Tom Mattson

Not necessarily. Very smart physicists thought that the old velocity addition formula was correct for centuries. What is "the observed portion"? It's the same for every inertial frame.

FYI' date=' in relativity light and matter do follow the same velocity addition rule. But this rule is not the one you cite, as has been noted already. The rule is given by the following equation. [math']u^{\prime}=\frac{u+v}{1+\frac{uv}{c^2}}[/math] Note that when [imath]uv[/imath] is much less than [imath]c^2[/imath] the term [imath]\frac{uv}{c^2}[/imath] is much less than 1, and the equation is approximately given by the following. [math]u^{\prime}=u+v[/math] This is the simple rule you stated.

Can you show your work? That would make it easier to check!

You aren't just supposed to flip reactions and/or add them together, you also have to multiply by appropriate coeffcients. For instance you have to multiply the first reaction by 1/2 so that you get a "1" in front of C2H2.

In both cases you've got the electron spinning like a top' date=' but this view of spin was abandoned long ago for some very good reasons. For one thing there is the behavior of spin 1/2 particles under [imath']2\pi[/imath] rotations: the wavefunction picks up a negative sign. You have to rotate an electron through [imath]4\pi[/imath] radians to get back to the initial state. That's clearly not ordinary rotation! The fact is that there is no acceptable description of electon spin that relates to our ordinary concept of motion. So the magnetic moment of the electron really is a counterexample to your claim that all of magnetism is just a consequence of moving charges. I know you know this for everyone else: An electric field in one frame is a magnetic field and an electric field in another frame. You can't transform away a pure electric field entirely, and neither can you transform a pure magnetic field entirely.

OK, I can add to this... They may not be the top 5 of all time, but they're my favorites. GoodFellas Shaun of the Dead Pulp Fiction Rocky The Empire Strikes Back

GoodFellas. Period.

No, it's not a dilemma. It's trivial. Do long division, as you would with any improper rational function. Then integrate the polynomial and the proper rational function and you're done.

Yeah, but it's an easy problem. You do it by parts: Let [imath]u=\ln(x)[/imath] and [imath]dv=x^5dx[/imath]. Same type of thing, you do it by parts. Let [imath]u=\ln(x+3)[/imath] and [imath]dv=x^5dx[/imath].

I bet the thing you're missing is the fact that [imath]\ln(3x)=\ln(x)+\ln(3)[/imath]. If you split the natural log up that way you get one integral that can be done by the power rule and another one that can be done by parts.

This exact same question could be asked of any career choice. People don't get jobs to be entertained, they do it to earn a living. And if you're lucky enough to earn a living doing something you enjoy, then more power to you. But no one enjoys everything about their work, no matter what they do. But you grow up, and suck it up, just like everyone else.

Yes, write down a sinusoidal function to describe each oscillator and add them up.

Well there's certainly an easier way. The basis [imath]\{\sin(x)' date='\cos(x)\}[/imath'] spans the exact same vector space as the basis [imath]\{exp(ix),exp(-ix)\}[/imath]. Convert your trig functions to complex exponentials and have a ball.

Ha. When I read the title of this thread I thought that an "excuse for massiveness" was something along the lines of I'm not fat, I'm just big boned.

None of these sets are orthogonal. Set (a): The inner product isn't defined on [imath](-\infty,\infty)[/imath] for any pair of these functions. Set (b): All of the functions in this set have mixed symmetry (except the first one, which has even symmetry). Set ©: The functions [imath]f(x)=x[/imath] and [imath]f(x)=4x^3[/imath] are both orthogonal to [imath]f(x)=e^{-ax^2}[/imath] on [imath](-\infty,\infty)[/imath], but they aren't orthogonal to each other.

The inner product for those two functions on [imath](-\infty' date='\infty)[/imath'] isn't even defined, because the limit you obtain when you evaluate the resulting integral doesn't exist. Is this question from the same instructor as the Wronskian question? Well then they aren't orthogonal! The inner product of two orthogonal functions is zero. I wouldn't expect them to be orthogonal anyway because their product has mixed symmetry (or no symmetry, if you prefer).

You've already done it. That's just what I would have done. How did you get that for [imath]W(f,g)[/imath]? The Wronskian clearly vanishes on [imath]-1<x<0[/imath] and on [imath]0<x<1[/imath]. Having said that, the Wronskian doesn't exist at [imath]x=0[/imath] because [imath]f[/imath] is not differentiable there. So the problem statement is wrong when it claims that [imath]W(f,g)\equiv 0[/imath] everywhere in [imath][-1,1][/imath].

I do agree that this has been overcomplicated' date=' but I don't see how your suggestion helps. It isn't. All you have to do is let [imath]u=\sqrt[6]{x}[/imath] in the original integral. Then you can do a partial fraction expansion, as caseclosed had initially surmised. That solution seems a whole lot more obvious to me, but to each his own...

It comes from you entering the expression into your calculator wrong. You put [imath]x^3+2[/imath] in for the numerator.

Not sure how this came about. In your previous step you should have inserted [imath]\cot^2(x)=\csc^2(x)-1[/imath] to obtain the following. [math]\int\csc^3(x)dx=-\csc(x)\cot(x)-\int\csc^3(x)dx+\int\csc(x)dx[/math] It all works out straightforwardly from there.

That is the correct method, but you should not have gotten 0. You must have made a small sign error.

Try integrating by parts. Let [imath]u=\csc(x)[/imath] and let [imath]du=\csc^2(x)dx[/imath].

A few years ago I saw the band Peaceful Knievel at Tobacco Road in Hell's Kitchen. For those of you who don't know that band is made up of singer Mike Farris and remnants of the Black Crowes. I bumped into Audley Freed at the bar. It was crowded but he turned around and looked right at me. Not wanting to look uncool, I deftly blamed the collision on someone else.