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Background:
I'm trying to transform the gaussian distribution from flat space to curved space. I start with the flat, 1D gaussian distribution in the form
[tex]\[{\textstyle{1 \over {{{(\pi {\Delta ^2})}^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}}}}{e^{{{ - {{({x_1} - {x_0})}^2}} \mathord{\left/
{\vphantom {{ - {{({x_1} - {x_0})}^2}} {{\Delta ^2}}}} \right.
\kern-\nulldelimiterspace} {{\Delta ^2}}}}}\][/tex]
And this has the property that
[tex]\[\int_{ - \infty }^{ + \infty } {{\textstyle{1 \over {{{(\pi {\Delta ^2})}^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}}}}{e^{{{ - {{({x_1} - {x_0})}^2}} \mathord{\left/
{\vphantom {{ - {{({x_1} - {x_0})}^2}} {{\Delta ^2}}}} \right.
\kern-\nulldelimiterspace} {{\Delta ^2}}}}}} d{x_1}\,\, = \,\,1\][/tex]
But the distance in the exponential, [tex]\[{{x_1} - {x_0}}\][/tex], does not tranform simply in curved space as, [tex]\[{q^i} - q_0^i\][/tex],
where [tex]\[{x^j} = {x^j}({q^i})\][/tex], and the [tex]\[{q^i}\][/tex] are the general curved space coordinates.
So I consider the exponential term in the form
[tex]\[{e^{{{ - {{({x_1} - {x_0})}^2}} \mathord{\left/
{\vphantom {{ - {{({x_1} - {x_0})}^2}} {{\Delta ^2}}}} \right.
\kern-\nulldelimiterspace} {{\Delta ^2}}}}}\, = \,{e^{{{ - {{(\int_{{x_0}}^{{x_1}} {dx} )}^2}} \mathord{\left/
{\vphantom {{ - {{(\int_{{x_0}}^{{x_1}} {dx} )}^2}} {{\Delta ^2}}}} \right.
\kern-\nulldelimiterspace} {{\Delta ^2}}})}}\][/tex],
Then the integral in the exponential can be transformed differentially to the curved space coordinates.
the 1D, flat gaussian generalizes to 3D flat space as,
[tex]\[{\textstyle{1 \over {{{(\pi {\Delta ^2})}^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}}}}{e^{{{ - ({{({x_1} - {x_0})}^2} + {{({y_1} - {y_0})}^2} + {{({z_1} - {z_0})}^2})} \mathord{\left/
{\vphantom {{ - ({{({x_1} - {x_0})}^2} + {{({y_1} - {y_0})}^2} + {{({z_1} - {z_0})}^2})} {{\Delta ^2}}}} \right.
\kern-\nulldelimiterspace} {{\Delta ^2}}}}}\][/tex]
And the differential distance of some line segment in 3D flat space is,
[tex]\[ds = {({(dx)^2} + {(dy)^2} + {(dz)^2})^{1/2}} = {({(\frac{{dx}}{{dt}})^2} + {(\frac{{dy}}{{dt}})^2} + {(\frac{{dz}}{{dt}})^2})^{1/2}} \cdot dt = {({\delta _{ij}}d{x^i}d{x^j})^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}\][/tex]
which generalizes to curved q-space as
[tex]\[ds = {({g_{ij}}d{q^i}d{q^j})^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}} = {({g_{ij}}\frac{{d{q^i}}}{{dt}}\frac{{d{q^j}}}{{dt}})^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}} \cdot dt\][/tex]
So that the exponential of the integral transforms as
[tex]\[{e^{ - {{(\int_{{x_0}}^{{x_1}} {dx} )}^2}/{\Delta ^2}}} \to \,\,\,\,\,{e^{ - {{(\int_{{t_0}}^t {{{({g_{ij}}d{q^i}d{q^j})}^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}} )}^2}/{\Delta ^2}}} = \,\,\,{e^{ - {{(\int_{{t_0}}^t {{{({g_{ij}}\frac{{d{q^i}}}{{dt'}}\frac{{d{q^j}}}{{dt'}})}^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}} \cdot dt'} )}^2}/{\Delta ^2}}}\][/tex]
But the square of the integral looks to be quite messy. I will not know the [tex]\[{q^i}(t)\][/tex] to begin with. And I will end up integrating everything that's already in the exponent. So I'm looking for a Taylor series expansion for the squared integral. I won't want to get an expansion for the integral squared since that will only give me copies of the integral. So I want an expansion of only the integral, and I will then square it for the leading terms in [tex]\[d{q^i}d{q^j}\][/tex]. Has anyone ever seen a Taylor series expansion for the general distance integral? Thank you.
I'm trying to transform the gaussian distribution from flat space to curved space. I start with the flat, 1D gaussian distribution in the form
[tex]\[{\textstyle{1 \over {{{(\pi {\Delta ^2})}^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}}}}{e^{{{ - {{({x_1} - {x_0})}^2}} \mathord{\left/
{\vphantom {{ - {{({x_1} - {x_0})}^2}} {{\Delta ^2}}}} \right.
\kern-\nulldelimiterspace} {{\Delta ^2}}}}}\][/tex]
And this has the property that
[tex]\[\int_{ - \infty }^{ + \infty } {{\textstyle{1 \over {{{(\pi {\Delta ^2})}^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}}}}{e^{{{ - {{({x_1} - {x_0})}^2}} \mathord{\left/
{\vphantom {{ - {{({x_1} - {x_0})}^2}} {{\Delta ^2}}}} \right.
\kern-\nulldelimiterspace} {{\Delta ^2}}}}}} d{x_1}\,\, = \,\,1\][/tex]
But the distance in the exponential, [tex]\[{{x_1} - {x_0}}\][/tex], does not tranform simply in curved space as, [tex]\[{q^i} - q_0^i\][/tex],
where [tex]\[{x^j} = {x^j}({q^i})\][/tex], and the [tex]\[{q^i}\][/tex] are the general curved space coordinates.
So I consider the exponential term in the form
[tex]\[{e^{{{ - {{({x_1} - {x_0})}^2}} \mathord{\left/
{\vphantom {{ - {{({x_1} - {x_0})}^2}} {{\Delta ^2}}}} \right.
\kern-\nulldelimiterspace} {{\Delta ^2}}}}}\, = \,{e^{{{ - {{(\int_{{x_0}}^{{x_1}} {dx} )}^2}} \mathord{\left/
{\vphantom {{ - {{(\int_{{x_0}}^{{x_1}} {dx} )}^2}} {{\Delta ^2}}}} \right.
\kern-\nulldelimiterspace} {{\Delta ^2}}})}}\][/tex],
Then the integral in the exponential can be transformed differentially to the curved space coordinates.
the 1D, flat gaussian generalizes to 3D flat space as,
[tex]\[{\textstyle{1 \over {{{(\pi {\Delta ^2})}^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}}}}{e^{{{ - ({{({x_1} - {x_0})}^2} + {{({y_1} - {y_0})}^2} + {{({z_1} - {z_0})}^2})} \mathord{\left/
{\vphantom {{ - ({{({x_1} - {x_0})}^2} + {{({y_1} - {y_0})}^2} + {{({z_1} - {z_0})}^2})} {{\Delta ^2}}}} \right.
\kern-\nulldelimiterspace} {{\Delta ^2}}}}}\][/tex]
And the differential distance of some line segment in 3D flat space is,
[tex]\[ds = {({(dx)^2} + {(dy)^2} + {(dz)^2})^{1/2}} = {({(\frac{{dx}}{{dt}})^2} + {(\frac{{dy}}{{dt}})^2} + {(\frac{{dz}}{{dt}})^2})^{1/2}} \cdot dt = {({\delta _{ij}}d{x^i}d{x^j})^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}\][/tex]
which generalizes to curved q-space as
[tex]\[ds = {({g_{ij}}d{q^i}d{q^j})^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}} = {({g_{ij}}\frac{{d{q^i}}}{{dt}}\frac{{d{q^j}}}{{dt}})^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}} \cdot dt\][/tex]
So that the exponential of the integral transforms as
[tex]\[{e^{ - {{(\int_{{x_0}}^{{x_1}} {dx} )}^2}/{\Delta ^2}}} \to \,\,\,\,\,{e^{ - {{(\int_{{t_0}}^t {{{({g_{ij}}d{q^i}d{q^j})}^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}} )}^2}/{\Delta ^2}}} = \,\,\,{e^{ - {{(\int_{{t_0}}^t {{{({g_{ij}}\frac{{d{q^i}}}{{dt'}}\frac{{d{q^j}}}{{dt'}})}^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}} \cdot dt'} )}^2}/{\Delta ^2}}}\][/tex]
But the square of the integral looks to be quite messy. I will not know the [tex]\[{q^i}(t)\][/tex] to begin with. And I will end up integrating everything that's already in the exponent. So I'm looking for a Taylor series expansion for the squared integral. I won't want to get an expansion for the integral squared since that will only give me copies of the integral. So I want an expansion of only the integral, and I will then square it for the leading terms in [tex]\[d{q^i}d{q^j}\][/tex]. Has anyone ever seen a Taylor series expansion for the general distance integral? Thank you.