Functional Equation -

ok

found it in my note book... it says x+y

what will it be if its xy ?

mirka cum on gmail

hey c post 1 i did na ............

woh rite hai na :-S

Well is it mentioned that the function is differentiable at other points than x=0?

f(x+y)=xf(y)+yf(x)

put y=-x

f(0)=xf(-x)-xf(x)

put x=y=0

f(0)=0

thus we get

f(-x)=f(x)

put y=-2x

f(-x)=xf(-2x)-2xf(x)

but f(-x)=f(x) and f(-2x)=f(2x)=2xf(x)

so we get

f(x)=2x²f(x)-2xf(x)

which gives f(x)=0

But then f'(0)=0 opposing to the condition given in question

f^' (x)=f(x+h)-f(x)/h where h tends to 0

{using first principle}

=xf(h)+hf(x)-f(x)/h
=f(x)+ [xf(h)-f(x)]/h
but f(x)=xf(0) using given condition
so we get
f'(x)=f(x) + [x{f(h)-f(0)}]/h where h tends to 0

the second part has 0/0 form

.. so we can apply l'hospital so

f'(x) =f(x) + x f'(h) putting h=0
we get f'(X)=f(x)+xf'(0) or

f'(x)=f(x)+x as f'(0)=1

put x=y

f(2x)=2xf(x)

differentiate

f'(2x)=xf'(x)+f(x)

put x=0

we get f(0)=1

this is in contradicton when we put x=y=0 in org. equn,by which we get f(0)=0

so,i think, no function satisfies the given condition !!!

cant we do this by first principle?

edited ....