Taylor series of a complex Gaussian function












0












$begingroup$


The first few terms of Taylor series at $a=0$ of the Gaussian function $e^{-(x-a)^2}$ when $x$ is real are:



$$e^{-x^2}+aleft(2xe^{-x^2}right)+a^2(2x^2-1)e^{-x^2}+...$$



I am wondering if one can take the "Taylor series" at $a=0$ for the complex Gaussian function $e^{-|x-a|^2}$. Will it have the following form?



$$e^{-|x|^2}+aleft(2xe^{-|x|^2}right)+|a|^2(2|x|^2-1)e^{-|x|^2}+...$$



I'm trying to approximate complex Gaussian around $a=0$, but am really stumbling around in the dark here, having never taken the course on complex analysis. I read about the relevant part of the wiki article on Taylor series, but am confused. Any help would be appreciated.






complex-analysis taylor-expansion







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Do you really mean $|x-a|^2$ instead of $(x-a)^2$ here?
    $endgroup$
    – kimchi lover
    Jan 8 at 3:45










  • $begingroup$
    Taylor series require analytic functions. In general, absolute values lead to non-analytic functions in the complex plane. In particular, $e^{-|x-a|^2}$ is always real, and no analytic functions are real except constants.
    $endgroup$
    – Robert Israel
    Jan 8 at 4:07










  • $begingroup$
    @RobertIsrael Hmm, I see -- thanks for clarification on analytic functions. I think instead of what I wrote I need a Taylor series of a bivariate function $e^{-(x_i-a_i)^2-(x_q-a_q)^2}$ where $x=x_i+ix_q$ and $a=a_a+ia_q$ (subscripts "i" and "q" denote in-phase and quadrature components of complex numbers).
    $endgroup$
    – HellRazor
    Jan 8 at 5:31










  • $begingroup$
    Again, this is always real, and thus non-analytic as a function of the complex variable $a$. On the other hand, you could regard $a_i$ and $a_q$ as separate variables, and do a multivariate Taylor expansion as a function of them.
    $endgroup$
    – Robert Israel
    Jan 8 at 16:22












  • $begingroup$
    $$ {{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2}}}+2,{{rm e}^{-{x_{{i}}}^{2}-{ x_{{q}}}^{2}}}x_{{i}}a_{{i}}+2,{{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2} }}x_{{q}}a_{{q}}+{{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2}}} left( 2,{x _{{i}}}^{2}-1 right) {a_{{i}}}^{2}+4,{{rm e}^{-{x_{{i}}}^{2}-{x_{{q }}}^{2}}}x_{{i}}x_{{q}}a_{{q}}a_{{i}}+{{rm e}^{-{x_{{i}}}^{2}-{x_{{q} }}^{2}}} left( 2,{x_{{q}}}^{2}-1 right) {a_{{q}}}^{2} +ldots$$
    $endgroup$
    – Robert Israel
    Jan 8 at 16:26
















0












$begingroup$


The first few terms of Taylor series at $a=0$ of the Gaussian function $e^{-(x-a)^2}$ when $x$ is real are:



$$e^{-x^2}+aleft(2xe^{-x^2}right)+a^2(2x^2-1)e^{-x^2}+...$$



I am wondering if one can take the "Taylor series" at $a=0$ for the complex Gaussian function $e^{-|x-a|^2}$. Will it have the following form?



$$e^{-|x|^2}+aleft(2xe^{-|x|^2}right)+|a|^2(2|x|^2-1)e^{-|x|^2}+...$$



I'm trying to approximate complex Gaussian around $a=0$, but am really stumbling around in the dark here, having never taken the course on complex analysis. I read about the relevant part of the wiki article on Taylor series, but am confused. Any help would be appreciated.






complex-analysis taylor-expansion







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Do you really mean $|x-a|^2$ instead of $(x-a)^2$ here?
    $endgroup$
    – kimchi lover
    Jan 8 at 3:45










  • $begingroup$
    Taylor series require analytic functions. In general, absolute values lead to non-analytic functions in the complex plane. In particular, $e^{-|x-a|^2}$ is always real, and no analytic functions are real except constants.
    $endgroup$
    – Robert Israel
    Jan 8 at 4:07










  • $begingroup$
    @RobertIsrael Hmm, I see -- thanks for clarification on analytic functions. I think instead of what I wrote I need a Taylor series of a bivariate function $e^{-(x_i-a_i)^2-(x_q-a_q)^2}$ where $x=x_i+ix_q$ and $a=a_a+ia_q$ (subscripts "i" and "q" denote in-phase and quadrature components of complex numbers).
    $endgroup$
    – HellRazor
    Jan 8 at 5:31










  • $begingroup$
    Again, this is always real, and thus non-analytic as a function of the complex variable $a$. On the other hand, you could regard $a_i$ and $a_q$ as separate variables, and do a multivariate Taylor expansion as a function of them.
    $endgroup$
    – Robert Israel
    Jan 8 at 16:22












  • $begingroup$
    $$ {{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2}}}+2,{{rm e}^{-{x_{{i}}}^{2}-{ x_{{q}}}^{2}}}x_{{i}}a_{{i}}+2,{{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2} }}x_{{q}}a_{{q}}+{{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2}}} left( 2,{x _{{i}}}^{2}-1 right) {a_{{i}}}^{2}+4,{{rm e}^{-{x_{{i}}}^{2}-{x_{{q }}}^{2}}}x_{{i}}x_{{q}}a_{{q}}a_{{i}}+{{rm e}^{-{x_{{i}}}^{2}-{x_{{q} }}^{2}}} left( 2,{x_{{q}}}^{2}-1 right) {a_{{q}}}^{2} +ldots$$
    $endgroup$
    – Robert Israel
    Jan 8 at 16:26














0












0








0





$begingroup$


The first few terms of Taylor series at $a=0$ of the Gaussian function $e^{-(x-a)^2}$ when $x$ is real are:



$$e^{-x^2}+aleft(2xe^{-x^2}right)+a^2(2x^2-1)e^{-x^2}+...$$



I am wondering if one can take the "Taylor series" at $a=0$ for the complex Gaussian function $e^{-|x-a|^2}$. Will it have the following form?



$$e^{-|x|^2}+aleft(2xe^{-|x|^2}right)+|a|^2(2|x|^2-1)e^{-|x|^2}+...$$



I'm trying to approximate complex Gaussian around $a=0$, but am really stumbling around in the dark here, having never taken the course on complex analysis. I read about the relevant part of the wiki article on Taylor series, but am confused. Any help would be appreciated.






complex-analysis taylor-expansion







share|cite|improve this question









$endgroup$




The first few terms of Taylor series at $a=0$ of the Gaussian function $e^{-(x-a)^2}$ when $x$ is real are:



$$e^{-x^2}+aleft(2xe^{-x^2}right)+a^2(2x^2-1)e^{-x^2}+...$$



I am wondering if one can take the "Taylor series" at $a=0$ for the complex Gaussian function $e^{-|x-a|^2}$. Will it have the following form?



$$e^{-|x|^2}+aleft(2xe^{-|x|^2}right)+|a|^2(2|x|^2-1)e^{-|x|^2}+...$$



I'm trying to approximate complex Gaussian around $a=0$, but am really stumbling around in the dark here, having never taken the course on complex analysis. I read about the relevant part of the wiki article on Taylor series, but am confused. Any help would be appreciated.






complex-analysis taylor-expansion




complex-analysis taylor-expansion






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 8 at 3:41









HellRazorHellRazor

16619




16619








  • 3




    $begingroup$
    Do you really mean $|x-a|^2$ instead of $(x-a)^2$ here?
    $endgroup$
    – kimchi lover
    Jan 8 at 3:45










  • $begingroup$
    Taylor series require analytic functions. In general, absolute values lead to non-analytic functions in the complex plane. In particular, $e^{-|x-a|^2}$ is always real, and no analytic functions are real except constants.
    $endgroup$
    – Robert Israel
    Jan 8 at 4:07










  • $begingroup$
    @RobertIsrael Hmm, I see -- thanks for clarification on analytic functions. I think instead of what I wrote I need a Taylor series of a bivariate function $e^{-(x_i-a_i)^2-(x_q-a_q)^2}$ where $x=x_i+ix_q$ and $a=a_a+ia_q$ (subscripts "i" and "q" denote in-phase and quadrature components of complex numbers).
    $endgroup$
    – HellRazor
    Jan 8 at 5:31










  • $begingroup$
    Again, this is always real, and thus non-analytic as a function of the complex variable $a$. On the other hand, you could regard $a_i$ and $a_q$ as separate variables, and do a multivariate Taylor expansion as a function of them.
    $endgroup$
    – Robert Israel
    Jan 8 at 16:22












  • $begingroup$
    $$ {{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2}}}+2,{{rm e}^{-{x_{{i}}}^{2}-{ x_{{q}}}^{2}}}x_{{i}}a_{{i}}+2,{{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2} }}x_{{q}}a_{{q}}+{{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2}}} left( 2,{x _{{i}}}^{2}-1 right) {a_{{i}}}^{2}+4,{{rm e}^{-{x_{{i}}}^{2}-{x_{{q }}}^{2}}}x_{{i}}x_{{q}}a_{{q}}a_{{i}}+{{rm e}^{-{x_{{i}}}^{2}-{x_{{q} }}^{2}}} left( 2,{x_{{q}}}^{2}-1 right) {a_{{q}}}^{2} +ldots$$
    $endgroup$
    – Robert Israel
    Jan 8 at 16:26














  • 3




    $begingroup$
    Do you really mean $|x-a|^2$ instead of $(x-a)^2$ here?
    $endgroup$
    – kimchi lover
    Jan 8 at 3:45










  • $begingroup$
    Taylor series require analytic functions. In general, absolute values lead to non-analytic functions in the complex plane. In particular, $e^{-|x-a|^2}$ is always real, and no analytic functions are real except constants.
    $endgroup$
    – Robert Israel
    Jan 8 at 4:07










  • $begingroup$
    @RobertIsrael Hmm, I see -- thanks for clarification on analytic functions. I think instead of what I wrote I need a Taylor series of a bivariate function $e^{-(x_i-a_i)^2-(x_q-a_q)^2}$ where $x=x_i+ix_q$ and $a=a_a+ia_q$ (subscripts "i" and "q" denote in-phase and quadrature components of complex numbers).
    $endgroup$
    – HellRazor
    Jan 8 at 5:31










  • $begingroup$
    Again, this is always real, and thus non-analytic as a function of the complex variable $a$. On the other hand, you could regard $a_i$ and $a_q$ as separate variables, and do a multivariate Taylor expansion as a function of them.
    $endgroup$
    – Robert Israel
    Jan 8 at 16:22












  • $begingroup$
    $$ {{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2}}}+2,{{rm e}^{-{x_{{i}}}^{2}-{ x_{{q}}}^{2}}}x_{{i}}a_{{i}}+2,{{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2} }}x_{{q}}a_{{q}}+{{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2}}} left( 2,{x _{{i}}}^{2}-1 right) {a_{{i}}}^{2}+4,{{rm e}^{-{x_{{i}}}^{2}-{x_{{q }}}^{2}}}x_{{i}}x_{{q}}a_{{q}}a_{{i}}+{{rm e}^{-{x_{{i}}}^{2}-{x_{{q} }}^{2}}} left( 2,{x_{{q}}}^{2}-1 right) {a_{{q}}}^{2} +ldots$$
    $endgroup$
    – Robert Israel
    Jan 8 at 16:26








3




3




$begingroup$
Do you really mean $|x-a|^2$ instead of $(x-a)^2$ here?
$endgroup$
– kimchi lover
Jan 8 at 3:45




$begingroup$
Do you really mean $|x-a|^2$ instead of $(x-a)^2$ here?
$endgroup$
– kimchi lover
Jan 8 at 3:45












$begingroup$
Taylor series require analytic functions. In general, absolute values lead to non-analytic functions in the complex plane. In particular, $e^{-|x-a|^2}$ is always real, and no analytic functions are real except constants.
$endgroup$
– Robert Israel
Jan 8 at 4:07




$begingroup$
Taylor series require analytic functions. In general, absolute values lead to non-analytic functions in the complex plane. In particular, $e^{-|x-a|^2}$ is always real, and no analytic functions are real except constants.
$endgroup$
– Robert Israel
Jan 8 at 4:07












$begingroup$
@RobertIsrael Hmm, I see -- thanks for clarification on analytic functions. I think instead of what I wrote I need a Taylor series of a bivariate function $e^{-(x_i-a_i)^2-(x_q-a_q)^2}$ where $x=x_i+ix_q$ and $a=a_a+ia_q$ (subscripts "i" and "q" denote in-phase and quadrature components of complex numbers).
$endgroup$
– HellRazor
Jan 8 at 5:31




$begingroup$
@RobertIsrael Hmm, I see -- thanks for clarification on analytic functions. I think instead of what I wrote I need a Taylor series of a bivariate function $e^{-(x_i-a_i)^2-(x_q-a_q)^2}$ where $x=x_i+ix_q$ and $a=a_a+ia_q$ (subscripts "i" and "q" denote in-phase and quadrature components of complex numbers).
$endgroup$
– HellRazor
Jan 8 at 5:31












$begingroup$
Again, this is always real, and thus non-analytic as a function of the complex variable $a$. On the other hand, you could regard $a_i$ and $a_q$ as separate variables, and do a multivariate Taylor expansion as a function of them.
$endgroup$
– Robert Israel
Jan 8 at 16:22






$begingroup$
Again, this is always real, and thus non-analytic as a function of the complex variable $a$. On the other hand, you could regard $a_i$ and $a_q$ as separate variables, and do a multivariate Taylor expansion as a function of them.
$endgroup$
– Robert Israel
Jan 8 at 16:22














$begingroup$
$$ {{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2}}}+2,{{rm e}^{-{x_{{i}}}^{2}-{ x_{{q}}}^{2}}}x_{{i}}a_{{i}}+2,{{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2} }}x_{{q}}a_{{q}}+{{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2}}} left( 2,{x _{{i}}}^{2}-1 right) {a_{{i}}}^{2}+4,{{rm e}^{-{x_{{i}}}^{2}-{x_{{q }}}^{2}}}x_{{i}}x_{{q}}a_{{q}}a_{{i}}+{{rm e}^{-{x_{{i}}}^{2}-{x_{{q} }}^{2}}} left( 2,{x_{{q}}}^{2}-1 right) {a_{{q}}}^{2} +ldots$$
$endgroup$
– Robert Israel
Jan 8 at 16:26




$begingroup$
$$ {{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2}}}+2,{{rm e}^{-{x_{{i}}}^{2}-{ x_{{q}}}^{2}}}x_{{i}}a_{{i}}+2,{{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2} }}x_{{q}}a_{{q}}+{{rm e}^{-{x_{{i}}}^{2}-{x_{{q}}}^{2}}} left( 2,{x _{{i}}}^{2}-1 right) {a_{{i}}}^{2}+4,{{rm e}^{-{x_{{i}}}^{2}-{x_{{q }}}^{2}}}x_{{i}}x_{{q}}a_{{q}}a_{{i}}+{{rm e}^{-{x_{{i}}}^{2}-{x_{{q} }}^{2}}} left( 2,{x_{{q}}}^{2}-1 right) {a_{{q}}}^{2} +ldots$$
$endgroup$
– Robert Israel
Jan 8 at 16:26










0






active

oldest

votes












Your Answer








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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065781%2ftaylor-series-of-a-complex-gaussian-function%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065781%2ftaylor-series-of-a-complex-gaussian-function%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Wiesbaden

Image
Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
Read more

Marschland

Image
Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
Read more

Dieringhausen

Image
Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
Read more