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Determine $f$ when $f(t)+f(t^-1) = g(t+t^-1)$
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up vote
1
down vote
favorite
I have the following equation:
$f(t)+f(t^-1) = g(t+t^-1)$,
where $g$ is a known function. The objective is to determine $f$. Also, assume that $t>0$ and $f,g geq 0$. My intuition says that $f$ should be
$frac12g(t+t^-1)$, i.e. $f(t)=f(t^-1)$.
functions
asked Jul 27 at 18:55
ToniAz
897
add a comment |Â
up vote
1
down vote
favorite
I have the following equation:
$f(t)+f(t^-1) = g(t+t^-1)$,
where $g$ is a known function. The objective is to determine $f$. Also, assume that $t>0$ and $f,g geq 0$. My intuition says that $f$ should be
$frac12g(t+t^-1)$, i.e. $f(t)=f(t^-1)$.
functions
asked Jul 27 at 18:55
ToniAz
897
What are the domains and codomains of your functions?
– Eric Wofsey
Jul 27 at 18:57
Suppose that $tinmathbbR$ and $f,g geq 0$.
– ToniAz
Jul 27 at 18:58
Presumably you mean to require $t$ to be nonzero as well?
– Eric Wofsey
Jul 27 at 18:59
@EricWofsey Ah yes, let me correct that. Thank you. I'm mostly concerned with $t>0$ specifically.
– ToniAz
Jul 27 at 19:00
1
Counterexamples: if $g(x) = x$ then $f(x) = x$ or $f(x)=frac1x$ will work; if $g(x)=x^2$ then $f(x)=x^2+1$ will work.
– Daniel Schepler
Jul 27 at 19:01
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have the following equation:
$f(t)+f(t^-1) = g(t+t^-1)$,
where $g$ is a known function. The objective is to determine $f$. Also, assume that $t>0$ and $f,g geq 0$. My intuition says that $f$ should be
$frac12g(t+t^-1)$, i.e. $f(t)=f(t^-1)$.
functions
asked Jul 27 at 18:55
ToniAz
897
I have the following equation:
$f(t)+f(t^-1) = g(t+t^-1)$,
where $g$ is a known function. The objective is to determine $f$. Also, assume that $t>0$ and $f,g geq 0$. My intuition says that $f$ should be
$frac12g(t+t^-1)$, i.e. $f(t)=f(t^-1)$.
functions
asked Jul 27 at 18:55
ToniAz
897
edited Jul 31 at 19:46
asked Jul 27 at 18:55
ToniAz
897
asked Jul 27 at 18:55
ToniAz
897
asked Jul 27 at 18:55
ToniAz
897
897
What are the domains and codomains of your functions?
– Eric Wofsey
Jul 27 at 18:57
Suppose that $tinmathbbR$ and $f,g geq 0$.
– ToniAz
Jul 27 at 18:58
Presumably you mean to require $t$ to be nonzero as well?
– Eric Wofsey
Jul 27 at 18:59
@EricWofsey Ah yes, let me correct that. Thank you. I'm mostly concerned with $t>0$ specifically.
– ToniAz
Jul 27 at 19:00
1
Counterexamples: if $g(x) = x$ then $f(x) = x$ or $f(x)=frac1x$ will work; if $g(x)=x^2$ then $f(x)=x^2+1$ will work.
– Daniel Schepler
Jul 27 at 19:01
add a comment |Â
What are the domains and codomains of your functions?
– Eric Wofsey
Jul 27 at 18:57
Suppose that $tinmathbbR$ and $f,g geq 0$.
– ToniAz
Jul 27 at 18:58
Presumably you mean to require $t$ to be nonzero as well?
– Eric Wofsey
Jul 27 at 18:59
@EricWofsey Ah yes, let me correct that. Thank you. I'm mostly concerned with $t>0$ specifically.
– ToniAz
Jul 27 at 19:00
1
Counterexamples: if $g(x) = x$ then $f(x) = x$ or $f(x)=frac1x$ will work; if $g(x)=x^2$ then $f(x)=x^2+1$ will work.
– Daniel Schepler
Jul 27 at 19:01
What are the domains and codomains of your functions?
– Eric Wofsey
Jul 27 at 18:57
What are the domains and codomains of your functions?
– Eric Wofsey
Jul 27 at 18:57
Suppose that $tinmathbbR$ and $f,g geq 0$.
– ToniAz
Jul 27 at 18:58
Suppose that $tinmathbbR$ and $f,g geq 0$.
– ToniAz
Jul 27 at 18:58
Presumably you mean to require $t$ to be nonzero as well?
– Eric Wofsey
Jul 27 at 18:59
Presumably you mean to require $t$ to be nonzero as well?
– Eric Wofsey
Jul 27 at 18:59
@EricWofsey Ah yes, let me correct that. Thank you. I'm mostly concerned with $t>0$ specifically.
– ToniAz
Jul 27 at 19:00
@EricWofsey Ah yes, let me correct that. Thank you. I'm mostly concerned with $t>0$ specifically.
– ToniAz
Jul 27 at 19:00
1
1
Counterexamples: if $g(x) = x$ then $f(x) = x$ or $f(x)=frac1x$ will work; if $g(x)=x^2$ then $f(x)=x^2+1$ will work.
– Daniel Schepler
Jul 27 at 19:01
Counterexamples: if $g(x) = x$ then $f(x) = x$ or $f(x)=frac1x$ will work; if $g(x)=x^2$ then $f(x)=x^2+1$ will work.
– Daniel Schepler
Jul 27 at 19:01
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
It turns out that $f$ is not uniquely determined by this functional equation. For example, your proposed solution
$$f(t)=fracg(t+t^-1)2$$
does satisfy the desired functional equation, but so does the function
$$f(t)=fracg(t+t^-1)2+o(t-t^-1)$$
...where $o$ is any odd function.
answered Jul 27 at 19:05
Frpzzd
16.9k63489
add a comment |Â
up vote
2
down vote
This equation does not give enough information to determine $f$. For every $t$ such that $tneq t^-1$, $f(t)$ and $f(t^-1)$ could be any pair of numbers at all whose sum is $g(t+t^-1)$. (Note that the restrictions given by different values of $t$ are essentially independent of each other, since the only time they would overlap is in the case of $t$ and $t^-1$ but then they give the exact same restriction.)
answered Jul 27 at 19:00
Eric Wofsey
162k12188299
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
It turns out that $f$ is not uniquely determined by this functional equation. For example, your proposed solution
$$f(t)=fracg(t+t^-1)2$$
does satisfy the desired functional equation, but so does the function
$$f(t)=fracg(t+t^-1)2+o(t-t^-1)$$
...where $o$ is any odd function.
answered Jul 27 at 19:05
Frpzzd
16.9k63489
add a comment |Â
up vote
4
down vote
accepted
It turns out that $f$ is not uniquely determined by this functional equation. For example, your proposed solution
$$f(t)=fracg(t+t^-1)2$$
does satisfy the desired functional equation, but so does the function
$$f(t)=fracg(t+t^-1)2+o(t-t^-1)$$
...where $o$ is any odd function.
answered Jul 27 at 19:05
Frpzzd
16.9k63489
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
It turns out that $f$ is not uniquely determined by this functional equation. For example, your proposed solution
$$f(t)=fracg(t+t^-1)2$$
does satisfy the desired functional equation, but so does the function
$$f(t)=fracg(t+t^-1)2+o(t-t^-1)$$
...where $o$ is any odd function.
answered Jul 27 at 19:05
Frpzzd
16.9k63489
It turns out that $f$ is not uniquely determined by this functional equation. For example, your proposed solution
$$f(t)=fracg(t+t^-1)2$$
does satisfy the desired functional equation, but so does the function
$$f(t)=fracg(t+t^-1)2+o(t-t^-1)$$
...where $o$ is any odd function.
answered Jul 27 at 19:05
Frpzzd
16.9k63489
answered Jul 27 at 19:05
Frpzzd
16.9k63489
answered Jul 27 at 19:05
Frpzzd
16.9k63489
answered Jul 27 at 19:05
Frpzzd
16.9k63489
16.9k63489
add a comment |Â
add a comment |Â
up vote
2
down vote
This equation does not give enough information to determine $f$. For every $t$ such that $tneq t^-1$, $f(t)$ and $f(t^-1)$ could be any pair of numbers at all whose sum is $g(t+t^-1)$. (Note that the restrictions given by different values of $t$ are essentially independent of each other, since the only time they would overlap is in the case of $t$ and $t^-1$ but then they give the exact same restriction.)
answered Jul 27 at 19:00
Eric Wofsey
162k12188299
add a comment |Â
up vote
2
down vote
This equation does not give enough information to determine $f$. For every $t$ such that $tneq t^-1$, $f(t)$ and $f(t^-1)$ could be any pair of numbers at all whose sum is $g(t+t^-1)$. (Note that the restrictions given by different values of $t$ are essentially independent of each other, since the only time they would overlap is in the case of $t$ and $t^-1$ but then they give the exact same restriction.)
answered Jul 27 at 19:00
Eric Wofsey
162k12188299
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This equation does not give enough information to determine $f$. For every $t$ such that $tneq t^-1$, $f(t)$ and $f(t^-1)$ could be any pair of numbers at all whose sum is $g(t+t^-1)$. (Note that the restrictions given by different values of $t$ are essentially independent of each other, since the only time they would overlap is in the case of $t$ and $t^-1$ but then they give the exact same restriction.)
answered Jul 27 at 19:00
Eric Wofsey
162k12188299
This equation does not give enough information to determine $f$. For every $t$ such that $tneq t^-1$, $f(t)$ and $f(t^-1)$ could be any pair of numbers at all whose sum is $g(t+t^-1)$. (Note that the restrictions given by different values of $t$ are essentially independent of each other, since the only time they would overlap is in the case of $t$ and $t^-1$ but then they give the exact same restriction.)
answered Jul 27 at 19:00
Eric Wofsey
162k12188299
answered Jul 27 at 19:00
Eric Wofsey
162k12188299
answered Jul 27 at 19:00
Eric Wofsey
162k12188299
answered Jul 27 at 19:00
Eric Wofsey
162k12188299
162k12188299
add a comment |Â
add a comment |Â
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What are the domains and codomains of your functions?
– Eric Wofsey
Jul 27 at 18:57
Suppose that $tinmathbbR$ and $f,g geq 0$.
– ToniAz
Jul 27 at 18:58
Presumably you mean to require $t$ to be nonzero as well?
– Eric Wofsey
Jul 27 at 18:59
@EricWofsey Ah yes, let me correct that. Thank you. I'm mostly concerned with $t>0$ specifically.
– ToniAz
Jul 27 at 19:00
1
Counterexamples: if $g(x) = x$ then $f(x) = x$ or $f(x)=frac1x$ will work; if $g(x)=x^2$ then $f(x)=x^2+1$ will work.
– Daniel Schepler
Jul 27 at 19:01