologies.
So I was thinking about real valued continuous non zero functions (from here I'll just refer to them as functions) and that when they are periodic of the same period (from here Ill just say periodic) we observe that the product of two periodic functions is periodic as is the sum of two periodic functions. After some easy work, I found that under the real definition of addition and multiplication that they form a Field with the Zero Element $0(x) = 0$ and the One Element $1(x) = 1$ and additive and multiplicative inverses defined as for real values, i.e. $-a(x) = -1 times a(x)$ and $a(x)^-1 = frac1a(x)$
Anyways, it got me thinking about how structures that have periodic and non periodic functions could be defined. I may be wrong, but given that the sum of two periodic functions is periodic and the product of a periodic with a non periodic is periodic that we can define a Vector Space $(V,+,times)$ over the Field $F$ where
$V$ = set of periodic functions
$F$ = set of all non-zero functions
Vector addition is the real addition of two periodic functions. Scalar Multiplication is the real multiplication of a non-zero function with a periodic function.
I believe this may be OK, but I'm concerned that the Scalar Field contains Vectors - i.e. a periodic function is a function and thus in F. I suppose I could circumnavigate this by defining F as non periodic non-zero functions but I'm not sure if I'm allowed to have it as it currently is defined.
Are there any pointers/references people can suggest.
Thanks,
David
Ps - Another form of this would be if we took integers divisible by a given integer k (addition of two divisible by k values and scalar mult of any integer with one divisible by k).
Edit _ as pointed out if can’t occur as the function set do not form a field.
abstract-algebra vector-spaces
asked Jul 29 at 11:59
user3053801
1008
This is not clear. Are you fixing the period? If not, then neither the sum nor product have to be periodic. And if your function has zeroes then it's multiplicative inverse can't be continuous.
– lulu
Jul 29 at 12:05
@lulu - Cheers. Yes, sorry I am fixing the period and only considering non-zero functions. I will edit now to reflect.
– user3053801
Jul 29 at 12:19
Actually, sorry, why does it need to have the same period? Is it not true that the sum/product of two periodic functions is also periodic. The condition that is required (I believe) is that they both have to be comeasurable. Here both are real valued continuous so I believe satisfy that. But 100% re non-zero functions. Thanks again for your comment.
– user3053801
Jul 29 at 12:28
Well, if the period of one function is a multiple of the period of the other, then sure. But then of course both are periodic with the same period (the larger one). But, really, it is your definition. Just write out in detail what you want and verify that it works (if verification is required).
– lulu
Jul 29 at 12:33
Sorry, I was confused when you said 'then neither the sum nor product have to be periodic' - my bad.
– user3053801
Jul 29 at 12:38
 |Â
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Preface - I'm very new to Abstract Algebra, so if my terminology/interpretations are off, my apologies.
So I was thinking about real valued continuous non zero functions (from here I'll just refer to them as functions) and that when they are periodic of the same period (from here Ill just say periodic) we observe that the product of two periodic functions is periodic as is the sum of two periodic functions. After some easy work, I found that under the real definition of addition and multiplication that they form a Field with the Zero Element $0(x) = 0$ and the One Element $1(x) = 1$ and additive and multiplicative inverses defined as for real values, i.e. $-a(x) = -1 times a(x)$ and $a(x)^-1 = frac1a(x)$
Anyways, it got me thinking about how structures that have periodic and non periodic functions could be defined. I may be wrong, but given that the sum of two periodic functions is periodic and the product of a periodic with a non periodic is periodic that we can define a Vector Space $(V,+,times)$ over the Field $F$ where
$V$ = set of periodic functions
$F$ = set of all non-zero functions
Vector addition is the real addition of two periodic functions. Scalar Multiplication is the real multiplication of a non-zero function with a periodic function.
I believe this may be OK, but I'm concerned that the Scalar Field contains Vectors - i.e. a periodic function is a function and thus in F. I suppose I could circumnavigate this by defining F as non periodic non-zero functions but I'm not sure if I'm allowed to have it as it currently is defined.
Are there any pointers/references people can suggest.
Thanks,
David
Ps - Another form of this would be if we took integers divisible by a given integer k (addition of two divisible by k values and scalar mult of any integer with one divisible by k).
Edit _ as pointed out if can’t occur as the function set do not form a field.
abstract-algebra vector-spaces
asked Jul 29 at 11:59
user3053801
1008
Preface - I'm very new to Abstract Algebra, so if my terminology/interpretations are off, my apologies.
So I was thinking about real valued continuous non zero functions (from here I'll just refer to them as functions) and that when they are periodic of the same period (from here Ill just say periodic) we observe that the product of two periodic functions is periodic as is the sum of two periodic functions. After some easy work, I found that under the real definition of addition and multiplication that they form a Field with the Zero Element $0(x) = 0$ and the One Element $1(x) = 1$ and additive and multiplicative inverses defined as for real values, i.e. $-a(x) = -1 times a(x)$ and $a(x)^-1 = frac1a(x)$
Anyways, it got me thinking about how structures that have periodic and non periodic functions could be defined. I may be wrong, but given that the sum of two periodic functions is periodic and the product of a periodic with a non periodic is periodic that we can define a Vector Space $(V,+,times)$ over the Field $F$ where
$V$ = set of periodic functions
$F$ = set of all non-zero functions
Vector addition is the real addition of two periodic functions. Scalar Multiplication is the real multiplication of a non-zero function with a periodic function.
I believe this may be OK, but I'm concerned that the Scalar Field contains Vectors - i.e. a periodic function is a function and thus in F. I suppose I could circumnavigate this by defining F as non periodic non-zero functions but I'm not sure if I'm allowed to have it as it currently is defined.
Are there any pointers/references people can suggest.
Thanks,
David
Ps - Another form of this would be if we took integers divisible by a given integer k (addition of two divisible by k values and scalar mult of any integer with one divisible by k).
Edit _ as pointed out if can’t occur as the function set do not form a field.
abstract-algebra vector-spaces
asked Jul 29 at 11:59
user3053801
1008
edited Jul 29 at 13:04
asked Jul 29 at 11:59
user3053801
1008
asked Jul 29 at 11:59
user3053801
1008
asked Jul 29 at 11:59
user3053801
1008
1008
This is not clear. Are you fixing the period? If not, then neither the sum nor product have to be periodic. And if your function has zeroes then it's multiplicative inverse can't be continuous.
– lulu
Jul 29 at 12:05
@lulu - Cheers. Yes, sorry I am fixing the period and only considering non-zero functions. I will edit now to reflect.
– user3053801
Jul 29 at 12:19
Actually, sorry, why does it need to have the same period? Is it not true that the sum/product of two periodic functions is also periodic. The condition that is required (I believe) is that they both have to be comeasurable. Here both are real valued continuous so I believe satisfy that. But 100% re non-zero functions. Thanks again for your comment.
– user3053801
Jul 29 at 12:28
Well, if the period of one function is a multiple of the period of the other, then sure. But then of course both are periodic with the same period (the larger one). But, really, it is your definition. Just write out in detail what you want and verify that it works (if verification is required).
– lulu
Jul 29 at 12:33
Sorry, I was confused when you said 'then neither the sum nor product have to be periodic' - my bad.
– user3053801
Jul 29 at 12:38
 |Â
show 2 more comments
This is not clear. Are you fixing the period? If not, then neither the sum nor product have to be periodic. And if your function has zeroes then it's multiplicative inverse can't be continuous.
– lulu
Jul 29 at 12:05
@lulu - Cheers. Yes, sorry I am fixing the period and only considering non-zero functions. I will edit now to reflect.
– user3053801
Jul 29 at 12:19
Actually, sorry, why does it need to have the same period? Is it not true that the sum/product of two periodic functions is also periodic. The condition that is required (I believe) is that they both have to be comeasurable. Here both are real valued continuous so I believe satisfy that. But 100% re non-zero functions. Thanks again for your comment.
– user3053801
Jul 29 at 12:28
Well, if the period of one function is a multiple of the period of the other, then sure. But then of course both are periodic with the same period (the larger one). But, really, it is your definition. Just write out in detail what you want and verify that it works (if verification is required).
– lulu
Jul 29 at 12:33
Sorry, I was confused when you said 'then neither the sum nor product have to be periodic' - my bad.
– user3053801
Jul 29 at 12:38
This is not clear. Are you fixing the period? If not, then neither the sum nor product have to be periodic. And if your function has zeroes then it's multiplicative inverse can't be continuous.
– lulu
Jul 29 at 12:05
This is not clear. Are you fixing the period? If not, then neither the sum nor product have to be periodic. And if your function has zeroes then it's multiplicative inverse can't be continuous.
– lulu
Jul 29 at 12:05
@lulu - Cheers. Yes, sorry I am fixing the period and only considering non-zero functions. I will edit now to reflect.
– user3053801
Jul 29 at 12:19
@lulu - Cheers. Yes, sorry I am fixing the period and only considering non-zero functions. I will edit now to reflect.
– user3053801
Jul 29 at 12:19
Actually, sorry, why does it need to have the same period? Is it not true that the sum/product of two periodic functions is also periodic. The condition that is required (I believe) is that they both have to be comeasurable. Here both are real valued continuous so I believe satisfy that. But 100% re non-zero functions. Thanks again for your comment.
– user3053801
Jul 29 at 12:28
Actually, sorry, why does it need to have the same period? Is it not true that the sum/product of two periodic functions is also periodic. The condition that is required (I believe) is that they both have to be comeasurable. Here both are real valued continuous so I believe satisfy that. But 100% re non-zero functions. Thanks again for your comment.
– user3053801
Jul 29 at 12:28
Well, if the period of one function is a multiple of the period of the other, then sure. But then of course both are periodic with the same period (the larger one). But, really, it is your definition. Just write out in detail what you want and verify that it works (if verification is required).
– lulu
Jul 29 at 12:33
Well, if the period of one function is a multiple of the period of the other, then sure. But then of course both are periodic with the same period (the larger one). But, really, it is your definition. Just write out in detail what you want and verify that it works (if verification is required).
– lulu
Jul 29 at 12:33
Sorry, I was confused when you said 'then neither the sum nor product have to be periodic' - my bad.
– user3053801
Jul 29 at 12:38
Sorry, I was confused when you said 'then neither the sum nor product have to be periodic' - my bad.
– user3053801
Jul 29 at 12:38
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You can't use the set of functions as scalars since they don't form a field. Functions that have zeroes somewhere don't have inverses. This has nothing to do with periodicity.
Restricting the scalars to functions that are never $0$ won't fix things. The sum of two such functions can have a root, so the set of those functions aren't closed under addition.
Your last PS is confusing, but hints at using the field of integers modulo a prime $p$ for the scalars. That will work: see the wikipedia entry on
finite fields
.
answered Jul 29 at 12:08
Ethan Bolker
35.7k54199
Cheers for your comment. What if we restricted to non-zero functions?
– user3053801
Jul 29 at 12:15
I have edited to reflect your comment. Thanks again for pointing it out. Sorry for the confusing PS part.
– user3053801
Jul 29 at 12:22
See my edit. Your suggestion doesn't fix things because they are essentially unfixable. The functions form a ring but not a field.
– Ethan Bolker
Jul 29 at 12:41
Even non zero functions?
– user3053801
Jul 29 at 12:49
Suppose $f(2) = 4$ and $g(2) = -4$. What can you say about $f+g$ at $2$?
– Ethan Bolker
Jul 29 at 12:55
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You can't use the set of functions as scalars since they don't form a field. Functions that have zeroes somewhere don't have inverses. This has nothing to do with periodicity.
Restricting the scalars to functions that are never $0$ won't fix things. The sum of two such functions can have a root, so the set of those functions aren't closed under addition.
Your last PS is confusing, but hints at using the field of integers modulo a prime $p$ for the scalars. That will work: see the wikipedia entry on
finite fields
.
answered Jul 29 at 12:08
Ethan Bolker
35.7k54199
Cheers for your comment. What if we restricted to non-zero functions?
– user3053801
Jul 29 at 12:15
I have edited to reflect your comment. Thanks again for pointing it out. Sorry for the confusing PS part.
– user3053801
Jul 29 at 12:22
See my edit. Your suggestion doesn't fix things because they are essentially unfixable. The functions form a ring but not a field.
– Ethan Bolker
Jul 29 at 12:41
Even non zero functions?
– user3053801
Jul 29 at 12:49
Suppose $f(2) = 4$ and $g(2) = -4$. What can you say about $f+g$ at $2$?
– Ethan Bolker
Jul 29 at 12:55
 |Â
show 4 more comments
up vote
1
down vote
accepted
You can't use the set of functions as scalars since they don't form a field. Functions that have zeroes somewhere don't have inverses. This has nothing to do with periodicity.
Restricting the scalars to functions that are never $0$ won't fix things. The sum of two such functions can have a root, so the set of those functions aren't closed under addition.
Your last PS is confusing, but hints at using the field of integers modulo a prime $p$ for the scalars. That will work: see the wikipedia entry on
finite fields
.
answered Jul 29 at 12:08
Ethan Bolker
35.7k54199
Cheers for your comment. What if we restricted to non-zero functions?
– user3053801
Jul 29 at 12:15
I have edited to reflect your comment. Thanks again for pointing it out. Sorry for the confusing PS part.
– user3053801
Jul 29 at 12:22
See my edit. Your suggestion doesn't fix things because they are essentially unfixable. The functions form a ring but not a field.
– Ethan Bolker
Jul 29 at 12:41
Even non zero functions?
– user3053801
Jul 29 at 12:49
Suppose $f(2) = 4$ and $g(2) = -4$. What can you say about $f+g$ at $2$?
– Ethan Bolker
Jul 29 at 12:55
 |Â
show 4 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You can't use the set of functions as scalars since they don't form a field. Functions that have zeroes somewhere don't have inverses. This has nothing to do with periodicity.
Restricting the scalars to functions that are never $0$ won't fix things. The sum of two such functions can have a root, so the set of those functions aren't closed under addition.
Your last PS is confusing, but hints at using the field of integers modulo a prime $p$ for the scalars. That will work: see the wikipedia entry on
finite fields
.
answered Jul 29 at 12:08
Ethan Bolker
35.7k54199
You can't use the set of functions as scalars since they don't form a field. Functions that have zeroes somewhere don't have inverses. This has nothing to do with periodicity.
Restricting the scalars to functions that are never $0$ won't fix things. The sum of two such functions can have a root, so the set of those functions aren't closed under addition.
Your last PS is confusing, but hints at using the field of integers modulo a prime $p$ for the scalars. That will work: see the wikipedia entry on
finite fields
.
answered Jul 29 at 12:08
Ethan Bolker
35.7k54199
edited Jul 29 at 12:40
answered Jul 29 at 12:08
Ethan Bolker
35.7k54199
answered Jul 29 at 12:08
Ethan Bolker
35.7k54199
answered Jul 29 at 12:08
Ethan Bolker
35.7k54199
35.7k54199
Cheers for your comment. What if we restricted to non-zero functions?
– user3053801
Jul 29 at 12:15
I have edited to reflect your comment. Thanks again for pointing it out. Sorry for the confusing PS part.
– user3053801
Jul 29 at 12:22
See my edit. Your suggestion doesn't fix things because they are essentially unfixable. The functions form a ring but not a field.
– Ethan Bolker
Jul 29 at 12:41
Even non zero functions?
– user3053801
Jul 29 at 12:49
Suppose $f(2) = 4$ and $g(2) = -4$. What can you say about $f+g$ at $2$?
– Ethan Bolker
Jul 29 at 12:55
 |Â
show 4 more comments
Cheers for your comment. What if we restricted to non-zero functions?
– user3053801
Jul 29 at 12:15
I have edited to reflect your comment. Thanks again for pointing it out. Sorry for the confusing PS part.
– user3053801
Jul 29 at 12:22
See my edit. Your suggestion doesn't fix things because they are essentially unfixable. The functions form a ring but not a field.
– Ethan Bolker
Jul 29 at 12:41
Even non zero functions?
– user3053801
Jul 29 at 12:49
Suppose $f(2) = 4$ and $g(2) = -4$. What can you say about $f+g$ at $2$?
– Ethan Bolker
Jul 29 at 12:55
Cheers for your comment. What if we restricted to non-zero functions?
– user3053801
Jul 29 at 12:15
Cheers for your comment. What if we restricted to non-zero functions?
– user3053801
Jul 29 at 12:15
I have edited to reflect your comment. Thanks again for pointing it out. Sorry for the confusing PS part.
– user3053801
Jul 29 at 12:22
I have edited to reflect your comment. Thanks again for pointing it out. Sorry for the confusing PS part.
– user3053801
Jul 29 at 12:22
See my edit. Your suggestion doesn't fix things because they are essentially unfixable. The functions form a ring but not a field.
– Ethan Bolker
Jul 29 at 12:41
See my edit. Your suggestion doesn't fix things because they are essentially unfixable. The functions form a ring but not a field.
– Ethan Bolker
Jul 29 at 12:41
Even non zero functions?
– user3053801
Jul 29 at 12:49
Even non zero functions?
– user3053801
Jul 29 at 12:49
Suppose $f(2) = 4$ and $g(2) = -4$. What can you say about $f+g$ at $2$?
– Ethan Bolker
Jul 29 at 12:55
Suppose $f(2) = 4$ and $g(2) = -4$. What can you say about $f+g$ at $2$?
– Ethan Bolker
Jul 29 at 12:55
 |Â
show 4 more comments
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This is not clear. Are you fixing the period? If not, then neither the sum nor product have to be periodic. And if your function has zeroes then it's multiplicative inverse can't be continuous.
– lulu
Jul 29 at 12:05
@lulu - Cheers. Yes, sorry I am fixing the period and only considering non-zero functions. I will edit now to reflect.
– user3053801
Jul 29 at 12:19
Actually, sorry, why does it need to have the same period? Is it not true that the sum/product of two periodic functions is also periodic. The condition that is required (I believe) is that they both have to be comeasurable. Here both are real valued continuous so I believe satisfy that. But 100% re non-zero functions. Thanks again for your comment.
– user3053801
Jul 29 at 12:28
Well, if the period of one function is a multiple of the period of the other, then sure. But then of course both are periodic with the same period (the larger one). But, really, it is your definition. Just write out in detail what you want and verify that it works (if verification is required).
– lulu
Jul 29 at 12:33
Sorry, I was confused when you said 'then neither the sum nor product have to be periodic' - my bad.
– user3053801
Jul 29 at 12:38