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Let R\mathbb{R}R be the set of real numbers. Determine all functions f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R that satisfy the equation
f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)
for all real numbers xxx and yyy.
Note by Sualeh Asif 5 years, 7 months ago
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Let P(x,y)P(x,y)P(x,y) be the above FE.
P(0,0) ⟹ f(f(0))+f(0)=f(0) ⟹ f(f(0))=0P(0,0)\implies f(f(0))+f(0)=f(0)\implies f(f(0))=0P(0,0)⟹f(f(0))+f(0)=f(0)⟹f(f(0))=0 P(0,f(0)) ⟹ 2f(0)=f(f(0))+f(0)2 ⟹ f(0)2−2f(0)=0 ⟹ f(0)=0,2P(0, f(0))\implies 2f(0)=f(f(0))+f(0)^2\implies f(0)^2-2f(0)=0\implies f(0)=0, 2P(0,f(0))⟹2f(0)=f(f(0))+f(0)2⟹f(0)2−2f(0)=0⟹f(0)=0,2 Case 1: f(0)=0f(0)=0f(0)=0. Then P(0,y) ⟹ f(f(y))+f(0)=f(y)+yf(0) ⟹ f(f(y))=f(y)P(0, y)\implies f(f(y)) + f(0)=f(y)+yf(0)\implies f(f(y))=f(y)P(0,y)⟹f(f(y))+f(0)=f(y)+yf(0)⟹f(f(y))=f(y) Since the range of f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R is R\mathbb{R}R, then f(x)=x\boxed{f(x)=x}f(x)=x.
Case 2: f(0)=2f(0)=2f(0)=2. Then let g(x)=f(x)−2g(x)=f(x)-2g(x)=f(x)−2 which changes the FE to P(x):=g(g(x+y)+x+2)+f(xy)+2=x+g(x+y)+yg(x)+2yP(x):= g(g(x+y)+x+2)+f(xy)+2=x+g(x+y)+yg(x)+2yP(x):=g(g(x+y)+x+2)+f(xy)+2=x+g(x+y)+yg(x)+2y [I'll finish later]
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In case 1, you must prove that fff is surjective to make that conclusion because R\mathbb RR is the codomain of fff, which is not necessarily the range.
Oh right. I'll think about this problem some more.
It does look like f(x)=2−x f(x) = 2-x f(x)=2−x satisfies the functional equation. My guess would be f(x)=x f(x) = x f(x)=x or f(x)=2−x f(x) = 2-x f(x)=2−x, but I do agree with the comment that case 1 still has some work to be done to get to f(x)=x f(x) = x f(x)=x.
Yes you are right @Patrick Corn !
f(x)=xf(x)= xf(x)=x or 2−x∀x2-x\quad \forall x2−x∀x
This problem is quite a bit harder than I perceived it to be. I don't think I can find a solution.
@Daniel Liu – Daniel Liu The problem is one of those bash types on the IMO where if you keep bashing long enough you do end up on a solution! You were on the right track! Here is a solution from AOPS (by wanwan4343) which bashed in the same way! I essentially also took the same path too and ended up on the functions! (with a somewhat incomplete proof):
Let P(x,y)P(x,y)P(x,y) be the assertion f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)
P(0,0)⇒f(f(0))=0P(0,f(0))⇒2f(0)=f(0)2⇒f(0)=0or2P(0,0)\Rightarrow f(f(0))=0 \\P(0,f(0))\Rightarrow 2f(0)=f(0)^2\Rightarrow f(0)=0 \text{or} 2P(0,0)⇒f(f(0))=0P(0,f(0))⇒2f(0)=f(0)2⇒f(0)=0or2
case 1.
f(0)=2⇒f(2)=0P(x,1)⇒f(x+f(x+1))=x+f(x+1)P(0,f(x+1)+x)⇒f(x+1)+x+2=f(x+1)+x+2(f(x+1)+x)⇒f(x)=2−x∀∈Rf(0)=2\Rightarrow f(2)=0 \\ P(x,1)\Rightarrow f(x+f(x+1))=x+f(x+1) \\ P(0,f(x+1)+x)\Rightarrow f(x+1)+x+2=f(x+1)+x+2(f(x+1)+x) \\ \Rightarrow f(x)=2-x\forall \in\mathbb{R}f(0)=2⇒f(2)=0P(x,1)⇒f(x+f(x+1))=x+f(x+1)P(0,f(x+1)+x)⇒f(x+1)+x+2=f(x+1)+x+2(f(x+1)+x)⇒f(x)=2−x∀∈R
Case 2.f(0)=0f(0)=0f(0)=0
P(x,0)⇒f(x+f(x))=x+f(x)P(x,1)⇒f(x+f(x+1))=x+1+f(x)P(1,f(x+1)+x)⇒f(1+f(1+x+f(x+1)))+f(x+f(x+1))=1+f(x+1+f(x+1))+f(x+1)+x⇒f(f(x)+x+1)=f(x)+x+1P(x,−1)⇒f(x+f(x−1))+f(−x)=x+f(x−1)−f(x)⇒−f(x)=f(−x)P(x,−x)⇒f(x)+f(−x2)=x−xf(x)P(−x,x)⇒f(−x)+f(−x2)=−x+x(−x)⇒f(x)−f(−x)=2x−x(f(x)+f(−x))⇒f(x)=x∀x∈RP(x,0)\Rightarrow f(x+f(x))=x+f(x) \\P(x,1)\Rightarrow f(x+f(x+1))=x+1+f(x) \\ P(1,f(x+1)+x)\Rightarrow f(1+f(1+x+f(x+1)))+f(x+f(x+1))=1+f(x+1+f(x+1))+f(x+1)+x \\ \Rightarrow f(f(x)+x+1)=f(x)+x+1 \\ P(x,-1)\Rightarrow f(x+f(x-1))+f(-x)=x+f(x-1)-f(x) \\ \Rightarrow -f(x)=f(-x)\\ P(x,-x)\Rightarrow f(x)+f(-x^2)=x-xf(x)\\ P(-x,x)\Rightarrow f(-x)+f(-x^2)=-x+x(-x) \\ \Rightarrow f(x)-f(-x)=2x-x(f(x)+f(-x)) \\ \Rightarrow f(x)=x \forall x\in\mathbb{R}P(x,0)⇒f(x+f(x))=x+f(x)P(x,1)⇒f(x+f(x+1))=x+1+f(x)P(1,f(x+1)+x)⇒f(1+f(1+x+f(x+1)))+f(x+f(x+1))=1+f(x+1+f(x+1))+f(x+1)+x⇒f(f(x)+x+1)=f(x)+x+1P(x,−1)⇒f(x+f(x−1))+f(−x)=x+f(x−1)−f(x)⇒−f(x)=f(−x)P(x,−x)⇒f(x)+f(−x2)=x−xf(x)P(−x,x)⇒f(−x)+f(−x2)=−x+x(−x)⇒f(x)−f(−x)=2x−x(f(x)+f(−x))⇒f(x)=x∀x∈R
hence, f(x)=2−xf(x)=2-xf(x)=2−x and f(x)=xf(x)=xf(x)=x are solutions.
@Sualeh Asif – Oh dang, I never thought of putting something as unusual as f(x+1)+xf(x+1)+xf(x+1)+x in the FE.
@Daniel Liu – That is what most people never thought! @Daniel Liu and that is why this is problem 5 of the IMO!
@Sualeh Asif – Okay, so Both problems #2 and #5 were bashy. :(
@Daniel Liu – Yeah that is what I also saw! A bashing Algebra and NT problem!
However problems #1, #4 and #6 have been generally appreciated
It is not very hard to see that f(x) = x and f(x) = 2 - x are solutions. The difficult part is to show that they are the only solutions. In my eyes, Sualeh Asif demonstrates that f(0) = 2 implies that f(x) = x - 2. My compliments! But I am not convinced by his reasoning that f(0) = 0 implies that f(x) = x for all real x. In particular, I do not see the consequence of P(x,1) on line 2 of case 2. Neither the consequence of P(1,f(x+1)+x) on line 4 of case 2.
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Top NewestLet P(x,y) be the above FE.
P(0,0)⟹f(f(0))+f(0)=f(0)⟹f(f(0))=0 P(0,f(0))⟹2f(0)=f(f(0))+f(0)2⟹f(0)2−2f(0)=0⟹f(0)=0,2 Case 1: f(0)=0. Then P(0,y)⟹f(f(y))+f(0)=f(y)+yf(0)⟹f(f(y))=f(y) Since the range of f:R→R is R, then f(x)=x.
Case 2: f(0)=2. Then let g(x)=f(x)−2 which changes the FE to P(x):=g(g(x+y)+x+2)+f(xy)+2=x+g(x+y)+yg(x)+2y [I'll finish later]
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In case 1, you must prove that f is surjective to make that conclusion because R is the codomain of f, which is not necessarily the range.
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Oh right. I'll think about this problem some more.
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It does look like f(x)=2−x satisfies the functional equation. My guess would be f(x)=x or f(x)=2−x, but I do agree with the comment that case 1 still has some work to be done to get to f(x)=x.
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Yes you are right @Patrick Corn !
f(x)=x or 2−x∀x
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This problem is quite a bit harder than I perceived it to be. I don't think I can find a solution.
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Let P(x,y) be the assertion f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)
P(0,0)⇒f(f(0))=0P(0,f(0))⇒2f(0)=f(0)2⇒f(0)=0or2
case 1.
f(0)=2⇒f(2)=0P(x,1)⇒f(x+f(x+1))=x+f(x+1)P(0,f(x+1)+x)⇒f(x+1)+x+2=f(x+1)+x+2(f(x+1)+x)⇒f(x)=2−x∀∈R
Case 2.f(0)=0
P(x,0)⇒f(x+f(x))=x+f(x)P(x,1)⇒f(x+f(x+1))=x+1+f(x)P(1,f(x+1)+x)⇒f(1+f(1+x+f(x+1)))+f(x+f(x+1))=1+f(x+1+f(x+1))+f(x+1)+x⇒f(f(x)+x+1)=f(x)+x+1P(x,−1)⇒f(x+f(x−1))+f(−x)=x+f(x−1)−f(x)⇒−f(x)=f(−x)P(x,−x)⇒f(x)+f(−x2)=x−xf(x)P(−x,x)⇒f(−x)+f(−x2)=−x+x(−x)⇒f(x)−f(−x)=2x−x(f(x)+f(−x))⇒f(x)=x∀x∈R
hence, f(x)=2−x and f(x)=x are solutions.
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f(x+1)+x in the FE.
Oh dang, I never thought of putting something as unusual asLog in to reply
@Daniel Liu and that is why this is problem 5 of the IMO!
That is what most people never thought!Log in to reply
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However problems #1, #4 and #6 have been generally appreciated
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It is not very hard to see that f(x) = x and f(x) = 2 - x are solutions. The difficult part is to show that they are the only solutions. In my eyes, Sualeh Asif demonstrates that f(0) = 2 implies that f(x) = x - 2. My compliments! But I am not convinced by his reasoning that f(0) = 0 implies that f(x) = x for all real x. In particular, I do not see the consequence of P(x,1) on line 2 of case 2. Neither the consequence of P(1,f(x+1)+x) on line 4 of case 2.
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