Find all$$f:R→R$$such that$$ f(x^2+f(y))=(f(x))^2+y,∀ x,y∈R$$ Help me to understand this type of questions.I am new to this pattern. Thanks in advance!
@rational
@Michele_Laino @rational please help :)
is the answer (x+4)/3?
hello..
answer is$$ f(x)=x$$
ok..
@Michele_Laino hey will you please help
here I'm
yes will you please wait for sometime Michele please :)
ok!
okay I am here :) Thanks sister :) @rvc
Thanks @Michele_Laino for waiting so patiently :)
@Michele_Laino will the function be an indentity function.
I think that your equation can be viewed as an operational equation
namely a functional equation
yes it is from the olympiad prep materials. And a question from the functional equation section.
there exist an operator, say T, such that, your equation can be written as below: \[T\left( f \right) = ...\] so, we have to know who is our operator T
taking x=0 we have $$\begin{align} \forall x,y\in \Bbb R^2&& f\left(f\left(y\right)\right)=y+f(0)^2=y\tag2\end{align}$$ which means that f is one one and onto. can it be ???
I'm not sure
for y=o we will have$$ f(x^2)=(f(x))^2$$
why f(0)=0?
please wait a moment, someone is calling me to my phone
here I'm
using the Taylor expansion, at the left side, we can write: \[\Large f\left( {{x^2} + f\left( y \right)} \right) = f\left( {{x^2}} \right) + f\left( y \right)\frac{{df}}{{dx}}\]
sorry, that formula is wrong
I guess so ! I am confused with the step.
for example one solution of your problem is the identity function, as you stated before.
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