Find all $f:mathbb Rtomathbb R$ such that $forall x,yinmathbb R$ the given equality holds:...












1












$begingroup$



Find all $f:mathbb Rto mathbb R$ such that $forall x,yinmathbb R$ the given equality holds: $$xf(y)+yf(x)=(x+y)f(x)f(y)$$





My try:



whenever $y=0$, we have $$xcdot f(0)cdot(1-f(x))=0$$



So the solutions can be either generated by: $$f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$$



where $c$ generate an infinite amount of solutions, or it can be any function satisfying $f(0)=0$.



However, I can't seem to find a way to check such a solution on the initial equation (we have to check it since we've simply eliminated the impossible solutions and we've only found that the solutions can be somehow generated using the way I've shown).



Note that this generates the trivial solutions $f(x)=0$ and $f(x)=1$ as well.




Is my reasoning correct?



If so, how can I check whether all the generated functions satisfy the initial equation?




Off-topic, but do $f^2(x)$ and $f(x)^2$ denote the same concept? My teacher uses the former while my book uses the latter (for squaring $f(x)$). Which one is correct, or are both of them correct?






functions functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Note that if you plug $y=0$ into your original equation you have $xf(0) = xf(x)f(0) forall x$, so if $f(0)neq 0$ then your solutions are much more tightly constrained.
    $endgroup$
    – Steven Stadnicki
    Jun 3 '14 at 18:22










  • $begingroup$
    Yes, you are right. But, you haven't proved that you already covered all functions.
    $endgroup$
    – Jlamprong
    Jun 3 '14 at 18:22










  • $begingroup$
    @Jlamprong I'm not sure what you mean. I have proved that whatever the solutions the initial equation has, they can be generated using the formula I've given.
    $endgroup$
    – user26486
    Jun 3 '14 at 18:25










  • $begingroup$
    @StevenStadnicki It follows that the solutions can be generated by $$f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$$ or it can be any function satisfying $f(0)=0$, if my reasoning is correct.
    $endgroup$
    – user26486
    Jun 3 '14 at 18:29












  • $begingroup$
    @StevenStadnicki I've edited the question.
    $endgroup$
    – user26486
    Jun 3 '14 at 18:38
















1












$begingroup$



Find all $f:mathbb Rto mathbb R$ such that $forall x,yinmathbb R$ the given equality holds: $$xf(y)+yf(x)=(x+y)f(x)f(y)$$





My try:



whenever $y=0$, we have $$xcdot f(0)cdot(1-f(x))=0$$



So the solutions can be either generated by: $$f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$$



where $c$ generate an infinite amount of solutions, or it can be any function satisfying $f(0)=0$.



However, I can't seem to find a way to check such a solution on the initial equation (we have to check it since we've simply eliminated the impossible solutions and we've only found that the solutions can be somehow generated using the way I've shown).



Note that this generates the trivial solutions $f(x)=0$ and $f(x)=1$ as well.




Is my reasoning correct?



If so, how can I check whether all the generated functions satisfy the initial equation?




Off-topic, but do $f^2(x)$ and $f(x)^2$ denote the same concept? My teacher uses the former while my book uses the latter (for squaring $f(x)$). Which one is correct, or are both of them correct?






functions functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Note that if you plug $y=0$ into your original equation you have $xf(0) = xf(x)f(0) forall x$, so if $f(0)neq 0$ then your solutions are much more tightly constrained.
    $endgroup$
    – Steven Stadnicki
    Jun 3 '14 at 18:22










  • $begingroup$
    Yes, you are right. But, you haven't proved that you already covered all functions.
    $endgroup$
    – Jlamprong
    Jun 3 '14 at 18:22










  • $begingroup$
    @Jlamprong I'm not sure what you mean. I have proved that whatever the solutions the initial equation has, they can be generated using the formula I've given.
    $endgroup$
    – user26486
    Jun 3 '14 at 18:25










  • $begingroup$
    @StevenStadnicki It follows that the solutions can be generated by $$f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$$ or it can be any function satisfying $f(0)=0$, if my reasoning is correct.
    $endgroup$
    – user26486
    Jun 3 '14 at 18:29












  • $begingroup$
    @StevenStadnicki I've edited the question.
    $endgroup$
    – user26486
    Jun 3 '14 at 18:38














1












1








1


1



$begingroup$



Find all $f:mathbb Rto mathbb R$ such that $forall x,yinmathbb R$ the given equality holds: $$xf(y)+yf(x)=(x+y)f(x)f(y)$$





My try:



whenever $y=0$, we have $$xcdot f(0)cdot(1-f(x))=0$$



So the solutions can be either generated by: $$f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$$



where $c$ generate an infinite amount of solutions, or it can be any function satisfying $f(0)=0$.



However, I can't seem to find a way to check such a solution on the initial equation (we have to check it since we've simply eliminated the impossible solutions and we've only found that the solutions can be somehow generated using the way I've shown).



Note that this generates the trivial solutions $f(x)=0$ and $f(x)=1$ as well.




Is my reasoning correct?



If so, how can I check whether all the generated functions satisfy the initial equation?




Off-topic, but do $f^2(x)$ and $f(x)^2$ denote the same concept? My teacher uses the former while my book uses the latter (for squaring $f(x)$). Which one is correct, or are both of them correct?






functions functional-equations







share|cite|improve this question











$endgroup$





Find all $f:mathbb Rto mathbb R$ such that $forall x,yinmathbb R$ the given equality holds: $$xf(y)+yf(x)=(x+y)f(x)f(y)$$





My try:



whenever $y=0$, we have $$xcdot f(0)cdot(1-f(x))=0$$



So the solutions can be either generated by: $$f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$$



where $c$ generate an infinite amount of solutions, or it can be any function satisfying $f(0)=0$.



However, I can't seem to find a way to check such a solution on the initial equation (we have to check it since we've simply eliminated the impossible solutions and we've only found that the solutions can be somehow generated using the way I've shown).



Note that this generates the trivial solutions $f(x)=0$ and $f(x)=1$ as well.




Is my reasoning correct?



If so, how can I check whether all the generated functions satisfy the initial equation?




Off-topic, but do $f^2(x)$ and $f(x)^2$ denote the same concept? My teacher uses the former while my book uses the latter (for squaring $f(x)$). Which one is correct, or are both of them correct?






functions functional-equations




functions functional-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 3 '14 at 18:35







user26486

















asked Jun 3 '14 at 18:16









user26486user26486

9,28221948




9,28221948












  • $begingroup$
    Note that if you plug $y=0$ into your original equation you have $xf(0) = xf(x)f(0) forall x$, so if $f(0)neq 0$ then your solutions are much more tightly constrained.
    $endgroup$
    – Steven Stadnicki
    Jun 3 '14 at 18:22










  • $begingroup$
    Yes, you are right. But, you haven't proved that you already covered all functions.
    $endgroup$
    – Jlamprong
    Jun 3 '14 at 18:22










  • $begingroup$
    @Jlamprong I'm not sure what you mean. I have proved that whatever the solutions the initial equation has, they can be generated using the formula I've given.
    $endgroup$
    – user26486
    Jun 3 '14 at 18:25










  • $begingroup$
    @StevenStadnicki It follows that the solutions can be generated by $$f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$$ or it can be any function satisfying $f(0)=0$, if my reasoning is correct.
    $endgroup$
    – user26486
    Jun 3 '14 at 18:29












  • $begingroup$
    @StevenStadnicki I've edited the question.
    $endgroup$
    – user26486
    Jun 3 '14 at 18:38


















  • $begingroup$
    Note that if you plug $y=0$ into your original equation you have $xf(0) = xf(x)f(0) forall x$, so if $f(0)neq 0$ then your solutions are much more tightly constrained.
    $endgroup$
    – Steven Stadnicki
    Jun 3 '14 at 18:22










  • $begingroup$
    Yes, you are right. But, you haven't proved that you already covered all functions.
    $endgroup$
    – Jlamprong
    Jun 3 '14 at 18:22










  • $begingroup$
    @Jlamprong I'm not sure what you mean. I have proved that whatever the solutions the initial equation has, they can be generated using the formula I've given.
    $endgroup$
    – user26486
    Jun 3 '14 at 18:25










  • $begingroup$
    @StevenStadnicki It follows that the solutions can be generated by $$f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$$ or it can be any function satisfying $f(0)=0$, if my reasoning is correct.
    $endgroup$
    – user26486
    Jun 3 '14 at 18:29












  • $begingroup$
    @StevenStadnicki I've edited the question.
    $endgroup$
    – user26486
    Jun 3 '14 at 18:38
















$begingroup$
Note that if you plug $y=0$ into your original equation you have $xf(0) = xf(x)f(0) forall x$, so if $f(0)neq 0$ then your solutions are much more tightly constrained.
$endgroup$
– Steven Stadnicki
Jun 3 '14 at 18:22




$begingroup$
Note that if you plug $y=0$ into your original equation you have $xf(0) = xf(x)f(0) forall x$, so if $f(0)neq 0$ then your solutions are much more tightly constrained.
$endgroup$
– Steven Stadnicki
Jun 3 '14 at 18:22












$begingroup$
Yes, you are right. But, you haven't proved that you already covered all functions.
$endgroup$
– Jlamprong
Jun 3 '14 at 18:22




$begingroup$
Yes, you are right. But, you haven't proved that you already covered all functions.
$endgroup$
– Jlamprong
Jun 3 '14 at 18:22












$begingroup$
@Jlamprong I'm not sure what you mean. I have proved that whatever the solutions the initial equation has, they can be generated using the formula I've given.
$endgroup$
– user26486
Jun 3 '14 at 18:25




$begingroup$
@Jlamprong I'm not sure what you mean. I have proved that whatever the solutions the initial equation has, they can be generated using the formula I've given.
$endgroup$
– user26486
Jun 3 '14 at 18:25












$begingroup$
@StevenStadnicki It follows that the solutions can be generated by $$f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$$ or it can be any function satisfying $f(0)=0$, if my reasoning is correct.
$endgroup$
– user26486
Jun 3 '14 at 18:29






$begingroup$
@StevenStadnicki It follows that the solutions can be generated by $$f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$$ or it can be any function satisfying $f(0)=0$, if my reasoning is correct.
$endgroup$
– user26486
Jun 3 '14 at 18:29














$begingroup$
@StevenStadnicki I've edited the question.
$endgroup$
– user26486
Jun 3 '14 at 18:38




$begingroup$
@StevenStadnicki I've edited the question.
$endgroup$
– user26486
Jun 3 '14 at 18:38










2 Answers
2






active

oldest

votes


















2












$begingroup$

You're most of the way there; now just show that your piecewise $f$ works for arbitrary $x$ and $y$. You may find it easiest to break the analysis into four cases: $x,yneq 0$; $x=0, yneq 0$; $xneq 0, y=0$; and $x, y=0$. For instance, for the last case, you have $xf(y)+yf(x) = 0cdot c+0cdot c = 0 = (0+0)cdot ccdot c = (x+y)cdot f(x)cdot f(y)$. You should find similar results in the other cases.



But note that not every function satisfying $f(0)=0$ satisfies the equation; what happens if you take e.g. $f(x) = sin x$, $x=y=fracpi4$? What you've shown is that every function with $f(0)=0$ satisfies the equation when we choose one of $x,y$ to be $0$, but the equation has to hold for arbitrary $x$ and $y$. You may find it useful to set $x=y$ and see what the resultant restriction on $f(x)$ is.



Incidentally, the approach that you're taking is the typical way of solving these questions: find special values of $x$ or $y$ that simplify the equation considerably and turn it into a one-parameter equation. For instance, taking $x=0$ here (obviously) simplifies the equation considerably; so does taking $x=y$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What I meant was if the solution can't be generated by $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$, then $f(0)=0$.
    $endgroup$
    – user26486
    Jun 3 '14 at 19:09










  • $begingroup$
    I haven't fully understood the solution yet. When $f(0)neq0$, it's clear that $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$ generates all the solutions. How to go on with the case $f(0)=0$?
    $endgroup$
    – user26486
    Jun 3 '14 at 20:12










  • $begingroup$
    We have that the solution $f(x)=0$ isn't generated by $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$, so there has to be something about the case $f(0)=0$ that creates such a solution.
    $endgroup$
    – user26486
    Jun 3 '14 at 20:21










  • $begingroup$
    @mathh If $f(a)=0$ for some $a not =0$, then putting $y=a$ gives $af(x)=0 , forall x$ so $f(x)=0 , forall x$. Otherwise $f(x) not =0$ for $x not =0$. Putting $x=y$ gives $2xf(x)=2xf(x)^2$. For $x not =0$ we have $xf(x) not =0$ by above so $f(x)=1$. Thus $f(x)=begin{cases} c , c in mathbb{R} , x=0 \ 1 , x not =0 end{cases}$.
    $endgroup$
    – Ivan Loh
    Jun 3 '14 at 20:26












  • $begingroup$
    @IvanLoh It seems like our problem was that we analysed the case $y=0implies xf(0)(1-f(x))=0$ and got lost when we were left with the case $f(0)=0$, which was never able to provide us the rest of the solutions. However, when we analysed the case $x=y$, we could check all the cases and finish our solution. I think that was the problem.
    $endgroup$
    – user26486
    Jun 3 '14 at 20:48





















1












$begingroup$

Your solution is not complete. You have only identified a necessary condition. It is not obvious, not true, that they are sufficient conditions.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That's why I emphasized: "However, I can't seem to find a way to check such a solution on the initial equation (we have to check it since we've simply eliminated the impossible solutions and we've only found that the solutions can be somehow generated using the way I've shown)." You shold've read the entire description before commenting.
    $endgroup$
    – user26486
    Jun 4 '14 at 10:55










  • $begingroup$
    Ivan Loh has posted the full solution in the comments. The final solutions are the functions $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$ and $f(x)=0$.
    $endgroup$
    – user26486
    Jun 4 '14 at 11:17










  • $begingroup$
    One of the main questions that were highlighted in brown was "If so, how can I check whether all the generated functions satisfy the initial equation?"
    $endgroup$
    – user26486
    Jun 4 '14 at 12:40











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

You're most of the way there; now just show that your piecewise $f$ works for arbitrary $x$ and $y$. You may find it easiest to break the analysis into four cases: $x,yneq 0$; $x=0, yneq 0$; $xneq 0, y=0$; and $x, y=0$. For instance, for the last case, you have $xf(y)+yf(x) = 0cdot c+0cdot c = 0 = (0+0)cdot ccdot c = (x+y)cdot f(x)cdot f(y)$. You should find similar results in the other cases.



But note that not every function satisfying $f(0)=0$ satisfies the equation; what happens if you take e.g. $f(x) = sin x$, $x=y=fracpi4$? What you've shown is that every function with $f(0)=0$ satisfies the equation when we choose one of $x,y$ to be $0$, but the equation has to hold for arbitrary $x$ and $y$. You may find it useful to set $x=y$ and see what the resultant restriction on $f(x)$ is.



Incidentally, the approach that you're taking is the typical way of solving these questions: find special values of $x$ or $y$ that simplify the equation considerably and turn it into a one-parameter equation. For instance, taking $x=0$ here (obviously) simplifies the equation considerably; so does taking $x=y$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What I meant was if the solution can't be generated by $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$, then $f(0)=0$.
    $endgroup$
    – user26486
    Jun 3 '14 at 19:09










  • $begingroup$
    I haven't fully understood the solution yet. When $f(0)neq0$, it's clear that $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$ generates all the solutions. How to go on with the case $f(0)=0$?
    $endgroup$
    – user26486
    Jun 3 '14 at 20:12










  • $begingroup$
    We have that the solution $f(x)=0$ isn't generated by $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$, so there has to be something about the case $f(0)=0$ that creates such a solution.
    $endgroup$
    – user26486
    Jun 3 '14 at 20:21










  • $begingroup$
    @mathh If $f(a)=0$ for some $a not =0$, then putting $y=a$ gives $af(x)=0 , forall x$ so $f(x)=0 , forall x$. Otherwise $f(x) not =0$ for $x not =0$. Putting $x=y$ gives $2xf(x)=2xf(x)^2$. For $x not =0$ we have $xf(x) not =0$ by above so $f(x)=1$. Thus $f(x)=begin{cases} c , c in mathbb{R} , x=0 \ 1 , x not =0 end{cases}$.
    $endgroup$
    – Ivan Loh
    Jun 3 '14 at 20:26












  • $begingroup$
    @IvanLoh It seems like our problem was that we analysed the case $y=0implies xf(0)(1-f(x))=0$ and got lost when we were left with the case $f(0)=0$, which was never able to provide us the rest of the solutions. However, when we analysed the case $x=y$, we could check all the cases and finish our solution. I think that was the problem.
    $endgroup$
    – user26486
    Jun 3 '14 at 20:48


















2












$begingroup$

You're most of the way there; now just show that your piecewise $f$ works for arbitrary $x$ and $y$. You may find it easiest to break the analysis into four cases: $x,yneq 0$; $x=0, yneq 0$; $xneq 0, y=0$; and $x, y=0$. For instance, for the last case, you have $xf(y)+yf(x) = 0cdot c+0cdot c = 0 = (0+0)cdot ccdot c = (x+y)cdot f(x)cdot f(y)$. You should find similar results in the other cases.



But note that not every function satisfying $f(0)=0$ satisfies the equation; what happens if you take e.g. $f(x) = sin x$, $x=y=fracpi4$? What you've shown is that every function with $f(0)=0$ satisfies the equation when we choose one of $x,y$ to be $0$, but the equation has to hold for arbitrary $x$ and $y$. You may find it useful to set $x=y$ and see what the resultant restriction on $f(x)$ is.



Incidentally, the approach that you're taking is the typical way of solving these questions: find special values of $x$ or $y$ that simplify the equation considerably and turn it into a one-parameter equation. For instance, taking $x=0$ here (obviously) simplifies the equation considerably; so does taking $x=y$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What I meant was if the solution can't be generated by $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$, then $f(0)=0$.
    $endgroup$
    – user26486
    Jun 3 '14 at 19:09










  • $begingroup$
    I haven't fully understood the solution yet. When $f(0)neq0$, it's clear that $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$ generates all the solutions. How to go on with the case $f(0)=0$?
    $endgroup$
    – user26486
    Jun 3 '14 at 20:12










  • $begingroup$
    We have that the solution $f(x)=0$ isn't generated by $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$, so there has to be something about the case $f(0)=0$ that creates such a solution.
    $endgroup$
    – user26486
    Jun 3 '14 at 20:21










  • $begingroup$
    @mathh If $f(a)=0$ for some $a not =0$, then putting $y=a$ gives $af(x)=0 , forall x$ so $f(x)=0 , forall x$. Otherwise $f(x) not =0$ for $x not =0$. Putting $x=y$ gives $2xf(x)=2xf(x)^2$. For $x not =0$ we have $xf(x) not =0$ by above so $f(x)=1$. Thus $f(x)=begin{cases} c , c in mathbb{R} , x=0 \ 1 , x not =0 end{cases}$.
    $endgroup$
    – Ivan Loh
    Jun 3 '14 at 20:26












  • $begingroup$
    @IvanLoh It seems like our problem was that we analysed the case $y=0implies xf(0)(1-f(x))=0$ and got lost when we were left with the case $f(0)=0$, which was never able to provide us the rest of the solutions. However, when we analysed the case $x=y$, we could check all the cases and finish our solution. I think that was the problem.
    $endgroup$
    – user26486
    Jun 3 '14 at 20:48
















2












2








2





$begingroup$

You're most of the way there; now just show that your piecewise $f$ works for arbitrary $x$ and $y$. You may find it easiest to break the analysis into four cases: $x,yneq 0$; $x=0, yneq 0$; $xneq 0, y=0$; and $x, y=0$. For instance, for the last case, you have $xf(y)+yf(x) = 0cdot c+0cdot c = 0 = (0+0)cdot ccdot c = (x+y)cdot f(x)cdot f(y)$. You should find similar results in the other cases.



But note that not every function satisfying $f(0)=0$ satisfies the equation; what happens if you take e.g. $f(x) = sin x$, $x=y=fracpi4$? What you've shown is that every function with $f(0)=0$ satisfies the equation when we choose one of $x,y$ to be $0$, but the equation has to hold for arbitrary $x$ and $y$. You may find it useful to set $x=y$ and see what the resultant restriction on $f(x)$ is.



Incidentally, the approach that you're taking is the typical way of solving these questions: find special values of $x$ or $y$ that simplify the equation considerably and turn it into a one-parameter equation. For instance, taking $x=0$ here (obviously) simplifies the equation considerably; so does taking $x=y$.






share|cite|improve this answer









$endgroup$



You're most of the way there; now just show that your piecewise $f$ works for arbitrary $x$ and $y$. You may find it easiest to break the analysis into four cases: $x,yneq 0$; $x=0, yneq 0$; $xneq 0, y=0$; and $x, y=0$. For instance, for the last case, you have $xf(y)+yf(x) = 0cdot c+0cdot c = 0 = (0+0)cdot ccdot c = (x+y)cdot f(x)cdot f(y)$. You should find similar results in the other cases.



But note that not every function satisfying $f(0)=0$ satisfies the equation; what happens if you take e.g. $f(x) = sin x$, $x=y=fracpi4$? What you've shown is that every function with $f(0)=0$ satisfies the equation when we choose one of $x,y$ to be $0$, but the equation has to hold for arbitrary $x$ and $y$. You may find it useful to set $x=y$ and see what the resultant restriction on $f(x)$ is.



Incidentally, the approach that you're taking is the typical way of solving these questions: find special values of $x$ or $y$ that simplify the equation considerably and turn it into a one-parameter equation. For instance, taking $x=0$ here (obviously) simplifies the equation considerably; so does taking $x=y$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jun 3 '14 at 18:57









Steven StadnickiSteven Stadnicki

41.1k867122




41.1k867122












  • $begingroup$
    What I meant was if the solution can't be generated by $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$, then $f(0)=0$.
    $endgroup$
    – user26486
    Jun 3 '14 at 19:09










  • $begingroup$
    I haven't fully understood the solution yet. When $f(0)neq0$, it's clear that $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$ generates all the solutions. How to go on with the case $f(0)=0$?
    $endgroup$
    – user26486
    Jun 3 '14 at 20:12










  • $begingroup$
    We have that the solution $f(x)=0$ isn't generated by $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$, so there has to be something about the case $f(0)=0$ that creates such a solution.
    $endgroup$
    – user26486
    Jun 3 '14 at 20:21










  • $begingroup$
    @mathh If $f(a)=0$ for some $a not =0$, then putting $y=a$ gives $af(x)=0 , forall x$ so $f(x)=0 , forall x$. Otherwise $f(x) not =0$ for $x not =0$. Putting $x=y$ gives $2xf(x)=2xf(x)^2$. For $x not =0$ we have $xf(x) not =0$ by above so $f(x)=1$. Thus $f(x)=begin{cases} c , c in mathbb{R} , x=0 \ 1 , x not =0 end{cases}$.
    $endgroup$
    – Ivan Loh
    Jun 3 '14 at 20:26












  • $begingroup$
    @IvanLoh It seems like our problem was that we analysed the case $y=0implies xf(0)(1-f(x))=0$ and got lost when we were left with the case $f(0)=0$, which was never able to provide us the rest of the solutions. However, when we analysed the case $x=y$, we could check all the cases and finish our solution. I think that was the problem.
    $endgroup$
    – user26486
    Jun 3 '14 at 20:48




















  • $begingroup$
    What I meant was if the solution can't be generated by $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$, then $f(0)=0$.
    $endgroup$
    – user26486
    Jun 3 '14 at 19:09










  • $begingroup$
    I haven't fully understood the solution yet. When $f(0)neq0$, it's clear that $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$ generates all the solutions. How to go on with the case $f(0)=0$?
    $endgroup$
    – user26486
    Jun 3 '14 at 20:12










  • $begingroup$
    We have that the solution $f(x)=0$ isn't generated by $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$, so there has to be something about the case $f(0)=0$ that creates such a solution.
    $endgroup$
    – user26486
    Jun 3 '14 at 20:21










  • $begingroup$
    @mathh If $f(a)=0$ for some $a not =0$, then putting $y=a$ gives $af(x)=0 , forall x$ so $f(x)=0 , forall x$. Otherwise $f(x) not =0$ for $x not =0$. Putting $x=y$ gives $2xf(x)=2xf(x)^2$. For $x not =0$ we have $xf(x) not =0$ by above so $f(x)=1$. Thus $f(x)=begin{cases} c , c in mathbb{R} , x=0 \ 1 , x not =0 end{cases}$.
    $endgroup$
    – Ivan Loh
    Jun 3 '14 at 20:26












  • $begingroup$
    @IvanLoh It seems like our problem was that we analysed the case $y=0implies xf(0)(1-f(x))=0$ and got lost when we were left with the case $f(0)=0$, which was never able to provide us the rest of the solutions. However, when we analysed the case $x=y$, we could check all the cases and finish our solution. I think that was the problem.
    $endgroup$
    – user26486
    Jun 3 '14 at 20:48


















$begingroup$
What I meant was if the solution can't be generated by $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$, then $f(0)=0$.
$endgroup$
– user26486
Jun 3 '14 at 19:09




$begingroup$
What I meant was if the solution can't be generated by $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$, then $f(0)=0$.
$endgroup$
– user26486
Jun 3 '14 at 19:09












$begingroup$
I haven't fully understood the solution yet. When $f(0)neq0$, it's clear that $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$ generates all the solutions. How to go on with the case $f(0)=0$?
$endgroup$
– user26486
Jun 3 '14 at 20:12




$begingroup$
I haven't fully understood the solution yet. When $f(0)neq0$, it's clear that $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$ generates all the solutions. How to go on with the case $f(0)=0$?
$endgroup$
– user26486
Jun 3 '14 at 20:12












$begingroup$
We have that the solution $f(x)=0$ isn't generated by $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$, so there has to be something about the case $f(0)=0$ that creates such a solution.
$endgroup$
– user26486
Jun 3 '14 at 20:21




$begingroup$
We have that the solution $f(x)=0$ isn't generated by $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$, so there has to be something about the case $f(0)=0$ that creates such a solution.
$endgroup$
– user26486
Jun 3 '14 at 20:21












$begingroup$
@mathh If $f(a)=0$ for some $a not =0$, then putting $y=a$ gives $af(x)=0 , forall x$ so $f(x)=0 , forall x$. Otherwise $f(x) not =0$ for $x not =0$. Putting $x=y$ gives $2xf(x)=2xf(x)^2$. For $x not =0$ we have $xf(x) not =0$ by above so $f(x)=1$. Thus $f(x)=begin{cases} c , c in mathbb{R} , x=0 \ 1 , x not =0 end{cases}$.
$endgroup$
– Ivan Loh
Jun 3 '14 at 20:26






$begingroup$
@mathh If $f(a)=0$ for some $a not =0$, then putting $y=a$ gives $af(x)=0 , forall x$ so $f(x)=0 , forall x$. Otherwise $f(x) not =0$ for $x not =0$. Putting $x=y$ gives $2xf(x)=2xf(x)^2$. For $x not =0$ we have $xf(x) not =0$ by above so $f(x)=1$. Thus $f(x)=begin{cases} c , c in mathbb{R} , x=0 \ 1 , x not =0 end{cases}$.
$endgroup$
– Ivan Loh
Jun 3 '14 at 20:26














$begingroup$
@IvanLoh It seems like our problem was that we analysed the case $y=0implies xf(0)(1-f(x))=0$ and got lost when we were left with the case $f(0)=0$, which was never able to provide us the rest of the solutions. However, when we analysed the case $x=y$, we could check all the cases and finish our solution. I think that was the problem.
$endgroup$
– user26486
Jun 3 '14 at 20:48






$begingroup$
@IvanLoh It seems like our problem was that we analysed the case $y=0implies xf(0)(1-f(x))=0$ and got lost when we were left with the case $f(0)=0$, which was never able to provide us the rest of the solutions. However, when we analysed the case $x=y$, we could check all the cases and finish our solution. I think that was the problem.
$endgroup$
– user26486
Jun 3 '14 at 20:48













1












$begingroup$

Your solution is not complete. You have only identified a necessary condition. It is not obvious, not true, that they are sufficient conditions.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That's why I emphasized: "However, I can't seem to find a way to check such a solution on the initial equation (we have to check it since we've simply eliminated the impossible solutions and we've only found that the solutions can be somehow generated using the way I've shown)." You shold've read the entire description before commenting.
    $endgroup$
    – user26486
    Jun 4 '14 at 10:55










  • $begingroup$
    Ivan Loh has posted the full solution in the comments. The final solutions are the functions $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$ and $f(x)=0$.
    $endgroup$
    – user26486
    Jun 4 '14 at 11:17










  • $begingroup$
    One of the main questions that were highlighted in brown was "If so, how can I check whether all the generated functions satisfy the initial equation?"
    $endgroup$
    – user26486
    Jun 4 '14 at 12:40
















1












$begingroup$

Your solution is not complete. You have only identified a necessary condition. It is not obvious, not true, that they are sufficient conditions.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That's why I emphasized: "However, I can't seem to find a way to check such a solution on the initial equation (we have to check it since we've simply eliminated the impossible solutions and we've only found that the solutions can be somehow generated using the way I've shown)." You shold've read the entire description before commenting.
    $endgroup$
    – user26486
    Jun 4 '14 at 10:55










  • $begingroup$
    Ivan Loh has posted the full solution in the comments. The final solutions are the functions $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$ and $f(x)=0$.
    $endgroup$
    – user26486
    Jun 4 '14 at 11:17










  • $begingroup$
    One of the main questions that were highlighted in brown was "If so, how can I check whether all the generated functions satisfy the initial equation?"
    $endgroup$
    – user26486
    Jun 4 '14 at 12:40














1












1








1





$begingroup$

Your solution is not complete. You have only identified a necessary condition. It is not obvious, not true, that they are sufficient conditions.






share|cite|improve this answer









$endgroup$



Your solution is not complete. You have only identified a necessary condition. It is not obvious, not true, that they are sufficient conditions.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jun 4 '14 at 1:16









Calvin LinCalvin Lin

36.3k349114




36.3k349114












  • $begingroup$
    That's why I emphasized: "However, I can't seem to find a way to check such a solution on the initial equation (we have to check it since we've simply eliminated the impossible solutions and we've only found that the solutions can be somehow generated using the way I've shown)." You shold've read the entire description before commenting.
    $endgroup$
    – user26486
    Jun 4 '14 at 10:55










  • $begingroup$
    Ivan Loh has posted the full solution in the comments. The final solutions are the functions $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$ and $f(x)=0$.
    $endgroup$
    – user26486
    Jun 4 '14 at 11:17










  • $begingroup$
    One of the main questions that were highlighted in brown was "If so, how can I check whether all the generated functions satisfy the initial equation?"
    $endgroup$
    – user26486
    Jun 4 '14 at 12:40


















  • $begingroup$
    That's why I emphasized: "However, I can't seem to find a way to check such a solution on the initial equation (we have to check it since we've simply eliminated the impossible solutions and we've only found that the solutions can be somehow generated using the way I've shown)." You shold've read the entire description before commenting.
    $endgroup$
    – user26486
    Jun 4 '14 at 10:55










  • $begingroup$
    Ivan Loh has posted the full solution in the comments. The final solutions are the functions $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$ and $f(x)=0$.
    $endgroup$
    – user26486
    Jun 4 '14 at 11:17










  • $begingroup$
    One of the main questions that were highlighted in brown was "If so, how can I check whether all the generated functions satisfy the initial equation?"
    $endgroup$
    – user26486
    Jun 4 '14 at 12:40
















$begingroup$
That's why I emphasized: "However, I can't seem to find a way to check such a solution on the initial equation (we have to check it since we've simply eliminated the impossible solutions and we've only found that the solutions can be somehow generated using the way I've shown)." You shold've read the entire description before commenting.
$endgroup$
– user26486
Jun 4 '14 at 10:55




$begingroup$
That's why I emphasized: "However, I can't seem to find a way to check such a solution on the initial equation (we have to check it since we've simply eliminated the impossible solutions and we've only found that the solutions can be somehow generated using the way I've shown)." You shold've read the entire description before commenting.
$endgroup$
– user26486
Jun 4 '14 at 10:55












$begingroup$
Ivan Loh has posted the full solution in the comments. The final solutions are the functions $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$ and $f(x)=0$.
$endgroup$
– user26486
Jun 4 '14 at 11:17




$begingroup$
Ivan Loh has posted the full solution in the comments. The final solutions are the functions $f(x)=begin{cases}c & cinmathbb R & x=0\ 1 & xneq 0end{cases}$ and $f(x)=0$.
$endgroup$
– user26486
Jun 4 '14 at 11:17












$begingroup$
One of the main questions that were highlighted in brown was "If so, how can I check whether all the generated functions satisfy the initial equation?"
$endgroup$
– user26486
Jun 4 '14 at 12:40




$begingroup$
One of the main questions that were highlighted in brown was "If so, how can I check whether all the generated functions satisfy the initial equation?"
$endgroup$
– user26486
Jun 4 '14 at 12:40


















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