Taylor series for the general distance integral

[friend]

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

\[{\textstyle{1 \over {{{(\pi {\Delta ^2})}^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}}}}{e^{{{ - {{({x_1} - {x_0})}^2}} \mathord{\left/<br /> {\vphantom {{ - {{({x_1} - {x_0})}^2}} {{\Delta ^2}}}} \right.<br /> \kern-\nulldelimiterspace} {{\Delta ^2}}}}}\]

And this has the property that

\[\int_{ - \infty }^{ + \infty } {{\textstyle{1 \over {{{(\pi {\Delta ^2})}^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}}}}{e^{{{ - {{({x_1} - {x_0})}^2}} \mathord{\left/<br /> {\vphantom {{ - {{({x_1} - {x_0})}^2}} {{\Delta ^2}}}} \right.<br /> \kern-\nulldelimiterspace} {{\Delta ^2}}}}}} d{x_1}\,\, = \,\,1\]

But the distance in the exponential, \[{{x_1} - {x_0}}\], does not tranform simply in curved space as, \[{q^i} - q_0^i\],

where \[{x^j} = {x^j}({q^i})\], and the \[{q^i}\] are the general curved space coordinates.



So I consider the exponential term in the form

\[{e^{{{ - {{({x_1} - {x_0})}^2}} \mathord{\left/<br /> {\vphantom {{ - {{({x_1} - {x_0})}^2}} {{\Delta ^2}}}} \right.<br /> \kern-\nulldelimiterspace} {{\Delta ^2}}}}}\, = \,{e^{{{ - {{(\int_{{x_0}}^{{x_1}} {dx} )}^2}} \mathord{\left/<br /> {\vphantom {{ - {{(\int_{{x_0}}^{{x_1}} {dx} )}^2}} {{\Delta ^2}}}} \right.<br /> \kern-\nulldelimiterspace} {{\Delta ^2}}})}}\],

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,

\[{\textstyle{1 \over {{{(\pi {\Delta ^2})}^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}}}}{e^{{{ - ({{({x_1} - {x_0})}^2} + {{({y_1} - {y_0})}^2} + {{({z_1} - {z_0})}^2})} \mathord{\left/<br /> {\vphantom {{ - ({{({x_1} - {x_0})}^2} + {{({y_1} - {y_0})}^2} + {{({z_1} - {z_0})}^2})} {{\Delta ^2}}}} \right.<br /> \kern-\nulldelimiterspace} {{\Delta ^2}}}}}\]

And the differential distance of some line segment in 3D flat space is,

\[ds = {({(dx)^2} + {(dy)^2} + {(dz)^2})^{1/2}} = {({(\frac{{dx}}{{dt}})^2} + {(\frac{{dy}}{{dt}})^2} + {(\frac{{dz}}{{dt}})^2})^{1/2}} \cdot dt = {({\delta _{ij}}d{x^i}d{x^j})^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}\]

which generalizes to curved q-space as

\[ds = {({g_{ij}}d{q^i}d{q^j})^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}} = {({g_{ij}}\frac{{d{q^i}}}{{dt}}\frac{{d{q^j}}}{{dt}})^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}} \cdot dt\]

So that the exponential of the integral transforms as

\[{e^{ - {{(\int_{{x_0}}^{{x_1}} {dx} )}^2}/{\Delta ^2}}} \to \,\,\,\,\,{e^{ - {{(\int_{{t_0}}^t {{{({g_{ij}}d{q^i}d{q^j})}^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}} )}^2}/{\Delta ^2}}} = \,\,\,{e^{ - {{(\int_{{t_0}}^t {{{({g_{ij}}\frac{{d{q^i}}}{{dt&#039;}}\frac{{d{q^j}}}{{dt&#039;}})}^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}} \cdot dt&#039;} )}^2}/{\Delta ^2}}}\]

But the square of the integral looks to be quite messy. I will not know the \[{q^i}(t)\] 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 \[d{q^i}d{q^j}\]. Has anyone ever seen a Taylor series expansion for the general distance integral? Thank you.

 

[friend]

So the question is what is the Taylor series of

\[f(t) = \int_{{t_0}}^t {{{({g_{ij}}d{q^i}d{q^j})}^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}} = \int_{{t_0}}^t {{{({g_{ij}}\frac{{d{q^i}}}{{dt&#039;}}\frac{{d{q^j}}}{{dt&#039;}})}^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}} \cdot dt&#039;} \]

Generally, the Taylor series expansion is

\[f(t) = \sum\limits_{n = 0}^\infty {{f^{(n)}}({t_0}) \cdot {{(t - {t_0})}^n}/n!\,\,\, = } \,\,f({t_0}) + f&#039;({t_0})(t - {t_0}) + f&#039;&#039;({t_0}){(t - {t_0})^2}/2! + f&#039;&#039;&#039;({t_0}){(t - {t_0})^3}/3! + ...\]

If f(t) as defined by the integral above is expanded about t₀,

\[f({t_0}) = 0\] since \[\int_{{t_0}}^{{t_0}} {...\,dt} = 0\]

And,

\[f&#039;({t_0}) = {({g_{ij}}\frac{{d{q^i}}}{{dt}}\frac{{d{q^j}}}{{dt}})^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}{|_{t = {t_0}}}\] since \[\frac{d}{{dt}}\int_{{t_0}}^t {F(t&#039;)dt&#039; = F(t)} \]

And,

\[f&#039;&#039;({t_0}) = {\textstyle{1 \over 2}}{({g_{ij}}\frac{{d{q^i}}}{{dt}}\frac{{d{q^j}}}{{dt}})^{ - {\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}({{g&#039;}_{ij}}\frac{{d{q^i}}}{{dt}}\frac{{d{q^j}}}{{dt}} + {g_{ij}}\frac{{{d^2}{q^i}}}{{d{t^2}}}\frac{{d{q^j}}}{{dt}} + {g_{ij}}\frac{{d{q^i}}}{{dt}}\frac{{{d^2}{q^j}}}{{d{t^2}}}){|_{t = {t_0}}}\]

So a linear approximate of f(t) is

\[f(t) \approx {({g_{ij}}\frac{{d{q^i}}}{{dt}}\frac{{d{q^j}}}{{dt}})^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}{|_{t = {t_0}}} \cdot (t - {t_0})\]

My question is do I include higher or terms, \[f&#039;&#039;({t_0}){(t - {t_0})^2}/2!\] and higher?