Is the set of periodic functions from $mathbb R$ to $mathbb R$a subspace of $mathbb R^{mathbb R}$?
$begingroup$
A function $f:mathbb R→mathbb R$ is called periodic if there exists a positive number $p$ such that $f(x)=f(x+p)$ for all $xinmathbb R$. Is the set of periodic functions from $mathbb{R}$ to $mathbb{R}$ a subspace of $mathbb{R}^mathbb{R}$? Explain.
The question has been asked here
I still don't quite understand why the ratio of two periods should not be irrational.
For example if $beta/alpha=r/s$ and $r/s = sqrt{2}$, the following equation still works?
$$h(x+sbeta)=f(x+ralpha)+g(x+sbeta)=f(x)+g(x)=h(x)$$
$$h(x+beta)=f(x+sqrt{2}alpha)+g(x+beta)=f(x)+g(x)=h(x)$$
linear-algebra proof-explanation
edited Dec 25 '18 at 0:41
Chris Custer
14.2k3827
asked Dec 25 '18 at 0:16
JOHN JOHN
4209
$endgroup$
add a comment |
$begingroup$
A function $f:mathbb R→mathbb R$ is called periodic if there exists a positive number $p$ such that $f(x)=f(x+p)$ for all $xinmathbb R$. Is the set of periodic functions from $mathbb{R}$ to $mathbb{R}$ a subspace of $mathbb{R}^mathbb{R}$? Explain.
The question has been asked here
I still don't quite understand why the ratio of two periods should not be irrational.
For example if $beta/alpha=r/s$ and $r/s = sqrt{2}$, the following equation still works?
$$h(x+sbeta)=f(x+ralpha)+g(x+sbeta)=f(x)+g(x)=h(x)$$
$$h(x+beta)=f(x+sqrt{2}alpha)+g(x+beta)=f(x)+g(x)=h(x)$$
linear-algebra proof-explanation
edited Dec 25 '18 at 0:41
Chris Custer
14.2k3827
asked Dec 25 '18 at 0:16
JOHN JOHN
4209
$endgroup$
add a comment |
$begingroup$
A function $f:mathbb R→mathbb R$ is called periodic if there exists a positive number $p$ such that $f(x)=f(x+p)$ for all $xinmathbb R$. Is the set of periodic functions from $mathbb{R}$ to $mathbb{R}$ a subspace of $mathbb{R}^mathbb{R}$? Explain.
The question has been asked here
I still don't quite understand why the ratio of two periods should not be irrational.
For example if $beta/alpha=r/s$ and $r/s = sqrt{2}$, the following equation still works?
$$h(x+sbeta)=f(x+ralpha)+g(x+sbeta)=f(x)+g(x)=h(x)$$
$$h(x+beta)=f(x+sqrt{2}alpha)+g(x+beta)=f(x)+g(x)=h(x)$$
linear-algebra proof-explanation
edited Dec 25 '18 at 0:41
Chris Custer
14.2k3827
asked Dec 25 '18 at 0:16
JOHN JOHN
4209
$endgroup$
A function $f:mathbb R→mathbb R$ is called periodic if there exists a positive number $p$ such that $f(x)=f(x+p)$ for all $xinmathbb R$. Is the set of periodic functions from $mathbb{R}$ to $mathbb{R}$ a subspace of $mathbb{R}^mathbb{R}$? Explain.
The question has been asked here
I still don't quite understand why the ratio of two periods should not be irrational.
For example if $beta/alpha=r/s$ and $r/s = sqrt{2}$, the following equation still works?
$$h(x+sbeta)=f(x+ralpha)+g(x+sbeta)=f(x)+g(x)=h(x)$$
$$h(x+beta)=f(x+sqrt{2}alpha)+g(x+beta)=f(x)+g(x)=h(x)$$
linear-algebra proof-explanation
linear-algebra proof-explanation
edited Dec 25 '18 at 0:41
Chris Custer
14.2k3827
asked Dec 25 '18 at 0:16
JOHN JOHN
4209
edited Dec 25 '18 at 0:41
Chris Custer
14.2k3827
asked Dec 25 '18 at 0:16
JOHN JOHN
4209
edited Dec 25 '18 at 0:41
Chris Custer
14.2k3827
edited Dec 25 '18 at 0:41
Chris Custer
14.2k3827
edited Dec 25 '18 at 0:41
Chris Custer
14.2k3827
14.2k3827
asked Dec 25 '18 at 0:16
JOHN JOHN
4209
asked Dec 25 '18 at 0:16
JOHN JOHN
4209
asked Dec 25 '18 at 0:16
JOHN JOHN
4209
4209
add a comment |
add a comment |
2 Answers
2
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$begingroup$
If $m(x)$ is periodic with $p$, then it is also periodic with integral multiples of $p$, because $m(x)=m(x+p)=m(x+p+p)...$, but we can't say about non-integral multiples of $p$.
Note that you can only write $f(x+ralpha)=f(x),g(x+sbeta)=g(x)$ in general if $r,s$ are integers. If $r/s$ is irrational, at-least one of $r,s$ is non-integral, and the equality doesn't work in general. As an example, take $f(x)=sin(sqrt2 x),g(x)=sin(sqrt3x)$.
answered Dec 25 '18 at 0:35
Shubham JohriShubham Johri
5,204718
$endgroup$
1
$begingroup$
An even better example might be $f(x)=cos(sqrt2 x)$ and $g(x)=cos(sqrt3 x)$ -- where there is a particularly direct argument that $f+g$ cannot be periodic with any period, namely that $f(x)+g(x)=2$ has exactly one solution, $x=0$.
$endgroup$
– Henning Makholm
Dec 25 '18 at 3:03
add a comment |
$begingroup$
$f$ has period $p$ if $f(x+p)=f(x)$ for all $x$. (Equivalently, $f(x+np)=f(x)$ for all $x$ and for all integers $n$). If $r$ is a real number and $f(x+rp)=f(x)$ for all $x$ we cannot say $f$ is periodic. In the quoted proof $r$ and $s$ are integers and that makes $frac r s$ rational.
answered Dec 25 '18 at 0:20
Kavi Rama MurthyKavi Rama Murthy
65.8k42867
$endgroup$
add a comment |
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2 Answers
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oldest
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2 Answers
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$begingroup$
If $m(x)$ is periodic with $p$, then it is also periodic with integral multiples of $p$, because $m(x)=m(x+p)=m(x+p+p)...$, but we can't say about non-integral multiples of $p$.
Note that you can only write $f(x+ralpha)=f(x),g(x+sbeta)=g(x)$ in general if $r,s$ are integers. If $r/s$ is irrational, at-least one of $r,s$ is non-integral, and the equality doesn't work in general. As an example, take $f(x)=sin(sqrt2 x),g(x)=sin(sqrt3x)$.
answered Dec 25 '18 at 0:35
Shubham JohriShubham Johri
5,204718
$endgroup$
1
$begingroup$
An even better example might be $f(x)=cos(sqrt2 x)$ and $g(x)=cos(sqrt3 x)$ -- where there is a particularly direct argument that $f+g$ cannot be periodic with any period, namely that $f(x)+g(x)=2$ has exactly one solution, $x=0$.
$endgroup$
– Henning Makholm
Dec 25 '18 at 3:03
add a comment |
$begingroup$
If $m(x)$ is periodic with $p$, then it is also periodic with integral multiples of $p$, because $m(x)=m(x+p)=m(x+p+p)...$, but we can't say about non-integral multiples of $p$.
Note that you can only write $f(x+ralpha)=f(x),g(x+sbeta)=g(x)$ in general if $r,s$ are integers. If $r/s$ is irrational, at-least one of $r,s$ is non-integral, and the equality doesn't work in general. As an example, take $f(x)=sin(sqrt2 x),g(x)=sin(sqrt3x)$.
answered Dec 25 '18 at 0:35
Shubham JohriShubham Johri
5,204718
$endgroup$
1
$begingroup$
An even better example might be $f(x)=cos(sqrt2 x)$ and $g(x)=cos(sqrt3 x)$ -- where there is a particularly direct argument that $f+g$ cannot be periodic with any period, namely that $f(x)+g(x)=2$ has exactly one solution, $x=0$.
$endgroup$
– Henning Makholm
Dec 25 '18 at 3:03
add a comment |
$begingroup$
If $m(x)$ is periodic with $p$, then it is also periodic with integral multiples of $p$, because $m(x)=m(x+p)=m(x+p+p)...$, but we can't say about non-integral multiples of $p$.
Note that you can only write $f(x+ralpha)=f(x),g(x+sbeta)=g(x)$ in general if $r,s$ are integers. If $r/s$ is irrational, at-least one of $r,s$ is non-integral, and the equality doesn't work in general. As an example, take $f(x)=sin(sqrt2 x),g(x)=sin(sqrt3x)$.
answered Dec 25 '18 at 0:35
Shubham JohriShubham Johri
5,204718
$endgroup$
If $m(x)$ is periodic with $p$, then it is also periodic with integral multiples of $p$, because $m(x)=m(x+p)=m(x+p+p)...$, but we can't say about non-integral multiples of $p$.
Note that you can only write $f(x+ralpha)=f(x),g(x+sbeta)=g(x)$ in general if $r,s$ are integers. If $r/s$ is irrational, at-least one of $r,s$ is non-integral, and the equality doesn't work in general. As an example, take $f(x)=sin(sqrt2 x),g(x)=sin(sqrt3x)$.
answered Dec 25 '18 at 0:35
Shubham JohriShubham Johri
5,204718
answered Dec 25 '18 at 0:35
Shubham JohriShubham Johri
5,204718
answered Dec 25 '18 at 0:35
Shubham JohriShubham Johri
5,204718
answered Dec 25 '18 at 0:35
Shubham JohriShubham Johri
5,204718
5,204718
1
$begingroup$
An even better example might be $f(x)=cos(sqrt2 x)$ and $g(x)=cos(sqrt3 x)$ -- where there is a particularly direct argument that $f+g$ cannot be periodic with any period, namely that $f(x)+g(x)=2$ has exactly one solution, $x=0$.
$endgroup$
– Henning Makholm
Dec 25 '18 at 3:03
add a comment |
1
$begingroup$
An even better example might be $f(x)=cos(sqrt2 x)$ and $g(x)=cos(sqrt3 x)$ -- where there is a particularly direct argument that $f+g$ cannot be periodic with any period, namely that $f(x)+g(x)=2$ has exactly one solution, $x=0$.
$endgroup$
– Henning Makholm
Dec 25 '18 at 3:03
1
1
$begingroup$
An even better example might be $f(x)=cos(sqrt2 x)$ and $g(x)=cos(sqrt3 x)$ -- where there is a particularly direct argument that $f+g$ cannot be periodic with any period, namely that $f(x)+g(x)=2$ has exactly one solution, $x=0$.
$endgroup$
– Henning Makholm
Dec 25 '18 at 3:03
$begingroup$
An even better example might be $f(x)=cos(sqrt2 x)$ and $g(x)=cos(sqrt3 x)$ -- where there is a particularly direct argument that $f+g$ cannot be periodic with any period, namely that $f(x)+g(x)=2$ has exactly one solution, $x=0$.
$endgroup$
– Henning Makholm
Dec 25 '18 at 3:03
add a comment |
$begingroup$
$f$ has period $p$ if $f(x+p)=f(x)$ for all $x$. (Equivalently, $f(x+np)=f(x)$ for all $x$ and for all integers $n$). If $r$ is a real number and $f(x+rp)=f(x)$ for all $x$ we cannot say $f$ is periodic. In the quoted proof $r$ and $s$ are integers and that makes $frac r s$ rational.
answered Dec 25 '18 at 0:20
Kavi Rama MurthyKavi Rama Murthy
65.8k42867
$endgroup$
add a comment |
$begingroup$
$f$ has period $p$ if $f(x+p)=f(x)$ for all $x$. (Equivalently, $f(x+np)=f(x)$ for all $x$ and for all integers $n$). If $r$ is a real number and $f(x+rp)=f(x)$ for all $x$ we cannot say $f$ is periodic. In the quoted proof $r$ and $s$ are integers and that makes $frac r s$ rational.
answered Dec 25 '18 at 0:20
Kavi Rama MurthyKavi Rama Murthy
65.8k42867
$endgroup$
add a comment |
$begingroup$
$f$ has period $p$ if $f(x+p)=f(x)$ for all $x$. (Equivalently, $f(x+np)=f(x)$ for all $x$ and for all integers $n$). If $r$ is a real number and $f(x+rp)=f(x)$ for all $x$ we cannot say $f$ is periodic. In the quoted proof $r$ and $s$ are integers and that makes $frac r s$ rational.
answered Dec 25 '18 at 0:20
Kavi Rama MurthyKavi Rama Murthy
65.8k42867
$endgroup$
$f$ has period $p$ if $f(x+p)=f(x)$ for all $x$. (Equivalently, $f(x+np)=f(x)$ for all $x$ and for all integers $n$). If $r$ is a real number and $f(x+rp)=f(x)$ for all $x$ we cannot say $f$ is periodic. In the quoted proof $r$ and $s$ are integers and that makes $frac r s$ rational.
answered Dec 25 '18 at 0:20
Kavi Rama MurthyKavi Rama Murthy
65.8k42867
answered Dec 25 '18 at 0:20
Kavi Rama MurthyKavi Rama Murthy
65.8k42867
answered Dec 25 '18 at 0:20
Kavi Rama MurthyKavi Rama Murthy
65.8k42867
answered Dec 25 '18 at 0:20
Kavi Rama MurthyKavi Rama Murthy
65.8k42867
65.8k42867
add a comment |
add a comment |
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