Mixing
Unique solution to an Algebraic equation
$begingroup$
$$f(x_{2}-x_{1})= f(x_{2}-y)f(y-x_{1})$$ I need to find a unique solution to this algebraic equation. Any hints as to how to proceed. The Exponential function works in this case, but I explicitly need to see the details. Thanks in advance.
algebra-precalculus multivariable-calculus
asked Jan 22 at 11:29
SUMIT DEYSUMIT DEY
61
$endgroup$
add a comment |
$begingroup$
$$f(x_{2}-x_{1})= f(x_{2}-y)f(y-x_{1})$$ I need to find a unique solution to this algebraic equation. Any hints as to how to proceed. The Exponential function works in this case, but I explicitly need to see the details. Thanks in advance.
algebra-precalculus multivariable-calculus
asked Jan 22 at 11:29
SUMIT DEYSUMIT DEY
61
$endgroup$
$begingroup$
There is no unique solution, as $f(x)=b^x$ works for any $bin(0,1)cup(1,infty)$.
$endgroup$
– Ben W
Jan 22 at 11:33
$begingroup$
@BenW Why do you exclude the case $b=1$?
$endgroup$
– Jose Brox
Jan 22 at 11:39
$begingroup$
Are $x_1,x_2$ supposed to be fixed real numbers, or is the equation supposed to hold for all $x_1,x_2inmathbb{R}$?
$endgroup$
– Jose Brox
Jan 22 at 11:44
$begingroup$
@JoseBrox just habit
$endgroup$
– Ben W
Jan 22 at 16:39
add a comment |
$begingroup$
$$f(x_{2}-x_{1})= f(x_{2}-y)f(y-x_{1})$$ I need to find a unique solution to this algebraic equation. Any hints as to how to proceed. The Exponential function works in this case, but I explicitly need to see the details. Thanks in advance.
algebra-precalculus multivariable-calculus
asked Jan 22 at 11:29
SUMIT DEYSUMIT DEY
61
$endgroup$
$$f(x_{2}-x_{1})= f(x_{2}-y)f(y-x_{1})$$ I need to find a unique solution to this algebraic equation. Any hints as to how to proceed. The Exponential function works in this case, but I explicitly need to see the details. Thanks in advance.
algebra-precalculus multivariable-calculus
algebra-precalculus multivariable-calculus
asked Jan 22 at 11:29
SUMIT DEYSUMIT DEY
61
asked Jan 22 at 11:29
SUMIT DEYSUMIT DEY
61
asked Jan 22 at 11:29
SUMIT DEYSUMIT DEY
61
asked Jan 22 at 11:29
SUMIT DEYSUMIT DEY
61
asked Jan 22 at 11:29
SUMIT DEYSUMIT DEY
61
61
$begingroup$
There is no unique solution, as $f(x)=b^x$ works for any $bin(0,1)cup(1,infty)$.
$endgroup$
– Ben W
Jan 22 at 11:33
$begingroup$
@BenW Why do you exclude the case $b=1$?
$endgroup$
– Jose Brox
Jan 22 at 11:39
$begingroup$
Are $x_1,x_2$ supposed to be fixed real numbers, or is the equation supposed to hold for all $x_1,x_2inmathbb{R}$?
$endgroup$
– Jose Brox
Jan 22 at 11:44
$begingroup$
@JoseBrox just habit
$endgroup$
– Ben W
Jan 22 at 16:39
add a comment |
$begingroup$
There is no unique solution, as $f(x)=b^x$ works for any $bin(0,1)cup(1,infty)$.
$endgroup$
– Ben W
Jan 22 at 11:33
$begingroup$
@BenW Why do you exclude the case $b=1$?
$endgroup$
– Jose Brox
Jan 22 at 11:39
$begingroup$
Are $x_1,x_2$ supposed to be fixed real numbers, or is the equation supposed to hold for all $x_1,x_2inmathbb{R}$?
$endgroup$
– Jose Brox
Jan 22 at 11:44
$begingroup$
@JoseBrox just habit
$endgroup$
– Ben W
Jan 22 at 16:39
$begingroup$
There is no unique solution, as $f(x)=b^x$ works for any $bin(0,1)cup(1,infty)$.
$endgroup$
– Ben W
Jan 22 at 11:33
$begingroup$
There is no unique solution, as $f(x)=b^x$ works for any $bin(0,1)cup(1,infty)$.
$endgroup$
– Ben W
Jan 22 at 11:33
$begingroup$
@BenW Why do you exclude the case $b=1$?
$endgroup$
– Jose Brox
Jan 22 at 11:39
$begingroup$
@BenW Why do you exclude the case $b=1$?
$endgroup$
– Jose Brox
Jan 22 at 11:39
$begingroup$
Are $x_1,x_2$ supposed to be fixed real numbers, or is the equation supposed to hold for all $x_1,x_2inmathbb{R}$?
$endgroup$
– Jose Brox
Jan 22 at 11:44
$begingroup$
Are $x_1,x_2$ supposed to be fixed real numbers, or is the equation supposed to hold for all $x_1,x_2inmathbb{R}$?
$endgroup$
– Jose Brox
Jan 22 at 11:44
$begingroup$
@JoseBrox just habit
$endgroup$
– Ben W
Jan 22 at 16:39
$begingroup$
@JoseBrox just habit
$endgroup$
– Ben W
Jan 22 at 16:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I'm going to suppose that $x_1,x_2inmathbb{R}$ are variables and not fixed real numbers, and that $f$ is differentiable.
First, if $y=x_2$ then $f(x_2-x_1)=f(0)f(x_2-x_1)$, hence $f(0)=1$ unless $f(x_2-x_1)=0$ for every $x_1,x_2$, i.e., unless $f(x)=0$ in all of $mathbb{R}$.
Now suppose $fneq0$ and let us differentiate the equation with respect to $y$:
$0=-f'(x_2-y)f(y-x_1)+f(x_2-y)f'(x_1-y)$,
so $$f'(x_2-y)f(y-x_1)=f(x_2-y)f'(x_1-y)$$
and substituting $y=x_1$ we get
$$f'(x_2-x_1)=f'(0)f(x_2-x_1)$$
and therefore
$$f'(x)=alpha_f f(x) ,,,, (1)$$
for some constant $alpha_f:=f'(0)inmathbb{R}$.
Since the derivative of $f$ is proportional to $f$, the unique family of solutions for (1) when $fneq0$ is $f(x)=a^x$ for $ain(0,infty)$.
Finally we check that $f(x)=0$ and $f(x)=a^x$, $a>1$ are solutions to the original equation, so they are the only ones.
answered Jan 22 at 11:57
Jose BroxJose Brox
3,15711128
$endgroup$
$begingroup$
OP is not assuming smoothness.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 12:05
$begingroup$
@KaviRamaMurthy Oh, yes, I forgot to add that explicitly. I'll edit, thanks!
$endgroup$
– Jose Brox
Jan 22 at 12:14
add a comment |
$begingroup$
The equation is equivalent to $f(a+b)=f(a)f(b)$ which is closely related to $g(a+b)=g(a)+g(b)$. It is well know that there are pathological examples of functions satisfying the equation which are not even measurable. However the only continuous solutions are the exponential functions mentioned in the comments. [Note that $f(a)=0$ for some $a$ implies $f equiv 0$. If $f$ is continuous and never vanishes then it is always positive and we can take logarithms to reduce the given equation to $g(a+b)=g(a)+g(b)$ where $g=log , f$. In this case $g(x)=cx$ for some $c$ and $f(x)=e^{cx}$].
answered Jan 22 at 12:09
Kavi Rama MurthyKavi Rama Murthy
65.9k42867
$endgroup$
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
$begingroup$
I'm going to suppose that $x_1,x_2inmathbb{R}$ are variables and not fixed real numbers, and that $f$ is differentiable.
First, if $y=x_2$ then $f(x_2-x_1)=f(0)f(x_2-x_1)$, hence $f(0)=1$ unless $f(x_2-x_1)=0$ for every $x_1,x_2$, i.e., unless $f(x)=0$ in all of $mathbb{R}$.
Now suppose $fneq0$ and let us differentiate the equation with respect to $y$:
$0=-f'(x_2-y)f(y-x_1)+f(x_2-y)f'(x_1-y)$,
so $$f'(x_2-y)f(y-x_1)=f(x_2-y)f'(x_1-y)$$
and substituting $y=x_1$ we get
$$f'(x_2-x_1)=f'(0)f(x_2-x_1)$$
and therefore
$$f'(x)=alpha_f f(x) ,,,, (1)$$
for some constant $alpha_f:=f'(0)inmathbb{R}$.
Since the derivative of $f$ is proportional to $f$, the unique family of solutions for (1) when $fneq0$ is $f(x)=a^x$ for $ain(0,infty)$.
Finally we check that $f(x)=0$ and $f(x)=a^x$, $a>1$ are solutions to the original equation, so they are the only ones.
answered Jan 22 at 11:57
Jose BroxJose Brox
3,15711128
$endgroup$
$begingroup$
OP is not assuming smoothness.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 12:05
$begingroup$
@KaviRamaMurthy Oh, yes, I forgot to add that explicitly. I'll edit, thanks!
$endgroup$
– Jose Brox
Jan 22 at 12:14
add a comment |
$begingroup$
I'm going to suppose that $x_1,x_2inmathbb{R}$ are variables and not fixed real numbers, and that $f$ is differentiable.
First, if $y=x_2$ then $f(x_2-x_1)=f(0)f(x_2-x_1)$, hence $f(0)=1$ unless $f(x_2-x_1)=0$ for every $x_1,x_2$, i.e., unless $f(x)=0$ in all of $mathbb{R}$.
Now suppose $fneq0$ and let us differentiate the equation with respect to $y$:
$0=-f'(x_2-y)f(y-x_1)+f(x_2-y)f'(x_1-y)$,
so $$f'(x_2-y)f(y-x_1)=f(x_2-y)f'(x_1-y)$$
and substituting $y=x_1$ we get
$$f'(x_2-x_1)=f'(0)f(x_2-x_1)$$
and therefore
$$f'(x)=alpha_f f(x) ,,,, (1)$$
for some constant $alpha_f:=f'(0)inmathbb{R}$.
Since the derivative of $f$ is proportional to $f$, the unique family of solutions for (1) when $fneq0$ is $f(x)=a^x$ for $ain(0,infty)$.
Finally we check that $f(x)=0$ and $f(x)=a^x$, $a>1$ are solutions to the original equation, so they are the only ones.
answered Jan 22 at 11:57
Jose BroxJose Brox
3,15711128
$endgroup$
$begingroup$
OP is not assuming smoothness.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 12:05
$begingroup$
@KaviRamaMurthy Oh, yes, I forgot to add that explicitly. I'll edit, thanks!
$endgroup$
– Jose Brox
Jan 22 at 12:14
add a comment |
$begingroup$
I'm going to suppose that $x_1,x_2inmathbb{R}$ are variables and not fixed real numbers, and that $f$ is differentiable.
First, if $y=x_2$ then $f(x_2-x_1)=f(0)f(x_2-x_1)$, hence $f(0)=1$ unless $f(x_2-x_1)=0$ for every $x_1,x_2$, i.e., unless $f(x)=0$ in all of $mathbb{R}$.
Now suppose $fneq0$ and let us differentiate the equation with respect to $y$:
$0=-f'(x_2-y)f(y-x_1)+f(x_2-y)f'(x_1-y)$,
so $$f'(x_2-y)f(y-x_1)=f(x_2-y)f'(x_1-y)$$
and substituting $y=x_1$ we get
$$f'(x_2-x_1)=f'(0)f(x_2-x_1)$$
and therefore
$$f'(x)=alpha_f f(x) ,,,, (1)$$
for some constant $alpha_f:=f'(0)inmathbb{R}$.
Since the derivative of $f$ is proportional to $f$, the unique family of solutions for (1) when $fneq0$ is $f(x)=a^x$ for $ain(0,infty)$.
Finally we check that $f(x)=0$ and $f(x)=a^x$, $a>1$ are solutions to the original equation, so they are the only ones.
answered Jan 22 at 11:57
Jose BroxJose Brox
3,15711128
$endgroup$
I'm going to suppose that $x_1,x_2inmathbb{R}$ are variables and not fixed real numbers, and that $f$ is differentiable.
First, if $y=x_2$ then $f(x_2-x_1)=f(0)f(x_2-x_1)$, hence $f(0)=1$ unless $f(x_2-x_1)=0$ for every $x_1,x_2$, i.e., unless $f(x)=0$ in all of $mathbb{R}$.
Now suppose $fneq0$ and let us differentiate the equation with respect to $y$:
$0=-f'(x_2-y)f(y-x_1)+f(x_2-y)f'(x_1-y)$,
so $$f'(x_2-y)f(y-x_1)=f(x_2-y)f'(x_1-y)$$
and substituting $y=x_1$ we get
$$f'(x_2-x_1)=f'(0)f(x_2-x_1)$$
and therefore
$$f'(x)=alpha_f f(x) ,,,, (1)$$
for some constant $alpha_f:=f'(0)inmathbb{R}$.
Since the derivative of $f$ is proportional to $f$, the unique family of solutions for (1) when $fneq0$ is $f(x)=a^x$ for $ain(0,infty)$.
Finally we check that $f(x)=0$ and $f(x)=a^x$, $a>1$ are solutions to the original equation, so they are the only ones.
answered Jan 22 at 11:57
Jose BroxJose Brox
3,15711128
edited Jan 22 at 12:15
answered Jan 22 at 11:57
Jose BroxJose Brox
3,15711128
answered Jan 22 at 11:57
Jose BroxJose Brox
3,15711128
answered Jan 22 at 11:57
Jose BroxJose Brox
3,15711128
3,15711128
$begingroup$
OP is not assuming smoothness.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 12:05
$begingroup$
@KaviRamaMurthy Oh, yes, I forgot to add that explicitly. I'll edit, thanks!
$endgroup$
– Jose Brox
Jan 22 at 12:14
add a comment |
$begingroup$
OP is not assuming smoothness.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 12:05
$begingroup$
@KaviRamaMurthy Oh, yes, I forgot to add that explicitly. I'll edit, thanks!
$endgroup$
– Jose Brox
Jan 22 at 12:14
$begingroup$
OP is not assuming smoothness.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 12:05
$begingroup$
OP is not assuming smoothness.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 12:05
$begingroup$
@KaviRamaMurthy Oh, yes, I forgot to add that explicitly. I'll edit, thanks!
$endgroup$
– Jose Brox
Jan 22 at 12:14
$begingroup$
@KaviRamaMurthy Oh, yes, I forgot to add that explicitly. I'll edit, thanks!
$endgroup$
– Jose Brox
Jan 22 at 12:14
add a comment |
$begingroup$
The equation is equivalent to $f(a+b)=f(a)f(b)$ which is closely related to $g(a+b)=g(a)+g(b)$. It is well know that there are pathological examples of functions satisfying the equation which are not even measurable. However the only continuous solutions are the exponential functions mentioned in the comments. [Note that $f(a)=0$ for some $a$ implies $f equiv 0$. If $f$ is continuous and never vanishes then it is always positive and we can take logarithms to reduce the given equation to $g(a+b)=g(a)+g(b)$ where $g=log , f$. In this case $g(x)=cx$ for some $c$ and $f(x)=e^{cx}$].
answered Jan 22 at 12:09
Kavi Rama MurthyKavi Rama Murthy
65.9k42867
$endgroup$
add a comment |
$begingroup$
The equation is equivalent to $f(a+b)=f(a)f(b)$ which is closely related to $g(a+b)=g(a)+g(b)$. It is well know that there are pathological examples of functions satisfying the equation which are not even measurable. However the only continuous solutions are the exponential functions mentioned in the comments. [Note that $f(a)=0$ for some $a$ implies $f equiv 0$. If $f$ is continuous and never vanishes then it is always positive and we can take logarithms to reduce the given equation to $g(a+b)=g(a)+g(b)$ where $g=log , f$. In this case $g(x)=cx$ for some $c$ and $f(x)=e^{cx}$].
answered Jan 22 at 12:09
Kavi Rama MurthyKavi Rama Murthy
65.9k42867
$endgroup$
add a comment |
$begingroup$
The equation is equivalent to $f(a+b)=f(a)f(b)$ which is closely related to $g(a+b)=g(a)+g(b)$. It is well know that there are pathological examples of functions satisfying the equation which are not even measurable. However the only continuous solutions are the exponential functions mentioned in the comments. [Note that $f(a)=0$ for some $a$ implies $f equiv 0$. If $f$ is continuous and never vanishes then it is always positive and we can take logarithms to reduce the given equation to $g(a+b)=g(a)+g(b)$ where $g=log , f$. In this case $g(x)=cx$ for some $c$ and $f(x)=e^{cx}$].
answered Jan 22 at 12:09
Kavi Rama MurthyKavi Rama Murthy
65.9k42867
$endgroup$
The equation is equivalent to $f(a+b)=f(a)f(b)$ which is closely related to $g(a+b)=g(a)+g(b)$. It is well know that there are pathological examples of functions satisfying the equation which are not even measurable. However the only continuous solutions are the exponential functions mentioned in the comments. [Note that $f(a)=0$ for some $a$ implies $f equiv 0$. If $f$ is continuous and never vanishes then it is always positive and we can take logarithms to reduce the given equation to $g(a+b)=g(a)+g(b)$ where $g=log , f$. In this case $g(x)=cx$ for some $c$ and $f(x)=e^{cx}$].
answered Jan 22 at 12:09
Kavi Rama MurthyKavi Rama Murthy
65.9k42867
edited Jan 22 at 12:16
answered Jan 22 at 12:09
Kavi Rama MurthyKavi Rama Murthy
65.9k42867
answered Jan 22 at 12:09
Kavi Rama MurthyKavi Rama Murthy
65.9k42867
answered Jan 22 at 12:09
Kavi Rama MurthyKavi Rama Murthy
65.9k42867
65.9k42867
add a comment |
add a comment |
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$begingroup$
There is no unique solution, as $f(x)=b^x$ works for any $bin(0,1)cup(1,infty)$.
$endgroup$
– Ben W
Jan 22 at 11:33
$begingroup$
@BenW Why do you exclude the case $b=1$?
$endgroup$
– Jose Brox
Jan 22 at 11:39
$begingroup$
Are $x_1,x_2$ supposed to be fixed real numbers, or is the equation supposed to hold for all $x_1,x_2inmathbb{R}$?
$endgroup$
– Jose Brox
Jan 22 at 11:44
$begingroup$
@JoseBrox just habit
$endgroup$
– Ben W
Jan 22 at 16:39