Inverse of a number-valued function on a curve
$begingroup$
$S={(x_1,x_2) | forall xinmathbb{R}colon x_1=x,x_2=exp(x)}$ be a set, and
$f$ be the function $fcolon Sto mathbb{R},(x1,x2)mapsto x_1+x_2$.
How can I show if $f$ is bijective?
How can the inverse of $f$ be calculated?
$$$$
I already know that $Sneqemptyset$, and $S$ is an open subset of $mathbb{R}$. Together with dimension invariance theorem follows that $f$ is bijective and has an inverse therefore.
In multivariate analysis, inverses of n-vector-valued n-vector functions are calculated. But it seems we have now a number-valued vector function.
abstract-algebra algebra-precalculus analysis closed-form
asked Dec 19 '18 at 21:52
IV_IV_
1,389525
$endgroup$
add a comment |
$begingroup$
$S={(x_1,x_2) | forall xinmathbb{R}colon x_1=x,x_2=exp(x)}$ be a set, and
$f$ be the function $fcolon Sto mathbb{R},(x1,x2)mapsto x_1+x_2$.
How can I show if $f$ is bijective?
How can the inverse of $f$ be calculated?
$$$$
I already know that $Sneqemptyset$, and $S$ is an open subset of $mathbb{R}$. Together with dimension invariance theorem follows that $f$ is bijective and has an inverse therefore.
In multivariate analysis, inverses of n-vector-valued n-vector functions are calculated. But it seems we have now a number-valued vector function.
abstract-algebra algebra-precalculus analysis closed-form
asked Dec 19 '18 at 21:52
IV_IV_
1,389525
$endgroup$
$begingroup$
Isn't $S$ a closed subset of $Bbb R^2$?
$endgroup$
– Shubham Johri
Dec 19 '18 at 22:04
$begingroup$
You can write the set more simply as $$ S = {(x, exp(x)) : xinmathbb R}. $$
$endgroup$
– Math1000
Dec 19 '18 at 22:08
$begingroup$
HINT: $f$ is indeed bijective. This follows from bijectiveness of $t mapsto t+ e^t$. However, its inverse cannot be written in terms of elementary functions (i.e. there is no formula for it). You need to use injectivity and surjectivity (use calculus).
$endgroup$
– Crostul
Dec 19 '18 at 22:13
$begingroup$
The inverse of $f$ can be written in terms of the Lambert W function, but this isn't really something you can compute analytically.
$endgroup$
– Math1000
Dec 19 '18 at 22:15
$begingroup$
I'm interested in calculating a closed-form expression of the inverse only from the given conditions and the inverses of the component functions. I'm looking for hints why the inverses of bijective number-valued vector functions cannot be calculated on this way. I hope, such hints could lead to a proof someday.
$endgroup$
– IV_
Dec 20 '18 at 16:59
add a comment |
$begingroup$
$S={(x_1,x_2) | forall xinmathbb{R}colon x_1=x,x_2=exp(x)}$ be a set, and
$f$ be the function $fcolon Sto mathbb{R},(x1,x2)mapsto x_1+x_2$.
How can I show if $f$ is bijective?
How can the inverse of $f$ be calculated?
$$$$
I already know that $Sneqemptyset$, and $S$ is an open subset of $mathbb{R}$. Together with dimension invariance theorem follows that $f$ is bijective and has an inverse therefore.
In multivariate analysis, inverses of n-vector-valued n-vector functions are calculated. But it seems we have now a number-valued vector function.
abstract-algebra algebra-precalculus analysis closed-form
asked Dec 19 '18 at 21:52
IV_IV_
1,389525
$endgroup$
$S={(x_1,x_2) | forall xinmathbb{R}colon x_1=x,x_2=exp(x)}$ be a set, and
$f$ be the function $fcolon Sto mathbb{R},(x1,x2)mapsto x_1+x_2$.
How can I show if $f$ is bijective?
How can the inverse of $f$ be calculated?
$$$$
I already know that $Sneqemptyset$, and $S$ is an open subset of $mathbb{R}$. Together with dimension invariance theorem follows that $f$ is bijective and has an inverse therefore.
In multivariate analysis, inverses of n-vector-valued n-vector functions are calculated. But it seems we have now a number-valued vector function.
abstract-algebra algebra-precalculus analysis closed-form
abstract-algebra algebra-precalculus analysis closed-form
asked Dec 19 '18 at 21:52
IV_IV_
1,389525
asked Dec 19 '18 at 21:52
IV_IV_
1,389525
edited Dec 20 '18 at 16:48
IV_
asked Dec 19 '18 at 21:52
IV_IV_
1,389525
asked Dec 19 '18 at 21:52
IV_IV_
1,389525
asked Dec 19 '18 at 21:52
IV_IV_
1,389525
1,389525
$begingroup$
Isn't $S$ a closed subset of $Bbb R^2$?
$endgroup$
– Shubham Johri
Dec 19 '18 at 22:04
$begingroup$
You can write the set more simply as $$ S = {(x, exp(x)) : xinmathbb R}. $$
$endgroup$
– Math1000
Dec 19 '18 at 22:08
$begingroup$
HINT: $f$ is indeed bijective. This follows from bijectiveness of $t mapsto t+ e^t$. However, its inverse cannot be written in terms of elementary functions (i.e. there is no formula for it). You need to use injectivity and surjectivity (use calculus).
$endgroup$
– Crostul
Dec 19 '18 at 22:13
$begingroup$
The inverse of $f$ can be written in terms of the Lambert W function, but this isn't really something you can compute analytically.
$endgroup$
– Math1000
Dec 19 '18 at 22:15
$begingroup$
I'm interested in calculating a closed-form expression of the inverse only from the given conditions and the inverses of the component functions. I'm looking for hints why the inverses of bijective number-valued vector functions cannot be calculated on this way. I hope, such hints could lead to a proof someday.
$endgroup$
– IV_
Dec 20 '18 at 16:59
add a comment |
$begingroup$
Isn't $S$ a closed subset of $Bbb R^2$?
$endgroup$
– Shubham Johri
Dec 19 '18 at 22:04
$begingroup$
You can write the set more simply as $$ S = {(x, exp(x)) : xinmathbb R}. $$
$endgroup$
– Math1000
Dec 19 '18 at 22:08
$begingroup$
HINT: $f$ is indeed bijective. This follows from bijectiveness of $t mapsto t+ e^t$. However, its inverse cannot be written in terms of elementary functions (i.e. there is no formula for it). You need to use injectivity and surjectivity (use calculus).
$endgroup$
– Crostul
Dec 19 '18 at 22:13
$begingroup$
The inverse of $f$ can be written in terms of the Lambert W function, but this isn't really something you can compute analytically.
$endgroup$
– Math1000
Dec 19 '18 at 22:15
$begingroup$
I'm interested in calculating a closed-form expression of the inverse only from the given conditions and the inverses of the component functions. I'm looking for hints why the inverses of bijective number-valued vector functions cannot be calculated on this way. I hope, such hints could lead to a proof someday.
$endgroup$
– IV_
Dec 20 '18 at 16:59
$begingroup$
Isn't $S$ a closed subset of $Bbb R^2$?
$endgroup$
– Shubham Johri
Dec 19 '18 at 22:04
$begingroup$
Isn't $S$ a closed subset of $Bbb R^2$?
$endgroup$
– Shubham Johri
Dec 19 '18 at 22:04
$begingroup$
You can write the set more simply as $$ S = {(x, exp(x)) : xinmathbb R}. $$
$endgroup$
– Math1000
Dec 19 '18 at 22:08
$begingroup$
You can write the set more simply as $$ S = {(x, exp(x)) : xinmathbb R}. $$
$endgroup$
– Math1000
Dec 19 '18 at 22:08
$begingroup$
HINT: $f$ is indeed bijective. This follows from bijectiveness of $t mapsto t+ e^t$. However, its inverse cannot be written in terms of elementary functions (i.e. there is no formula for it). You need to use injectivity and surjectivity (use calculus).
$endgroup$
– Crostul
Dec 19 '18 at 22:13
$begingroup$
HINT: $f$ is indeed bijective. This follows from bijectiveness of $t mapsto t+ e^t$. However, its inverse cannot be written in terms of elementary functions (i.e. there is no formula for it). You need to use injectivity and surjectivity (use calculus).
$endgroup$
– Crostul
Dec 19 '18 at 22:13
$begingroup$
The inverse of $f$ can be written in terms of the Lambert W function, but this isn't really something you can compute analytically.
$endgroup$
– Math1000
Dec 19 '18 at 22:15
$begingroup$
The inverse of $f$ can be written in terms of the Lambert W function, but this isn't really something you can compute analytically.
$endgroup$
– Math1000
Dec 19 '18 at 22:15
$begingroup$
I'm interested in calculating a closed-form expression of the inverse only from the given conditions and the inverses of the component functions. I'm looking for hints why the inverses of bijective number-valued vector functions cannot be calculated on this way. I hope, such hints could lead to a proof someday.
$endgroup$
– IV_
Dec 20 '18 at 16:59
$begingroup$
I'm interested in calculating a closed-form expression of the inverse only from the given conditions and the inverses of the component functions. I'm looking for hints why the inverses of bijective number-valued vector functions cannot be calculated on this way. I hope, such hints could lead to a proof someday.
$endgroup$
– IV_
Dec 20 '18 at 16:59
add a comment |
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$begingroup$
Isn't $S$ a closed subset of $Bbb R^2$?
$endgroup$
– Shubham Johri
Dec 19 '18 at 22:04
$begingroup$
You can write the set more simply as $$ S = {(x, exp(x)) : xinmathbb R}. $$
$endgroup$
– Math1000
Dec 19 '18 at 22:08
$begingroup$
HINT: $f$ is indeed bijective. This follows from bijectiveness of $t mapsto t+ e^t$. However, its inverse cannot be written in terms of elementary functions (i.e. there is no formula for it). You need to use injectivity and surjectivity (use calculus).
$endgroup$
– Crostul
Dec 19 '18 at 22:13
$begingroup$
The inverse of $f$ can be written in terms of the Lambert W function, but this isn't really something you can compute analytically.
$endgroup$
– Math1000
Dec 19 '18 at 22:15
$begingroup$
I'm interested in calculating a closed-form expression of the inverse only from the given conditions and the inverses of the component functions. I'm looking for hints why the inverses of bijective number-valued vector functions cannot be calculated on this way. I hope, such hints could lead to a proof someday.
$endgroup$
– IV_
Dec 20 '18 at 16:59