Cap'n Refsmmat

I think the old server's default timezone was Central time, whereas the new uses British Standard Time. (Unless the script specifically asks for GMT/UTC, it gets the default timezone). You'll also notice that times on the forum are no longer 17 minutes slow.

Yeah. They should catch up fairly quickly -- the most anyone should cache the DNS information is a day or two (even though I explicitly set the maximum cache time to 1 hour, some ISPs ignore it). I'm surprised that blog stats are in the tank though. I think the trouble is that a lot of the bots (like Google and other companies that crawl blogs) cache DNS for a reaaaally long time to save their servers some work, so it'll take them a little while. If stats are still low in a few days I'd start getting worried.

If you're running Windows, try downloading SpeedFan to see how hot the notebook is actually getting inside.

For C, I'd recommend Programming in C.

Yeah, sorry about that. I just couldn't find a good way to make it long and painful like everyone was hoping for.

It's the same problem we had before, so I made the same change I made before. Tell me if it gives more errors.

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I have a nasty tendency to spontaneously remember three or four notes from a song I've heard at some point in the past, which then of course means I have to sit around trying to figure out what song they belong to. Very irritating.

If you steal energy from it, you also steal mass.

The thing is that we don't understand the causes very well. There are various treatments that can attack the plaque buildups in the brain, but we don't know if the plaque buildups are a cause or a symptom of something else. Whereas with HIV we understand the virus and have no good way to fight it, with Alzheimer's we have an incomplete understanding and possible ways to fight it. It's a matter of time, really.

Remember Hawking radiation. Hawking radiation works by way of "virtual" particles -- particles and antiparticles that pop into existence for a tiny fraction of a second and then annihilate, leaving no overall change to the system. (It takes energy to make them and they give off the same energy when they annihilate.) It's possible that half of a particle/antiparticle pair could get sucked in to the black hole, leaving a particle behind with nothing to annihilate it. The energy it took to make that particle is extracted from the black hole's gravitational energy. At least that's how I understand it.

Black holes do not "eject" matter. Nothing can escape. The streams of particles you see exiting a black hole actually come from the accretion disk (which is outside the black hole's event horizon).

The world may have once been warmer, but that doesn't mean that the present warning isn't going to screw us over. That's my problem with the "it's happened before" argument: it might have happened millions of years ago, but it's still bad.

Lesson 6: The Chain Rule The Chain Rule helps you solve another important type of equation. This kind: [math]g(x) = 4(x^2 - 7)^6[/math] You have a choice: you could expand the equation out (which would take a very long time) and apply the other rules of differentiation, or you could use the chain rule. Let's break that above equation into two separate functions, a and b: [math]a(x) = 4x^6[/math] [math]b(x) = x^2 - 7[/math] That means we can redefine g(x) like this: [math]g(x) = a(b(x))[/math] For those of you who don't see how g(x) can be a(b(x)), try it. a(b(x)) is a(x) with b(x) stuck in wherever there's an x, like this: [math]g(x) = a(b(x)) = 4(b(x))^6 = 4(x^2 - 7)^6[/math] We simply inserted x2 - 7 where there was an x. How do you find the derivative? The rule says it's this: [math]\frac{d}{dx} g(x) = a'(b(x)) \cdot b'(x)[/math] So it's as simple as finding the derivatives of a(x) and b(x) using all the rules we learned before and substituting them back into the problem. [math]a'(x) = 24x^5[/math] [math]b'(x) = 2x[/math] Then we substitute those back in: [math]\frac{d}{dx} g(x) = 24(b(x))^5 \cdot 2x[/math] [math]\frac{d}{dx} g(x) = 24(x^2 - 7)^5 \cdot 2x[/math] From there you can simplify the equation any way you'd like. So the chain rule is as simple as breaking the equation down into parts. Try a few below. The answers are at the end of this post. [math]g(x) = 2(x + 4)^3 + 7x[/math] [math]q(x) = ((x + 4)^4)^2[/math] Answers: You could split g(x) into [imath]a(x) = x + 4[/imath] and [imath]b(x) = 2x^3[/imath], making [imath]g(x) = b(a(x)) + 7x[/imath]. You can safely leave the 7x sitting around and derive it by itself because it's being added, not multiplied. q(x) should be split into [imath]a(x) = x + 4[/imath], [imath]b(x) = x^4[/imath], and [imath]c(x) = x^2[/imath]. That makes [imath]q(x) = c(b(a(x)))[/imath]. How do you solve that? Easy. [imath]q'(x) = c'(b(a(x))) \cdot b'(a(x)) \cdot a'(x)[/imath]. Remember, it's "the derivative of the outer function, times the derivative of the inner function." You just have to apply the chain rule to the inner function to find its derivative. Remember, if you need help understanding any of this, you can just ask in our calculus forum.

There you go The second derivative of position is acceleration, so d''(t) = a(t) = 0.

I'm asking you as a Socratic question -- what do you think the second derivative of position is?

Hold up. If d(t) represents position, what would d''(t) (the second derivative) represent? Velocity, acceleration, or something else? That's very important for your answer. That looks right. I can't speak for the Gaussian elimination problem as I'm not familiar with it. Good luck.

Take a look at the Wikipedia article on the subject of Hawking radiation, which is believed to be the way black holes could "evaporate".

At least in the US, power companies are government-sanctioned monopolies, so they won't gain or lose market share.

What sort of water source do you use? (I have no idea what a lab might use.) If tap water is used at any point, remember that some municipalities add extra chlorine and such to their water during the summer to keep it from stagnating in the heat.

And here's the key point. My thought was that with sufficiently mainstream electric cars, all charging up at night, night-time will no longer be off-peak -- there'll be as much demand then as any other time. That clearly would not be the case unless there were a very large number of electric cars. (I'm not entirely sure how the demand from charging automobiles compares to the usual night-time demand, so I may just be making crap up. But that's my job.)

Because if electric cars become mainstream, overnight hours are no longer off-peak.

The question is whether our electrical distribution systems can handle the load.

Did you even read the entire article?