Variational Calculus: When Is dg(r=r+) ≠ dg(r=r++)?

PeteSampras
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Homework Statement



Question:

If ##r_+ \neq r_{++}## and ## g(r=r_+) \neq g(r=r_{++}) ##

When is it fulfilled that ## d g (r=r_+) \neq d g (r=r_+) ## ?

Homework Equations


##r_+ \neq r_{++}##
## g(r=r_+) \neq g(r=r_{++}) ##

The Attempt at a Solution



I tried computing ## dg(r_+) = \frac{\partial g}{\partial r} \Big |_{r=r_+} dr_+##, but i am very confused. This is all the information that i have of this problem.
 
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Can you define #r_{+} , r_{++}## for us? Do it through a forehand or backhand, either is OK ;).
 
WWGD said:
Can you define #r_{+} , r_{++}## for us? Do it through a forehand or backhand, either is OK ;).
I don't understand your assumption. ##r=r_+## and ##r=r_{++}## are two differents points such that ##r_{++}>r_+##
 
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PeteSampras said:
I don't understand your assumption. ##r=r_+## and ##r=r_{++}## are two differents points such that ##r_{++}>r_+##
Sorry, a tennis joke; guess you're tired of them ;).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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