Integral of a function which equals division of its values at integral boundaries
I am doing some derivations and stuck at some point.
Given a $C^0$ function $f(x): mathbb{R} to mathbb{R}$ where $f(x) > 0$ $forall x$, is it possible to define an integral over an interval $[a,b]$ such that
$$
int_a^b g[f(x)],dx = frac{f(b)}{f(a)}
$$
where $g$ is a functional of $f$?
If so, what is $g$?
If it makes it easier, the condition could also be
$$
int_a^b g[f(x)],dx = hleft(frac{f(b)}{f(a)}right)
$$
where $h$ is an invertible function.
calculus
asked Nov 27 at 11:50
osolmaz
11414
|
show 1 more comment
I am doing some derivations and stuck at some point.
Given a $C^0$ function $f(x): mathbb{R} to mathbb{R}$ where $f(x) > 0$ $forall x$, is it possible to define an integral over an interval $[a,b]$ such that
$$
int_a^b g[f(x)],dx = frac{f(b)}{f(a)}
$$
where $g$ is a functional of $f$?
If so, what is $g$?
If it makes it easier, the condition could also be
$$
int_a^b g[f(x)],dx = hleft(frac{f(b)}{f(a)}right)
$$
where $h$ is an invertible function.
calculus
asked Nov 27 at 11:50
osolmaz
11414
What do you do when $f(a)=0$?
– Ingix
Nov 27 at 11:55
In my case, $f(x) > 0 ,forall x$.
– osolmaz
Nov 27 at 12:00
$a$ and $b$ can very in some interval, I assume?
– Ingix
Nov 27 at 12:01
$a,bin mathbb{R}$ if that is what you are asking.
– osolmaz
Nov 27 at 12:03
What I don't understand is why $g(x) = frac{f(b)}{(b-a)f(a)}$ for $x in mathbb R$ isn't a solution. Or do you want to find a function $g: mathbb R to mathbb R$ (so you have to determine $g(x)$ for each $x$ by itself, rather than a functional $g: C^0 to C^?$ that allows you to determine $g(f)$ knowing all of $f$ (as I did above).
– Ingix
Nov 27 at 12:08
|
show 1 more comment
I am doing some derivations and stuck at some point.
Given a $C^0$ function $f(x): mathbb{R} to mathbb{R}$ where $f(x) > 0$ $forall x$, is it possible to define an integral over an interval $[a,b]$ such that
$$
int_a^b g[f(x)],dx = frac{f(b)}{f(a)}
$$
where $g$ is a functional of $f$?
If so, what is $g$?
If it makes it easier, the condition could also be
$$
int_a^b g[f(x)],dx = hleft(frac{f(b)}{f(a)}right)
$$
where $h$ is an invertible function.
calculus
asked Nov 27 at 11:50
osolmaz
11414
I am doing some derivations and stuck at some point.
Given a $C^0$ function $f(x): mathbb{R} to mathbb{R}$ where $f(x) > 0$ $forall x$, is it possible to define an integral over an interval $[a,b]$ such that
$$
int_a^b g[f(x)],dx = frac{f(b)}{f(a)}
$$
where $g$ is a functional of $f$?
If so, what is $g$?
If it makes it easier, the condition could also be
$$
int_a^b g[f(x)],dx = hleft(frac{f(b)}{f(a)}right)
$$
where $h$ is an invertible function.
calculus
calculus
asked Nov 27 at 11:50
osolmaz
11414
asked Nov 27 at 11:50
osolmaz
11414
edited Nov 27 at 12:01
asked Nov 27 at 11:50
osolmaz
11414
asked Nov 27 at 11:50
osolmaz
11414
asked Nov 27 at 11:50
osolmaz
11414
11414
What do you do when $f(a)=0$?
– Ingix
Nov 27 at 11:55
In my case, $f(x) > 0 ,forall x$.
– osolmaz
Nov 27 at 12:00
$a$ and $b$ can very in some interval, I assume?
– Ingix
Nov 27 at 12:01
$a,bin mathbb{R}$ if that is what you are asking.
– osolmaz
Nov 27 at 12:03
What I don't understand is why $g(x) = frac{f(b)}{(b-a)f(a)}$ for $x in mathbb R$ isn't a solution. Or do you want to find a function $g: mathbb R to mathbb R$ (so you have to determine $g(x)$ for each $x$ by itself, rather than a functional $g: C^0 to C^?$ that allows you to determine $g(f)$ knowing all of $f$ (as I did above).
– Ingix
Nov 27 at 12:08
|
show 1 more comment
What do you do when $f(a)=0$?
– Ingix
Nov 27 at 11:55
In my case, $f(x) > 0 ,forall x$.
– osolmaz
Nov 27 at 12:00
$a$ and $b$ can very in some interval, I assume?
– Ingix
Nov 27 at 12:01
$a,bin mathbb{R}$ if that is what you are asking.
– osolmaz
Nov 27 at 12:03
What I don't understand is why $g(x) = frac{f(b)}{(b-a)f(a)}$ for $x in mathbb R$ isn't a solution. Or do you want to find a function $g: mathbb R to mathbb R$ (so you have to determine $g(x)$ for each $x$ by itself, rather than a functional $g: C^0 to C^?$ that allows you to determine $g(f)$ knowing all of $f$ (as I did above).
– Ingix
Nov 27 at 12:08
What do you do when $f(a)=0$?
– Ingix
Nov 27 at 11:55
What do you do when $f(a)=0$?
– Ingix
Nov 27 at 11:55
In my case, $f(x) > 0 ,forall x$.
– osolmaz
Nov 27 at 12:00
In my case, $f(x) > 0 ,forall x$.
– osolmaz
Nov 27 at 12:00
$a$ and $b$ can very in some interval, I assume?
– Ingix
Nov 27 at 12:01
$a$ and $b$ can very in some interval, I assume?
– Ingix
Nov 27 at 12:01
$a,bin mathbb{R}$ if that is what you are asking.
– osolmaz
Nov 27 at 12:03
$a,bin mathbb{R}$ if that is what you are asking.
– osolmaz
Nov 27 at 12:03
What I don't understand is why $g(x) = frac{f(b)}{(b-a)f(a)}$ for $x in mathbb R$ isn't a solution. Or do you want to find a function $g: mathbb R to mathbb R$ (so you have to determine $g(x)$ for each $x$ by itself, rather than a functional $g: C^0 to C^?$ that allows you to determine $g(f)$ knowing all of $f$ (as I did above).
– Ingix
Nov 27 at 12:08
What I don't understand is why $g(x) = frac{f(b)}{(b-a)f(a)}$ for $x in mathbb R$ isn't a solution. Or do you want to find a function $g: mathbb R to mathbb R$ (so you have to determine $g(x)$ for each $x$ by itself, rather than a functional $g: C^0 to C^?$ that allows you to determine $g(f)$ knowing all of $f$ (as I did above).
– Ingix
Nov 27 at 12:08
|
show 1 more comment
1 Answer
1
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Such a well defined and bounded function independent from $f(x)$ and the bounds of integral $a,b$ does not exist because otherwise we should have $$int_a^ag[f(x)]dx=0$$and while $int_a^ag[f(x)]dx={f(a)over f(x)}=1$ we must have $$0=1$$ which is a contradiction. Therefore such $g$ doesn't exist.
answered Nov 27 at 13:16
Mostafa Ayaz
13.7k3836
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
Such a well defined and bounded function independent from $f(x)$ and the bounds of integral $a,b$ does not exist because otherwise we should have $$int_a^ag[f(x)]dx=0$$and while $int_a^ag[f(x)]dx={f(a)over f(x)}=1$ we must have $$0=1$$ which is a contradiction. Therefore such $g$ doesn't exist.
answered Nov 27 at 13:16
Mostafa Ayaz
13.7k3836
add a comment |
Such a well defined and bounded function independent from $f(x)$ and the bounds of integral $a,b$ does not exist because otherwise we should have $$int_a^ag[f(x)]dx=0$$and while $int_a^ag[f(x)]dx={f(a)over f(x)}=1$ we must have $$0=1$$ which is a contradiction. Therefore such $g$ doesn't exist.
answered Nov 27 at 13:16
Mostafa Ayaz
13.7k3836
add a comment |
Such a well defined and bounded function independent from $f(x)$ and the bounds of integral $a,b$ does not exist because otherwise we should have $$int_a^ag[f(x)]dx=0$$and while $int_a^ag[f(x)]dx={f(a)over f(x)}=1$ we must have $$0=1$$ which is a contradiction. Therefore such $g$ doesn't exist.
answered Nov 27 at 13:16
Mostafa Ayaz
13.7k3836
Such a well defined and bounded function independent from $f(x)$ and the bounds of integral $a,b$ does not exist because otherwise we should have $$int_a^ag[f(x)]dx=0$$and while $int_a^ag[f(x)]dx={f(a)over f(x)}=1$ we must have $$0=1$$ which is a contradiction. Therefore such $g$ doesn't exist.
answered Nov 27 at 13:16
Mostafa Ayaz
13.7k3836
answered Nov 27 at 13:16
Mostafa Ayaz
13.7k3836
answered Nov 27 at 13:16
Mostafa Ayaz
13.7k3836
answered Nov 27 at 13:16
Mostafa Ayaz
13.7k3836
13.7k3836
add a comment |
add a comment |
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What do you do when $f(a)=0$?
– Ingix
Nov 27 at 11:55
In my case, $f(x) > 0 ,forall x$.
– osolmaz
Nov 27 at 12:00
$a$ and $b$ can very in some interval, I assume?
– Ingix
Nov 27 at 12:01
$a,bin mathbb{R}$ if that is what you are asking.
– osolmaz
Nov 27 at 12:03
What I don't understand is why $g(x) = frac{f(b)}{(b-a)f(a)}$ for $x in mathbb R$ isn't a solution. Or do you want to find a function $g: mathbb R to mathbb R$ (so you have to determine $g(x)$ for each $x$ by itself, rather than a functional $g: C^0 to C^?$ that allows you to determine $g(f)$ knowing all of $f$ (as I did above).
– Ingix
Nov 27 at 12:08