Find all solutions of the following functional equation: f(x2+y+f(y))=2y+f(x)2.
Problem
Source: Iran TST 2007, Day 1
Tags: algebra proposed, algebra
05.05.2007 21:36
See here http://www.mathlinks.ro/Forum/viewtopic.php?p=18289#18289
29.09.2009 12:58
Here's my Ugly Proof... Let given functional eq as T-eq We know f is surjective cuz for given x, 2y moves all in Real number. Let f(a)=0 for some a, take y=0 in T-eq then f(x2+f(0))=f(x)2, Thus f(x)2=f(x2), so f(−a)=0 Take x=0,y=a,−a, we get f(a)=2a+f(0)2, f(−a)=−2a+f(0)2 so a=0 Here, We get f(x)=0⟺x=0 Now put y=−f(x)22, we get f(x2−f(x)22+f(−f(x)22))=0 which means x2−f(x)22+f(−f(x)22)=0 Thus f(p)=f(q)⟹p2=q2 If f(p)=f(−p)=r, where p>0 put y=0 in T-eq, f(x2)=f(x)2 Thus f(x)≥0 when x≥0, Thus r≥0 put y=p,−p in T-eq, we get f(x2+p+r)=2p+f(x2), f(x2−p+r)=−2p+f(x2) Put x=0 in first one, f(p+r)=2p, put x=√p+r in Second one, we get f(2r)=−2p+f(p+r)=0 so r=0 means p=0 Therefore, f(p)=−f(−p) Thus f(p)=f(q) means p2=q2 so p=q which means f is injective Now, take x=0 in T-eq, we get f(y+f(y))=2y. We know f is bijective. Thus y+f(y) moves all of real numbers. So we can set y+f(y)=t with ∀t∈R Let's reconsider about T-eq, we get f(x+t)=f(x)+f(t) with x≥0 Then f(x+y)=f(|x|+x+y−|x|)=f(|x|−x)+f(y−|x|)=f(|x|)+f(x)−f(|x|−y) =f(x)+f(y) Thus f(x+y)=f(x)+f(y) Thus f(q)=kq for q∈Q Then ∀α≥0, f(x+α)=f(x)+f(√α)2≥f(x) which means f in increasing function. ∀x∈R, let two rational sequence (an),(bn) which are converge to x with an<x, bn>x for all ninN Then f(an)=kan≤f(x)≤f(bn)=kbn, so f(x)=kx for all x∈R cuz f(x2)=f(x)2, f(1)=1. so we get f(x)=x for all x∈R
29.04.2014 18:15
we have f is surjective and f(0)=0, f(y+f(y))=2y and f(x2)=f(x)2. We can f(a)=0 ⇒ a=0. So g(x)=f(x)+x is surjective in g:(−∞,0]→(−∞,0]. For all z≤0, x∈R, we have f(x+z)=f(x)+f(z) and f(x)=−f(−x) - odd function. So g is surjective in all R. For all x,y≥0 we get that f(x+y)=f(x)+f(y) and f is monotone increasing. Answer: f(x)=x.
07.07.2014 07:48
P(x,y) and P(−x,y) give us f(x)2=f(−x)2 So f(x)=f(−x) or f(x)=−f(−x) Suppose that f(0)≠0, If there exist a t satisfy f(t)=−f(−t) then P(0,t) and P(0,−t) give us f(0)=0, a condradiction. Since f(x)2=f(−x)2, f is even. If there exist p s.t. f(p)≤0 and p≠0 We can get that p=0 from P(√−f(p),p) and P(√−f(p),−p) So f(x)>0 for all non-zero real number x a contracdition with P(x,−y) for a sufficient large y So f(0)=0, and it's easy to get f(x2)=f(x)2 and f(x+y)=f(x)+f(y) Hence f(x)=x
19.11.2017 21:29
Here is a solution by my friend and teammate RAMİL JABİYEV (İ like his fe solutions that is why I post here) Claim 1:f(a)=0 ⟹ a=0 Proof:Just see P(0,a) and P(0,−a) (Note:Here we used that f(x)2=f(−x)2) Then done... Claim 2:f(f(x))+f(x)=2x Proof:Firstly lets look at P(x,−f(x)22) which gives us x2−f(x)22−f(−f(x)22)=0.Then P(0,−f(x)22) yields that f is odd.Now P(x,−x2) gives the desired relation . Claim 3:f(2x)=2f(x) Proof:Using the above claim it is nothing Claim 3:f(x)=x is a solution Proof: P(x,f(y)) and using the fact that f is odd we conclude that f is additive and since f is bounded in some interval we get f(x)=x for any x∈R.
19.11.2017 21:35
tenplusten wrote: ... and since f is bounded we get f(x)=x for any x∈R. Quite funny "since f(x) is bounded ...", we get ... an unbounded function .... Do you think this sentence is quite consistant ?
19.11.2017 21:47
f(x2)=f(x)2≥0 .So f is bounded in some interval
20.11.2017 11:46
tenplusten wrote: f(x2)=f(x)2≥0 .So f is bounded in some interval According to me, each word is important in maths statements. And "f is bounded" is not equivalent to "f is lowerbounded over R+" With this last sentence, I do agree.
29.11.2022 00:30
Alternative ending: After we establish that 2x=f(x)+f(f(x)) we set an=nth iteration of f. Then, we get that an+an−1=an−2 and solving the recurrence we find that an=l+m(−2)n. But f is lowerbounded in positive reals and therefore we must have m=0 and an=l which is constant. Thus, f(f(x))=f(x) and due to surjectivity we have f(x)=x in positive reals. Now choose x large enough s.t. x2+y+f(y)>0 and from the initial relation we get that f(y)=y,∀y∈R