Recent content by klite

b^2 = a^2 + c^2 - 2ac(cos B)

It depends on what you like. Observational Astrophysicists: extracts physical information from astronomical observations which can be directly compared with the models, and uses theoretical models to suggest unambiguous observational tests. Theoretical Astrophysicists: use theoretical...

What leads us to believe that there is a black hole in the center of the milky way? If there is one, why is the center of galaxies always depicted as being a bright sphere? Is it a rotating black hole, and how would've it formed?

You mean a black hole or the singularity? A black hole: the more mass it has, the more space it takes up (I think you mean how much space it takes up). For example, a black hole with the mass of our sun would be about 2 miles in radius. A typical black hole has a ten solar mass, and therefore a...

I remember reading that somewhere, you're right. Sorry for misinforming :blushing:

To answer both your first question, and this one: Yes, you can work for NASA while being an astrophysicist/theoretical physicist. They employ people from a variety of professions, not just aerospace engineers and astronauts. Look: http://astrophysics.gsfc.nasa.gov/...

Even if you are just under c, you will still be unable to escape (the escape velocity of a black hole is c+). Hover there as long as you want, but you will still slowly be hovering toward the black hole. Even if you slow down the time it takes to get to the singularity, you'd still reach it...

I appreciate your help, thank you!

I was looking at an older incorrect step of the problem...it's another mistake. I meant: \lim_{n \to \infty} \frac{2}{n} \left( \sum_{i=1}^n 1+\frac{4}{n^2} \sum_{i=1}^n i^2+\frac{4}{n} \sum_{i=1}^n i \right) \lim_{n \to \infty} \frac{2}{n} \left( n + \frac{4}{n^2}...

Alright, thank you! I noticed the mistake - wrong sums: \lim_{n \to \infty} \frac{2}{n} \left( \sum_{i=1}^n 1+\frac{4}{n^2} \sum_{i=1}^n i^2+\frac{4}{n} \sum_{i=1}^n i \right) = \lim_{n \to \infty} \frac{2}{n} \left( n + \frac{4}{n^2} \frac{2n(n+1)(2n+1)}{6} + \frac{4}{n}...

\lim_{n \to \infty} \frac{2}{n} \left( \sum_{i=1}^n 1+\frac{4}{n^2} \sum_{i=1}^n i^2+\frac{4}{n} \sum_{i=1}^n i \right) = \lim_{n \to \infty} \frac{2}{n} \left( n+\frac{2n(n+1)}{n^2} + \frac{2n(n+1)(2n+1)}{3n} \right) = \lim_{n \to \infty} \frac{2}{n} \left( n+2+2n+\frac{2n(2n^2+3n+1)}{3n}...

Oh, I see. Therefore the correct sum is (n(n+1)/2)^2? We were writing them down in class, and what I got from my teacher was: \sum_{i=1}^n i^2 = n(n+1)(2n+1)/6 \sum_{i=1}^n i^3 = (n(n+1)/2)^2 = n^2(n+1)^2/4 So where would I go from \lim_{n \to \infty} \frac{2}{n} \left(...

I took 2/n out in front, after which I took the denominators n and n^2 out in front, and multiplied them by 2/n, yielding 2n^2 out in front. Then I took both 4s out from the sums, leaving \sum_{i=1}^n 1 4\sum_{i=1}^n i and 4\sum_{i=1}^n i^2 inside the brackets, and 2n^2 outside the...

I have something similar to this... \lim_{n \to \infty} n^2 [\sum_{i=1}^n (1) + 4\sum_{i=1}^n (i) + 4\sum_{i^2}i^2] But your steps make it much more clearer, thank you. Afterwards I just use these equations for 1, i, and i^2, \sum_{i=1}^n 1 = 1*n = n \sum_{i=1}^n i = n(n+1)/n...

Ohh...I have no idea how to use the symbols, I'm new here :-p \lim_{n \to \infty} \sum_{i=1}^n (1+\frac{2i}{n})^2 \frac{2}{n} = \lim_{n \to \infty} 2/n [\sum_{i=1}^n (1+\frac{4i}{n})^2 \frac{4i^2}{n^2} = \lim_{n \to \infty} 2/n [\sum_{i=1}^n (1) + 4\sum_{i=1}^n (i) + 4\sum_{i^2} =...