User:Guardian of Light

User at Wikipedia.

Major Contributions
Some pages I've invested a lot in:


 * 1) Differentiable manifold
 * 2) Proofs of Fermat's little theorem
 * 3) John Derbyshire
 * 4) Ayla (Chrono Trigger)
 * 5) Frog (Chrono Trigger)
 * 6) Sonic Heroes

Mathematical

 * 1) Trigonometric function
 * 2) Hyperbolic function
 * 3) Taylor series
 * 4) Taylor's theorem
 * 5) Table of integrals
 * 6) Derivative
 * 7) Partial fraction
 * 8) Metamathematics
 * 9) Coversine
 * 10) Critical line theorem
 * 11) Fermat's little theorem
 * 12) Limit (mathematics)
 * 13) Domain (mathematics)
 * 14) Laplace transform
 * 15) 173 (number)
 * 16) Amortization (business)

Scientific

 * 1) Carl Friedrich Gauss
 * 2) Chaos theory
 * 3) Wave equation
 * 4) P-branes
 * 5) Great White Shark

Miscellaneous

 * 1) Chrono Trigger
 * 2) Melchior (Chrono Trigger)
 * 3) English Language
 * 4) Crono
 * 5) Serge (Chrono Cross)
 * 6) Kid (Chrono Cross)
 * 7) Lynx (Chrono Cross)
 * 8) Luccia
 * 9) Harle

Creations
The pages I myself have made (from ye olde scratch) are--in chronological order:


 * 1) Haversine
 * 2) This Page
 * 3) Heliotrope
 * 4) Magnometer
 * 5) Belthasar (Chrono Trigger)
 * 6) 12000 B.C. (Chrono Trigger)

Problems
To show the probability that two integers chosen at random are relatively prime is $${6\over \pi^2}$$.

Proof: It is sufficient to show $$\sum_{n=1}^{\infty}{1\over n^2} = {\pi^2\over 6}$$. When we have a polynomial with constant term one, we may rewrite it in factored form as follows: If $$\alpha_1,\alpha_2,...,\alpha_r\,$$ are the roots of a polynomial p(z), then we may write $$p(z)=\left(1-{z\over\alpha_1}\right)...\left(1-{z\over \alpha_r}\right)$$.

Now examine the power series for the function sin(z)/z. $${\sin z\over z}=1-{z^2\over 3!}+{z^4\over 5!}+...+{(-1)^nz^{2n}\over (2n+1)!}$$

Well we also know we can rewrite sin(z)/z in terms of its roots to be:

$$\left(1-{z\over \pi}\right)\left(1+{z\over \pi}\right)\left(1-{z\over 2\pi}\right)\left(1+{z\over 2\pi}\right)\left(1-{z\over 3\pi}\right)\left(1+{z\over 3\pi}\right)...\left(1-{z\over k\pi}\right)\left(1+{z\over k\pi}\right)...$$

If we examine the quadratic term in each we find that:

$${1\over 3!} = {1\over \pi^2}\sum_{n = 1}^{\infty}{1\over n^2}\rightarrow {\pi^2\over 6}=\sum_{n = 1}^{\infty}{1\over n^2}\text{Q.E.D.}$$