Recent content by kumusta

Changing the base of the logarithm from ##~x~## to that of natural logarithm, ##{~~~~}{ \small { \rm {log} }_x~N } = \frac { { \rm {log} }_e~N } { { \rm {log} }_e~x } = \frac { { \rm {ln} }~N } { { \rm {ln} }~x }~\Rightarrow~{ \small { \rm {ln} }~x~{\rm {log} }_x~N } = { \small { \rm {ln} }~N...

This is how I look at the solution of the problem. First, by replacing the ##~k^2-2k-6~## by ##~r~## in the first given quadratic equation with real roots and then solving for ##~x~## in terms of ##~k~## and ##~k~##, one gets the condition ##~k~\geq \pm\sqrt {r}## for the roots of ##~x~## to be...

I don't understand this thread. According to the OP Since ##{~~~~}x = \left \{ \begin{align} & \text {variable of integration } \nonumber \\ &~~ \text {and is therefore the} \nonumber \\ & ~\text {independent variable} \nonumber \end{align} \right \}~~\Rightarrow~## ##\left \{ \begin{align} &...

@chawala Thanks for the explanation.

I find this thread confusing. First, in post #1: The ##~x^{1/2}+x^{1/3}=12~## became ##~x^{1/6}=12x^{1/3}-1~##. Then, in post #4, the ##~x^{1/6}=12x^{1/3}-1~## became ##x^{1/6}=12x^{-1/3}-1## I do not know which of the two is the actual equation of interest.

If the problem is to find a real root of the equation (1)##\qquad~\sqrt {x+3} + \sqrt {x-3} - 2√(x+5) = 0## then clearly, ##~x~\geq~3##. But if the aim is to find the general root of (1), not just the real root, then obviously complex roots must be admissible..

If the equation referred to above is (1)##\qquad–~\sqrt {z + 3}~+ \sqrt {z - 3} = 2\sqrt {z + 5}~## that could have only real roots ##~z = x~\geq~3~##, then why does ##~z = x = 4~\gt~3~##give for the left-hand side of (1) ##\qquad{~~~~}–~\sqrt {4 + 3}~+ \sqrt {4 - 3} = –\sqrt 7 + 1 = –2.6 + 1...

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...

I don't know how and where to start. I'll just think for a minute.

I found the mistake in my solution. I don't know if you're interested in knowing where the mistake is.

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)...

I wonder if this drawing of the moving platform seen fom the top would help solve the problem in this thread.

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...