How to deal with a problem of surjective functions?
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Here is a problem right here: find all the surjective functions $f :Bbb RtoBbb R$ such as
for each $x$ and $y inBbb R$: $f(xf(y)+y^2)=f((x+y)^2)-xf(x)$.
I don't understand really well the principle of surjectivity and I was wondering if my solution is right . And if it's wrong, can you please explain to me the mistake?
Here is my solution:
As $f$ is surjective for each $yinBbb R$ , there exists $xinBbb R$ such as $f(x)=y$ so there exists $yinBbb R$ such as $f(y)=x+2y$.
Hence $xf(x)=0$ so $x=0$ ( which means that $f(y)=2y$) or $f(x)=0$
And the functions $f(x)=0$ and $f(x)=2x$ verify the equation so they are the solutions
And thank you for your help
functions
edited Dec 30 '18 at 16:52
Bernard
124k742117
asked Dec 30 '18 at 16:42
Mariam Mariam
11
$endgroup$
add a comment |
$begingroup$
Here is a problem right here: find all the surjective functions $f :Bbb RtoBbb R$ such as
for each $x$ and $y inBbb R$: $f(xf(y)+y^2)=f((x+y)^2)-xf(x)$.
I don't understand really well the principle of surjectivity and I was wondering if my solution is right . And if it's wrong, can you please explain to me the mistake?
Here is my solution:
As $f$ is surjective for each $yinBbb R$ , there exists $xinBbb R$ such as $f(x)=y$ so there exists $yinBbb R$ such as $f(y)=x+2y$.
Hence $xf(x)=0$ so $x=0$ ( which means that $f(y)=2y$) or $f(x)=0$
And the functions $f(x)=0$ and $f(x)=2x$ verify the equation so they are the solutions
And thank you for your help
functions
edited Dec 30 '18 at 16:52
Bernard
124k742117
asked Dec 30 '18 at 16:42
Mariam Mariam
11
$endgroup$
$begingroup$
Please see math.meta.stackexchange.com/questions/5020 for formatting tips.
$endgroup$
– Lord Shark the Unknown
Dec 30 '18 at 16:45
$begingroup$
The logic seems unclear, or at least the quantifiers are unclear. As I understand it, you fix $x$ and seek a $y$ such that $f(y)=x+2y$, yes? But why should that exist? If, say, $f(z)=2z$ for all $z$ then $f$ is surjective then $f(y)=x+2y$ only has a solution if $x=0$. Also note: the function which is identically $0$ is not surjective.
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– lulu
Dec 30 '18 at 17:24
$begingroup$
Ok thanks for the clarification
$endgroup$
– Mariam
Dec 30 '18 at 19:06
add a comment |
$begingroup$
Here is a problem right here: find all the surjective functions $f :Bbb RtoBbb R$ such as
for each $x$ and $y inBbb R$: $f(xf(y)+y^2)=f((x+y)^2)-xf(x)$.
I don't understand really well the principle of surjectivity and I was wondering if my solution is right . And if it's wrong, can you please explain to me the mistake?
Here is my solution:
As $f$ is surjective for each $yinBbb R$ , there exists $xinBbb R$ such as $f(x)=y$ so there exists $yinBbb R$ such as $f(y)=x+2y$.
Hence $xf(x)=0$ so $x=0$ ( which means that $f(y)=2y$) or $f(x)=0$
And the functions $f(x)=0$ and $f(x)=2x$ verify the equation so they are the solutions
And thank you for your help
functions
edited Dec 30 '18 at 16:52
Bernard
124k742117
asked Dec 30 '18 at 16:42
Mariam Mariam
11
$endgroup$
Here is a problem right here: find all the surjective functions $f :Bbb RtoBbb R$ such as
for each $x$ and $y inBbb R$: $f(xf(y)+y^2)=f((x+y)^2)-xf(x)$.
I don't understand really well the principle of surjectivity and I was wondering if my solution is right . And if it's wrong, can you please explain to me the mistake?
Here is my solution:
As $f$ is surjective for each $yinBbb R$ , there exists $xinBbb R$ such as $f(x)=y$ so there exists $yinBbb R$ such as $f(y)=x+2y$.
Hence $xf(x)=0$ so $x=0$ ( which means that $f(y)=2y$) or $f(x)=0$
And the functions $f(x)=0$ and $f(x)=2x$ verify the equation so they are the solutions
And thank you for your help
functions
functions
edited Dec 30 '18 at 16:52
Bernard
124k742117
asked Dec 30 '18 at 16:42
Mariam Mariam
11
edited Dec 30 '18 at 16:52
Bernard
124k742117
asked Dec 30 '18 at 16:42
Mariam Mariam
11
edited Dec 30 '18 at 16:52
Bernard
124k742117
edited Dec 30 '18 at 16:52
Bernard
124k742117
edited Dec 30 '18 at 16:52
Bernard
124k742117
124k742117
asked Dec 30 '18 at 16:42
Mariam Mariam
11
asked Dec 30 '18 at 16:42
Mariam Mariam
11
asked Dec 30 '18 at 16:42
Mariam Mariam
11
11
$begingroup$
Please see math.meta.stackexchange.com/questions/5020 for formatting tips.
$endgroup$
– Lord Shark the Unknown
Dec 30 '18 at 16:45
$begingroup$
The logic seems unclear, or at least the quantifiers are unclear. As I understand it, you fix $x$ and seek a $y$ such that $f(y)=x+2y$, yes? But why should that exist? If, say, $f(z)=2z$ for all $z$ then $f$ is surjective then $f(y)=x+2y$ only has a solution if $x=0$. Also note: the function which is identically $0$ is not surjective.
$endgroup$
– lulu
Dec 30 '18 at 17:24
$begingroup$
Ok thanks for the clarification
$endgroup$
– Mariam
Dec 30 '18 at 19:06
add a comment |
$begingroup$
Please see math.meta.stackexchange.com/questions/5020 for formatting tips.
$endgroup$
– Lord Shark the Unknown
Dec 30 '18 at 16:45
$begingroup$
The logic seems unclear, or at least the quantifiers are unclear. As I understand it, you fix $x$ and seek a $y$ such that $f(y)=x+2y$, yes? But why should that exist? If, say, $f(z)=2z$ for all $z$ then $f$ is surjective then $f(y)=x+2y$ only has a solution if $x=0$. Also note: the function which is identically $0$ is not surjective.
$endgroup$
– lulu
Dec 30 '18 at 17:24
$begingroup$
Ok thanks for the clarification
$endgroup$
– Mariam
Dec 30 '18 at 19:06
$begingroup$
Please see math.meta.stackexchange.com/questions/5020 for formatting tips.
$endgroup$
– Lord Shark the Unknown
Dec 30 '18 at 16:45
$begingroup$
Please see math.meta.stackexchange.com/questions/5020 for formatting tips.
$endgroup$
– Lord Shark the Unknown
Dec 30 '18 at 16:45
$begingroup$
The logic seems unclear, or at least the quantifiers are unclear. As I understand it, you fix $x$ and seek a $y$ such that $f(y)=x+2y$, yes? But why should that exist? If, say, $f(z)=2z$ for all $z$ then $f$ is surjective then $f(y)=x+2y$ only has a solution if $x=0$. Also note: the function which is identically $0$ is not surjective.
$endgroup$
– lulu
Dec 30 '18 at 17:24
$begingroup$
The logic seems unclear, or at least the quantifiers are unclear. As I understand it, you fix $x$ and seek a $y$ such that $f(y)=x+2y$, yes? But why should that exist? If, say, $f(z)=2z$ for all $z$ then $f$ is surjective then $f(y)=x+2y$ only has a solution if $x=0$. Also note: the function which is identically $0$ is not surjective.
$endgroup$
– lulu
Dec 30 '18 at 17:24
$begingroup$
Ok thanks for the clarification
$endgroup$
– Mariam
Dec 30 '18 at 19:06
$begingroup$
Ok thanks for the clarification
$endgroup$
– Mariam
Dec 30 '18 at 19:06
add a comment |
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$begingroup$
Please see math.meta.stackexchange.com/questions/5020 for formatting tips.
$endgroup$
– Lord Shark the Unknown
Dec 30 '18 at 16:45
$begingroup$
The logic seems unclear, or at least the quantifiers are unclear. As I understand it, you fix $x$ and seek a $y$ such that $f(y)=x+2y$, yes? But why should that exist? If, say, $f(z)=2z$ for all $z$ then $f$ is surjective then $f(y)=x+2y$ only has a solution if $x=0$. Also note: the function which is identically $0$ is not surjective.
$endgroup$
– lulu
Dec 30 '18 at 17:24
$begingroup$
Ok thanks for the clarification
$endgroup$
– Mariam
Dec 30 '18 at 19:06