The given equation is correct:
##\qquad\qquad(\vec A \times \vec B)\times(\vec C \times \vec D) = [ \vec A \cdot (\vec B \times \vec D) ] \vec C - [ \vec A \cdot (\vec B \times \vec C) ] \vec D##
The result I got was
##\qquad\qquad(\vec A \times \vec B)\times(\vec C \times \vec D) = [ \vec A...
The equation to be proven had been improperly written because the vectors in the third and fourth terms had not been properly grouped. Replacing the lower-case-letter vectors with upper case letters, we should have $$\vec \nabla (\vec A\cdot \vec B)=(\vec A \cdot \vec \nabla)\vec B + (\vec B...
Show that $$(1){~~~~~~~~~~~~~~~~~~~~~~~~~}\vec \nabla\cdot(\vec a \times \vec b) = \vec b \cdot (\vec \nabla \times \vec a) - \vec a \cdot (\vec \nabla \times \vec b){~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}$$ Using suffix notation, we get for the left hand side of eq. (1)...
The OP posted this problem-to-prove question:
As I already said in post #11, the correct simplification should lead to$$\vec{\nabla}(\vec{r} \cdot \vec{u}) = \vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + [~(\vec r \times \vec{\nabla}) \times \vec{u}~]$$But I was wondering if the identity to...
I find the use of brute-force method and suffix notations both tedious and cumbersome. Why don't you try using first useful vector identities to simplify your equation. The BAC minus CAB rule is one such useful identity.
But I noticed in your OP where you wrote:
that you didn't properly group...
I think you're right there. Just two steps away from the answer.
Using ln, just doing two computations won't give you the answer yet. You'll need to do at least three; first the value of ln##~y~##, then the value of (ln##~y##)/##n~##, then that of ##exp[{\frac {\ln y}{n}}]~## before you finally...
@kuruman ... You know fairly well what we were talking about. We were not arguing about the acceptability of settling a $75,000 debt by paying back only $15,000 , which you mentioned in post #26 and that seem to imply, that according to my own point of view, should be acceptable because they...
@jbriggs444 ...
Of course I wasn't talking about any two numbers having the same order of magnitude in general. I was talking to kuruman about these numbers given by the OP
in post #1:
and these numbers that I wrote in post #20:
I know that they are not equal but for the purposes of my...
@kuruman again ... Your post #9 said:
There you mentioned nothing about ##|β|~,## that ##β## can be ##\pm~.## Then, suddenly in post #19, only after 10 messages had already been posted, you surprisingly come up with the idea that ##β~,## in the formula you used in #9 , can be ##\pm~.## Why...
@kuruman ... Of course they are, they both have the same order of magnitude
There are minimum and maximum values of the frequency for any given color. I used the average value.