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

$$\[{e^{ - {{(\int_{{x_0}}^{{x_1}} {dx} )}^2}/{\Delta ^2}}} \to \,\,\,\,\,{e^{ - {{(\int_{{t_0}}^t {{{({g_{ij}}d{q^i}d{q^j})}^{{\raise0.5ex\hbox{\scriptstyle 1}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{\scriptstyle 2}}}}} )}^2}/{\Delta ^2}}} = \,\,\,{e^{ - {{(\int_{{t_0}}^t {{{({g_{ij}}\frac{{d{q^i}}}{{dt'}}\frac{{d{q^j}}}{{dt'}})}^{{\raise0.5ex\hbox{\scriptstyle 1}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{\scriptstyle 2}}}} \cdot dt'} )}^2}/{\Delta ^2}}}\]$$

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

$$\[f(t) = \int_{{t_0}}^t {{{({g_{ij}}d{q^i}d{q^j})}^{{\raise0.5ex\hbox{\scriptstyle 1}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{\scriptstyle 2}}}}} = \int_{{t_0}}^t {{{({g_{ij}}\frac{{d{q^i}}}{{dt'}}\frac{{d{q^j}}}{{dt'}})}^{{\raise0.5ex\hbox{\scriptstyle 1}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{\scriptstyle 2}}}} \cdot dt'} \]$$

Generally, the Taylor series expansion is

$$\[f(t) = \sum\limits_{n = 0}^\infty {{f^{(n)}}({t_0}) \cdot {{(t - {t_0})}^n}/n!\,\,\, = } \,\,f({t_0}) + f'({t_0})(t - {t_0}) + f''({t_0}){(t - {t_0})^2}/2! + f'''({t_0}){(t - {t_0})^3}/3! + ...\]$$

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,

$$\[f'({t_0}) = {({g_{ij}}\frac{{d{q^i}}}{{dt}}\frac{{d{q^j}}}{{dt}})^{{\raise0.5ex\hbox{\scriptstyle 1}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{\scriptstyle 2}}}}{|_{t = {t_0}}}\]$$ since $$\[\frac{d}{{dt}}\int_{{t_0}}^t {F(t')dt' = F(t)} \]$$

And,

$$\[f''({t_0}) = {\textstyle{1 \over 2}}{({g_{ij}}\frac{{d{q^i}}}{{dt}}\frac{{d{q^j}}}{{dt}})^{ - {\raise0.5ex\hbox{\scriptstyle 1}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{\scriptstyle 2}}}}({{g'}_{ij}}\frac{{d{q^i}}}{{dt}}\frac{{d{q^j}}}{{dt}} + {g_{ij}}\frac{{{d^2}{q^i}}}{{d{t^2}}}\frac{{d{q^j}}}{{dt}} + {g_{ij}}\frac{{d{q^i}}}{{dt}}\frac{{{d^2}{q^j}}}{{d{t^2}}}){|_{t = {t_0}}}\]$$

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, $$\[f''({t_0}){(t - {t_0})^2}/2!\]$$ and higher?