Recent content by Nexus[Free-DC]

There's an old problem concerning an old drunken person who is so sloshed he can't walk properly: he's got a 50/50 chance of stepping a metre forwards; otherwise he moves a metre backwards. It can be shown that he winds up where he started infinitely many times and that his average displacement...

Spot on, Cragwolf.

Well, you learn something new every day.

Yes, it is. Chemically it's identical to water and is not radioactive.

I wonder what the Jehovah's Witnesses' policy is on storing your own blood in case you ever need it for a transfusion.

That zillion-digit prime is one of a specific type of prime, called a Mersenne prime. Mersenne numbers are of the form 2^k-1 where p is itself some other prime number, and it's easier to verify the primeness of a Mersenne number than it is be to check other numbers of similar magnitude. Mersenne...

There's a number of reasons why dividing anything by zero is absurd, but here's a simple argument: Suppose there was such a number as 1/0, call it k. Now suppose we find 1/(0+0). This must be equal to k/2 since the denominator in the second fraction is twice the denominator of the first...

The major influence on Physics will be what is has always been: Mathematics.

Let's look at this logically. We have postulated one thing; that God can do anything. Thus he can lift all rocks. It follows that the uber-heavy rock in question is not in the set of all rocks. So for God to create this rock would require the breaking of the laws of logic. Now we've got two...

Hey there Ygmaince241. With the union and intersection symbols it's sometimes a bit hard to remember which is which. I always think of it this way: \cupnion and i\captersection.

I dunno. Sometimes if you have a polynomial you can spot one factor real easy, but the others are hard to find. It becomes easier if you divide through by the factor you've found.

Okay, suppose x=a/b, where a,b are integers. Assume that the fraction is reduced, ie. gcd(a,b)=1. Then x^x= (a/b)^(a/b)=(\frac{a^a}{b^a})^\frac{1}{b} But gcd(a^a,b^a)=1, and therefore (\frac{a^a}{b^a})^\frac{1}{b} is irrational. It follows that if x^x is an integer, then either x is an...

Okay, this is a combination of the chain rule and implicit differentiation. The first thing to do is let a=u(t)+iv(t) Now let f(x)=\sqrt{a^2} and the derivative becomes \frac{df}{dt}=\frac{df}{da}\frac{da}{dt} You should be able to proceed from there. If not, yell out.

It's the limits of integration that count here. Say you're trying to integrate a hard function, but there's a neat little trick for working out the integral from zero to infinity. That trick probably won't help you if you're integrating from, say, 1 to 5.729.

We exist. Deal with it.