What are the tensor and series questions in this homework?

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SUMMARY

The forum discussion focuses on solving tensor and series questions from a homework assignment in vector calculus and series convergence. The correct tensor index notation for the divergence and curl operations is established as ##\vec{\triangledown}\cdot\vec{\triangledown}\times\vec{A} = \epsilon_{ijk}\partial_{i}\partial_{j}A_{k}## and ##\vec{\triangledown}\times\vec{\triangledown}\phi = \epsilon_{ijk}\partial_{j}\partial_{k}\phi##. The spherical coordinate transformations are clarified, with corrections made to the expression for ##dy##. The series convergence analysis reveals that the first series converges while the second diverges, leading to a final conclusion that the overall result is divergence.

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  • Understanding of vector calculus, specifically tensor notation in ##R^3##.
  • Knowledge of spherical coordinates and their transformations.
  • Familiarity with series convergence tests, including the D'Alembert ratio test.
  • Proficiency in mathematical notation and manipulation of limits.
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  • Study tensor calculus applications in physics and engineering.
  • Learn about spherical coordinate systems and their applications in multivariable calculus.
  • Explore various series convergence tests beyond the D'Alembert ratio test.
  • Review the properties of geometric series and their convergence criteria.
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Students in advanced mathematics, particularly those studying vector calculus, physics majors, and anyone involved in mathematical analysis of series and tensor operations.

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Homework Statement


i have a few homework question and want to be sure if I have solved them right.
Q1) Write ##\vec{\triangledown}\cdot\vec{\triangledown}\times\vec{A}## and ##\vec{\triangledown}\times\vec{\triangledown}\phi## in tensor index notation in ##R^3##

Q2) the spherical coordinates
##x=r sin\theta cos\phi##
##y=r sin\theta sin\phi##
##z=r cos\theta##
what is the relataion of ##dx, dy, dz## in terms of ## dr , d\theta , d\phi , ##

Q3) Determine whether the following series converges
##\sum \left(\frac{2}{5^{k+1} }+\frac{(2k)!}{3^k}\right)##
this problem has no summation startpoint. I thought such question must have a start point and go to infinity. like k=0 or k=2

The Attempt at a Solution


Q1) ##\vec{\triangledown}\cdot\vec{\triangledown}\times\vec{A} = \epsilon_{ijk}\partial_{i}\partial_{j}A_{k}##
##\vec{\triangledown}\times\vec{\triangledown}\phi = \epsilon_{ijk}\partial_{j}\partial_{k}\phi##

Q2) ##dx=sin\theta cos\phi dr + r cos\theta cos\phi d\theta - r sin\theta sin\phi dphi##
##dy=sin\theta sin\phi dr + r cos\theta sin\phi d\phi + r cos\theta cos\phi d\phi##
## dz=cos\theta dr - r sin\theta d\theta##

Q3)
##\sum \frac{2}{5^{k+1} }=\frac{2}{5}+\frac{2}{25}+\frac{2}{125}+...##
from geometric series
##\lim_{k \rightarrow \infty}S_n=\frac{a}{1-r}##

##a=\frac{2}{5}## ##r=\frac{1}{5}##

since ##\mid r\mid<1## the series converges
##\lim_{k \rightarrow \infty}S_n=\frac{\frac{2}{5}}{1-\frac{1}{5}}=\frac{1}{2}##

##\sum \frac{(2k)!}{3^k}##

using D'Alembert ratio test
##\rho=\lim_{k \rightarrow \infty} \frac{u_{k+1}}{u_{k}}##

##\rho=\lim_{k \rightarrow \infty} \frac{\frac{[2(k+1)]!}{3^{k+1}}}{\frac{(2k)!}{3^{k}}}####\rho=\lim_{k \rightarrow \infty} \frac{(2k+2)!3^{k}}{(2k)!3^{k+1}}####\rho=\lim_{k \rightarrow \infty} \frac{(2k+2)(2k+1)}{3}##

this gives infinity and therefore this series diverges
 
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Q1) The first expression is correct. The second is missing a basis vector and has a mismatched free index on the right hand side.

Q2) Double check the last term of dy.

Q3) ok
 
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Orodruin said:
Q1) The first expression is correct. The second is missing a basis vector and has a mismatched free index on the right hand side.

Q2) Double check the last term of dy.

Q3) ok

Q1) the second expression. I think ##\phi## is a scalar and therefore it should not have an index. Right?

Q2) yes i made a mistak. It should be
##dy=sin\theta sin\phi dr + r cos\theta sin\phi d\theta + r sin\theta cos\phi d\phi##

Q3) so the final answer is
Converge + Diverge = Diverge.
And is it normal that the Summation has no startpoint? I thought there is a trick.
 

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