Mixing
Inverse function in integral form
Here is my question:
Suppose my function is defined in terms of Riemann integral in the following form
$$z=g(y)= int_{}^{}f(x,y)dx$$.
Is there any explicit formula of inverse function $h(z)$; $y=h(z)=h(g(y))$.
I'm sorry if this is a stupid question.
Thank you.
inverse-function
asked Nov 21 '18 at 19:37
kolobokishkolobokish
40438
|
show 1 more comment
Here is my question:
Suppose my function is defined in terms of Riemann integral in the following form
$$z=g(y)= int_{}^{}f(x,y)dx$$.
Is there any explicit formula of inverse function $h(z)$; $y=h(z)=h(g(y))$.
I'm sorry if this is a stupid question.
Thank you.
inverse-function
asked Nov 21 '18 at 19:37
kolobokishkolobokish
40438
1
In general there is no reason for $g$ to be invertible
– Federico
Nov 21 '18 at 19:41
Yes, i mean where $g(y)$ is strictly increasing function.
– kolobokish
Nov 21 '18 at 19:43
And even if g was invertible, the inverse may not have a clean, closed form.
– rubikscube09
Nov 21 '18 at 19:43
And given that $g$ is expressed as an integral, $g$ itself might not have a clean, closed form.
– Federico
Nov 21 '18 at 19:45
Maybe you can provide a concrete example?
– Federico
Nov 21 '18 at 19:45
|
show 1 more comment
Here is my question:
Suppose my function is defined in terms of Riemann integral in the following form
$$z=g(y)= int_{}^{}f(x,y)dx$$.
Is there any explicit formula of inverse function $h(z)$; $y=h(z)=h(g(y))$.
I'm sorry if this is a stupid question.
Thank you.
inverse-function
asked Nov 21 '18 at 19:37
kolobokishkolobokish
40438
Here is my question:
Suppose my function is defined in terms of Riemann integral in the following form
$$z=g(y)= int_{}^{}f(x,y)dx$$.
Is there any explicit formula of inverse function $h(z)$; $y=h(z)=h(g(y))$.
I'm sorry if this is a stupid question.
Thank you.
inverse-function
inverse-function
asked Nov 21 '18 at 19:37
kolobokishkolobokish
40438
asked Nov 21 '18 at 19:37
kolobokishkolobokish
40438
asked Nov 21 '18 at 19:37
kolobokishkolobokish
40438
asked Nov 21 '18 at 19:37
kolobokishkolobokish
40438
asked Nov 21 '18 at 19:37
kolobokishkolobokish
40438
40438
1
In general there is no reason for $g$ to be invertible
– Federico
Nov 21 '18 at 19:41
Yes, i mean where $g(y)$ is strictly increasing function.
– kolobokish
Nov 21 '18 at 19:43
And even if g was invertible, the inverse may not have a clean, closed form.
– rubikscube09
Nov 21 '18 at 19:43
And given that $g$ is expressed as an integral, $g$ itself might not have a clean, closed form.
– Federico
Nov 21 '18 at 19:45
Maybe you can provide a concrete example?
– Federico
Nov 21 '18 at 19:45
|
show 1 more comment
1
In general there is no reason for $g$ to be invertible
– Federico
Nov 21 '18 at 19:41
Yes, i mean where $g(y)$ is strictly increasing function.
– kolobokish
Nov 21 '18 at 19:43
And even if g was invertible, the inverse may not have a clean, closed form.
– rubikscube09
Nov 21 '18 at 19:43
And given that $g$ is expressed as an integral, $g$ itself might not have a clean, closed form.
– Federico
Nov 21 '18 at 19:45
Maybe you can provide a concrete example?
– Federico
Nov 21 '18 at 19:45
1
1
In general there is no reason for $g$ to be invertible
– Federico
Nov 21 '18 at 19:41
In general there is no reason for $g$ to be invertible
– Federico
Nov 21 '18 at 19:41
Yes, i mean where $g(y)$ is strictly increasing function.
– kolobokish
Nov 21 '18 at 19:43
Yes, i mean where $g(y)$ is strictly increasing function.
– kolobokish
Nov 21 '18 at 19:43
And even if g was invertible, the inverse may not have a clean, closed form.
– rubikscube09
Nov 21 '18 at 19:43
And even if g was invertible, the inverse may not have a clean, closed form.
– rubikscube09
Nov 21 '18 at 19:43
And given that $g$ is expressed as an integral, $g$ itself might not have a clean, closed form.
– Federico
Nov 21 '18 at 19:45
And given that $g$ is expressed as an integral, $g$ itself might not have a clean, closed form.
– Federico
Nov 21 '18 at 19:45
Maybe you can provide a concrete example?
– Federico
Nov 21 '18 at 19:45
Maybe you can provide a concrete example?
– Federico
Nov 21 '18 at 19:45
|
show 1 more comment
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1
In general there is no reason for $g$ to be invertible
– Federico
Nov 21 '18 at 19:41
Yes, i mean where $g(y)$ is strictly increasing function.
– kolobokish
Nov 21 '18 at 19:43
And even if g was invertible, the inverse may not have a clean, closed form.
– rubikscube09
Nov 21 '18 at 19:43
And given that $g$ is expressed as an integral, $g$ itself might not have a clean, closed form.
– Federico
Nov 21 '18 at 19:45
Maybe you can provide a concrete example?
– Federico
Nov 21 '18 at 19:45