- #1
apostolosdt
- 160
- 187
My field is physics and I'm very cautious about the "math describing the Nature" attitude, but I can't help admiring the deep richness of mathematics!
The other day, I was checking about isotropic tensors. An isotropic tensor keeps its components in all coordinated systems transformed under rotation. Then, unexpectedly, I came across some beautiful remarks on the topic. Here are some of them.
How many isotropic tensors exist? There are only single rank-0, rank-2, and rank-3 tensors, respectively, a scalar, the Kronecker ##\delta^{ij}##, and the permutation symbol ##\epsilon_{ijk}##. There are no rank-1 isotropic tensors, that is, vectors.
Now, if one attempts to enumerate all the isotropic tensors, starting with the ones above and going to higher ranks, one gets the sequence:
$$1, 0, 1, 1, 3, 6, 15, 36, 91, 232, {\rm etc.} $$
These numbers are called ##{\it Motzkin\,sum\,numbers}## and obey a recurrence relation:
$$a_n = {n-1\over n+1}\left(2a_{n-1} + 3a_{n-2}\right)$$
with ##a(0) =1, a(1) = 0##.
There is also a generating function:
$$G(x) = {1\over 2x}\left(1 -\sqrt{1-3x\over 1+x}\right)$$
that produces the following polynomial series:
$$\sum_{n=0}^\infty a_n\, x^n = 1 + x^2 + x^3 + 3x^4 + 6x^5 + 15x^6 + \cdots$$
No need to draw your attention to the coefficients of the powers! Plain beautiful!
The other day, I was checking about isotropic tensors. An isotropic tensor keeps its components in all coordinated systems transformed under rotation. Then, unexpectedly, I came across some beautiful remarks on the topic. Here are some of them.
How many isotropic tensors exist? There are only single rank-0, rank-2, and rank-3 tensors, respectively, a scalar, the Kronecker ##\delta^{ij}##, and the permutation symbol ##\epsilon_{ijk}##. There are no rank-1 isotropic tensors, that is, vectors.
Now, if one attempts to enumerate all the isotropic tensors, starting with the ones above and going to higher ranks, one gets the sequence:
$$1, 0, 1, 1, 3, 6, 15, 36, 91, 232, {\rm etc.} $$
These numbers are called ##{\it Motzkin\,sum\,numbers}## and obey a recurrence relation:
$$a_n = {n-1\over n+1}\left(2a_{n-1} + 3a_{n-2}\right)$$
with ##a(0) =1, a(1) = 0##.
There is also a generating function:
$$G(x) = {1\over 2x}\left(1 -\sqrt{1-3x\over 1+x}\right)$$
that produces the following polynomial series:
$$\sum_{n=0}^\infty a_n\, x^n = 1 + x^2 + x^3 + 3x^4 + 6x^5 + 15x^6 + \cdots$$
No need to draw your attention to the coefficients of the powers! Plain beautiful!
Last edited: