What functions $g$ satisfy $int_{-L}^{L} frac{f(x)}{1 + g(x)}:dx = int_{0}^{L} f(x):dx$ for every even...
$begingroup$
As has been covered in a number of questions on this site, there is a well know property of single variable real continuous even functions $f(x)$:
begin{equation}
int_{-L}^{L} frac{f(x)}{1 + e^x}:dx = int_{0}^{L} f(x):dx
end{equation}
for $L in mathbb{R}^+$ being either finite or infinite.
When you evaluate the proof, there is a fundamental property of $g(x) = e^x$ that allows for this to occur and that is:
begin{equation}
g(-x) = frac{1}{g(x)}
end{equation}
We see this holds not only for $e$ but for any $a in mathbb{R}^+$
My question: outside of $a^x$ are there any real valued functions the satisfy this condition?
real-analysis integration definite-integrals
edited Dec 17 '18 at 10:47
Did
248k23224463
asked Dec 17 '18 at 5:09
DavidGDavidG
2,2271724
$endgroup$
|
show 3 more comments
$begingroup$
As has been covered in a number of questions on this site, there is a well know property of single variable real continuous even functions $f(x)$:
begin{equation}
int_{-L}^{L} frac{f(x)}{1 + e^x}:dx = int_{0}^{L} f(x):dx
end{equation}
for $L in mathbb{R}^+$ being either finite or infinite.
When you evaluate the proof, there is a fundamental property of $g(x) = e^x$ that allows for this to occur and that is:
begin{equation}
g(-x) = frac{1}{g(x)}
end{equation}
We see this holds not only for $e$ but for any $a in mathbb{R}^+$
My question: outside of $a^x$ are there any real valued functions the satisfy this condition?
real-analysis integration definite-integrals
edited Dec 17 '18 at 10:47
Did
248k23224463
asked Dec 17 '18 at 5:09
DavidGDavidG
2,2271724
$endgroup$
1
$begingroup$
Function $f(x)$ should be even.
$endgroup$
– Kemono Chen
Dec 17 '18 at 5:21
$begingroup$
Yes, thanks, I forgot that in my question. I will amend now.
$endgroup$
– DavidG
Dec 17 '18 at 5:21
$begingroup$
There are many. How about $e^{x^{3}}$? You can start with all kinds of functions on $[0,infty)$ with $g(0)=1$ an define $g(x)=frac 1 {g(-x)}$ for $x<0$.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 5:52
$begingroup$
@KaviRamaMurthy - I'm curious in compiling a list of functions (hence the 'Big List' tag). If you could please post up families of functions that satisfy the condition spoken to, I would be very appreciative.
$endgroup$
– DavidG
Dec 17 '18 at 5:54
$begingroup$
Assume $gin C^omega(mathbb{R})$, we can expand $g(-x)g(x)=1$ at $x=0$ and get the series representation of $g$ with some free variables.
$endgroup$
– Kemono Chen
Dec 17 '18 at 5:56
|
show 3 more comments
$begingroup$
As has been covered in a number of questions on this site, there is a well know property of single variable real continuous even functions $f(x)$:
begin{equation}
int_{-L}^{L} frac{f(x)}{1 + e^x}:dx = int_{0}^{L} f(x):dx
end{equation}
for $L in mathbb{R}^+$ being either finite or infinite.
When you evaluate the proof, there is a fundamental property of $g(x) = e^x$ that allows for this to occur and that is:
begin{equation}
g(-x) = frac{1}{g(x)}
end{equation}
We see this holds not only for $e$ but for any $a in mathbb{R}^+$
My question: outside of $a^x$ are there any real valued functions the satisfy this condition?
real-analysis integration definite-integrals
edited Dec 17 '18 at 10:47
Did
248k23224463
asked Dec 17 '18 at 5:09
DavidGDavidG
2,2271724
$endgroup$
As has been covered in a number of questions on this site, there is a well know property of single variable real continuous even functions $f(x)$:
begin{equation}
int_{-L}^{L} frac{f(x)}{1 + e^x}:dx = int_{0}^{L} f(x):dx
end{equation}
for $L in mathbb{R}^+$ being either finite or infinite.
When you evaluate the proof, there is a fundamental property of $g(x) = e^x$ that allows for this to occur and that is:
begin{equation}
g(-x) = frac{1}{g(x)}
end{equation}
We see this holds not only for $e$ but for any $a in mathbb{R}^+$
My question: outside of $a^x$ are there any real valued functions the satisfy this condition?
real-analysis integration definite-integrals
real-analysis integration definite-integrals
edited Dec 17 '18 at 10:47
Did
248k23224463
asked Dec 17 '18 at 5:09
DavidGDavidG
2,2271724
edited Dec 17 '18 at 10:47
Did
248k23224463
asked Dec 17 '18 at 5:09
DavidGDavidG
2,2271724
edited Dec 17 '18 at 10:47
Did
248k23224463
edited Dec 17 '18 at 10:47
Did
248k23224463
edited Dec 17 '18 at 10:47
Did
248k23224463
248k23224463
asked Dec 17 '18 at 5:09
DavidGDavidG
2,2271724
asked Dec 17 '18 at 5:09
DavidGDavidG
2,2271724
asked Dec 17 '18 at 5:09
DavidGDavidG
2,2271724
2,2271724
1
$begingroup$
Function $f(x)$ should be even.
$endgroup$
– Kemono Chen
Dec 17 '18 at 5:21
$begingroup$
Yes, thanks, I forgot that in my question. I will amend now.
$endgroup$
– DavidG
Dec 17 '18 at 5:21
$begingroup$
There are many. How about $e^{x^{3}}$? You can start with all kinds of functions on $[0,infty)$ with $g(0)=1$ an define $g(x)=frac 1 {g(-x)}$ for $x<0$.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 5:52
$begingroup$
@KaviRamaMurthy - I'm curious in compiling a list of functions (hence the 'Big List' tag). If you could please post up families of functions that satisfy the condition spoken to, I would be very appreciative.
$endgroup$
– DavidG
Dec 17 '18 at 5:54
$begingroup$
Assume $gin C^omega(mathbb{R})$, we can expand $g(-x)g(x)=1$ at $x=0$ and get the series representation of $g$ with some free variables.
$endgroup$
– Kemono Chen
Dec 17 '18 at 5:56
|
show 3 more comments
1
$begingroup$
Function $f(x)$ should be even.
$endgroup$
– Kemono Chen
Dec 17 '18 at 5:21
$begingroup$
Yes, thanks, I forgot that in my question. I will amend now.
$endgroup$
– DavidG
Dec 17 '18 at 5:21
$begingroup$
There are many. How about $e^{x^{3}}$? You can start with all kinds of functions on $[0,infty)$ with $g(0)=1$ an define $g(x)=frac 1 {g(-x)}$ for $x<0$.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 5:52
$begingroup$
@KaviRamaMurthy - I'm curious in compiling a list of functions (hence the 'Big List' tag). If you could please post up families of functions that satisfy the condition spoken to, I would be very appreciative.
$endgroup$
– DavidG
Dec 17 '18 at 5:54
$begingroup$
Assume $gin C^omega(mathbb{R})$, we can expand $g(-x)g(x)=1$ at $x=0$ and get the series representation of $g$ with some free variables.
$endgroup$
– Kemono Chen
Dec 17 '18 at 5:56
1
1
$begingroup$
Function $f(x)$ should be even.
$endgroup$
– Kemono Chen
Dec 17 '18 at 5:21
$begingroup$
Function $f(x)$ should be even.
$endgroup$
– Kemono Chen
Dec 17 '18 at 5:21
$begingroup$
Yes, thanks, I forgot that in my question. I will amend now.
$endgroup$
– DavidG
Dec 17 '18 at 5:21
$begingroup$
Yes, thanks, I forgot that in my question. I will amend now.
$endgroup$
– DavidG
Dec 17 '18 at 5:21
$begingroup$
There are many. How about $e^{x^{3}}$? You can start with all kinds of functions on $[0,infty)$ with $g(0)=1$ an define $g(x)=frac 1 {g(-x)}$ for $x<0$.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 5:52
$begingroup$
There are many. How about $e^{x^{3}}$? You can start with all kinds of functions on $[0,infty)$ with $g(0)=1$ an define $g(x)=frac 1 {g(-x)}$ for $x<0$.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 5:52
$begingroup$
@KaviRamaMurthy - I'm curious in compiling a list of functions (hence the 'Big List' tag). If you could please post up families of functions that satisfy the condition spoken to, I would be very appreciative.
$endgroup$
– DavidG
Dec 17 '18 at 5:54
$begingroup$
@KaviRamaMurthy - I'm curious in compiling a list of functions (hence the 'Big List' tag). If you could please post up families of functions that satisfy the condition spoken to, I would be very appreciative.
$endgroup$
– DavidG
Dec 17 '18 at 5:54
$begingroup$
Assume $gin C^omega(mathbb{R})$, we can expand $g(-x)g(x)=1$ at $x=0$ and get the series representation of $g$ with some free variables.
$endgroup$
– Kemono Chen
Dec 17 '18 at 5:56
$begingroup$
Assume $gin C^omega(mathbb{R})$, we can expand $g(-x)g(x)=1$ at $x=0$ and get the series representation of $g$ with some free variables.
$endgroup$
– Kemono Chen
Dec 17 '18 at 5:56
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Assume that $g: Bbb R to Bbb R setminus { -1 }$ is a continuous function with
$$ tag 1
int_{-L}^{L} frac{f(x)}{1 + g(x)},dx = int_{0}^{L} f(x),dx
$$
for all $L > 0$ and all even continuous functions $f: [-L, L]to Bbb R$.
Then in particular (choosing $f(x) = 1$)
$$
int_{-L}^{L} frac{1}{1 + g(x)},dx = L
$$
for all $L > 0$, and differentiating this with respect to $L$ gives
$$
frac{1}{1 + g(L)} + frac{1}{1 + g(-L)} = 1
iff g(L) g(-L) = 1 , .
$$
Therefore
$$ tag 2
g(x) g(-x) = 1
$$ must hold for all $x in Bbb R$.
It is clear now that $g$ can have no zeros. Also $g(0)^2 = 1$ and $g(0) ne -1$, therefore $g(0) = 1$. Since we assumed $g$ to be continuous, $g(x)> 0$ for all $x in Bbb R$ follows.
So we can define $h(x) = log g(x)$. Substituting this in $(2)$ gives
$$
h(x) + h(-x) = 0
$$
so that
$$ tag 3 g(x) = e^{h(x)} text{ for some odd continuous function $h$.}$$
On the other hand, every function $g$ defined by $(3)$ satisfies $(2)$, and consequently $(1)$, so that is the most general (continuous) solution.
answered Dec 17 '18 at 10:17
Martin RMartin R
29.3k33558
$endgroup$
1
$begingroup$
Is this grosso modo the perfect answer? I believe so. (+1)
$endgroup$
– Did
Dec 17 '18 at 10:25
$begingroup$
Fantastic solution. Thanks very much.
$endgroup$
– DavidG
Dec 17 '18 at 10:45
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assume that $g: Bbb R to Bbb R setminus { -1 }$ is a continuous function with
$$ tag 1
int_{-L}^{L} frac{f(x)}{1 + g(x)},dx = int_{0}^{L} f(x),dx
$$
for all $L > 0$ and all even continuous functions $f: [-L, L]to Bbb R$.
Then in particular (choosing $f(x) = 1$)
$$
int_{-L}^{L} frac{1}{1 + g(x)},dx = L
$$
for all $L > 0$, and differentiating this with respect to $L$ gives
$$
frac{1}{1 + g(L)} + frac{1}{1 + g(-L)} = 1
iff g(L) g(-L) = 1 , .
$$
Therefore
$$ tag 2
g(x) g(-x) = 1
$$ must hold for all $x in Bbb R$.
It is clear now that $g$ can have no zeros. Also $g(0)^2 = 1$ and $g(0) ne -1$, therefore $g(0) = 1$. Since we assumed $g$ to be continuous, $g(x)> 0$ for all $x in Bbb R$ follows.
So we can define $h(x) = log g(x)$. Substituting this in $(2)$ gives
$$
h(x) + h(-x) = 0
$$
so that
$$ tag 3 g(x) = e^{h(x)} text{ for some odd continuous function $h$.}$$
On the other hand, every function $g$ defined by $(3)$ satisfies $(2)$, and consequently $(1)$, so that is the most general (continuous) solution.
answered Dec 17 '18 at 10:17
Martin RMartin R
29.3k33558
$endgroup$
1
$begingroup$
Is this grosso modo the perfect answer? I believe so. (+1)
$endgroup$
– Did
Dec 17 '18 at 10:25
$begingroup$
Fantastic solution. Thanks very much.
$endgroup$
– DavidG
Dec 17 '18 at 10:45
add a comment |
$begingroup$
Assume that $g: Bbb R to Bbb R setminus { -1 }$ is a continuous function with
$$ tag 1
int_{-L}^{L} frac{f(x)}{1 + g(x)},dx = int_{0}^{L} f(x),dx
$$
for all $L > 0$ and all even continuous functions $f: [-L, L]to Bbb R$.
Then in particular (choosing $f(x) = 1$)
$$
int_{-L}^{L} frac{1}{1 + g(x)},dx = L
$$
for all $L > 0$, and differentiating this with respect to $L$ gives
$$
frac{1}{1 + g(L)} + frac{1}{1 + g(-L)} = 1
iff g(L) g(-L) = 1 , .
$$
Therefore
$$ tag 2
g(x) g(-x) = 1
$$ must hold for all $x in Bbb R$.
It is clear now that $g$ can have no zeros. Also $g(0)^2 = 1$ and $g(0) ne -1$, therefore $g(0) = 1$. Since we assumed $g$ to be continuous, $g(x)> 0$ for all $x in Bbb R$ follows.
So we can define $h(x) = log g(x)$. Substituting this in $(2)$ gives
$$
h(x) + h(-x) = 0
$$
so that
$$ tag 3 g(x) = e^{h(x)} text{ for some odd continuous function $h$.}$$
On the other hand, every function $g$ defined by $(3)$ satisfies $(2)$, and consequently $(1)$, so that is the most general (continuous) solution.
answered Dec 17 '18 at 10:17
Martin RMartin R
29.3k33558
$endgroup$
1
$begingroup$
Is this grosso modo the perfect answer? I believe so. (+1)
$endgroup$
– Did
Dec 17 '18 at 10:25
$begingroup$
Fantastic solution. Thanks very much.
$endgroup$
– DavidG
Dec 17 '18 at 10:45
add a comment |
$begingroup$
Assume that $g: Bbb R to Bbb R setminus { -1 }$ is a continuous function with
$$ tag 1
int_{-L}^{L} frac{f(x)}{1 + g(x)},dx = int_{0}^{L} f(x),dx
$$
for all $L > 0$ and all even continuous functions $f: [-L, L]to Bbb R$.
Then in particular (choosing $f(x) = 1$)
$$
int_{-L}^{L} frac{1}{1 + g(x)},dx = L
$$
for all $L > 0$, and differentiating this with respect to $L$ gives
$$
frac{1}{1 + g(L)} + frac{1}{1 + g(-L)} = 1
iff g(L) g(-L) = 1 , .
$$
Therefore
$$ tag 2
g(x) g(-x) = 1
$$ must hold for all $x in Bbb R$.
It is clear now that $g$ can have no zeros. Also $g(0)^2 = 1$ and $g(0) ne -1$, therefore $g(0) = 1$. Since we assumed $g$ to be continuous, $g(x)> 0$ for all $x in Bbb R$ follows.
So we can define $h(x) = log g(x)$. Substituting this in $(2)$ gives
$$
h(x) + h(-x) = 0
$$
so that
$$ tag 3 g(x) = e^{h(x)} text{ for some odd continuous function $h$.}$$
On the other hand, every function $g$ defined by $(3)$ satisfies $(2)$, and consequently $(1)$, so that is the most general (continuous) solution.
answered Dec 17 '18 at 10:17
Martin RMartin R
29.3k33558
$endgroup$
Assume that $g: Bbb R to Bbb R setminus { -1 }$ is a continuous function with
$$ tag 1
int_{-L}^{L} frac{f(x)}{1 + g(x)},dx = int_{0}^{L} f(x),dx
$$
for all $L > 0$ and all even continuous functions $f: [-L, L]to Bbb R$.
Then in particular (choosing $f(x) = 1$)
$$
int_{-L}^{L} frac{1}{1 + g(x)},dx = L
$$
for all $L > 0$, and differentiating this with respect to $L$ gives
$$
frac{1}{1 + g(L)} + frac{1}{1 + g(-L)} = 1
iff g(L) g(-L) = 1 , .
$$
Therefore
$$ tag 2
g(x) g(-x) = 1
$$ must hold for all $x in Bbb R$.
It is clear now that $g$ can have no zeros. Also $g(0)^2 = 1$ and $g(0) ne -1$, therefore $g(0) = 1$. Since we assumed $g$ to be continuous, $g(x)> 0$ for all $x in Bbb R$ follows.
So we can define $h(x) = log g(x)$. Substituting this in $(2)$ gives
$$
h(x) + h(-x) = 0
$$
so that
$$ tag 3 g(x) = e^{h(x)} text{ for some odd continuous function $h$.}$$
On the other hand, every function $g$ defined by $(3)$ satisfies $(2)$, and consequently $(1)$, so that is the most general (continuous) solution.
answered Dec 17 '18 at 10:17
Martin RMartin R
29.3k33558
edited Dec 17 '18 at 10:56
answered Dec 17 '18 at 10:17
Martin RMartin R
29.3k33558
answered Dec 17 '18 at 10:17
Martin RMartin R
29.3k33558
answered Dec 17 '18 at 10:17
Martin RMartin R
29.3k33558
29.3k33558
1
$begingroup$
Is this grosso modo the perfect answer? I believe so. (+1)
$endgroup$
– Did
Dec 17 '18 at 10:25
$begingroup$
Fantastic solution. Thanks very much.
$endgroup$
– DavidG
Dec 17 '18 at 10:45
add a comment |
1
$begingroup$
Is this grosso modo the perfect answer? I believe so. (+1)
$endgroup$
– Did
Dec 17 '18 at 10:25
$begingroup$
Fantastic solution. Thanks very much.
$endgroup$
– DavidG
Dec 17 '18 at 10:45
1
1
$begingroup$
Is this grosso modo the perfect answer? I believe so. (+1)
$endgroup$
– Did
Dec 17 '18 at 10:25
$begingroup$
Is this grosso modo the perfect answer? I believe so. (+1)
$endgroup$
– Did
Dec 17 '18 at 10:25
$begingroup$
Fantastic solution. Thanks very much.
$endgroup$
– DavidG
Dec 17 '18 at 10:45
$begingroup$
Fantastic solution. Thanks very much.
$endgroup$
– DavidG
Dec 17 '18 at 10:45
add a comment |
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1
$begingroup$
Function $f(x)$ should be even.
$endgroup$
– Kemono Chen
Dec 17 '18 at 5:21
$begingroup$
Yes, thanks, I forgot that in my question. I will amend now.
$endgroup$
– DavidG
Dec 17 '18 at 5:21
$begingroup$
There are many. How about $e^{x^{3}}$? You can start with all kinds of functions on $[0,infty)$ with $g(0)=1$ an define $g(x)=frac 1 {g(-x)}$ for $x<0$.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 5:52
$begingroup$
@KaviRamaMurthy - I'm curious in compiling a list of functions (hence the 'Big List' tag). If you could please post up families of functions that satisfy the condition spoken to, I would be very appreciative.
$endgroup$
– DavidG
Dec 17 '18 at 5:54
$begingroup$
Assume $gin C^omega(mathbb{R})$, we can expand $g(-x)g(x)=1$ at $x=0$ and get the series representation of $g$ with some free variables.
$endgroup$
– Kemono Chen
Dec 17 '18 at 5:56