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Finally got it! Thanks a lot for the help

That would result in: A_{jilk} + A_{jkil} = A_{lkji} + A_{likj} The only thing it reminds me is of the third symmetry again, but if I use it I end up with a meaningless result: A_{jlki} = A_{ljik} Which translates in the first two symmetries.

New attempt, got further but still missing something, hope this was what you meant. From the third property: A_{ijkl} + A_{iklj} + A_{iljk} = 0 A_{klij} + A_{kijl} + A_{kjli} = 0 Therefore: A_{ijkl} + A_{iklj} + A_{iljk} = A_{klij} + A_{kijl} + A_{kjli} Since the first two properties...

Homework Statement Let Aijkl be a rank 4 square tensor with the following symmetries: A_{ijkl} = -A_{jikl}, \qquad A_{ijkl} = - A_{ijlk}, \qquad A_{ijkl} + A_{iklj} + A_{iljk} = 0, Prove that A_{ijkl} = A_{klij} Homework EquationsThe Attempt at a Solution From the first two properties...

I think I do, because when taking the limit, the only 2 possibilities that satisfy that relation are: - the limits do not exist (which would be the case of a sine or cosine) - the limit is 0, which is the point that prevents the function from oscillating That's a nice example, but taking...

Homework Statement Given the differential equation : \frac{dy}{dt} + a(t)*y = f(t) in which a and f are continuous functions in ℝ and verify: a(t) > c > 0 \forallt , lim f(t) = 0 Show that any solution of the differential equation...