From Wikipedia, the free encyclopedia
User at Wikipedia.
Major Contributions
Some pages I've invested a lot in:
Differentiable manifold
Proofs of Fermat's little theorem
John Derbyshire
Ayla (Chrono Trigger)
Frog (Chrono Trigger)
Sonic Heroes
Minor Contributions
Mathematical
Trigonometric function
Hyperbolic function
Taylor series
Taylor's theorem
Table of integrals
Derivative
Partial fraction
Metamathematics
Coversine
Critical line theorem
Fermat's little theorem
Limit (mathematics)
Domain (mathematics)
Laplace transform
173 (number)
Amortization (business)
Scientific
Carl Friedrich Gauss
Chaos theory
Wave equation
P-branes
Great White Shark
Miscellaneous
Chrono Trigger
Melchior (Chrono Trigger)
English Language
Crono
Serge (Chrono Cross)
Kid (Chrono Cross)
Lynx (Chrono Cross)
Luccia
Harle
Creations
The pages I myself have made (from ye olde scratch) are--in chronological order:
Haversine
This Page
Heliotrope
Magnometer
Belthasar (Chrono Trigger)
12000 B.C. (Chrono Trigger)
Problems
To show the probability that two integers chosen at random are relatively prime is
6
π
2
{\displaystyle {6 \over \pi ^{2}}}
.
Proof: It is sufficient to show
∑
n
=
1
∞
1
n
2
=
π
2
6
{\displaystyle \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
α
1
,
α
2
,
.
.
.
,
α
r
{\displaystyle \alpha _{1},\alpha _{2},...,\alpha _{r}\,}
are the roots of a polynomial p(z), then we may write
p
(
z
)
=
(
1
−
z
α
1
)
.
.
.
(
1
−
z
α
r
)
{\displaystyle 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
z
=
1
−
z
2
3
!
+
z
4
5
!
+
.
.
.
+
(
−
1
)
n
z
2
n
(
2
n
+
1
)
!
{\displaystyle {\sin z \over z}=1-{z^{2} \over 3!}+{z^{4} \over 5!}+...+{(-1)^{n}z^{2n} \over (2n+1)!}}
Well we also know we can rewrite sin(z)/z in terms of its roots to be:
(
1
−
z
π
)
(
1
+
z
π
)
(
1
−
z
2
π
)
(
1
+
z
2
π
)
(
1
−
z
3
π
)
(
1
+
z
3
π
)
.
.
.
(
1
−
z
k
π
)
(
1
+
z
k
π
)
.
.
.
{\displaystyle \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
3
!
=
1
π
2
∑
n
=
1
∞
1
n
2
→
π
2
6
=
∑
n
=
1
∞
1
n
2
Q.E.D.
{\displaystyle {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.}}}