How to prove vector identities WITHOUT using levi civita ?

darksilence
Messages
6
Reaction score
1
Mentor note: Thread moved from homework sections as being a better fit in the math technical section.
Multiplying components of both sides are also off limits.
I am trying to derive vector identities on introduction chapters in various EMT books. For example : (AXB).(CXD) = (A.C)(B.D) - (A.D)(B.C)
After a few hours i noticed B.(CXD) = C.(DXB) and replaced B's with AXB's its Done.
AX(BX(CXD)) was even simpler didnt take any time at all.
I want to do that to
∇. (AXB) = B.(∇xA) - A.(∇xB)
∇x(AxB) = ...
∇(A.B) = ...
∇x(∇xA) = ∇(∇.A) - ∇2A etc

So far last 2 days after solving the first two of them just looking them and hoping to see it. What i should do to improve my ability to see them fast ? (I also have to finish half the book in 2-3 weeks before exam so i am hoping to solve this problem in a few days at most.)
 
Last edited by a moderator:
Mathematics news on Phys.org
Thank you. It wasnt what i wanted at all but some way i didnt imagine they were still helpfull. Instead of working on my weakness i will go on with my strengths.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top