Recent content by renrenbin

yeaa, it is a right way,absolutely.L'Hopital's Rule is for 0/0 and inf/inf.

I made it! Another form of this equation is: \nabla ( \vec{A} \cdot \vec{B} ) = (\vec{B} \cdot \nabla ) \vec{A} + ( \vec{A} \cdot \nabla) \vec{B} + \vec{A} \times ( \nabla \times \vec{B}) + \vec{B} \times ( \nabla \times \vec{A}) So, if I prove this form of it ,the proving process will be very...

ok ,I will try again

well,the editing drives me crazy,the 6th and 7th line are not printed properly,but it seems not very vital,the main process starts from the 8th line. thank you everyone! I was puzzelled by this problem for almost 2 days.

left=\nabla(\vec{a} \cdot \vec{b}) =\partial_{i}\vec{e}_{i}(a_{j}b_{j}) =\vec{e}_{i}(b_{j}\partial_{i}a_{j}+a_{j}\partial_{i}b_{j}) =\vec{e}_{i}b_{j}\partial_{i}a_{j}+\vec{e}_{i}a_{j}\partial_{i}b_{j}...

sorry,what I want to prove is '(b dot nabla) a=b (nabla dot a)'.It is clearly that they are not in a same direction , but when I was using Einstein summation convention , they have to be the same,which confused me a lot .I would show you my procedure of proving.Please wait.

please prove : nabla(a dot b)=(b dot nabla)a+(a dot nabla)b+b cross nabla cross a+a cross nabla cross b; (a and b are all vectors) when I was proving it,I found it impossible to go from right side to left side. I don't know whether '(b dot nabla)a = b (nabla a)' is right or not , when...