Mixing
Finding functions $f: Bbb R_*^+ to Bbb R_*^+$ with certain properties
$begingroup$
Find the functions $f:Bbb R_*^+ to Bbb R_*^+$ such that:
$$f(x)f left(frac{1}{x}right)=1$$
functions functional-equations
edited Oct 24 '12 at 15:35
Martin Sleziak
44.8k10118272
asked Oct 24 '12 at 15:02
LevanDokiteLevanDokite
789
$endgroup$
|
show 2 more comments
$begingroup$
Find the functions $f:Bbb R_*^+ to Bbb R_*^+$ such that:
$$f(x)f left(frac{1}{x}right)=1$$
functions functional-equations
edited Oct 24 '12 at 15:35
Martin Sleziak
44.8k10118272
asked Oct 24 '12 at 15:02
LevanDokiteLevanDokite
789
$endgroup$
$begingroup$
What is $R_ast^+$?
$endgroup$
– Asaf Karagila♦
Oct 24 '12 at 15:05
$begingroup$
I presume it's better known, and less ambiguously written, as $Bbb R_{> 0}$.
$endgroup$
– Lord_Farin
Oct 24 '12 at 15:06
$begingroup$
Am I missing something? $f(x) = x$.
$endgroup$
– copper.hat
Oct 24 '12 at 15:07
$begingroup$
Where's your proof that no other $f$ does the job?
$endgroup$
– Lord_Farin
Oct 24 '12 at 15:08
$begingroup$
None. $f(x) = 1$ also works. As does $f(x) = frac{1}{x}$. I guess the question is to characterize all such functions...
$endgroup$
– copper.hat
Oct 24 '12 at 15:08
|
show 2 more comments
$begingroup$
Find the functions $f:Bbb R_*^+ to Bbb R_*^+$ such that:
$$f(x)f left(frac{1}{x}right)=1$$
functions functional-equations
edited Oct 24 '12 at 15:35
Martin Sleziak
44.8k10118272
asked Oct 24 '12 at 15:02
LevanDokiteLevanDokite
789
$endgroup$
Find the functions $f:Bbb R_*^+ to Bbb R_*^+$ such that:
$$f(x)f left(frac{1}{x}right)=1$$
functions functional-equations
functions functional-equations
edited Oct 24 '12 at 15:35
Martin Sleziak
44.8k10118272
asked Oct 24 '12 at 15:02
LevanDokiteLevanDokite
789
edited Oct 24 '12 at 15:35
Martin Sleziak
44.8k10118272
asked Oct 24 '12 at 15:02
LevanDokiteLevanDokite
789
edited Oct 24 '12 at 15:35
Martin Sleziak
44.8k10118272
edited Oct 24 '12 at 15:35
Martin Sleziak
44.8k10118272
edited Oct 24 '12 at 15:35
Martin Sleziak
44.8k10118272
44.8k10118272
asked Oct 24 '12 at 15:02
LevanDokiteLevanDokite
789
asked Oct 24 '12 at 15:02
LevanDokiteLevanDokite
789
asked Oct 24 '12 at 15:02
LevanDokiteLevanDokite
789
789
$begingroup$
What is $R_ast^+$?
$endgroup$
– Asaf Karagila♦
Oct 24 '12 at 15:05
$begingroup$
I presume it's better known, and less ambiguously written, as $Bbb R_{> 0}$.
$endgroup$
– Lord_Farin
Oct 24 '12 at 15:06
$begingroup$
Am I missing something? $f(x) = x$.
$endgroup$
– copper.hat
Oct 24 '12 at 15:07
$begingroup$
Where's your proof that no other $f$ does the job?
$endgroup$
– Lord_Farin
Oct 24 '12 at 15:08
$begingroup$
None. $f(x) = 1$ also works. As does $f(x) = frac{1}{x}$. I guess the question is to characterize all such functions...
$endgroup$
– copper.hat
Oct 24 '12 at 15:08
|
show 2 more comments
$begingroup$
What is $R_ast^+$?
$endgroup$
– Asaf Karagila♦
Oct 24 '12 at 15:05
$begingroup$
I presume it's better known, and less ambiguously written, as $Bbb R_{> 0}$.
$endgroup$
– Lord_Farin
Oct 24 '12 at 15:06
$begingroup$
Am I missing something? $f(x) = x$.
$endgroup$
– copper.hat
Oct 24 '12 at 15:07
$begingroup$
Where's your proof that no other $f$ does the job?
$endgroup$
– Lord_Farin
Oct 24 '12 at 15:08
$begingroup$
None. $f(x) = 1$ also works. As does $f(x) = frac{1}{x}$. I guess the question is to characterize all such functions...
$endgroup$
– copper.hat
Oct 24 '12 at 15:08
$begingroup$
What is $R_ast^+$?
$endgroup$
– Asaf Karagila♦
Oct 24 '12 at 15:05
$begingroup$
What is $R_ast^+$?
$endgroup$
– Asaf Karagila♦
Oct 24 '12 at 15:05
$begingroup$
I presume it's better known, and less ambiguously written, as $Bbb R_{> 0}$.
$endgroup$
– Lord_Farin
Oct 24 '12 at 15:06
$begingroup$
I presume it's better known, and less ambiguously written, as $Bbb R_{> 0}$.
$endgroup$
– Lord_Farin
Oct 24 '12 at 15:06
$begingroup$
Am I missing something? $f(x) = x$.
$endgroup$
– copper.hat
Oct 24 '12 at 15:07
$begingroup$
Am I missing something? $f(x) = x$.
$endgroup$
– copper.hat
Oct 24 '12 at 15:07
$begingroup$
Where's your proof that no other $f$ does the job?
$endgroup$
– Lord_Farin
Oct 24 '12 at 15:08
$begingroup$
Where's your proof that no other $f$ does the job?
$endgroup$
– Lord_Farin
Oct 24 '12 at 15:08
$begingroup$
None. $f(x) = 1$ also works. As does $f(x) = frac{1}{x}$. I guess the question is to characterize all such functions...
$endgroup$
– copper.hat
Oct 24 '12 at 15:08
$begingroup$
None. $f(x) = 1$ also works. As does $f(x) = frac{1}{x}$. I guess the question is to characterize all such functions...
$endgroup$
– copper.hat
Oct 24 '12 at 15:08
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
In fact this functional equation belongs to the form of http://eqworld.ipmnet.ru/en/solutions/fe/fe2111.pdf.
The general solution is $f(x)=pm e^{Cleft(x,frac{1}{x}right)}$ , where $C(u,v)$ is any antisymmetric function.
answered Oct 26 '12 at 8:10
doraemonpauldoraemonpaul
12.6k31660
$endgroup$
add a comment |
$begingroup$
For any function $tilde{f}:[1,infty)rightarrow mathbb{R}_+^*$ such that $f(1)=1$, there is a unique extension $f:(0,infty)rightarrow mathbb{R}_+^*$ defined for $x<1$ by $f(x)=1/tilde{f}(1/x)$.
answered Oct 24 '12 at 15:21
Kevin VentulloKevin Ventullo
2,15411422
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In fact this functional equation belongs to the form of http://eqworld.ipmnet.ru/en/solutions/fe/fe2111.pdf.
The general solution is $f(x)=pm e^{Cleft(x,frac{1}{x}right)}$ , where $C(u,v)$ is any antisymmetric function.
answered Oct 26 '12 at 8:10
doraemonpauldoraemonpaul
12.6k31660
$endgroup$
add a comment |
$begingroup$
In fact this functional equation belongs to the form of http://eqworld.ipmnet.ru/en/solutions/fe/fe2111.pdf.
The general solution is $f(x)=pm e^{Cleft(x,frac{1}{x}right)}$ , where $C(u,v)$ is any antisymmetric function.
answered Oct 26 '12 at 8:10
doraemonpauldoraemonpaul
12.6k31660
$endgroup$
add a comment |
$begingroup$
In fact this functional equation belongs to the form of http://eqworld.ipmnet.ru/en/solutions/fe/fe2111.pdf.
The general solution is $f(x)=pm e^{Cleft(x,frac{1}{x}right)}$ , where $C(u,v)$ is any antisymmetric function.
answered Oct 26 '12 at 8:10
doraemonpauldoraemonpaul
12.6k31660
$endgroup$
In fact this functional equation belongs to the form of http://eqworld.ipmnet.ru/en/solutions/fe/fe2111.pdf.
The general solution is $f(x)=pm e^{Cleft(x,frac{1}{x}right)}$ , where $C(u,v)$ is any antisymmetric function.
answered Oct 26 '12 at 8:10
doraemonpauldoraemonpaul
12.6k31660
answered Oct 26 '12 at 8:10
doraemonpauldoraemonpaul
12.6k31660
answered Oct 26 '12 at 8:10
doraemonpauldoraemonpaul
12.6k31660
answered Oct 26 '12 at 8:10
doraemonpauldoraemonpaul
12.6k31660
12.6k31660
add a comment |
add a comment |
$begingroup$
For any function $tilde{f}:[1,infty)rightarrow mathbb{R}_+^*$ such that $f(1)=1$, there is a unique extension $f:(0,infty)rightarrow mathbb{R}_+^*$ defined for $x<1$ by $f(x)=1/tilde{f}(1/x)$.
answered Oct 24 '12 at 15:21
Kevin VentulloKevin Ventullo
2,15411422
$endgroup$
add a comment |
$begingroup$
For any function $tilde{f}:[1,infty)rightarrow mathbb{R}_+^*$ such that $f(1)=1$, there is a unique extension $f:(0,infty)rightarrow mathbb{R}_+^*$ defined for $x<1$ by $f(x)=1/tilde{f}(1/x)$.
answered Oct 24 '12 at 15:21
Kevin VentulloKevin Ventullo
2,15411422
$endgroup$
add a comment |
$begingroup$
For any function $tilde{f}:[1,infty)rightarrow mathbb{R}_+^*$ such that $f(1)=1$, there is a unique extension $f:(0,infty)rightarrow mathbb{R}_+^*$ defined for $x<1$ by $f(x)=1/tilde{f}(1/x)$.
answered Oct 24 '12 at 15:21
Kevin VentulloKevin Ventullo
2,15411422
$endgroup$
For any function $tilde{f}:[1,infty)rightarrow mathbb{R}_+^*$ such that $f(1)=1$, there is a unique extension $f:(0,infty)rightarrow mathbb{R}_+^*$ defined for $x<1$ by $f(x)=1/tilde{f}(1/x)$.
answered Oct 24 '12 at 15:21
Kevin VentulloKevin Ventullo
2,15411422
answered Oct 24 '12 at 15:21
Kevin VentulloKevin Ventullo
2,15411422
answered Oct 24 '12 at 15:21
Kevin VentulloKevin Ventullo
2,15411422
answered Oct 24 '12 at 15:21
Kevin VentulloKevin Ventullo
2,15411422
2,15411422
add a comment |
add a comment |
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$begingroup$
What is $R_ast^+$?
$endgroup$
– Asaf Karagila♦
Oct 24 '12 at 15:05
$begingroup$
I presume it's better known, and less ambiguously written, as $Bbb R_{> 0}$.
$endgroup$
– Lord_Farin
Oct 24 '12 at 15:06
$begingroup$
Am I missing something? $f(x) = x$.
$endgroup$
– copper.hat
Oct 24 '12 at 15:07
$begingroup$
Where's your proof that no other $f$ does the job?
$endgroup$
– Lord_Farin
Oct 24 '12 at 15:08
$begingroup$
None. $f(x) = 1$ also works. As does $f(x) = frac{1}{x}$. I guess the question is to characterize all such functions...
$endgroup$
– copper.hat
Oct 24 '12 at 15:08