Finding all and every continuous function $ mathbb{R}tomathbb{R} $ such that $ f(x+y)=f(x)cdot f(y) $...
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This question already has an answer here:
Is there a name for function with the exponential property $f(x+y)=f(x) cdot f(y)$?
5 answers
How to find all and every continuous function $ f $ from $ mathbb{R} $ in $ mathbb{R} $ that verifies $$ f(x+y)=f(x)cdot f(y)quadforall x,yinmathbb{R} $$
All I could find is $ f(x)=fleft(frac{1}{-x}right) $... but I don't like that answer at all... any hints please??
analysis functions continuity
edited Dec 2 '18 at 17:02
marco21
308211
asked Dec 2 '18 at 15:40
AbderrazzakAbderrazzak
12
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marked as duplicate by José Carlos Santos, Ross Millikan, Community♦ Dec 2 '18 at 18:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Is there a name for function with the exponential property $f(x+y)=f(x) cdot f(y)$?
5 answers
How to find all and every continuous function $ f $ from $ mathbb{R} $ in $ mathbb{R} $ that verifies $$ f(x+y)=f(x)cdot f(y)quadforall x,yinmathbb{R} $$
All I could find is $ f(x)=fleft(frac{1}{-x}right) $... but I don't like that answer at all... any hints please??
analysis functions continuity
edited Dec 2 '18 at 17:02
marco21
308211
asked Dec 2 '18 at 15:40
AbderrazzakAbderrazzak
12
$endgroup$
marked as duplicate by José Carlos Santos, Ross Millikan, Community♦ Dec 2 '18 at 18:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
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How about $e^x$?
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– lulu
Dec 2 '18 at 15:41
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Very logical I like it .. but what is the procedure to prove that?
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– Abderrazzak
Dec 2 '18 at 15:45
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@Abderrazzak The exponential $ exp_e:mathbb{R}tomathbb{R}_{>0} $ is usually defined as a monotonic function such that $ exp_e(x_1+x_2)=exp_e(x_1)exp_e(x_2) $ (and such that $ exp_e(1)=e $); at most you should check the (uniqueness and the) existence of such a function.
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– marco21
Dec 2 '18 at 16:26
add a comment |
$begingroup$
This question already has an answer here:
Is there a name for function with the exponential property $f(x+y)=f(x) cdot f(y)$?
5 answers
How to find all and every continuous function $ f $ from $ mathbb{R} $ in $ mathbb{R} $ that verifies $$ f(x+y)=f(x)cdot f(y)quadforall x,yinmathbb{R} $$
All I could find is $ f(x)=fleft(frac{1}{-x}right) $... but I don't like that answer at all... any hints please??
analysis functions continuity
edited Dec 2 '18 at 17:02
marco21
308211
asked Dec 2 '18 at 15:40
AbderrazzakAbderrazzak
12
$endgroup$
This question already has an answer here:
Is there a name for function with the exponential property $f(x+y)=f(x) cdot f(y)$?
5 answers
How to find all and every continuous function $ f $ from $ mathbb{R} $ in $ mathbb{R} $ that verifies $$ f(x+y)=f(x)cdot f(y)quadforall x,yinmathbb{R} $$
All I could find is $ f(x)=fleft(frac{1}{-x}right) $... but I don't like that answer at all... any hints please??
This question already has an answer here:
Is there a name for function with the exponential property $f(x+y)=f(x) cdot f(y)$?
5 answers
analysis functions continuity
analysis functions continuity
edited Dec 2 '18 at 17:02
marco21
308211
asked Dec 2 '18 at 15:40
AbderrazzakAbderrazzak
12
edited Dec 2 '18 at 17:02
marco21
308211
asked Dec 2 '18 at 15:40
AbderrazzakAbderrazzak
12
edited Dec 2 '18 at 17:02
marco21
308211
edited Dec 2 '18 at 17:02
marco21
308211
edited Dec 2 '18 at 17:02
marco21
308211
308211
asked Dec 2 '18 at 15:40
AbderrazzakAbderrazzak
12
asked Dec 2 '18 at 15:40
AbderrazzakAbderrazzak
12
asked Dec 2 '18 at 15:40
AbderrazzakAbderrazzak
12
12
marked as duplicate by José Carlos Santos, Ross Millikan, Community♦ Dec 2 '18 at 18:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by José Carlos Santos, Ross Millikan, Community♦ Dec 2 '18 at 18:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
$begingroup$
How about $e^x$?
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– lulu
Dec 2 '18 at 15:41
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Very logical I like it .. but what is the procedure to prove that?
$endgroup$
– Abderrazzak
Dec 2 '18 at 15:45
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@Abderrazzak The exponential $ exp_e:mathbb{R}tomathbb{R}_{>0} $ is usually defined as a monotonic function such that $ exp_e(x_1+x_2)=exp_e(x_1)exp_e(x_2) $ (and such that $ exp_e(1)=e $); at most you should check the (uniqueness and the) existence of such a function.
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– marco21
Dec 2 '18 at 16:26
add a comment |
3
$begingroup$
How about $e^x$?
$endgroup$
– lulu
Dec 2 '18 at 15:41
$begingroup$
Very logical I like it .. but what is the procedure to prove that?
$endgroup$
– Abderrazzak
Dec 2 '18 at 15:45
$begingroup$
@Abderrazzak The exponential $ exp_e:mathbb{R}tomathbb{R}_{>0} $ is usually defined as a monotonic function such that $ exp_e(x_1+x_2)=exp_e(x_1)exp_e(x_2) $ (and such that $ exp_e(1)=e $); at most you should check the (uniqueness and the) existence of such a function.
$endgroup$
– marco21
Dec 2 '18 at 16:26
3
3
$begingroup$
How about $e^x$?
$endgroup$
– lulu
Dec 2 '18 at 15:41
$begingroup$
How about $e^x$?
$endgroup$
– lulu
Dec 2 '18 at 15:41
$begingroup$
Very logical I like it .. but what is the procedure to prove that?
$endgroup$
– Abderrazzak
Dec 2 '18 at 15:45
$begingroup$
Very logical I like it .. but what is the procedure to prove that?
$endgroup$
– Abderrazzak
Dec 2 '18 at 15:45
$begingroup$
@Abderrazzak The exponential $ exp_e:mathbb{R}tomathbb{R}_{>0} $ is usually defined as a monotonic function such that $ exp_e(x_1+x_2)=exp_e(x_1)exp_e(x_2) $ (and such that $ exp_e(1)=e $); at most you should check the (uniqueness and the) existence of such a function.
$endgroup$
– marco21
Dec 2 '18 at 16:26
$begingroup$
@Abderrazzak The exponential $ exp_e:mathbb{R}tomathbb{R}_{>0} $ is usually defined as a monotonic function such that $ exp_e(x_1+x_2)=exp_e(x_1)exp_e(x_2) $ (and such that $ exp_e(1)=e $); at most you should check the (uniqueness and the) existence of such a function.
$endgroup$
– marco21
Dec 2 '18 at 16:26
add a comment |
1 Answer
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This is not the complete answer, but it finds all differentiable functions $f:mathbb Rrightarrowmathbb R$ satisfying $f(x+y)=f(x)cdot f(y), forall x,yinmathbb R$.
Since $f$ is differentiable,
$f'(x)=lim_{hto0}frac{f(x+h)-f(x)}h=lim_{hto0}frac{f(x)cdot f(h)-f(x)}h=f(x)cdotlim_{hto0}frac{f(h)-1}h=f(x)cdot f'(0), forall xinmathbb R$
$implies f'(x)=f(x)cdot f'(0)$
Integrate this equation to obtain $f(x)=ke^{f'(0)x}$.
Since $f(x+y)=f(x)cdot f(y), k=k^2implies k=0,1$. Thus, the differentiable solutions of this functional equation are $f(x)=0, e^{kx}; kinmathbb R$.
answered Dec 2 '18 at 16:18
Shubham JohriShubham Johri
4,992717
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I understand this part ... but is this function unique .. if so how can I prove that ?
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– Abderrazzak
Dec 6 '18 at 6:28
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is not the complete answer, but it finds all differentiable functions $f:mathbb Rrightarrowmathbb R$ satisfying $f(x+y)=f(x)cdot f(y), forall x,yinmathbb R$.
Since $f$ is differentiable,
$f'(x)=lim_{hto0}frac{f(x+h)-f(x)}h=lim_{hto0}frac{f(x)cdot f(h)-f(x)}h=f(x)cdotlim_{hto0}frac{f(h)-1}h=f(x)cdot f'(0), forall xinmathbb R$
$implies f'(x)=f(x)cdot f'(0)$
Integrate this equation to obtain $f(x)=ke^{f'(0)x}$.
Since $f(x+y)=f(x)cdot f(y), k=k^2implies k=0,1$. Thus, the differentiable solutions of this functional equation are $f(x)=0, e^{kx}; kinmathbb R$.
answered Dec 2 '18 at 16:18
Shubham JohriShubham Johri
4,992717
$endgroup$
$begingroup$
I understand this part ... but is this function unique .. if so how can I prove that ?
$endgroup$
– Abderrazzak
Dec 6 '18 at 6:28
add a comment |
$begingroup$
This is not the complete answer, but it finds all differentiable functions $f:mathbb Rrightarrowmathbb R$ satisfying $f(x+y)=f(x)cdot f(y), forall x,yinmathbb R$.
Since $f$ is differentiable,
$f'(x)=lim_{hto0}frac{f(x+h)-f(x)}h=lim_{hto0}frac{f(x)cdot f(h)-f(x)}h=f(x)cdotlim_{hto0}frac{f(h)-1}h=f(x)cdot f'(0), forall xinmathbb R$
$implies f'(x)=f(x)cdot f'(0)$
Integrate this equation to obtain $f(x)=ke^{f'(0)x}$.
Since $f(x+y)=f(x)cdot f(y), k=k^2implies k=0,1$. Thus, the differentiable solutions of this functional equation are $f(x)=0, e^{kx}; kinmathbb R$.
answered Dec 2 '18 at 16:18
Shubham JohriShubham Johri
4,992717
$endgroup$
$begingroup$
I understand this part ... but is this function unique .. if so how can I prove that ?
$endgroup$
– Abderrazzak
Dec 6 '18 at 6:28
add a comment |
$begingroup$
This is not the complete answer, but it finds all differentiable functions $f:mathbb Rrightarrowmathbb R$ satisfying $f(x+y)=f(x)cdot f(y), forall x,yinmathbb R$.
Since $f$ is differentiable,
$f'(x)=lim_{hto0}frac{f(x+h)-f(x)}h=lim_{hto0}frac{f(x)cdot f(h)-f(x)}h=f(x)cdotlim_{hto0}frac{f(h)-1}h=f(x)cdot f'(0), forall xinmathbb R$
$implies f'(x)=f(x)cdot f'(0)$
Integrate this equation to obtain $f(x)=ke^{f'(0)x}$.
Since $f(x+y)=f(x)cdot f(y), k=k^2implies k=0,1$. Thus, the differentiable solutions of this functional equation are $f(x)=0, e^{kx}; kinmathbb R$.
answered Dec 2 '18 at 16:18
Shubham JohriShubham Johri
4,992717
$endgroup$
This is not the complete answer, but it finds all differentiable functions $f:mathbb Rrightarrowmathbb R$ satisfying $f(x+y)=f(x)cdot f(y), forall x,yinmathbb R$.
Since $f$ is differentiable,
$f'(x)=lim_{hto0}frac{f(x+h)-f(x)}h=lim_{hto0}frac{f(x)cdot f(h)-f(x)}h=f(x)cdotlim_{hto0}frac{f(h)-1}h=f(x)cdot f'(0), forall xinmathbb R$
$implies f'(x)=f(x)cdot f'(0)$
Integrate this equation to obtain $f(x)=ke^{f'(0)x}$.
Since $f(x+y)=f(x)cdot f(y), k=k^2implies k=0,1$. Thus, the differentiable solutions of this functional equation are $f(x)=0, e^{kx}; kinmathbb R$.
answered Dec 2 '18 at 16:18
Shubham JohriShubham Johri
4,992717
answered Dec 2 '18 at 16:18
Shubham JohriShubham Johri
4,992717
answered Dec 2 '18 at 16:18
Shubham JohriShubham Johri
4,992717
answered Dec 2 '18 at 16:18
Shubham JohriShubham Johri
4,992717
4,992717
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I understand this part ... but is this function unique .. if so how can I prove that ?
$endgroup$
– Abderrazzak
Dec 6 '18 at 6:28
add a comment |
$begingroup$
I understand this part ... but is this function unique .. if so how can I prove that ?
$endgroup$
– Abderrazzak
Dec 6 '18 at 6:28
$begingroup$
I understand this part ... but is this function unique .. if so how can I prove that ?
$endgroup$
– Abderrazzak
Dec 6 '18 at 6:28
$begingroup$
I understand this part ... but is this function unique .. if so how can I prove that ?
$endgroup$
– Abderrazzak
Dec 6 '18 at 6:28
add a comment |
3
$begingroup$
How about $e^x$?
$endgroup$
– lulu
Dec 2 '18 at 15:41
$begingroup$
Very logical I like it .. but what is the procedure to prove that?
$endgroup$
– Abderrazzak
Dec 2 '18 at 15:45
$begingroup$
@Abderrazzak The exponential $ exp_e:mathbb{R}tomathbb{R}_{>0} $ is usually defined as a monotonic function such that $ exp_e(x_1+x_2)=exp_e(x_1)exp_e(x_2) $ (and such that $ exp_e(1)=e $); at most you should check the (uniqueness and the) existence of such a function.
$endgroup$
– marco21
Dec 2 '18 at 16:26