Mixing
Find all continuous function $f: mathbb R rightarrow mathbb R$
$begingroup$
Find all continuous function $f: mathbb R rightarrow mathbb R$ for which $f(3)=5$ and for every $x,y in mathbb R$ it is truth that $f(x+y)=2+f(x)+f(y)$.
I tried to find some dependence before $x$ and $y$ because I have for example $f(1+2)=2+f(1)+f(2)=5$ and then $f(1)+f(2)=3$. Hovewer I cannot find dependence for every $x,y$ so I don't know how I can do this task.
I also thought about create new function $g(x)$ which is dependent of $f(x)$ but I also don't have good idea to do this.
Can you get me some tips?
real-analysis functional-equations
edited Jan 14 at 23:33
Eric Wofsey
186k14215342
asked Jan 14 at 23:29
MP3129MP3129
35310
$endgroup$
add a comment |
$begingroup$
Find all continuous function $f: mathbb R rightarrow mathbb R$ for which $f(3)=5$ and for every $x,y in mathbb R$ it is truth that $f(x+y)=2+f(x)+f(y)$.
I tried to find some dependence before $x$ and $y$ because I have for example $f(1+2)=2+f(1)+f(2)=5$ and then $f(1)+f(2)=3$. Hovewer I cannot find dependence for every $x,y$ so I don't know how I can do this task.
I also thought about create new function $g(x)$ which is dependent of $f(x)$ but I also don't have good idea to do this.
Can you get me some tips?
real-analysis functional-equations
edited Jan 14 at 23:33
Eric Wofsey
186k14215342
asked Jan 14 at 23:29
MP3129MP3129
35310
$endgroup$
add a comment |
$begingroup$
Find all continuous function $f: mathbb R rightarrow mathbb R$ for which $f(3)=5$ and for every $x,y in mathbb R$ it is truth that $f(x+y)=2+f(x)+f(y)$.
I tried to find some dependence before $x$ and $y$ because I have for example $f(1+2)=2+f(1)+f(2)=5$ and then $f(1)+f(2)=3$. Hovewer I cannot find dependence for every $x,y$ so I don't know how I can do this task.
I also thought about create new function $g(x)$ which is dependent of $f(x)$ but I also don't have good idea to do this.
Can you get me some tips?
real-analysis functional-equations
edited Jan 14 at 23:33
Eric Wofsey
186k14215342
asked Jan 14 at 23:29
MP3129MP3129
35310
$endgroup$
Find all continuous function $f: mathbb R rightarrow mathbb R$ for which $f(3)=5$ and for every $x,y in mathbb R$ it is truth that $f(x+y)=2+f(x)+f(y)$.
I tried to find some dependence before $x$ and $y$ because I have for example $f(1+2)=2+f(1)+f(2)=5$ and then $f(1)+f(2)=3$. Hovewer I cannot find dependence for every $x,y$ so I don't know how I can do this task.
I also thought about create new function $g(x)$ which is dependent of $f(x)$ but I also don't have good idea to do this.
Can you get me some tips?
real-analysis functional-equations
real-analysis functional-equations
edited Jan 14 at 23:33
Eric Wofsey
186k14215342
asked Jan 14 at 23:29
MP3129MP3129
35310
edited Jan 14 at 23:33
Eric Wofsey
186k14215342
asked Jan 14 at 23:29
MP3129MP3129
35310
edited Jan 14 at 23:33
Eric Wofsey
186k14215342
edited Jan 14 at 23:33
Eric Wofsey
186k14215342
edited Jan 14 at 23:33
Eric Wofsey
186k14215342
186k14215342
asked Jan 14 at 23:29
MP3129MP3129
35310
asked Jan 14 at 23:29
MP3129MP3129
35310
asked Jan 14 at 23:29
MP3129MP3129
35310
35310
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As transformed by @Hagen von Eitzen into the Cauchy functional equation (https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation). The only continous solution to Cauchy's functional equation is $g(x)=kx$. Therefore, $f(x)+2=kx$. Now given that $f(3)=5$ so that $k=frac{7}{3}$. Thus the required function is $f(x)=frac{7}{3}x-2$.
answered Jan 15 at 0:10
ershersh
408113
$endgroup$
$begingroup$
ok, but why I can assume that's this function is additive? Does she have to be like that? How to prove it?
$endgroup$
– MP3129
Jan 15 at 10:24
1
$begingroup$
It is not additive homomorphism but we have transformed it to Cauchy's functional equation, which is additive.
$endgroup$
– ersh
Jan 15 at 15:25
add a comment |
$begingroup$
Hint: With $g(x):=f(x)+2$, we have$$ g(x+y)=g(x)+g(y).$$
answered Jan 14 at 23:32
Hagen von EitzenHagen von Eitzen
280k23272504
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
As transformed by @Hagen von Eitzen into the Cauchy functional equation (https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation). The only continous solution to Cauchy's functional equation is $g(x)=kx$. Therefore, $f(x)+2=kx$. Now given that $f(3)=5$ so that $k=frac{7}{3}$. Thus the required function is $f(x)=frac{7}{3}x-2$.
answered Jan 15 at 0:10
ershersh
408113
$endgroup$
$begingroup$
ok, but why I can assume that's this function is additive? Does she have to be like that? How to prove it?
$endgroup$
– MP3129
Jan 15 at 10:24
1
$begingroup$
It is not additive homomorphism but we have transformed it to Cauchy's functional equation, which is additive.
$endgroup$
– ersh
Jan 15 at 15:25
add a comment |
$begingroup$
As transformed by @Hagen von Eitzen into the Cauchy functional equation (https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation). The only continous solution to Cauchy's functional equation is $g(x)=kx$. Therefore, $f(x)+2=kx$. Now given that $f(3)=5$ so that $k=frac{7}{3}$. Thus the required function is $f(x)=frac{7}{3}x-2$.
answered Jan 15 at 0:10
ershersh
408113
$endgroup$
$begingroup$
ok, but why I can assume that's this function is additive? Does she have to be like that? How to prove it?
$endgroup$
– MP3129
Jan 15 at 10:24
1
$begingroup$
It is not additive homomorphism but we have transformed it to Cauchy's functional equation, which is additive.
$endgroup$
– ersh
Jan 15 at 15:25
add a comment |
$begingroup$
As transformed by @Hagen von Eitzen into the Cauchy functional equation (https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation). The only continous solution to Cauchy's functional equation is $g(x)=kx$. Therefore, $f(x)+2=kx$. Now given that $f(3)=5$ so that $k=frac{7}{3}$. Thus the required function is $f(x)=frac{7}{3}x-2$.
answered Jan 15 at 0:10
ershersh
408113
$endgroup$
As transformed by @Hagen von Eitzen into the Cauchy functional equation (https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation). The only continous solution to Cauchy's functional equation is $g(x)=kx$. Therefore, $f(x)+2=kx$. Now given that $f(3)=5$ so that $k=frac{7}{3}$. Thus the required function is $f(x)=frac{7}{3}x-2$.
answered Jan 15 at 0:10
ershersh
408113
answered Jan 15 at 0:10
ershersh
408113
answered Jan 15 at 0:10
ershersh
408113
answered Jan 15 at 0:10
ershersh
408113
408113
$begingroup$
ok, but why I can assume that's this function is additive? Does she have to be like that? How to prove it?
$endgroup$
– MP3129
Jan 15 at 10:24
1
$begingroup$
It is not additive homomorphism but we have transformed it to Cauchy's functional equation, which is additive.
$endgroup$
– ersh
Jan 15 at 15:25
add a comment |
$begingroup$
ok, but why I can assume that's this function is additive? Does she have to be like that? How to prove it?
$endgroup$
– MP3129
Jan 15 at 10:24
1
$begingroup$
It is not additive homomorphism but we have transformed it to Cauchy's functional equation, which is additive.
$endgroup$
– ersh
Jan 15 at 15:25
$begingroup$
ok, but why I can assume that's this function is additive? Does she have to be like that? How to prove it?
$endgroup$
– MP3129
Jan 15 at 10:24
$begingroup$
ok, but why I can assume that's this function is additive? Does she have to be like that? How to prove it?
$endgroup$
– MP3129
Jan 15 at 10:24
1
1
$begingroup$
It is not additive homomorphism but we have transformed it to Cauchy's functional equation, which is additive.
$endgroup$
– ersh
Jan 15 at 15:25
$begingroup$
It is not additive homomorphism but we have transformed it to Cauchy's functional equation, which is additive.
$endgroup$
– ersh
Jan 15 at 15:25
add a comment |
$begingroup$
Hint: With $g(x):=f(x)+2$, we have$$ g(x+y)=g(x)+g(y).$$
answered Jan 14 at 23:32
Hagen von EitzenHagen von Eitzen
280k23272504
$endgroup$
add a comment |
$begingroup$
Hint: With $g(x):=f(x)+2$, we have$$ g(x+y)=g(x)+g(y).$$
answered Jan 14 at 23:32
Hagen von EitzenHagen von Eitzen
280k23272504
$endgroup$
add a comment |
$begingroup$
Hint: With $g(x):=f(x)+2$, we have$$ g(x+y)=g(x)+g(y).$$
answered Jan 14 at 23:32
Hagen von EitzenHagen von Eitzen
280k23272504
$endgroup$
Hint: With $g(x):=f(x)+2$, we have$$ g(x+y)=g(x)+g(y).$$
answered Jan 14 at 23:32
Hagen von EitzenHagen von Eitzen
280k23272504
answered Jan 14 at 23:32
Hagen von EitzenHagen von Eitzen
280k23272504
answered Jan 14 at 23:32
Hagen von EitzenHagen von Eitzen
280k23272504
answered Jan 14 at 23:32
Hagen von EitzenHagen von Eitzen
280k23272504
280k23272504
add a comment |
add a comment |
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