Firefox3 RC1

It's coming in 1 week or 2.
And after that, there will be RC2, and then the golden release!

Shutting down computers

run -> command.com
shutdown -i
And then type in comment, timeout, etc.
You can shut down ANYBODY in the local network

HCE Class

Yeah, so we went to this teachingsexualhealth.ca website.
This is in the front page:



Lol.....

M$ Frontpage 2000

I to this day, still do not understand why the school teaches how to make website with it.
Frontpage provides horrible, non-standard compliant HTML code.
Its themes are really horrible too; crappy gif graphic and looser design.
It is based on IE5 and thus does NOT properly display CSS2.1 in the WYSIWYG editor.

In general, it sucks. Period.

Snowing in April?

I am surprised too.
Here, in Victoria, B.C.?
I mean, it hardly even snows throughout the winter!
Ugh stupid global warming + extreme climates.

Some boarder suspended

One of our fellow gr8 boarders got suspended for a few days.
Not much to say.
Reasons:
- theft
- major late homework in various subjects
- skipping classes (fake sickness and then lie to nurse)
- more

He will be coming back next Monday

Fibonacci Numbers

You know the sequence:
1, 1, 2, 3, 5, 8, 13, 21, 34, ...

Where each number is the sum of the 2 previous ones and the first 2 numbers are 1's.
So, I was curious if there is a formula for finding the nth Fibonacci number...
Here is what I found.

There is 1 rule that will guide you through the process (contact me if you want the prove)

if F(n) = a * F(n-1) + b * F(n-2)

and that c and d are 2 solutions to the quadratic equation x^2 = ax + b
then for any integer n, F(n) = r * c^n + s * d^n where r,s are real numbers

Applied to this case
(1 + √5)/2 and (1 - √5)/2 are solutions for x^2 = x + 1
so we have

F(n) = r * ((1 + √5)/2)^ n + s * ((1 + √5)/2)^n

since f(1) = f(2) = 1

r * (1 + √5)/2 + s * (1 -√5) / 2 = r * (3 + √5)/2 + s * (3 - √5)/2 = 1
r + s = 0 //subtract f(1) from f(2)
s = -r

r * (1 + √5)/2 - r * (1 - √5) / 2 = 1
r * (1 + √5 - 1 + √5)/2 = 1
r * √5 = 1
r = 1 / √5

Thus, the formula is:
F(n) = (1/√5)((1+√5)/2)^n - (1/√5)((1-√5)/2)^n

or to simplify things...

F(n) = (1/√5)(φ^n) - (1/√5)((1-φ)^n)
Where φ = Golden Ratio = (1+√5)/2