- #36
Vanilla Gorilla
- 78
- 24
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 $$\frac{\partial}{\partial x^i}=\frac{\partial q^j}{\partial x^i}\frac{\partial }{\partial q^j}$$
I think this holds, since, in Einstein Notation, we have the multivariable chain rule as $$\frac{\partial}{\partial x^i}=\frac{\partial q^j}{\partial x^i}\frac{\partial }{\partial q^j}$$