Series expansion of an expression that has productlog function about -1/e The 2019 Stack Overflow Developer Survey Results Are InSeries Expansion of $frac11-e^int$Taylor / Maclaurin series expansion origin.If I have a holomorphic function on the unit disc, do I know anything about the radius of convergence of its series expansion about zero?Find complex power series expansion for $int e^-w^2 dw$Power series expansion of $f(z)=frac13-z$ about the point $4i$Expansion in power series of $z=0$Laurent series expansion of a given functionPower series expansion questionPower series expansion involving Lambert-W functionFind the power series expansion of $f(x)= frace^x - 1x$
Deadlock Graph and Interpretation, solution to avoid
What is the meaning of Triage in Cybersec world?
Pristine Bit Checking
What is the steepest angle that a canal can be traversable without locks?
What does "sndry explns" mean in one of the Hitchhiker's guide books?
Access elements in std::string where positon of string is greater than its size
How are circuits which use complex ICs normally simulated?
Patience, young "Padovan"
Monty Hall variation
Realistic Alternatives to Dust: What Else Could Feed a Plankton Bloom?
Does light intensity oscillate really fast since it is a wave?
"To split hairs" vs "To be pedantic"
Unbreakable Formation vs. Cry of the Carnarium
What effect does the “loading” weapon property have in practical terms?
aging parents with no investments
Inline version of a function returns different value than non-inline version
What do hard-Brexiteers want with respect to the Irish border?
How can I create a character who can assume the widest possible range of creature sizes?
Time travel alters history but people keep saying nothing's changed
Manuscript was "unsubmitted" because the manuscript was deposited in Arxiv Preprints
Why is the maximum length of OpenWrt’s root password 8 characters?
A poker game description that does not feel gimmicky
Are USB sockets on wall outlets live all the time, even when the switch is off?
Why can Shazam do this?
Series expansion of an expression that has productlog function about -1/e
The 2019 Stack Overflow Developer Survey Results Are InSeries Expansion of $frac11-e^int$Taylor / Maclaurin series expansion origin.If I have a holomorphic function on the unit disc, do I know anything about the radius of convergence of its series expansion about zero?Find complex power series expansion for $int e^-w^2 dw$Power series expansion of $f(z)=frac13-z$ about the point $4i$Expansion in power series of $z=0$Laurent series expansion of a given functionPower series expansion questionPower series expansion involving Lambert-W functionFind the power series expansion of $f(x)= frace^x - 1x$
$begingroup$
I have this term: $-frac1pW(-fracpe)bigg[logbigg(-frac1pW(-fracpe)bigg)-1bigg]$.
In order to reduce complexity I've taken
$-fracpe=x$ which gives $f(x)=frac1xeW(x)bigg[logbigg(frac1xeW(x)bigg)-1bigg]$.
I need to do a power series expansion for this term at $x=-1/e$.
From the basics of Taylor expansion: $f(x)=f(-1/e)+fracdfdx(-1/e) (x+1/e)^1+fracd^2fdx^2(-1/e)(x+1/e)^2+.....$
The first term $f(-1/e)=-1$ but the second term $fracdfdx=-frac[W(x)]^2bigg[logbigg(fracW(x)xbigg)-1bigg]ex^2[w(x)+1]$ is not defined at $x=-1/e$ since $W(-1/e)=-1$ making the denominator to be $0$ also the numerator is $0$ as well. If the second term is undefined then the Taylor expansion does not work (to my understanding).
I'm not sure where I'm going wrong with this approach or if there's any other way to expand this term as a power series at $x=-1/e$. Any suggestions/help appreciated. Thanks!
power-series lambert-w
$endgroup$
add a comment |
$begingroup$
I have this term: $-frac1pW(-fracpe)bigg[logbigg(-frac1pW(-fracpe)bigg)-1bigg]$.
In order to reduce complexity I've taken
$-fracpe=x$ which gives $f(x)=frac1xeW(x)bigg[logbigg(frac1xeW(x)bigg)-1bigg]$.
I need to do a power series expansion for this term at $x=-1/e$.
From the basics of Taylor expansion: $f(x)=f(-1/e)+fracdfdx(-1/e) (x+1/e)^1+fracd^2fdx^2(-1/e)(x+1/e)^2+.....$
The first term $f(-1/e)=-1$ but the second term $fracdfdx=-frac[W(x)]^2bigg[logbigg(fracW(x)xbigg)-1bigg]ex^2[w(x)+1]$ is not defined at $x=-1/e$ since $W(-1/e)=-1$ making the denominator to be $0$ also the numerator is $0$ as well. If the second term is undefined then the Taylor expansion does not work (to my understanding).
I'm not sure where I'm going wrong with this approach or if there's any other way to expand this term as a power series at $x=-1/e$. Any suggestions/help appreciated. Thanks!
power-series lambert-w
$endgroup$
add a comment |
$begingroup$
I have this term: $-frac1pW(-fracpe)bigg[logbigg(-frac1pW(-fracpe)bigg)-1bigg]$.
In order to reduce complexity I've taken
$-fracpe=x$ which gives $f(x)=frac1xeW(x)bigg[logbigg(frac1xeW(x)bigg)-1bigg]$.
I need to do a power series expansion for this term at $x=-1/e$.
From the basics of Taylor expansion: $f(x)=f(-1/e)+fracdfdx(-1/e) (x+1/e)^1+fracd^2fdx^2(-1/e)(x+1/e)^2+.....$
The first term $f(-1/e)=-1$ but the second term $fracdfdx=-frac[W(x)]^2bigg[logbigg(fracW(x)xbigg)-1bigg]ex^2[w(x)+1]$ is not defined at $x=-1/e$ since $W(-1/e)=-1$ making the denominator to be $0$ also the numerator is $0$ as well. If the second term is undefined then the Taylor expansion does not work (to my understanding).
I'm not sure where I'm going wrong with this approach or if there's any other way to expand this term as a power series at $x=-1/e$. Any suggestions/help appreciated. Thanks!
power-series lambert-w
$endgroup$
I have this term: $-frac1pW(-fracpe)bigg[logbigg(-frac1pW(-fracpe)bigg)-1bigg]$.
In order to reduce complexity I've taken
$-fracpe=x$ which gives $f(x)=frac1xeW(x)bigg[logbigg(frac1xeW(x)bigg)-1bigg]$.
I need to do a power series expansion for this term at $x=-1/e$.
From the basics of Taylor expansion: $f(x)=f(-1/e)+fracdfdx(-1/e) (x+1/e)^1+fracd^2fdx^2(-1/e)(x+1/e)^2+.....$
The first term $f(-1/e)=-1$ but the second term $fracdfdx=-frac[W(x)]^2bigg[logbigg(fracW(x)xbigg)-1bigg]ex^2[w(x)+1]$ is not defined at $x=-1/e$ since $W(-1/e)=-1$ making the denominator to be $0$ also the numerator is $0$ as well. If the second term is undefined then the Taylor expansion does not work (to my understanding).
I'm not sure where I'm going wrong with this approach or if there's any other way to expand this term as a power series at $x=-1/e$. Any suggestions/help appreciated. Thanks!
power-series lambert-w
power-series lambert-w
edited Mar 26 at 4:05
anikfaisal
asked Mar 23 at 4:19
anikfaisalanikfaisal
266
266
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
According to R. M. Corless et al. if $p=+sqrt2(ez+1)$ and $mid p mid<sqrt2$ then $W(z)=-1+p-frac13p^2+frac1172p^3+....$
Also according to the definition of Lambert-W function $We^W=z$, taking logarithm on both sides of this equation yields: $ln W=ln z-W$ gives us a relatively easier way to break down the log term in our function.
Now getting back to the function we have, $-frac1yW(-fracye)bigg[lnbigg(-frac1yW(-fracye)bigg)-1bigg]$. Let's take $z=-fracye$ and so we get $p=sqrt2(1-y)$. Now we are able to do the power series expansion for our function:
$W(z)=-1+sqrt2(1-y)^1/2-frac23(1-y)+frac11 times 2^3/272(1-y)^3/2+....$
Now, using this expansion and after some very tedious multiplication the final expression for the expansion can be derived.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3158943%2fseries-expansion-of-an-expression-that-has-productlog-function-about-1-e%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
According to R. M. Corless et al. if $p=+sqrt2(ez+1)$ and $mid p mid<sqrt2$ then $W(z)=-1+p-frac13p^2+frac1172p^3+....$
Also according to the definition of Lambert-W function $We^W=z$, taking logarithm on both sides of this equation yields: $ln W=ln z-W$ gives us a relatively easier way to break down the log term in our function.
Now getting back to the function we have, $-frac1yW(-fracye)bigg[lnbigg(-frac1yW(-fracye)bigg)-1bigg]$. Let's take $z=-fracye$ and so we get $p=sqrt2(1-y)$. Now we are able to do the power series expansion for our function:
$W(z)=-1+sqrt2(1-y)^1/2-frac23(1-y)+frac11 times 2^3/272(1-y)^3/2+....$
Now, using this expansion and after some very tedious multiplication the final expression for the expansion can be derived.
$endgroup$
add a comment |
$begingroup$
According to R. M. Corless et al. if $p=+sqrt2(ez+1)$ and $mid p mid<sqrt2$ then $W(z)=-1+p-frac13p^2+frac1172p^3+....$
Also according to the definition of Lambert-W function $We^W=z$, taking logarithm on both sides of this equation yields: $ln W=ln z-W$ gives us a relatively easier way to break down the log term in our function.
Now getting back to the function we have, $-frac1yW(-fracye)bigg[lnbigg(-frac1yW(-fracye)bigg)-1bigg]$. Let's take $z=-fracye$ and so we get $p=sqrt2(1-y)$. Now we are able to do the power series expansion for our function:
$W(z)=-1+sqrt2(1-y)^1/2-frac23(1-y)+frac11 times 2^3/272(1-y)^3/2+....$
Now, using this expansion and after some very tedious multiplication the final expression for the expansion can be derived.
$endgroup$
add a comment |
$begingroup$
According to R. M. Corless et al. if $p=+sqrt2(ez+1)$ and $mid p mid<sqrt2$ then $W(z)=-1+p-frac13p^2+frac1172p^3+....$
Also according to the definition of Lambert-W function $We^W=z$, taking logarithm on both sides of this equation yields: $ln W=ln z-W$ gives us a relatively easier way to break down the log term in our function.
Now getting back to the function we have, $-frac1yW(-fracye)bigg[lnbigg(-frac1yW(-fracye)bigg)-1bigg]$. Let's take $z=-fracye$ and so we get $p=sqrt2(1-y)$. Now we are able to do the power series expansion for our function:
$W(z)=-1+sqrt2(1-y)^1/2-frac23(1-y)+frac11 times 2^3/272(1-y)^3/2+....$
Now, using this expansion and after some very tedious multiplication the final expression for the expansion can be derived.
$endgroup$
According to R. M. Corless et al. if $p=+sqrt2(ez+1)$ and $mid p mid<sqrt2$ then $W(z)=-1+p-frac13p^2+frac1172p^3+....$
Also according to the definition of Lambert-W function $We^W=z$, taking logarithm on both sides of this equation yields: $ln W=ln z-W$ gives us a relatively easier way to break down the log term in our function.
Now getting back to the function we have, $-frac1yW(-fracye)bigg[lnbigg(-frac1yW(-fracye)bigg)-1bigg]$. Let's take $z=-fracye$ and so we get $p=sqrt2(1-y)$. Now we are able to do the power series expansion for our function:
$W(z)=-1+sqrt2(1-y)^1/2-frac23(1-y)+frac11 times 2^3/272(1-y)^3/2+....$
Now, using this expansion and after some very tedious multiplication the final expression for the expansion can be derived.
answered Mar 26 at 0:09
anikfaisalanikfaisal
266
266
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3158943%2fseries-expansion-of-an-expression-that-has-productlog-function-about-1-e%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown