Fdtxjr

What is the difference between lands and mana?

How much theory knowledge is actually used while playing?

What to do when eye contact makes your coworker uncomfortable?

Shouldn’t conservatives embrace universal basic income?

How could a planet have erratic days?

Non-trope happy ending?

Will number of steps recorded on FitBit/any fitness tracker add up distance in PokemonGo?

Does the Linux kernel need a file system to run?

Doesn't the system of the Supreme Court oppose justice?

awk assign to multiple variables at once

Are Captain Marvel's powers affected by Thanos breaking the Tesseract and claiming the stone?

What is Cash Advance APR?

Why do Radio Buttons not fill the entire outer circle?

Is this part of the description of the Archfey warlock's Misty Escape feature redundant?

Why Shazam when there is already Superman?

Does the reader need to like the PoV character?

How would you translate "more" for use as an interface button?

Is this toilet slogan correct usage of the English language?

A Trivial Diagnosis

What fields between the rationals and the reals allow a good notion of 2D distance?

How does electrical safety system work on ISS?

Is there a nicer/politer/more positive alternative for "negates"?

Biological Blimps: Propulsion

Why does Carol not get rid of the Kree symbol on her suit when she changes its colours?

Show that taylor expansion

Troublesome functional derivative: second term of Euler-Lagrange equationTaylor Series of the Complex Log and Contour IntegrationTaylor theorem for f(x+h)Why is n-th Fréchet derivative symmetric?How do I show that this function is not smooth?$nabla^2 u = 0 $ and integral $u$ around $partial B_rho$Find the Taylor series for $sqrtx$ centered at 16Precisely, what is the polar coordinate system?Change of Variables in Surface Integralfirst order Taylor expansion term of a function multiplied by a dot product of gradients.

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

active

oldest

votes

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 | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3103478%2fshow-that-taylor-expansion%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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