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Show that taylor expansion

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2

$begingroup$

Let $M$ be a spherically symmetric $C^2$ manifold. Consider the open ball $B_R$, centered in $xin M$.
Let $yin B_R$, then we define by $rho$ the geodesic distance, beetwen $x$ and $y$.

I don't see the term $-frac12h_p$. Sorry if this is very imprecise, but my point is that I do not understand the indexes of the derivatives of $(*)$, but for me it is not correct that in the derivative the same argument of the function appears and even more multiplying (for instance $h_p(p,theta)p$). Thanks!!!!

multivariable-calculus taylor-expansion

share|cite|improve this question

edited Mar 15 at 20:09

Pablo_

asked Feb 7 at 6:08

Pablo_Pablo_

1647

$endgroup$

add a comment | 

2

$begingroup$

Let $M$ be a spherically symmetric $C^2$ manifold. Consider the open ball $B_R$, centered in $xin M$.
Let $yin B_R$, then we define by $rho$ the geodesic distance, beetwen $x$ and $y$.

I don't see the term $-frac12h_p$. Sorry if this is very imprecise, but my point is that I do not understand the indexes of the derivatives of $(*)$, but for me it is not correct that in the derivative the same argument of the function appears and even more multiplying (for instance $h_p(p,theta)p$). Thanks!!!!

multivariable-calculus taylor-expansion

share|cite|improve this question

edited Mar 15 at 20:09

Pablo_

asked Feb 7 at 6:08

Pablo_Pablo_

1647

$endgroup$

add a comment | 

$begingroup$

Let $M$ be a spherically symmetric $C^2$ manifold. Consider the open ball $B_R$, centered in $xin M$.
Let $yin B_R$, then we define by $rho$ the geodesic distance, beetwen $x$ and $y$.

I don't see the term $-frac12h_p$. Sorry if this is very imprecise, but my point is that I do not understand the indexes of the derivatives of $(*)$, but for me it is not correct that in the derivative the same argument of the function appears and even more multiplying (for instance $h_p(p,theta)p$). Thanks!!!!

multivariable-calculus taylor-expansion

share|cite|improve this question

edited Mar 15 at 20:09

Pablo_

asked Feb 7 at 6:08

Pablo_Pablo_

1647

$endgroup$

share|cite|improve this question

edited Mar 15 at 20:09

Pablo_

asked Feb 7 at 6:08

Pablo_Pablo_

1647

share|cite|improve this question

edited Mar 15 at 20:09

Pablo_

asked Feb 7 at 6:08

Pablo_Pablo_

1647

asked Feb 7 at 6:08

Pablo_Pablo_

1647

asked Feb 7 at 6:08

Pablo_Pablo_

1647

1 Answer
1

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1

$begingroup$

A full 3-dimensional power series, centered at $x$... that's not something you want to put into spherical coordinates (centered at $x$). The angle variables are singular there. The radius increases in every direction from there. The fundamental premise of the series requires an independent system of coordinates, and you can't have that when the series is centered at the polar origin.

So, what are we really doing here? We're treating this as a single-variable problem. That $h_rho$ isn't the derivative with respect to radius, it's a directional derivative in the (fixed) $rho$ direction. The series is based entirely on the values of the function along that line, and is used only to calculate values on that line. It is not a Taylor series for the full three-dimensional function.

share|cite|improve this answer

answered Feb 11 at 4:22

jmerryjmerry

15.3k1632

$endgroup$

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1

$begingroup$

A full 3-dimensional power series, centered at $x$... that's not something you want to put into spherical coordinates (centered at $x$). The angle variables are singular there. The radius increases in every direction from there. The fundamental premise of the series requires an independent system of coordinates, and you can't have that when the series is centered at the polar origin.

So, what are we really doing here? We're treating this as a single-variable problem. That $h_rho$ isn't the derivative with respect to radius, it's a directional derivative in the (fixed) $rho$ direction. The series is based entirely on the values of the function along that line, and is used only to calculate values on that line. It is not a Taylor series for the full three-dimensional function.

share|cite|improve this answer

answered Feb 11 at 4:22

jmerryjmerry

15.3k1632

$endgroup$

add a comment | 

1

$begingroup$

A full 3-dimensional power series, centered at $x$... that's not something you want to put into spherical coordinates (centered at $x$). The angle variables are singular there. The radius increases in every direction from there. The fundamental premise of the series requires an independent system of coordinates, and you can't have that when the series is centered at the polar origin.

So, what are we really doing here? We're treating this as a single-variable problem. That $h_rho$ isn't the derivative with respect to radius, it's a directional derivative in the (fixed) $rho$ direction. The series is based entirely on the values of the function along that line, and is used only to calculate values on that line. It is not a Taylor series for the full three-dimensional function.

share|cite|improve this answer

answered Feb 11 at 4:22

jmerryjmerry

15.3k1632

$endgroup$

add a comment | 

$begingroup$

A full 3-dimensional power series, centered at $x$... that's not something you want to put into spherical coordinates (centered at $x$). The angle variables are singular there. The radius increases in every direction from there. The fundamental premise of the series requires an independent system of coordinates, and you can't have that when the series is centered at the polar origin.

So, what are we really doing here? We're treating this as a single-variable problem. That $h_rho$ isn't the derivative with respect to radius, it's a directional derivative in the (fixed) $rho$ direction. The series is based entirely on the values of the function along that line, and is used only to calculate values on that line. It is not a Taylor series for the full three-dimensional function.

share|cite|improve this answer

answered Feb 11 at 4:22

jmerryjmerry

15.3k1632

$endgroup$

share|cite|improve this answer

answered Feb 11 at 4:22

jmerryjmerry

15.3k1632

answered Feb 11 at 4:22

jmerryjmerry

15.3k1632

answered Feb 11 at 4:22

jmerryjmerry

15.3k1632