Is it possible to have $f(x)f(y) = g(x)+g(y)$? The Next CEO of Stack OverflowSolution for a functional equationUniqueness of solution of functional equationFunctions minimized at the median of their argumentsFind all $f:mathbb Rtomathbb R$ such that $forall x,yinmathbb R$ the given equality holds: $xf(y)+yf(x)=(x+y)f(x)f(y)$.What can be said about a function with rotational symmetry of order other than 2?discontinuous solutions of functional equationQuestion regarding the Cauchy functional equationSolution of a function equation $f(x) + f(y) = f(x + y + 2f(xy))$Prove linear functions that are not multiplications by a constant are unbounded on every intervalFind all functions $f:mathbbNrightarrowmathbbN$ such that $varphi(f(x+y))=varphi(f(x))+varphi(f(y))quadforall x,yinmathbbN$Which functions satisfy $f^n(x) = f(x)^n$ for some $n ge 2$?
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Is it possible to have $f(x)f(y) = g(x)+g(y)$?
The Next CEO of Stack OverflowSolution for a functional equationUniqueness of solution of functional equationFunctions minimized at the median of their argumentsFind all $f:mathbb Rtomathbb R$ such that $forall x,yinmathbb R$ the given equality holds: $xf(y)+yf(x)=(x+y)f(x)f(y)$.What can be said about a function with rotational symmetry of order other than 2?discontinuous solutions of functional equationQuestion regarding the Cauchy functional equationSolution of a function equation $f(x) + f(y) = f(x + y + 2f(xy))$Prove linear functions that are not multiplications by a constant are unbounded on every intervalFind all functions $f:mathbbNrightarrowmathbbN$ such that $varphi(f(x+y))=varphi(f(x))+varphi(f(y))quadforall x,yinmathbbN$Which functions satisfy $f^n(x) = f(x)^n$ for some $n ge 2$?
$begingroup$
Inspired by this question I wondered whether there are any "notable" functions $f,g$ on (or on some subset $Omega$ of) $mathbb R$ or $mathbb C$ that satisfy
$$f(x)f(y) = g(x) + g(y) :forall x,y in Omega$$
By "notable" I mean nontrivial solutions (for example $f(x) = c, g(x) = fracc^22$ for some $c$ and all $x$ would be trivial, or also if you chose e.g. a one element domain $|Omega|=1$), that are sufficiently well behaved (e.g. continuous or even differentiable).
functional-equations
asked 2 days ago
flawrflawr
11.7k32546
$endgroup$
add a comment |
$begingroup$
Inspired by this question I wondered whether there are any "notable" functions $f,g$ on (or on some subset $Omega$ of) $mathbb R$ or $mathbb C$ that satisfy
$$f(x)f(y) = g(x) + g(y) :forall x,y in Omega$$
By "notable" I mean nontrivial solutions (for example $f(x) = c, g(x) = fracc^22$ for some $c$ and all $x$ would be trivial, or also if you chose e.g. a one element domain $|Omega|=1$), that are sufficiently well behaved (e.g. continuous or even differentiable).
functional-equations
asked 2 days ago
flawrflawr
11.7k32546
$endgroup$
add a comment |
$begingroup$
Inspired by this question I wondered whether there are any "notable" functions $f,g$ on (or on some subset $Omega$ of) $mathbb R$ or $mathbb C$ that satisfy
$$f(x)f(y) = g(x) + g(y) :forall x,y in Omega$$
By "notable" I mean nontrivial solutions (for example $f(x) = c, g(x) = fracc^22$ for some $c$ and all $x$ would be trivial, or also if you chose e.g. a one element domain $|Omega|=1$), that are sufficiently well behaved (e.g. continuous or even differentiable).
functional-equations
asked 2 days ago
flawrflawr
11.7k32546
$endgroup$
Inspired by this question I wondered whether there are any "notable" functions $f,g$ on (or on some subset $Omega$ of) $mathbb R$ or $mathbb C$ that satisfy
$$f(x)f(y) = g(x) + g(y) :forall x,y in Omega$$
By "notable" I mean nontrivial solutions (for example $f(x) = c, g(x) = fracc^22$ for some $c$ and all $x$ would be trivial, or also if you chose e.g. a one element domain $|Omega|=1$), that are sufficiently well behaved (e.g. continuous or even differentiable).
functional-equations
functional-equations
asked 2 days ago
flawrflawr
11.7k32546
asked 2 days ago
flawrflawr
11.7k32546
asked 2 days ago
flawrflawr
11.7k32546
asked 2 days ago
flawrflawr
11.7k32546
asked 2 days ago
flawrflawr
11.7k32546
11.7k32546
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Suppose that $0 in Omega$ and that $f(0) = c, g(0) = d$. We then have
$$cf(x) = g(x) + d$$
for all $x in Omega$. If $c = 0$, then $g(x)$ is also constant $0$.
Otherwise, we have $$f(x) = fracg(x) + dc.$$
Note that $x = 0$ gives us the equation
$$c = frac2dc implies c^2 = 2d.$$
Looking at the diagonal case $x = y$, we have
$$f(x)^2 = 2g(x)$$
which reduces us to
$$f(x) = fracfrac 1 2 f(x)^2 + fracf(0)^22f(0) = frac 1 2f(0) (f(x)^2 + f(0)^2).$$
There aren't so many solutions to this.
To be even more explicit, we have a quadratic equation
$$(f(x) - f(0))^2 = f(x)^2 - 2f(0) f(x) + f(0)^2 = 0$$
which implies that $f$ is constant.
answered 2 days ago
T. BongersT. Bongers
23.5k54762
$endgroup$
add a comment |
$begingroup$
Notice that deriving both side ($dx$):
$$f'(x)f(y)=g'(x)$$
And now both sides $dy$:
$$f'(x)f'(y)=0$$
So $f'(x)=0 Rightarrow f(x)=K $
And:
$$K^2=g(x)+g(y)$$
That has trivially as only solution $g(x)=fracK^22$ . These are the only solution if $f$ and $g$ are both derivable(you asked for "notable function").
answered 2 days ago
EurekaEureka
423112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
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active
oldest
votes
active
oldest
votes
$begingroup$
Suppose that $0 in Omega$ and that $f(0) = c, g(0) = d$. We then have
$$cf(x) = g(x) + d$$
for all $x in Omega$. If $c = 0$, then $g(x)$ is also constant $0$.
Otherwise, we have $$f(x) = fracg(x) + dc.$$
Note that $x = 0$ gives us the equation
$$c = frac2dc implies c^2 = 2d.$$
Looking at the diagonal case $x = y$, we have
$$f(x)^2 = 2g(x)$$
which reduces us to
$$f(x) = fracfrac 1 2 f(x)^2 + fracf(0)^22f(0) = frac 1 2f(0) (f(x)^2 + f(0)^2).$$
There aren't so many solutions to this.
To be even more explicit, we have a quadratic equation
$$(f(x) - f(0))^2 = f(x)^2 - 2f(0) f(x) + f(0)^2 = 0$$
which implies that $f$ is constant.
answered 2 days ago
T. BongersT. Bongers
23.5k54762
$endgroup$
add a comment |
$begingroup$
Suppose that $0 in Omega$ and that $f(0) = c, g(0) = d$. We then have
$$cf(x) = g(x) + d$$
for all $x in Omega$. If $c = 0$, then $g(x)$ is also constant $0$.
Otherwise, we have $$f(x) = fracg(x) + dc.$$
Note that $x = 0$ gives us the equation
$$c = frac2dc implies c^2 = 2d.$$
Looking at the diagonal case $x = y$, we have
$$f(x)^2 = 2g(x)$$
which reduces us to
$$f(x) = fracfrac 1 2 f(x)^2 + fracf(0)^22f(0) = frac 1 2f(0) (f(x)^2 + f(0)^2).$$
There aren't so many solutions to this.
To be even more explicit, we have a quadratic equation
$$(f(x) - f(0))^2 = f(x)^2 - 2f(0) f(x) + f(0)^2 = 0$$
which implies that $f$ is constant.
answered 2 days ago
T. BongersT. Bongers
23.5k54762
$endgroup$
add a comment |
$begingroup$
Suppose that $0 in Omega$ and that $f(0) = c, g(0) = d$. We then have
$$cf(x) = g(x) + d$$
for all $x in Omega$. If $c = 0$, then $g(x)$ is also constant $0$.
Otherwise, we have $$f(x) = fracg(x) + dc.$$
Note that $x = 0$ gives us the equation
$$c = frac2dc implies c^2 = 2d.$$
Looking at the diagonal case $x = y$, we have
$$f(x)^2 = 2g(x)$$
which reduces us to
$$f(x) = fracfrac 1 2 f(x)^2 + fracf(0)^22f(0) = frac 1 2f(0) (f(x)^2 + f(0)^2).$$
There aren't so many solutions to this.
To be even more explicit, we have a quadratic equation
$$(f(x) - f(0))^2 = f(x)^2 - 2f(0) f(x) + f(0)^2 = 0$$
which implies that $f$ is constant.
answered 2 days ago
T. BongersT. Bongers
23.5k54762
$endgroup$
Suppose that $0 in Omega$ and that $f(0) = c, g(0) = d$. We then have
$$cf(x) = g(x) + d$$
for all $x in Omega$. If $c = 0$, then $g(x)$ is also constant $0$.
Otherwise, we have $$f(x) = fracg(x) + dc.$$
Note that $x = 0$ gives us the equation
$$c = frac2dc implies c^2 = 2d.$$
Looking at the diagonal case $x = y$, we have
$$f(x)^2 = 2g(x)$$
which reduces us to
$$f(x) = fracfrac 1 2 f(x)^2 + fracf(0)^22f(0) = frac 1 2f(0) (f(x)^2 + f(0)^2).$$
There aren't so many solutions to this.
To be even more explicit, we have a quadratic equation
$$(f(x) - f(0))^2 = f(x)^2 - 2f(0) f(x) + f(0)^2 = 0$$
which implies that $f$ is constant.
answered 2 days ago
T. BongersT. Bongers
23.5k54762
edited 2 days ago
answered 2 days ago
T. BongersT. Bongers
23.5k54762
answered 2 days ago
T. BongersT. Bongers
23.5k54762
answered 2 days ago
T. BongersT. Bongers
23.5k54762
23.5k54762
add a comment |
add a comment |
$begingroup$
Notice that deriving both side ($dx$):
$$f'(x)f(y)=g'(x)$$
And now both sides $dy$:
$$f'(x)f'(y)=0$$
So $f'(x)=0 Rightarrow f(x)=K $
And:
$$K^2=g(x)+g(y)$$
That has trivially as only solution $g(x)=fracK^22$ . These are the only solution if $f$ and $g$ are both derivable(you asked for "notable function").
answered 2 days ago
EurekaEureka
423112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Notice that deriving both side ($dx$):
$$f'(x)f(y)=g'(x)$$
And now both sides $dy$:
$$f'(x)f'(y)=0$$
So $f'(x)=0 Rightarrow f(x)=K $
And:
$$K^2=g(x)+g(y)$$
That has trivially as only solution $g(x)=fracK^22$ . These are the only solution if $f$ and $g$ are both derivable(you asked for "notable function").
answered 2 days ago
EurekaEureka
423112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Notice that deriving both side ($dx$):
$$f'(x)f(y)=g'(x)$$
And now both sides $dy$:
$$f'(x)f'(y)=0$$
So $f'(x)=0 Rightarrow f(x)=K $
And:
$$K^2=g(x)+g(y)$$
That has trivially as only solution $g(x)=fracK^22$ . These are the only solution if $f$ and $g$ are both derivable(you asked for "notable function").
answered 2 days ago
EurekaEureka
423112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Notice that deriving both side ($dx$):
$$f'(x)f(y)=g'(x)$$
And now both sides $dy$:
$$f'(x)f'(y)=0$$
So $f'(x)=0 Rightarrow f(x)=K $
And:
$$K^2=g(x)+g(y)$$
That has trivially as only solution $g(x)=fracK^22$ . These are the only solution if $f$ and $g$ are both derivable(you asked for "notable function").
answered 2 days ago
EurekaEureka
423112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
EurekaEureka
423112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
EurekaEureka
423112
answered 2 days ago
EurekaEureka
423112
423112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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