Taylor series of a complex Gaussian function
$begingroup$
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
asked Jan 8 at 3:41
HellRazorHellRazor
16619
$endgroup$
add a comment |
$begingroup$
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
asked Jan 8 at 3:41
HellRazorHellRazor
16619
$endgroup$
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
add a comment |
$begingroup$
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
asked Jan 8 at 3:41
HellRazorHellRazor
16619
$endgroup$
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
asked Jan 8 at 3:41
HellRazorHellRazor
16619
asked Jan 8 at 3:41
HellRazorHellRazor
16619
asked Jan 8 at 3:41
HellRazorHellRazor
16619
asked Jan 8 at 3:41
HellRazorHellRazor
16619
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
add a comment |
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
add a comment |
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
});
}
});
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%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
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%2f3065781%2ftaylor-series-of-a-complex-gaussian-function%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
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