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You obtain the above relation by putting \$vec A=vec B =vec v\$ in the following relation: \$\$

abla(vec Acdot vec B)=(vec Acdot

abla)vec B(vec Bcdot

abla)vec Avec B times (

abla times vec A)vec A times (

abla times vec B)\$\$Now as you can see, the physical/geometrical/intuititve explanation for your relation is the same as that for this relation.And this can be easily done as L.H.S. denotes the gradient of the scalar field "the projection of \$vec A\$ on \$vec B\$" and hence on.

1. How to prove \$(F,)\$ and \$(Fsetminus 0,cdot)\$ aren't isomorphic, where \$(F,,cdot)\$ is an arbitrary field .

If \$operatornamechar(F)

eq 2\$ then \$(-1)\$ has order \$2\$ in \$(F^times,cdot)\$, but there is no element of order \$2\$ in \$(F,)\$.If \$operatornamechar(F)=2\$ then any element has order \$2\$ in \$(F,)\$ but no element has order \$2\$ in \$(F^times, cdot)\$ as \$\$x^2=1 Rightarrow (x-1)^2=0 Rightarrow x-1=0\$\$

2. Solve recurrence relation \$T(n) = ncdot T(n/2)^2 \$

We have \$\$4nT(n) = left(4fracn2T(n/2)

ight)^2\$\$ so that \$\$F(n) = F(n/2)^2 \$\$ where \$F(n) = 4nT(n).\$ Then take logs \$\$ln(F(n)) = 2ln(F(n/2)) \$\$so that we have \$\$ G(n) = 2G(n/2)\$\$ where \$G(n) = ln(F(n)).\$ The solution to this is obvious: when \$n\$ doubles, so does \$G. \$ Thus it is a linear function \$\$ G(n) = an.\$\$Unwinding the transformations gives \$F(n) = e^an\$ and \$T(n) = frace^an4n.\$As for your question about whether the master method works, I am only familiar with that as a theorem about asymptotics, but when you take logs of the initial equation it immediately becomes a recurrence of that type and you can conclude that the log of the solution is \$Theta(n). \$.

3. Why is small work done always taken as \$dW=F cdot dx\$ and not \$dW=x cdot dF\$?

There are already several good answers. In this answer, we will just highlight a geometric argument.Even if you are unswayed by physical arguments, you should think twice about introducing non-geometric quantities

4. How to deduce \$

ablacdot (Atimes B) = (

ablatimes A)cdot B - (

ablatimes B)cdot A \$ using differential forms?

The product rule you are using is wrong.The correct rule is:If \$alpha\$ is a \$p\$-form, and \$beta\$ is a \$q\$-form, then \$\$d(alphawedgebeta) = dalphawedgebeta (-1)^p alphawedge dbeta\$\$In particular, if both \$alpha\$ and \$beta\$ are \$1\$-forms, then \$\$d(alphawedgebeta) = dalphawedgebeta (-1)^1 alphawedge dbeta = dalphawedgebeta - alphawedge dbeta\$\$Note that, as seen in the image, this equation is also found in the hint to the problem.

5. Non-trigonometric proof: \$|AD|^2=|AB|cdot |AC|-|DB|cdot |DC|\$.

Note the second step (\$|BD|cdot|DC|=|AD|cdot|DE|\$ is a well-known theorem (Intersecting Chords Theorem), so you might just as well refer to it instead of proving it yourself. Other than that, this proof is perfectly valid, and being very short, I can not see how it could be improved

6. Using "\$cdot\$â€ as variable placeholder

In this particular instance, I can see no particular benefit, but this notation has some uses. For example, assume that \$f\$ is a function of two variables, and we want to compute the Laplace transform with respect to the second variable. Then \$\$mathcal L(f(x,cdot))\$\$ is a convenient way to write it.

7. \$ab=3\$ and \$a^2 b^2=7\$. Find \$acdot b\$

Although above solutions are preferable, there is always the obvious approach of setting \$a = 3 - b\$, and then solve the polynomial

8. Motivation for the ring product rule \$(a_1, a_2, a_3) cdot (b_1, b_2, b_3) = (a_1 cdot b_1, a_2 cdot b_2, a_1 cdot b_3 a_3 cdot b_2)\$

It is isomorphic to the ring of matrices\$\$ leftbeginbmatrixa_1&a_30&a_2endbmatrix,middle|,a_1, a_2,a_3in mathbb R

ight \$\$It's a semiprimary ring whose Jacobson radical is the subset with \$a_1=a_2=0\$. The Jacobson radical is nilpotent, and \$R/J(R)congmathbb Rtimesmathbb R\$. Here is a list of more properties of such a ring.This sort of ring is fairly famous, and has nice interpretations. One of them is that if you select a chain of subspaces \$0

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