Mixing
Properties of Inverse Lambert W function
$begingroup$
Starting with the two-branched Lambert W function (from Wikipedia):
Suppose we just flip it like this:
Is there a single power series for this $y=W^{-1}(x)$?
special-functions
edited Jan 1 at 10:46
Glorfindel
3,41981830
asked Aug 14 '11 at 22:01
deoxygerbedeoxygerbe
3,06711736
$endgroup$
add a comment |
$begingroup$
Starting with the two-branched Lambert W function (from Wikipedia):
Suppose we just flip it like this:
Is there a single power series for this $y=W^{-1}(x)$?
special-functions
edited Jan 1 at 10:46
Glorfindel
3,41981830
asked Aug 14 '11 at 22:01
deoxygerbedeoxygerbe
3,06711736
$endgroup$
add a comment |
$begingroup$
Starting with the two-branched Lambert W function (from Wikipedia):
Suppose we just flip it like this:
Is there a single power series for this $y=W^{-1}(x)$?
special-functions
edited Jan 1 at 10:46
Glorfindel
3,41981830
asked Aug 14 '11 at 22:01
deoxygerbedeoxygerbe
3,06711736
$endgroup$
Starting with the two-branched Lambert W function (from Wikipedia):
Suppose we just flip it like this:
Is there a single power series for this $y=W^{-1}(x)$?
special-functions
special-functions
edited Jan 1 at 10:46
Glorfindel
3,41981830
asked Aug 14 '11 at 22:01
deoxygerbedeoxygerbe
3,06711736
edited Jan 1 at 10:46
Glorfindel
3,41981830
asked Aug 14 '11 at 22:01
deoxygerbedeoxygerbe
3,06711736
edited Jan 1 at 10:46
Glorfindel
3,41981830
edited Jan 1 at 10:46
Glorfindel
3,41981830
edited Jan 1 at 10:46
Glorfindel
3,41981830
3,41981830
asked Aug 14 '11 at 22:01
deoxygerbedeoxygerbe
3,06711736
asked Aug 14 '11 at 22:01
deoxygerbedeoxygerbe
3,06711736
asked Aug 14 '11 at 22:01
deoxygerbedeoxygerbe
3,06711736
3,06711736
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Lambert W function, $y=W(x)$ is a solution for $y mathrm{e}^y = x$. Hence $W^{-1}(y) = y mathrm{e}^y$.
answered Aug 14 '11 at 22:15
SashaSasha
60.5k5107179
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Lambert W function, $y=W(x)$ is a solution for $y mathrm{e}^y = x$. Hence $W^{-1}(y) = y mathrm{e}^y$.
answered Aug 14 '11 at 22:15
SashaSasha
60.5k5107179
$endgroup$
add a comment |
$begingroup$
Lambert W function, $y=W(x)$ is a solution for $y mathrm{e}^y = x$. Hence $W^{-1}(y) = y mathrm{e}^y$.
answered Aug 14 '11 at 22:15
SashaSasha
60.5k5107179
$endgroup$
add a comment |
$begingroup$
Lambert W function, $y=W(x)$ is a solution for $y mathrm{e}^y = x$. Hence $W^{-1}(y) = y mathrm{e}^y$.
answered Aug 14 '11 at 22:15
SashaSasha
60.5k5107179
$endgroup$
Lambert W function, $y=W(x)$ is a solution for $y mathrm{e}^y = x$. Hence $W^{-1}(y) = y mathrm{e}^y$.
answered Aug 14 '11 at 22:15
SashaSasha
60.5k5107179
answered Aug 14 '11 at 22:15
SashaSasha
60.5k5107179
answered Aug 14 '11 at 22:15
SashaSasha
60.5k5107179
answered Aug 14 '11 at 22:15
SashaSasha
60.5k5107179
60.5k5107179
add a comment |
add a comment |
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