Taylor series of a complex Gaussian function
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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 |
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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