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Can the product of any two aperiodic functions which are defined on the entire number line be periodic?
Can the product of two monotone functions have more than one turning point?Find area of triangle which sides is limited by two functions and the x axisAre there any theorems linking periodic functions to the number of times they are differentiable?Are injective functions necessarily defined on their entire domain?Is there any way to compress two or more numbers which I can get back without any errors?The exponential integrals are the functions $E_n$ defined by $E_n(x)=int_1^infty(e^xtt^n)^-1dt$Are there any exceptions to the rule where a product of two odd functions is even?Functions which are the identity when applied repeatedlyCan any two identical functions necessarily be written in the same form?Relation between two functions defined by a line integral
$begingroup$
My book says that this case is possible, however, I'm unable to find any such example.
functions
edited 5 hours ago
David Mitra
64.1k6102167
asked 5 hours ago
Abhinav RawalAbhinav Rawal
104
$endgroup$
add a comment |
$begingroup$
My book says that this case is possible, however, I'm unable to find any such example.
functions
edited 5 hours ago
David Mitra
64.1k6102167
asked 5 hours ago
Abhinav RawalAbhinav Rawal
104
$endgroup$
add a comment |
$begingroup$
My book says that this case is possible, however, I'm unable to find any such example.
functions
edited 5 hours ago
David Mitra
64.1k6102167
asked 5 hours ago
Abhinav RawalAbhinav Rawal
104
$endgroup$
My book says that this case is possible, however, I'm unable to find any such example.
functions
functions
edited 5 hours ago
David Mitra
64.1k6102167
asked 5 hours ago
Abhinav RawalAbhinav Rawal
104
edited 5 hours ago
David Mitra
64.1k6102167
asked 5 hours ago
Abhinav RawalAbhinav Rawal
104
edited 5 hours ago
David Mitra
64.1k6102167
edited 5 hours ago
David Mitra
64.1k6102167
edited 5 hours ago
David Mitra
64.1k6102167
64.1k6102167
asked 5 hours ago
Abhinav RawalAbhinav Rawal
104
asked 5 hours ago
Abhinav RawalAbhinav Rawal
104
asked 5 hours ago
Abhinav RawalAbhinav Rawal
104
104
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$f(x)=x^2+1$
$g(x)=fracsin xx^2+1$
answered 5 hours ago
CY AriesCY Aries
19.5k11845
$endgroup$
$begingroup$
It only rests to prove that $g(x)$ is aperiodic
$endgroup$
– Tito Eliatron
5 hours ago
$begingroup$
$g(frac pi2)>frac12^2+1=frac 15$. But for $|x|>10$, $|g(x)|<frac 1100$. $g$ cannot be periodic.
$endgroup$
– CY Aries
13 mins ago
add a comment |
$begingroup$
Given any aperiodic function whose values are only $+1$ and $-1$, then the product of the function with itself is the constant $1$ which is periodic for all periods.
P.S. As Mark Bennet commented, you can also use any aperiodic function taking values only $0$ and $1$ and then the product of the function with its $1$-complement is the constant $0$ which is periodic for all periods. Thus it suffices to let $f(0)=1$ and $f(x)=0$ otherwise.
answered 5 hours ago
SomosSomos
15.9k11437
$endgroup$
1
$begingroup$
I was going to post the a very similar answer with values $0$ and $1$ and the product $f(1-f)$
$endgroup$
– Mark Bennet
5 hours ago
add a comment |
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2 Answers
2
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
$f(x)=x^2+1$
$g(x)=fracsin xx^2+1$
answered 5 hours ago
CY AriesCY Aries
19.5k11845
$endgroup$
$begingroup$
It only rests to prove that $g(x)$ is aperiodic
$endgroup$
– Tito Eliatron
5 hours ago
$begingroup$
$g(frac pi2)>frac12^2+1=frac 15$. But for $|x|>10$, $|g(x)|<frac 1100$. $g$ cannot be periodic.
$endgroup$
– CY Aries
13 mins ago
add a comment |
$begingroup$
$f(x)=x^2+1$
$g(x)=fracsin xx^2+1$
answered 5 hours ago
CY AriesCY Aries
19.5k11845
$endgroup$
$begingroup$
It only rests to prove that $g(x)$ is aperiodic
$endgroup$
– Tito Eliatron
5 hours ago
$begingroup$
$g(frac pi2)>frac12^2+1=frac 15$. But for $|x|>10$, $|g(x)|<frac 1100$. $g$ cannot be periodic.
$endgroup$
– CY Aries
13 mins ago
add a comment |
$begingroup$
$f(x)=x^2+1$
$g(x)=fracsin xx^2+1$
answered 5 hours ago
CY AriesCY Aries
19.5k11845
$endgroup$
$f(x)=x^2+1$
$g(x)=fracsin xx^2+1$
answered 5 hours ago
CY AriesCY Aries
19.5k11845
answered 5 hours ago
CY AriesCY Aries
19.5k11845
answered 5 hours ago
CY AriesCY Aries
19.5k11845
answered 5 hours ago
CY AriesCY Aries
19.5k11845
19.5k11845
$begingroup$
It only rests to prove that $g(x)$ is aperiodic
$endgroup$
– Tito Eliatron
5 hours ago
$begingroup$
$g(frac pi2)>frac12^2+1=frac 15$. But for $|x|>10$, $|g(x)|<frac 1100$. $g$ cannot be periodic.
$endgroup$
– CY Aries
13 mins ago
add a comment |
$begingroup$
It only rests to prove that $g(x)$ is aperiodic
$endgroup$
– Tito Eliatron
5 hours ago
$begingroup$
$g(frac pi2)>frac12^2+1=frac 15$. But for $|x|>10$, $|g(x)|<frac 1100$. $g$ cannot be periodic.
$endgroup$
– CY Aries
13 mins ago
$begingroup$
It only rests to prove that $g(x)$ is aperiodic
$endgroup$
– Tito Eliatron
5 hours ago
$begingroup$
It only rests to prove that $g(x)$ is aperiodic
$endgroup$
– Tito Eliatron
5 hours ago
$begingroup$
$g(frac pi2)>frac12^2+1=frac 15$. But for $|x|>10$, $|g(x)|<frac 1100$. $g$ cannot be periodic.
$endgroup$
– CY Aries
13 mins ago
$begingroup$
$g(frac pi2)>frac12^2+1=frac 15$. But for $|x|>10$, $|g(x)|<frac 1100$. $g$ cannot be periodic.
$endgroup$
– CY Aries
13 mins ago
add a comment |
$begingroup$
Given any aperiodic function whose values are only $+1$ and $-1$, then the product of the function with itself is the constant $1$ which is periodic for all periods.
P.S. As Mark Bennet commented, you can also use any aperiodic function taking values only $0$ and $1$ and then the product of the function with its $1$-complement is the constant $0$ which is periodic for all periods. Thus it suffices to let $f(0)=1$ and $f(x)=0$ otherwise.
answered 5 hours ago
SomosSomos
15.9k11437
$endgroup$
1
$begingroup$
I was going to post the a very similar answer with values $0$ and $1$ and the product $f(1-f)$
$endgroup$
– Mark Bennet
5 hours ago
add a comment |
$begingroup$
Given any aperiodic function whose values are only $+1$ and $-1$, then the product of the function with itself is the constant $1$ which is periodic for all periods.
P.S. As Mark Bennet commented, you can also use any aperiodic function taking values only $0$ and $1$ and then the product of the function with its $1$-complement is the constant $0$ which is periodic for all periods. Thus it suffices to let $f(0)=1$ and $f(x)=0$ otherwise.
answered 5 hours ago
SomosSomos
15.9k11437
$endgroup$
1
$begingroup$
I was going to post the a very similar answer with values $0$ and $1$ and the product $f(1-f)$
$endgroup$
– Mark Bennet
5 hours ago
add a comment |
$begingroup$
Given any aperiodic function whose values are only $+1$ and $-1$, then the product of the function with itself is the constant $1$ which is periodic for all periods.
P.S. As Mark Bennet commented, you can also use any aperiodic function taking values only $0$ and $1$ and then the product of the function with its $1$-complement is the constant $0$ which is periodic for all periods. Thus it suffices to let $f(0)=1$ and $f(x)=0$ otherwise.
answered 5 hours ago
SomosSomos
15.9k11437
$endgroup$
Given any aperiodic function whose values are only $+1$ and $-1$, then the product of the function with itself is the constant $1$ which is periodic for all periods.
P.S. As Mark Bennet commented, you can also use any aperiodic function taking values only $0$ and $1$ and then the product of the function with its $1$-complement is the constant $0$ which is periodic for all periods. Thus it suffices to let $f(0)=1$ and $f(x)=0$ otherwise.
answered 5 hours ago
SomosSomos
15.9k11437
edited 5 hours ago
answered 5 hours ago
SomosSomos
15.9k11437
answered 5 hours ago
SomosSomos
15.9k11437
answered 5 hours ago
SomosSomos
15.9k11437
15.9k11437
1
$begingroup$
I was going to post the a very similar answer with values $0$ and $1$ and the product $f(1-f)$
$endgroup$
– Mark Bennet
5 hours ago
add a comment |
1
$begingroup$
I was going to post the a very similar answer with values $0$ and $1$ and the product $f(1-f)$
$endgroup$
– Mark Bennet
5 hours ago
1
1
$begingroup$
I was going to post the a very similar answer with values $0$ and $1$ and the product $f(1-f)$
$endgroup$
– Mark Bennet
5 hours ago
$begingroup$
I was going to post the a very similar answer with values $0$ and $1$ and the product $f(1-f)$
$endgroup$
– Mark Bennet
5 hours ago
add a comment |
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