Recent content by CTS

http://mathworld.wolfram.com/Indeterminate.html

Induction is a pretty simple way to show that the sequence converges: A1 = 2^(1/2) < 2 Assume An < 2 A(n+1) = (An + 2)^(1/2) < (2 + 2)^(1/2) = 2 By the principle of mathematical induction, An < 2 for all n.

By the prime number theorem, the nth prime is about n*log(n).

The information in this thread is great! It's nice for us students to hear what being/becoming a mathematician is like from those who have had the experience. Thanks for giving your advice, stories, etc.

A google search popped up http://www.cut-the-knot.org/do_you_know/numbers.shtml , maybe some of that would be interesting.

Then to show that the almost all real numbers are irrational, show the rationals have measure zero. Start with rationals h/k with 1<h<k. Enclose them in the intervals (h/k - \epsilon/k^3, h/k + \epsilon/k^3). Because h and k are not necessarily relatively prime, we are duplicating many...

Math is definitely a fun subject to study and combines well with other majors. I'm currently in college majoring in math/economics and the combination is great. It also overlaps well with physics and computer science. The structure of a math degree is pretty much what you make of it...

Stopple's "A Primer of Analytic Number Theory" is a good introduction to analytic number theory for those without complex analysis. However, it would probably help to have a good background in elementary number theory before starting it.

Does looking at a generating function help at all? It's easy to show that the numbers of the sequence, a_n, are given by the coefficients (mod10) of x_n in the power series of \frac{3x^3-2x^2-2x+2}{1-x-x^2-x^3-x^4} = 2 + 0x + 0x^2 + 5x^3 + 7x^4 + 12x^5 + 24x^6 + 48x^7 + 91x^8 + 175x^9 +...

Doesn't 2 * 3 * 5 * 7 have \tau (2 * 3 * 5 * 7) = \tau (2) * \tau (3) * \tau (5) * \tau (7) = 2 * 2 * 2 * 2 = 16 positive factors? I think what you need to look at is the factors of 15: 1, 3, 5, and 15 15 = 1 * 15 15 = 3 * 5 The first case gives \tau (p^{1-1} *...

Although 0! = 1 by definition only, here's a reason it makes sense using the recursive definition of factorial: n! * (n + 1) = (n + 1)! n! = (n + 1)! / (n + 1) 0! = (0 + 1)! / (0 + 1) = 1! / 1 = 1

e^ix = cos x + i sin x

Consider the infinite sums: e^x = 1 + x/1! + x^2/2! + x^3/3! + ... sin(x) = x/1! - x^3/3! + x^5/5! -x^7/7! + ... cos(x) = 1 - x^2/2! + x^4/4! -x^6/6! + ... e^ix = 1 + ix/1! + (ix)^2/2! + (ix)^3/3!... Notice the patterns with the powers of i and rearrange the terms to see how they relate...

http://mathworld.wolfram.com/MillsConstant.html

1.) (10k + 5)^2 = 100k^2 + 100k + 25 = 100(k^2 + k) + 25 2.) (2n)^3 -4*(2n) = 8n^3 - 8n = 8n(n-1)(n+1) = 8(n-1)n(n+1) One of (n-1), n, (n+1) must be divisible by three, and at least one of them must be even, therefore 48|(2n)^3 -4*(2n) 3.) (2n + 1)^2 = 4n^2 + 4n + 1 = 4n(n + 1) + 1...