How's that FE?

[SANZEED]

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ which satisfies
$f(x^{2}-y^{2})=xf(x)-yf(y)$.

[Tahmid Hasan]

Nice problem.For the last part I used a technique which I learned form Bappa vai.

Let $P(x,y) \Rightarrow f(x^2-y^2)=xf(x)-yf(y)$
$P(x,x) \Rightarrow f(0)=0$
$P(x,0) \Rightarrow f(x^2)=xf(x)$
So $P(x,y)$ can be written as $f(x^2-y^2)=f(x^2)-f(y^2)$
Hence $f(x-y)=f(x)-f(y) \forall x \ge 0,y \ge 0$.Let it be $Q(x,y)$.
$Q(x+y,y) \Rightarrow f(x+y)=f(x)+f(y) \forall x \ge 0,y \ge 0$.Let it be $R(x,y)$
Now $\forall x<0, f(-x)=f(0-x)=f(0)-f(x)=-f(x)$
Let $y<0$,then $R(x,-y) \Rightarrow f(x-y)=f(x)-f(y)$
So $f(x-y)=f(x)-f(y) \forall x \ge 0,y \in \mathbb{R}$
Let $y<0$,then $Q(x,-y) \Rightarrow f(x+y)=f(x)+f(y)$
So $f(x+y)=f(x)+f(y) \forall x \ge 0,y \in \mathbb{R}$
Let $x<0$,then take any $y<0$;so $f(x+y)=-f(-x-y)=-(f(-x)+f(-y))=-(-f(x)-f(y))=f(x)+f(y)$
So we conclude $f(x+y)=f(x)+f(y) \forall x,y \in \mathbb{R}$
Now $2f(x)+f(1)=f(x)+(f(x)+f(1))=f(x)+f(x+1)=f(2x+1)=f((x+1)^2-x^2)$
$=(x+1)f(x+1)-xf(x)=(x+1)(f(x)+f(1))-xf(x)=f(x)+xf(1)+f(1)$
Hence $f(x)=xf(1) \forall x \in \mathbb{R}$ which indeed satisfies $P(x,y)$.

[SANZEED]

Alright. But I hope additionally I can post another problem under this very topic.
Find all functions $f:\mathbb{Q}\rightarrow \mathbb{Q}$ such that $f(x+y)=f(x)+f(y)+xy$.
And that was a nice solution Tahmid vai.

[SANZEED]

I am posting what I have done yet for the second problem under this topic.

Let us denote the sentence $f(x+y)=f(x)+f(y)+xy$ with $P(x,y)$.
Then, $P(0,0)\Rightarrow f(0)=2f(0)\Rightarrow f(0)=0$
$P(x,1)\Rightarrow f(x+1)=f(x)+f(1)+x$.
From the last one, by induction,we can say that $f(x)=\frac{x(x-1)}{2}+xf(1)\forall x\in \mathbb{N}$
Again,$P(x,y)\Rightarrow f(0)=f(x)+f(-x)-x^{2}$. Using the fact that $f(0)=0$,we can say,
$f(-x)=x^{2}-f(x)$. From this relationship,we can decide that
$f(z)=\frac{z(z-1)}{2}+zf(1)\forall z\in \mathbb{Z}$.
Now the flaw in my approach is,I don't know how to,and whether it can be,extended to $\mathbb{Q}$.

[Nadim Ul Abrar]

Umm what about this bro ?

Let $g(x)=f(x)-\frac{x(x-1)}{2}$
Now $g:Q \to Q$
and $g(x+y)=g(x)+g(y)$

Using Cauchy , $g(x)=ax=f(x)-\frac{x(x-1)}{2}$
So $f(x)=\frac{x(x-1)}{2}+ax$

Thanku Tahmid Vai ... <3

[Phlembac Adib Hasan]

$f(z)=\frac{z(z-1)}{2}+zf(1)\forall z\in \mathbb{Z}$.
Now the flaw in my approach is,I don't know how to,and whether it can be,extended to $\mathbb{Q}$.

(I'm using your notations)
Firstly take $P(1,m+1/x),m,x\in\mathbb {N}$ and use induction to find the value of $f(m+1/x)$.Next take $P(x,\displaystyle \frac {1}{x}), x\in \mathbb {N}$.Then we'll get the value of $f(\frac {1}{x})$.Now again use induction and take $P(n/x,1/x), n\in \mathbb {N}$ to finish the problem.

[Phlembac Adib Hasan]

Guys, this is a very similar f.e. which I've solved just one or two days before.Try this now!
http://www.matholympiad.org.bd/forum/vi ... =27&t=2411

[SANZEED]

$f(z)=\frac{z(z-1)}{2}+zf(1)\forall z\in \mathbb{Z}$.
Now the flaw in my approach is,I don't know how to,and whether it can be,extended to $\mathbb{Q}$.

(I'm using your notations)
Firstly take $P(1,m+1/x),m,x\in\mathbb {N}$ and use induction to find the value of $f(m+1/x)$.Next take $P(x,\displaystyle \frac {1}{x}), x\in \mathbb {N}$.Then we'll get the value of $f(\frac {1}{x})$.Now again use induction and take $P(n/x,1/x), n\in \mathbb {N}$ to finish the problem.

Adib,can you please do a little favour to me by explaining the first induction here to use?

[Phlembac Adib Hasan]

Adib,can you please do a little favour to me by explaining the first induction here to use?

$f(1+1/x)=f(1)+f(1/x)+1/x$
$f(1+(1+1/x))=f(1)+f(1+1/x)+1+1/x=2f(1)+f(1/x)+1+2/x$
I hope it's clear enough now.

[SANZEED]

(I'm using your notations)
Firstly take $P(1,m+1/x),m,x\in \mathbb N$ and use induction to find the value of $f(m+1/x)$ .Next take $P(x,1/x)$.Then we'll get the value of $f(1/x)$.Now again use induction and take $P(n/x,1/x),n\in \mathbb N$ to finish the problem.

Can anyone tell me how to get the value of $f(1/x)$ from here?