Find all functions satisfy an equality
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The question: Find all functions $f$ defined over $mathbb{R}$ satisfying the equality: $forall x,y in mathbb{R}$ $$f(y - f(x)) = f(x^{2002} - y) - 2001y f(x)$$
How do I approach (any hints) to solve the problem above?
real-analysis
asked Nov 30 '18 at 4:36
Dong LeDong Le
717
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add a comment |
$begingroup$
The question: Find all functions $f$ defined over $mathbb{R}$ satisfying the equality: $forall x,y in mathbb{R}$ $$f(y - f(x)) = f(x^{2002} - y) - 2001y f(x)$$
How do I approach (any hints) to solve the problem above?
real-analysis
asked Nov 30 '18 at 4:36
Dong LeDong Le
717
$endgroup$
add a comment |
$begingroup$
The question: Find all functions $f$ defined over $mathbb{R}$ satisfying the equality: $forall x,y in mathbb{R}$ $$f(y - f(x)) = f(x^{2002} - y) - 2001y f(x)$$
How do I approach (any hints) to solve the problem above?
real-analysis
asked Nov 30 '18 at 4:36
Dong LeDong Le
717
$endgroup$
The question: Find all functions $f$ defined over $mathbb{R}$ satisfying the equality: $forall x,y in mathbb{R}$ $$f(y - f(x)) = f(x^{2002} - y) - 2001y f(x)$$
How do I approach (any hints) to solve the problem above?
real-analysis
real-analysis
asked Nov 30 '18 at 4:36
Dong LeDong Le
717
asked Nov 30 '18 at 4:36
Dong LeDong Le
717
asked Nov 30 '18 at 4:36
Dong LeDong Le
717
asked Nov 30 '18 at 4:36
Dong LeDong Le
717
asked Nov 30 '18 at 4:36
Dong LeDong Le
717
717
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Set $y=f(x)$ so $f(0)=f(x^{2002}-f(x))-2001f(x)^2$. Set $y=x^{2002}$ to get $f(x^{2002}-f(x))=f(0)-2001x^{2002}f(x)=f(x^{2002}-f(x))-2001f(x)^2-2001x^{2002}f(x)$ and therefore $2001f(x)^2+2001x^{2002}f(x)=0$. Can you continue from here?
answered Nov 30 '18 at 5:42
Guacho PerezGuacho Perez
3,92911132
$endgroup$
$begingroup$
I see and appreciate very much your help!! I got $f(x) = 0$
$endgroup$
– Dong Le
Nov 30 '18 at 5:48
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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active
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active
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votes
$begingroup$
Set $y=f(x)$ so $f(0)=f(x^{2002}-f(x))-2001f(x)^2$. Set $y=x^{2002}$ to get $f(x^{2002}-f(x))=f(0)-2001x^{2002}f(x)=f(x^{2002}-f(x))-2001f(x)^2-2001x^{2002}f(x)$ and therefore $2001f(x)^2+2001x^{2002}f(x)=0$. Can you continue from here?
answered Nov 30 '18 at 5:42
Guacho PerezGuacho Perez
3,92911132
$endgroup$
$begingroup$
I see and appreciate very much your help!! I got $f(x) = 0$
$endgroup$
– Dong Le
Nov 30 '18 at 5:48
add a comment |
$begingroup$
Set $y=f(x)$ so $f(0)=f(x^{2002}-f(x))-2001f(x)^2$. Set $y=x^{2002}$ to get $f(x^{2002}-f(x))=f(0)-2001x^{2002}f(x)=f(x^{2002}-f(x))-2001f(x)^2-2001x^{2002}f(x)$ and therefore $2001f(x)^2+2001x^{2002}f(x)=0$. Can you continue from here?
answered Nov 30 '18 at 5:42
Guacho PerezGuacho Perez
3,92911132
$endgroup$
$begingroup$
I see and appreciate very much your help!! I got $f(x) = 0$
$endgroup$
– Dong Le
Nov 30 '18 at 5:48
add a comment |
$begingroup$
Set $y=f(x)$ so $f(0)=f(x^{2002}-f(x))-2001f(x)^2$. Set $y=x^{2002}$ to get $f(x^{2002}-f(x))=f(0)-2001x^{2002}f(x)=f(x^{2002}-f(x))-2001f(x)^2-2001x^{2002}f(x)$ and therefore $2001f(x)^2+2001x^{2002}f(x)=0$. Can you continue from here?
answered Nov 30 '18 at 5:42
Guacho PerezGuacho Perez
3,92911132
$endgroup$
Set $y=f(x)$ so $f(0)=f(x^{2002}-f(x))-2001f(x)^2$. Set $y=x^{2002}$ to get $f(x^{2002}-f(x))=f(0)-2001x^{2002}f(x)=f(x^{2002}-f(x))-2001f(x)^2-2001x^{2002}f(x)$ and therefore $2001f(x)^2+2001x^{2002}f(x)=0$. Can you continue from here?
answered Nov 30 '18 at 5:42
Guacho PerezGuacho Perez
3,92911132
answered Nov 30 '18 at 5:42
Guacho PerezGuacho Perez
3,92911132
answered Nov 30 '18 at 5:42
Guacho PerezGuacho Perez
3,92911132
answered Nov 30 '18 at 5:42
Guacho PerezGuacho Perez
3,92911132
3,92911132
$begingroup$
I see and appreciate very much your help!! I got $f(x) = 0$
$endgroup$
– Dong Le
Nov 30 '18 at 5:48
add a comment |
$begingroup$
I see and appreciate very much your help!! I got $f(x) = 0$
$endgroup$
– Dong Le
Nov 30 '18 at 5:48
$begingroup$
I see and appreciate very much your help!! I got $f(x) = 0$
$endgroup$
– Dong Le
Nov 30 '18 at 5:48
$begingroup$
I see and appreciate very much your help!! I got $f(x) = 0$
$endgroup$
– Dong Le
Nov 30 '18 at 5:48
add a comment |
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