High School Solving for the Nth divergence in any coordinate system

[Vanilla Gorilla]

Ok, let's try $n=5$ $$\nabla^{n} F = F_{r^{ n } } + \frac {1} {r} F_{r^{ n-1 }} + \frac {1} {r^{2}} F_{\theta^{ n } } = F_{r^{ 5 } } + \frac {1} {r} F_{r^{ 5-1 }} + \frac {1} {r^{2}} F_{\theta^{ 5 } } = F_{r^{ 5 } } + \frac {1} {r} F_{r^{ 4 }} + \frac {1} {r^{2}} F_{\theta^{ 5 } } $$

 

[Vanilla Gorilla]

I want to try and calculate if this formula is correct $$g^{ d_{1} d_{2} } \left[\prod\limits_{w = 3}^{n} g^{ c_{w} d_{w} } \right] f_{;d_{n}...d_{1}}$$ but I'm unsure of how to quickly calculate iterated covariant derivatives, like we have with $f_{;d_{n}...d_{1}}$

 

[anuttarasammyak]

Your $f_{;d_{n}...d_{1}}$ are all n-th derivative. As you calculated in #29

$\Delta$ or $\nabla^2$ contains RHS 2nd term, first derivative. How do you concilitate it ?

 

[Vanilla Gorilla]

I'm sorry, I don't know what you mean by "concilitate"

 

[anuttarasammyak]

My bad typo, conciliate. $\Delta$ has first derivative of r. I wonder your #32 goes with it.

 

[Vanilla Gorilla]

Ok, so I was rereading through your posts 19, 21, 23, and 25, and I think I understand them better now. I also think I MIGHT have found what I was looking for $$\nabla^{n} f = \sum_{i=1}^{m} \left [ \partial_{x_{i}}^{n} \left [ f \right ] \right ] = \sum_{i=1}^{m} \left [ \left [ \partial_{x_{i}} \right ]^{n} \left [ f \right ] \right ] = \sum_{i=1}^{m} \left [ \left [ \frac {\partial } {\partial {x_{i}}} \right ] ^{n} \left [ f \right ] \right ] = \sum_{i=1}^{m_{x}} \left ( \left [ \sum_{j=1}^{m_{q}} \left [ \frac{\partial q^j}{\partial x^i}\frac{\partial }{\partial q^j} \right ] \right ]^{n} \left [ f \right ] \right )$$ and the $\nabla^{n}$ operator itself $$\nabla^{n}= \sum_{i=1}^{m_{x}} \left ( \left [ \sum_{j=1}^{m_{q}} \left [ \frac{\partial q^j}{\partial x^i}\frac{\partial }{\partial q^j} \right ] \right ]^{n} \right )$$ where $x$ represents the Cartesian Basis, and $q$ represents the arbitrary Basis. There are $m_x$ dimensions in the Cartesian Basis and $m_q$ dimensions in the arbitrary Basis.
I think this holds, since, in Einstein Notation, we have the multivariable chain rule as $$\frac{\partial}{\partial x^i}=\frac{\partial q^j}{\partial x^i}\frac{\partial }{\partial q^j}$$

 

[Vanilla Gorilla]

Is the like meant to indicate that that is correct? :D

 

[anuttarasammyak]

I observe my #25 and your #36 coincide.

 

[Vanilla Gorilla]

Also, would it be necessary for $m_x=m_q$ for that equation $$\nabla^{n}= \sum_{i=1}^{m_{x}} \left ( \left [ \sum_{j=1}^{m_{q}} \left [ \frac{\partial q^j}{\partial x^i}\frac{\partial }{\partial q^j} \right ] \right ]^{n} \right )$$ to work?

 

[anuttarasammyak]

We need same number of parameters to describe same space.

 

[Vanilla Gorilla]

What do I do with mixed terms (I.e., more than 1 basis vector involved), such as $$-2\cos \phi\frac{\partial}{\partial r}\frac{\sin \phi}{r}\frac{\partial}{\partial \phi}$$ in my calculation? Do I just discard those?

 

[anuttarasammyak]

What do I do with mixed terms (I.e., more than 1 basis vector involved), such as −2cos⁡ϕ∂∂rsin⁡ϕr∂∂ϕ in my calculation? Do I just discard those?

-2\cos \phi\frac{\partial}{\partial r}\frac{\sin \phi}{r}\frac{\partial}{\partial \phi}
We have no problem on the leftest
-2\cos \phi. Applying produclt rule of differentiaion,
\frac{\partial}{\partial r}\frac{\sin \phi}{r}\frac{\partial}{\partial \phi}
=[\frac{\partial}{\partial r}\frac{\sin \phi}{r}]\frac{\partial}{\partial \phi}+\frac{\sin \phi}{r}\frac{\partial}{\partial r}\frac{\partial}{\partial \phi}
=-\frac{\sin \phi}{r^2}\frac{\partial}{\partial \phi}+\frac{\sin \phi}{r}\frac{\partial}{\partial r}\frac{\partial}{\partial \phi}

As for change of order of operators for an example
\frac{d}{dx}xA=A+x\frac{d}{dx}A
As operator we may delete A as
\frac{d}{dx}x=1+x\frac{d}{dx}
1=\frac{d}{dx}x-x\frac{d}{dx}=[\frac{d}{dx},x]
In general
[\frac{d}{dx},f(x)]=f'(x)