- #1
Karol
- 1,380
- 22
I have to prove the above equation.
I tried:
$$\bigl[ A\times ( B\times C) \bigr]_i=\sum_{jk} \varepsilon_{ijk}A_j\biggl[ \sum_{ij}\varepsilon_{ijk}B_iC_j\biggr]$$
In order to reach the k'th member of (BxC) but i tested it with i=1 and saw that in the member:
$$\biggl[ \sum_{ij}\varepsilon_{ijk}B_iC_j\biggr]=\biggl[ \sum_{1j}B_1C_j\biggr]$$
And it can't be that B1 is inside the sum since it has to receive also other values.
Then i tried:
$$\bigl[ A\times ( B\times C) \bigr]_i=\sum_{jk} \varepsilon_{ijk}A_j\biggl[ \sum_{ij}\varepsilon_{ijk}B_iC_j\biggr]_k$$
But it doesn't enable me to open the [] brackets, or does it? i am new to this
I tried:
$$\bigl[ A\times ( B\times C) \bigr]_i=\sum_{jk} \varepsilon_{ijk}A_j\biggl[ \sum_{ij}\varepsilon_{ijk}B_iC_j\biggr]$$
In order to reach the k'th member of (BxC) but i tested it with i=1 and saw that in the member:
$$\biggl[ \sum_{ij}\varepsilon_{ijk}B_iC_j\biggr]=\biggl[ \sum_{1j}B_1C_j\biggr]$$
And it can't be that B1 is inside the sum since it has to receive also other values.
Then i tried:
$$\bigl[ A\times ( B\times C) \bigr]_i=\sum_{jk} \varepsilon_{ijk}A_j\biggl[ \sum_{ij}\varepsilon_{ijk}B_iC_j\biggr]_k$$
But it doesn't enable me to open the [] brackets, or does it? i am new to this