Mixing
A particular Functional equation
$begingroup$
Bonjour,
Find all continuous functions, $f$, such that $f(x)-1999fbig(frac{2x}{1-x^2}big)=18$ for $|x|neq 1$.
My method: Putting $x=tan{h}, hin]frac{-pi}{2},frac{pi}{2}[ ,hneq pm pi/4$, cooking a little, and end up with a constant function. Is there someone here who can give a different way to handle this stuff?
real-analysis calculus functional-analysis functional-equations alternative-proof
edited Feb 3 at 21:31
Daniele Tampieri
2,5522922
asked Jan 26 at 14:43
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,18929
$endgroup$
add a comment |
$begingroup$
Bonjour,
Find all continuous functions, $f$, such that $f(x)-1999fbig(frac{2x}{1-x^2}big)=18$ for $|x|neq 1$.
My method: Putting $x=tan{h}, hin]frac{-pi}{2},frac{pi}{2}[ ,hneq pm pi/4$, cooking a little, and end up with a constant function. Is there someone here who can give a different way to handle this stuff?
real-analysis calculus functional-analysis functional-equations alternative-proof
edited Feb 3 at 21:31
Daniele Tampieri
2,5522922
asked Jan 26 at 14:43
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,18929
$endgroup$
$begingroup$
Please note that the tag "functional analysis" is not suitable for functional equations, it has a different meaning (see it's info page): math.stackexchange.com/tags/functional-analysis/info
$endgroup$
– pitariver
Jan 27 at 10:32
add a comment |
$begingroup$
Bonjour,
Find all continuous functions, $f$, such that $f(x)-1999fbig(frac{2x}{1-x^2}big)=18$ for $|x|neq 1$.
My method: Putting $x=tan{h}, hin]frac{-pi}{2},frac{pi}{2}[ ,hneq pm pi/4$, cooking a little, and end up with a constant function. Is there someone here who can give a different way to handle this stuff?
real-analysis calculus functional-analysis functional-equations alternative-proof
edited Feb 3 at 21:31
Daniele Tampieri
2,5522922
asked Jan 26 at 14:43
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,18929
$endgroup$
Bonjour,
Find all continuous functions, $f$, such that $f(x)-1999fbig(frac{2x}{1-x^2}big)=18$ for $|x|neq 1$.
My method: Putting $x=tan{h}, hin]frac{-pi}{2},frac{pi}{2}[ ,hneq pm pi/4$, cooking a little, and end up with a constant function. Is there someone here who can give a different way to handle this stuff?
real-analysis calculus functional-analysis functional-equations alternative-proof
real-analysis calculus functional-analysis functional-equations alternative-proof
edited Feb 3 at 21:31
Daniele Tampieri
2,5522922
asked Jan 26 at 14:43
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,18929
edited Feb 3 at 21:31
Daniele Tampieri
2,5522922
asked Jan 26 at 14:43
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,18929
edited Feb 3 at 21:31
Daniele Tampieri
2,5522922
edited Feb 3 at 21:31
Daniele Tampieri
2,5522922
edited Feb 3 at 21:31
Daniele Tampieri
2,5522922
2,5522922
asked Jan 26 at 14:43
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,18929
asked Jan 26 at 14:43
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,18929
asked Jan 26 at 14:43
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,18929
1,18929
$begingroup$
Please note that the tag "functional analysis" is not suitable for functional equations, it has a different meaning (see it's info page): math.stackexchange.com/tags/functional-analysis/info
$endgroup$
– pitariver
Jan 27 at 10:32
add a comment |
$begingroup$
Please note that the tag "functional analysis" is not suitable for functional equations, it has a different meaning (see it's info page): math.stackexchange.com/tags/functional-analysis/info
$endgroup$
– pitariver
Jan 27 at 10:32
$begingroup$
Please note that the tag "functional analysis" is not suitable for functional equations, it has a different meaning (see it's info page): math.stackexchange.com/tags/functional-analysis/info
$endgroup$
– pitariver
Jan 27 at 10:32
$begingroup$
Please note that the tag "functional analysis" is not suitable for functional equations, it has a different meaning (see it's info page): math.stackexchange.com/tags/functional-analysis/info
$endgroup$
– pitariver
Jan 27 at 10:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
My proof: taking $x=tan{h}$ leads to:
$f(tan{4h})-f(tan{2h})=frac{1}{1999}(f(tan{2h})-f(tan{h}))$. So by induction: $f(tan{h})-f(tan{frac{h}{2}})=frac{1}{1999^n}(f(tan{frac{h}{2^n}})-f(tan{frac{h}{2^{n+1}}}))$ for $n$ positive. Using continuity and limit we get: $f(tan{h})=f(tan{h/2})$. With same argument as before we end showing that $f(x)=f(0)$. Since $f(0)=-frac{1}{111}$ therefore $f(x)=-frac{1}{111}$
answered Jan 26 at 15:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,18929
$endgroup$
$begingroup$
Yes and the result is $-1/111$ which of course is the value of $f$
$endgroup$
– HAMIDINE SOUMARE
Jan 26 at 15:53
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
My proof: taking $x=tan{h}$ leads to:
$f(tan{4h})-f(tan{2h})=frac{1}{1999}(f(tan{2h})-f(tan{h}))$. So by induction: $f(tan{h})-f(tan{frac{h}{2}})=frac{1}{1999^n}(f(tan{frac{h}{2^n}})-f(tan{frac{h}{2^{n+1}}}))$ for $n$ positive. Using continuity and limit we get: $f(tan{h})=f(tan{h/2})$. With same argument as before we end showing that $f(x)=f(0)$. Since $f(0)=-frac{1}{111}$ therefore $f(x)=-frac{1}{111}$
answered Jan 26 at 15:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,18929
$endgroup$
$begingroup$
Yes and the result is $-1/111$ which of course is the value of $f$
$endgroup$
– HAMIDINE SOUMARE
Jan 26 at 15:53
add a comment |
$begingroup$
My proof: taking $x=tan{h}$ leads to:
$f(tan{4h})-f(tan{2h})=frac{1}{1999}(f(tan{2h})-f(tan{h}))$. So by induction: $f(tan{h})-f(tan{frac{h}{2}})=frac{1}{1999^n}(f(tan{frac{h}{2^n}})-f(tan{frac{h}{2^{n+1}}}))$ for $n$ positive. Using continuity and limit we get: $f(tan{h})=f(tan{h/2})$. With same argument as before we end showing that $f(x)=f(0)$. Since $f(0)=-frac{1}{111}$ therefore $f(x)=-frac{1}{111}$
answered Jan 26 at 15:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,18929
$endgroup$
$begingroup$
Yes and the result is $-1/111$ which of course is the value of $f$
$endgroup$
– HAMIDINE SOUMARE
Jan 26 at 15:53
add a comment |
$begingroup$
My proof: taking $x=tan{h}$ leads to:
$f(tan{4h})-f(tan{2h})=frac{1}{1999}(f(tan{2h})-f(tan{h}))$. So by induction: $f(tan{h})-f(tan{frac{h}{2}})=frac{1}{1999^n}(f(tan{frac{h}{2^n}})-f(tan{frac{h}{2^{n+1}}}))$ for $n$ positive. Using continuity and limit we get: $f(tan{h})=f(tan{h/2})$. With same argument as before we end showing that $f(x)=f(0)$. Since $f(0)=-frac{1}{111}$ therefore $f(x)=-frac{1}{111}$
answered Jan 26 at 15:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,18929
$endgroup$
My proof: taking $x=tan{h}$ leads to:
$f(tan{4h})-f(tan{2h})=frac{1}{1999}(f(tan{2h})-f(tan{h}))$. So by induction: $f(tan{h})-f(tan{frac{h}{2}})=frac{1}{1999^n}(f(tan{frac{h}{2^n}})-f(tan{frac{h}{2^{n+1}}}))$ for $n$ positive. Using continuity and limit we get: $f(tan{h})=f(tan{h/2})$. With same argument as before we end showing that $f(x)=f(0)$. Since $f(0)=-frac{1}{111}$ therefore $f(x)=-frac{1}{111}$
answered Jan 26 at 15:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,18929
edited Jan 26 at 16:52
answered Jan 26 at 15:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,18929
answered Jan 26 at 15:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,18929
answered Jan 26 at 15:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,18929
1,18929
$begingroup$
Yes and the result is $-1/111$ which of course is the value of $f$
$endgroup$
– HAMIDINE SOUMARE
Jan 26 at 15:53
add a comment |
$begingroup$
Yes and the result is $-1/111$ which of course is the value of $f$
$endgroup$
– HAMIDINE SOUMARE
Jan 26 at 15:53
$begingroup$
Yes and the result is $-1/111$ which of course is the value of $f$
$endgroup$
– HAMIDINE SOUMARE
Jan 26 at 15:53
$begingroup$
Yes and the result is $-1/111$ which of course is the value of $f$
$endgroup$
– HAMIDINE SOUMARE
Jan 26 at 15:53
add a comment |
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$begingroup$
Please note that the tag "functional analysis" is not suitable for functional equations, it has a different meaning (see it's info page): math.stackexchange.com/tags/functional-analysis/info
$endgroup$
– pitariver
Jan 27 at 10:32