Mixing
I was trying to find the function which satisfies $f(x)*f(y)+2=f(x)+f(y)+f(xy)$
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I was trying to find the function which satisfies $$f(x)*f(y)+2=f(x)+f(y)+f(xy) ,$$ where its given that $f'(0)=0$ and $f'(1)=2$.
So what I did was partially differentiated the relation wrt $x$.
So I got $f'(x)*f(y)=f'(x)+f'(xy)y$.
Putting $x=1$ we get
$f'(1)*f(y)= f'(1)+f'(y)y$.
Solving this diff equation I couldn't get the proper function.
limits functions
edited Jan 14 at 10:03
Christian Blatter
173k8114326
asked Jan 14 at 9:32
SANYUKTA DEOGADESANYUKTA DEOGADE
11
$endgroup$
add a comment |
$begingroup$
I was trying to find the function which satisfies $$f(x)*f(y)+2=f(x)+f(y)+f(xy) ,$$ where its given that $f'(0)=0$ and $f'(1)=2$.
So what I did was partially differentiated the relation wrt $x$.
So I got $f'(x)*f(y)=f'(x)+f'(xy)y$.
Putting $x=1$ we get
$f'(1)*f(y)= f'(1)+f'(y)y$.
Solving this diff equation I couldn't get the proper function.
limits functions
edited Jan 14 at 10:03
Christian Blatter
173k8114326
asked Jan 14 at 9:32
SANYUKTA DEOGADESANYUKTA DEOGADE
11
$endgroup$
$begingroup$
$f(0)f(0) + 2 = f(0)+f(0)+f(0)$ so $f(0)^2 + 2 = 3f(0)$ which can be factored as $(f(0)-1)(f(0)-2) = 0$. You have either $f(0) = 1$ or $f(0) = 2$, so there's probably something wrong with your problem statement.
$endgroup$
– RcnSc
Jan 14 at 9:41
add a comment |
$begingroup$
I was trying to find the function which satisfies $$f(x)*f(y)+2=f(x)+f(y)+f(xy) ,$$ where its given that $f'(0)=0$ and $f'(1)=2$.
So what I did was partially differentiated the relation wrt $x$.
So I got $f'(x)*f(y)=f'(x)+f'(xy)y$.
Putting $x=1$ we get
$f'(1)*f(y)= f'(1)+f'(y)y$.
Solving this diff equation I couldn't get the proper function.
limits functions
edited Jan 14 at 10:03
Christian Blatter
173k8114326
asked Jan 14 at 9:32
SANYUKTA DEOGADESANYUKTA DEOGADE
11
$endgroup$
I was trying to find the function which satisfies $$f(x)*f(y)+2=f(x)+f(y)+f(xy) ,$$ where its given that $f'(0)=0$ and $f'(1)=2$.
So what I did was partially differentiated the relation wrt $x$.
So I got $f'(x)*f(y)=f'(x)+f'(xy)y$.
Putting $x=1$ we get
$f'(1)*f(y)= f'(1)+f'(y)y$.
Solving this diff equation I couldn't get the proper function.
limits functions
limits functions
edited Jan 14 at 10:03
Christian Blatter
173k8114326
asked Jan 14 at 9:32
SANYUKTA DEOGADESANYUKTA DEOGADE
11
edited Jan 14 at 10:03
Christian Blatter
173k8114326
asked Jan 14 at 9:32
SANYUKTA DEOGADESANYUKTA DEOGADE
11
edited Jan 14 at 10:03
Christian Blatter
173k8114326
edited Jan 14 at 10:03
Christian Blatter
173k8114326
edited Jan 14 at 10:03
Christian Blatter
173k8114326
173k8114326
asked Jan 14 at 9:32
SANYUKTA DEOGADESANYUKTA DEOGADE
11
asked Jan 14 at 9:32
SANYUKTA DEOGADESANYUKTA DEOGADE
11
asked Jan 14 at 9:32
SANYUKTA DEOGADESANYUKTA DEOGADE
11
11
$begingroup$
$f(0)f(0) + 2 = f(0)+f(0)+f(0)$ so $f(0)^2 + 2 = 3f(0)$ which can be factored as $(f(0)-1)(f(0)-2) = 0$. You have either $f(0) = 1$ or $f(0) = 2$, so there's probably something wrong with your problem statement.
$endgroup$
– RcnSc
Jan 14 at 9:41
add a comment |
$begingroup$
$f(0)f(0) + 2 = f(0)+f(0)+f(0)$ so $f(0)^2 + 2 = 3f(0)$ which can be factored as $(f(0)-1)(f(0)-2) = 0$. You have either $f(0) = 1$ or $f(0) = 2$, so there's probably something wrong with your problem statement.
$endgroup$
– RcnSc
Jan 14 at 9:41
$begingroup$
$f(0)f(0) + 2 = f(0)+f(0)+f(0)$ so $f(0)^2 + 2 = 3f(0)$ which can be factored as $(f(0)-1)(f(0)-2) = 0$. You have either $f(0) = 1$ or $f(0) = 2$, so there's probably something wrong with your problem statement.
$endgroup$
– RcnSc
Jan 14 at 9:41
$begingroup$
$f(0)f(0) + 2 = f(0)+f(0)+f(0)$ so $f(0)^2 + 2 = 3f(0)$ which can be factored as $(f(0)-1)(f(0)-2) = 0$. You have either $f(0) = 1$ or $f(0) = 2$, so there's probably something wrong with your problem statement.
$endgroup$
– RcnSc
Jan 14 at 9:41
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $f(x) = g(x)+1$. Then the above equation becomes $g(x)g(y) = g(xy)$. Assuming continuity of $f(x)$ the solutions for this Cauchy Functional Equation are if the form $g(x) =|x|^c$. The given conditions imply $c=2$ so that $f(x)=x^2+1$
answered Jan 14 at 9:54
Hari ShankarHari Shankar
2,276139
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add a comment |
$begingroup$
From
$f(x)f(y)+2 = f(x) + f(y) + f(xy)$
setting $x=y=0$ we have
$f(0)^2 + 2 = 3f(0)$
Solving this quadratic gives $f(0)in {1,2 }$. So there is no solution with $f(0)=0$.
Similarly, setting $x=y=1$ gives $f(1)^2 + 2 = 3f(1)$ and so $f(1)in {1,2 }$.
If $f(0)=2$ then setting $x=0$ gives us
$2f(y)+2=f(y)+4\ Rightarrow f(y)=2 space forall y$
On the other hand, if $f(1)=1$ then setting $x=1$ gives us
$f(y)+2=2f(y)+1 \ Rightarrow f(y)=1 space forall y$
Note: the problem statement was changed after I posted this.
answered Jan 14 at 9:50
gandalf61gandalf61
8,761725
$endgroup$
add a comment |
$begingroup$
Of course you can solve it from an ODE! Here is a hint. Substitute $f(y)=y^2g(y)$ in your differential equation and find $g(y)$. From which then, you can find $f(y)$... plain and simple!!
answered Jan 14 at 10:13
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $f(x) = g(x)+1$. Then the above equation becomes $g(x)g(y) = g(xy)$. Assuming continuity of $f(x)$ the solutions for this Cauchy Functional Equation are if the form $g(x) =|x|^c$. The given conditions imply $c=2$ so that $f(x)=x^2+1$
answered Jan 14 at 9:54
Hari ShankarHari Shankar
2,276139
$endgroup$
add a comment |
$begingroup$
Let $f(x) = g(x)+1$. Then the above equation becomes $g(x)g(y) = g(xy)$. Assuming continuity of $f(x)$ the solutions for this Cauchy Functional Equation are if the form $g(x) =|x|^c$. The given conditions imply $c=2$ so that $f(x)=x^2+1$
answered Jan 14 at 9:54
Hari ShankarHari Shankar
2,276139
$endgroup$
add a comment |
$begingroup$
Let $f(x) = g(x)+1$. Then the above equation becomes $g(x)g(y) = g(xy)$. Assuming continuity of $f(x)$ the solutions for this Cauchy Functional Equation are if the form $g(x) =|x|^c$. The given conditions imply $c=2$ so that $f(x)=x^2+1$
answered Jan 14 at 9:54
Hari ShankarHari Shankar
2,276139
$endgroup$
Let $f(x) = g(x)+1$. Then the above equation becomes $g(x)g(y) = g(xy)$. Assuming continuity of $f(x)$ the solutions for this Cauchy Functional Equation are if the form $g(x) =|x|^c$. The given conditions imply $c=2$ so that $f(x)=x^2+1$
answered Jan 14 at 9:54
Hari ShankarHari Shankar
2,276139
answered Jan 14 at 9:54
Hari ShankarHari Shankar
2,276139
answered Jan 14 at 9:54
Hari ShankarHari Shankar
2,276139
answered Jan 14 at 9:54
Hari ShankarHari Shankar
2,276139
2,276139
add a comment |
add a comment |
$begingroup$
From
$f(x)f(y)+2 = f(x) + f(y) + f(xy)$
setting $x=y=0$ we have
$f(0)^2 + 2 = 3f(0)$
Solving this quadratic gives $f(0)in {1,2 }$. So there is no solution with $f(0)=0$.
Similarly, setting $x=y=1$ gives $f(1)^2 + 2 = 3f(1)$ and so $f(1)in {1,2 }$.
If $f(0)=2$ then setting $x=0$ gives us
$2f(y)+2=f(y)+4\ Rightarrow f(y)=2 space forall y$
On the other hand, if $f(1)=1$ then setting $x=1$ gives us
$f(y)+2=2f(y)+1 \ Rightarrow f(y)=1 space forall y$
Note: the problem statement was changed after I posted this.
answered Jan 14 at 9:50
gandalf61gandalf61
8,761725
$endgroup$
add a comment |
$begingroup$
From
$f(x)f(y)+2 = f(x) + f(y) + f(xy)$
setting $x=y=0$ we have
$f(0)^2 + 2 = 3f(0)$
Solving this quadratic gives $f(0)in {1,2 }$. So there is no solution with $f(0)=0$.
Similarly, setting $x=y=1$ gives $f(1)^2 + 2 = 3f(1)$ and so $f(1)in {1,2 }$.
If $f(0)=2$ then setting $x=0$ gives us
$2f(y)+2=f(y)+4\ Rightarrow f(y)=2 space forall y$
On the other hand, if $f(1)=1$ then setting $x=1$ gives us
$f(y)+2=2f(y)+1 \ Rightarrow f(y)=1 space forall y$
Note: the problem statement was changed after I posted this.
answered Jan 14 at 9:50
gandalf61gandalf61
8,761725
$endgroup$
add a comment |
$begingroup$
From
$f(x)f(y)+2 = f(x) + f(y) + f(xy)$
setting $x=y=0$ we have
$f(0)^2 + 2 = 3f(0)$
Solving this quadratic gives $f(0)in {1,2 }$. So there is no solution with $f(0)=0$.
Similarly, setting $x=y=1$ gives $f(1)^2 + 2 = 3f(1)$ and so $f(1)in {1,2 }$.
If $f(0)=2$ then setting $x=0$ gives us
$2f(y)+2=f(y)+4\ Rightarrow f(y)=2 space forall y$
On the other hand, if $f(1)=1$ then setting $x=1$ gives us
$f(y)+2=2f(y)+1 \ Rightarrow f(y)=1 space forall y$
Note: the problem statement was changed after I posted this.
answered Jan 14 at 9:50
gandalf61gandalf61
8,761725
$endgroup$
From
$f(x)f(y)+2 = f(x) + f(y) + f(xy)$
setting $x=y=0$ we have
$f(0)^2 + 2 = 3f(0)$
Solving this quadratic gives $f(0)in {1,2 }$. So there is no solution with $f(0)=0$.
Similarly, setting $x=y=1$ gives $f(1)^2 + 2 = 3f(1)$ and so $f(1)in {1,2 }$.
If $f(0)=2$ then setting $x=0$ gives us
$2f(y)+2=f(y)+4\ Rightarrow f(y)=2 space forall y$
On the other hand, if $f(1)=1$ then setting $x=1$ gives us
$f(y)+2=2f(y)+1 \ Rightarrow f(y)=1 space forall y$
Note: the problem statement was changed after I posted this.
answered Jan 14 at 9:50
gandalf61gandalf61
8,761725
edited Jan 14 at 10:04
answered Jan 14 at 9:50
gandalf61gandalf61
8,761725
answered Jan 14 at 9:50
gandalf61gandalf61
8,761725
answered Jan 14 at 9:50
gandalf61gandalf61
8,761725
8,761725
add a comment |
add a comment |
$begingroup$
Of course you can solve it from an ODE! Here is a hint. Substitute $f(y)=y^2g(y)$ in your differential equation and find $g(y)$. From which then, you can find $f(y)$... plain and simple!!
answered Jan 14 at 10:13
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
Of course you can solve it from an ODE! Here is a hint. Substitute $f(y)=y^2g(y)$ in your differential equation and find $g(y)$. From which then, you can find $f(y)$... plain and simple!!
answered Jan 14 at 10:13
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
Of course you can solve it from an ODE! Here is a hint. Substitute $f(y)=y^2g(y)$ in your differential equation and find $g(y)$. From which then, you can find $f(y)$... plain and simple!!
answered Jan 14 at 10:13
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
Of course you can solve it from an ODE! Here is a hint. Substitute $f(y)=y^2g(y)$ in your differential equation and find $g(y)$. From which then, you can find $f(y)$... plain and simple!!
answered Jan 14 at 10:13
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 14 at 10:13
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 14 at 10:13
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 14 at 10:13
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
add a comment |
add a comment |
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$begingroup$
$f(0)f(0) + 2 = f(0)+f(0)+f(0)$ so $f(0)^2 + 2 = 3f(0)$ which can be factored as $(f(0)-1)(f(0)-2) = 0$. You have either $f(0) = 1$ or $f(0) = 2$, so there's probably something wrong with your problem statement.
$endgroup$
– RcnSc
Jan 14 at 9:41