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The functional equation $f(xy)=f(x)f(y)$ [duplicate]
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This question already has an answer here:
If $f(xy)=f(x)f(y)$ then show that $f(x) = x^t$ for some t
3 answers
Let $f(x)$ be a function that satisfies this functional equation, $f(xy)=f(x)f(y)$.
With a little bit of intuition and luck one may come to a conclusion that these are perhaps the solutions of $f(x)$,
- $f(x)=x$
- $f(x)=1$
- $f(x)=0$
However, these solutions are family solutions of $f(x)=x^n$. What I meant by this is that, when $n=1$ you get the function $f(x)=x$. When $n=0$ you get $f(x)=1$ and when $x=0$ well you get $f(x)=0$.
So, it seems $f(x)=x^n$ is the genuine solution to that functional equation and when you're taking different values for $x$ and $n$ you're getting bunch of other functions of the same family.
Getting excited by this I tried to take different values for $x$, for instance when $x=2$, $x^n$ becomes $2^n$. So, now I'm expecting the function $f(x)=2^n$ to satisfy this functional equation $f(xy)=f(x)f(y)$. However, it doesn't. I don't know why it's not satisfying. May I get your explanation?
functional-equations
asked 2 days ago
user571036
25318
marked as duplicate by Jyrki Lahtonen, GEdgar, Claude Leibovici, rtybase, Arnaud Mortier 2 days ago
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.
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show 1 more comment
up vote
0
down vote
favorite
This question already has an answer here:
If $f(xy)=f(x)f(y)$ then show that $f(x) = x^t$ for some t
3 answers
Let $f(x)$ be a function that satisfies this functional equation, $f(xy)=f(x)f(y)$.
With a little bit of intuition and luck one may come to a conclusion that these are perhaps the solutions of $f(x)$,
- $f(x)=x$
- $f(x)=1$
- $f(x)=0$
However, these solutions are family solutions of $f(x)=x^n$. What I meant by this is that, when $n=1$ you get the function $f(x)=x$. When $n=0$ you get $f(x)=1$ and when $x=0$ well you get $f(x)=0$.
So, it seems $f(x)=x^n$ is the genuine solution to that functional equation and when you're taking different values for $x$ and $n$ you're getting bunch of other functions of the same family.
Getting excited by this I tried to take different values for $x$, for instance when $x=2$, $x^n$ becomes $2^n$. So, now I'm expecting the function $f(x)=2^n$ to satisfy this functional equation $f(xy)=f(x)f(y)$. However, it doesn't. I don't know why it's not satisfying. May I get your explanation?
functional-equations
asked 2 days ago
user571036
25318
marked as duplicate by Jyrki Lahtonen, GEdgar, Claude Leibovici, rtybase, Arnaud Mortier 2 days ago
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.
In the expression $f(x)=x^n$ the argument of the function is $x$, not $n$. $n$ is fixed (for any specific function of the type you like). you can't get a new solution by picking a value for $x$.
– lulu
2 days ago
Also, there are other solutions to your functional equation, though not any other nice ones. See, for example, this question and its answers.
– lulu
2 days ago
@lulu Then how do we get the solution $f(x)=0$ ?
– user571036
2 days ago
2
It simply isn't true that every solution is of the form $x^n$. for some $n$. $0$ isn't of that form and, as I mentioned, there are discontinuous solutions as well.
– lulu
2 days ago
I don't think this is a duplicate of @greedoid's proposed original. The original functional equation is the same, but the answers there don't (for good reason) address the particular confusion the OP has here.
– Henning Makholm
2 days ago
 |Â
show 1 more comment
up vote
0
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favorite
up vote
0
down vote
favorite
This question already has an answer here:
If $f(xy)=f(x)f(y)$ then show that $f(x) = x^t$ for some t
3 answers
Let $f(x)$ be a function that satisfies this functional equation, $f(xy)=f(x)f(y)$.
With a little bit of intuition and luck one may come to a conclusion that these are perhaps the solutions of $f(x)$,
- $f(x)=x$
- $f(x)=1$
- $f(x)=0$
However, these solutions are family solutions of $f(x)=x^n$. What I meant by this is that, when $n=1$ you get the function $f(x)=x$. When $n=0$ you get $f(x)=1$ and when $x=0$ well you get $f(x)=0$.
So, it seems $f(x)=x^n$ is the genuine solution to that functional equation and when you're taking different values for $x$ and $n$ you're getting bunch of other functions of the same family.
Getting excited by this I tried to take different values for $x$, for instance when $x=2$, $x^n$ becomes $2^n$. So, now I'm expecting the function $f(x)=2^n$ to satisfy this functional equation $f(xy)=f(x)f(y)$. However, it doesn't. I don't know why it's not satisfying. May I get your explanation?
functional-equations
asked 2 days ago
user571036
25318
This question already has an answer here:
If $f(xy)=f(x)f(y)$ then show that $f(x) = x^t$ for some t
3 answers
Let $f(x)$ be a function that satisfies this functional equation, $f(xy)=f(x)f(y)$.
With a little bit of intuition and luck one may come to a conclusion that these are perhaps the solutions of $f(x)$,
- $f(x)=x$
- $f(x)=1$
- $f(x)=0$
However, these solutions are family solutions of $f(x)=x^n$. What I meant by this is that, when $n=1$ you get the function $f(x)=x$. When $n=0$ you get $f(x)=1$ and when $x=0$ well you get $f(x)=0$.
So, it seems $f(x)=x^n$ is the genuine solution to that functional equation and when you're taking different values for $x$ and $n$ you're getting bunch of other functions of the same family.
Getting excited by this I tried to take different values for $x$, for instance when $x=2$, $x^n$ becomes $2^n$. So, now I'm expecting the function $f(x)=2^n$ to satisfy this functional equation $f(xy)=f(x)f(y)$. However, it doesn't. I don't know why it's not satisfying. May I get your explanation?
This question already has an answer here:
If $f(xy)=f(x)f(y)$ then show that $f(x) = x^t$ for some t
3 answers
functional-equations
asked 2 days ago
user571036
25318
edited 2 days ago
asked 2 days ago
user571036
25318
asked 2 days ago
user571036
25318
asked 2 days ago
user571036
25318
25318
marked as duplicate by Jyrki Lahtonen, GEdgar, Claude Leibovici, rtybase, Arnaud Mortier 2 days ago
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 Jyrki Lahtonen, GEdgar, Claude Leibovici, rtybase, Arnaud Mortier 2 days ago
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.
In the expression $f(x)=x^n$ the argument of the function is $x$, not $n$. $n$ is fixed (for any specific function of the type you like). you can't get a new solution by picking a value for $x$.
– lulu
2 days ago
Also, there are other solutions to your functional equation, though not any other nice ones. See, for example, this question and its answers.
– lulu
2 days ago
@lulu Then how do we get the solution $f(x)=0$ ?
– user571036
2 days ago
2
It simply isn't true that every solution is of the form $x^n$. for some $n$. $0$ isn't of that form and, as I mentioned, there are discontinuous solutions as well.
– lulu
2 days ago
I don't think this is a duplicate of @greedoid's proposed original. The original functional equation is the same, but the answers there don't (for good reason) address the particular confusion the OP has here.
– Henning Makholm
2 days ago
 |Â
show 1 more comment
In the expression $f(x)=x^n$ the argument of the function is $x$, not $n$. $n$ is fixed (for any specific function of the type you like). you can't get a new solution by picking a value for $x$.
– lulu
2 days ago
Also, there are other solutions to your functional equation, though not any other nice ones. See, for example, this question and its answers.
– lulu
2 days ago
@lulu Then how do we get the solution $f(x)=0$ ?
– user571036
2 days ago
2
It simply isn't true that every solution is of the form $x^n$. for some $n$. $0$ isn't of that form and, as I mentioned, there are discontinuous solutions as well.
– lulu
2 days ago
I don't think this is a duplicate of @greedoid's proposed original. The original functional equation is the same, but the answers there don't (for good reason) address the particular confusion the OP has here.
– Henning Makholm
2 days ago
In the expression $f(x)=x^n$ the argument of the function is $x$, not $n$. $n$ is fixed (for any specific function of the type you like). you can't get a new solution by picking a value for $x$.
– lulu
2 days ago
In the expression $f(x)=x^n$ the argument of the function is $x$, not $n$. $n$ is fixed (for any specific function of the type you like). you can't get a new solution by picking a value for $x$.
– lulu
2 days ago
Also, there are other solutions to your functional equation, though not any other nice ones. See, for example, this question and its answers.
– lulu
2 days ago
Also, there are other solutions to your functional equation, though not any other nice ones. See, for example, this question and its answers.
– lulu
2 days ago
@lulu Then how do we get the solution $f(x)=0$ ?
– user571036
2 days ago
@lulu Then how do we get the solution $f(x)=0$ ?
– user571036
2 days ago
2
2
It simply isn't true that every solution is of the form $x^n$. for some $n$. $0$ isn't of that form and, as I mentioned, there are discontinuous solutions as well.
– lulu
2 days ago
It simply isn't true that every solution is of the form $x^n$. for some $n$. $0$ isn't of that form and, as I mentioned, there are discontinuous solutions as well.
– lulu
2 days ago
I don't think this is a duplicate of @greedoid's proposed original. The original functional equation is the same, but the answers there don't (for good reason) address the particular confusion the OP has here.
– Henning Makholm
2 days ago
I don't think this is a duplicate of @greedoid's proposed original. The original functional equation is the same, but the answers there don't (for good reason) address the particular confusion the OP has here.
– Henning Makholm
2 days ago
 |Â
show 1 more comment
2 Answers
2
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up vote
0
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If you take the equation
$$ tag* f(x) = x^n $$
and "set $x=0$", what you get is not $f(x)=0$, but
$$ f(0) = 0^n $$
which does not define a function -- it only says what the function value must be at $0$ (and doesn't even do that until you decide what $n$ is).
As @lulu pointed out in comments, your assumption that every solution of the functional equation must have the form $text(*)$ is simply not true. $f(x)=0$ does not have this form, and there is an infinity of "wild" discontinuous solutions too (at least if we assume the Axiom of Choice).
answered 2 days ago
Henning Makholm
225k16289516
add a comment |Â
up vote
0
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We may prove that the non-identical function $f(x)$ which is continuous over $(0,+infty)$ and satisfies $$f(xy)=f(x)f(y)$$ is $$f(x)=x^alpha$$ only.
answered 2 days ago
mengdie1982
2,800216
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If you take the equation
$$ tag* f(x) = x^n $$
and "set $x=0$", what you get is not $f(x)=0$, but
$$ f(0) = 0^n $$
which does not define a function -- it only says what the function value must be at $0$ (and doesn't even do that until you decide what $n$ is).
As @lulu pointed out in comments, your assumption that every solution of the functional equation must have the form $text(*)$ is simply not true. $f(x)=0$ does not have this form, and there is an infinity of "wild" discontinuous solutions too (at least if we assume the Axiom of Choice).
answered 2 days ago
Henning Makholm
225k16289516
add a comment |Â
up vote
0
down vote
If you take the equation
$$ tag* f(x) = x^n $$
and "set $x=0$", what you get is not $f(x)=0$, but
$$ f(0) = 0^n $$
which does not define a function -- it only says what the function value must be at $0$ (and doesn't even do that until you decide what $n$ is).
As @lulu pointed out in comments, your assumption that every solution of the functional equation must have the form $text(*)$ is simply not true. $f(x)=0$ does not have this form, and there is an infinity of "wild" discontinuous solutions too (at least if we assume the Axiom of Choice).
answered 2 days ago
Henning Makholm
225k16289516
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If you take the equation
$$ tag* f(x) = x^n $$
and "set $x=0$", what you get is not $f(x)=0$, but
$$ f(0) = 0^n $$
which does not define a function -- it only says what the function value must be at $0$ (and doesn't even do that until you decide what $n$ is).
As @lulu pointed out in comments, your assumption that every solution of the functional equation must have the form $text(*)$ is simply not true. $f(x)=0$ does not have this form, and there is an infinity of "wild" discontinuous solutions too (at least if we assume the Axiom of Choice).
answered 2 days ago
Henning Makholm
225k16289516
If you take the equation
$$ tag* f(x) = x^n $$
and "set $x=0$", what you get is not $f(x)=0$, but
$$ f(0) = 0^n $$
which does not define a function -- it only says what the function value must be at $0$ (and doesn't even do that until you decide what $n$ is).
As @lulu pointed out in comments, your assumption that every solution of the functional equation must have the form $text(*)$ is simply not true. $f(x)=0$ does not have this form, and there is an infinity of "wild" discontinuous solutions too (at least if we assume the Axiom of Choice).
answered 2 days ago
Henning Makholm
225k16289516
answered 2 days ago
Henning Makholm
225k16289516
answered 2 days ago
Henning Makholm
225k16289516
answered 2 days ago
Henning Makholm
225k16289516
225k16289516
add a comment |Â
add a comment |Â
up vote
0
down vote
We may prove that the non-identical function $f(x)$ which is continuous over $(0,+infty)$ and satisfies $$f(xy)=f(x)f(y)$$ is $$f(x)=x^alpha$$ only.
answered 2 days ago
mengdie1982
2,800216
add a comment |Â
up vote
0
down vote
We may prove that the non-identical function $f(x)$ which is continuous over $(0,+infty)$ and satisfies $$f(xy)=f(x)f(y)$$ is $$f(x)=x^alpha$$ only.
answered 2 days ago
mengdie1982
2,800216
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We may prove that the non-identical function $f(x)$ which is continuous over $(0,+infty)$ and satisfies $$f(xy)=f(x)f(y)$$ is $$f(x)=x^alpha$$ only.
answered 2 days ago
mengdie1982
2,800216
We may prove that the non-identical function $f(x)$ which is continuous over $(0,+infty)$ and satisfies $$f(xy)=f(x)f(y)$$ is $$f(x)=x^alpha$$ only.
answered 2 days ago
mengdie1982
2,800216
edited 2 days ago
answered 2 days ago
mengdie1982
2,800216
answered 2 days ago
mengdie1982
2,800216
answered 2 days ago
mengdie1982
2,800216
2,800216
add a comment |Â
add a comment |Â
In the expression $f(x)=x^n$ the argument of the function is $x$, not $n$. $n$ is fixed (for any specific function of the type you like). you can't get a new solution by picking a value for $x$.
– lulu
2 days ago
Also, there are other solutions to your functional equation, though not any other nice ones. See, for example, this question and its answers.
– lulu
2 days ago
@lulu Then how do we get the solution $f(x)=0$ ?
– user571036
2 days ago
2
It simply isn't true that every solution is of the form $x^n$. for some $n$. $0$ isn't of that form and, as I mentioned, there are discontinuous solutions as well.
– lulu
2 days ago
I don't think this is a duplicate of @greedoid's proposed original. The original functional equation is the same, but the answers there don't (for good reason) address the particular confusion the OP has here.
– Henning Makholm
2 days ago