Recent content by cwbullivant

## F(x) = P(0 \leq x) = 1, x \geq 0, 0 otherwise##?

Sloppiness is exactly the issue here. Anyhow, I make the appropriate substitution, with u = nt, du = n dt, and come out to: $$ \int_{-infty}^{nx} \it{f}(u) du $$ This has eliminated the linear dependence on n from the original statement and reduced the problem to an integral with just a...

Homework Statement Let $$ \it{f}(x) $$ be a probability density function. Now let Xn have the density: $$ \it{f}_{n}(x) = n\it{f}(nx) $$ Determine whether or not Xn converges in distribution to zero. (this is the verbatim statement, there is no additional information given)Homework...

Of course; I should have noticed that sooner, I was a bit too fixated on the parentheses there. Thanks for the assistance.

I don't see where the ## \frac{d}{dt} u^2 ## term is in the initial expression. The initial one as written is: ## \frac{d}{dt}(\frac{m\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} = (\frac{m\vec{u}u\frac{du}{dt}}{c^2(1-\frac{u^2}{c^2})^{3/2}} +...

The fact that the one of the time derivatives is ## \frac{du}{dt} ##, and the other is ## \frac{d\vec{u}}{dt} ## doesn't make a difference with regard to pulling them out? I'd think the unit vector in the u direction wouldn't be time independent in general (so ## \frac{d\vec{u}}{dt} =...

I get a two term expression when attempting this. Namely: ## \frac{d}{dt}(\frac{m\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) = \frac{m\vec{u}u\frac{du}{dt}}{c^2(1-\frac{u^2}{c^2})^{3/2}} + \frac{m\frac{d\vec{u}}{dt}}{\sqrt{1-\frac{u^2}{c^2}}} ## Take the dot product with u on both terms and it...

Homework Statement ## \frac{d}{dt}\gamma(t)\vec{u(t)} ## Homework Equations See above The Attempt at a Solution This comes from trying to verify a claim in Chapter 12 of Griffiths Electrodynamics, 4th. edition (specifically Eq. 12.62 -> Eq. 12.63, if anyone has it on hand). I would have...

Homework Statement Using 26.40, show that a pseudovector p and antisymmetric second rank tensor (in three dimensions) A are related by: $$ {A}_{ij} = {\epsilon}_{ijk}{p}_{k} $$ Homework Equations 26.40: $$ {p}_{i} = \frac{1}{2}{\epsilon}_{ijk}{A}_{jk} $$ The Attempt at a Solution This...

The general form is given as: ## x(t) = x_0 + at ## ## y(t) = y_0 + bt ## ## z(t) = z_0 + ct ## And ## \vec R (t) = \langle x(t), y(t), z(t) \rangle ## Since the line starts at the origin, the initial points (x0, y0, z0) are all zero. The end points are (X, Y, Z). So then if t goes from 0 to...

I hit "enter" too soon. It's been edited to add the rest.

Typos abound in the above post, sorry about that. It was supposed to be: ## \vec R = <t, t, t>, \frac{d\vec R}{dt} = <1,1,1> ## So if the path is from (0,0,0) to (X,Y,Z), if I were to do it by integrating with dt, would I need to use bounds on t from 0 to, say, T?

I'm not sure exactly what would be an easier curve than a straight line, but here's how I set up the straight line using the formula you gave me: ## \vec F = <1,1,1> ## (since it's a straight line, x(t), y(t), z(t) are all t from 0 to 1) ##\frac{d\vec R}{dt} = <1,1,1> ## (the derivative of a...

That's true, and since F is conservative, it shouldn't matter which of those infinite paths I take. That's convincing for why I needed to choose a particular path, but after that, is the number that I got when using a straight line from the origin to (1,1,1) correct? Based on that, the gradient...

Quite right, I did mean the last component to be ## cy\hat y ##. When you say to specify a path, do you mean I need to specify a specific path, like (0,0,0)->(1,1,1), and can't do a generic path from the origin to some (x,y,z) (it's been awhile since I last did vector calc)? If so, say I...