Friday, February 16, 2007

Derivation of the most beautiful identity

I want to share this with you. The derivation of one of my favourite identites in mathematics so far:

Let f(x) = cos x = a + bx + cx^2 + dx^3 + ex^4 + fx^5 + ...

(don't mix up e in ex^4 with e = 2.7182821828...; in this case e is just any constant)

so

f(0) = cos 0 = 1 = a, thus a = 1

f'(x) = -sinx = b + 2cx + 3dx^2 + 4ex^3 + 5fx^4 + ...
f'(0) = -sin0 = 0 = b; thus b = 0

f''(x) = -cosx = (2*1)c + (3*2)d + (4*3)ex^2 + (5*4)fx^3 + ...
f''(0) = -cos0 = -1 = (2*1)c; thus c = -1/(2*1) = -1/2!

f'''(x) = sinx = (3*2*1)d + (4*3*2)ex + (5*4*3)fx^2 + ...
f'''(0) = sin0 = 0 = (3*2*1)d; thus d = 0

f''''(x) = cosx = (4*3*2*1)e + (5*4*3*2)fx + ...
f''''(0) = cos0 = 1 = (4*3*2*1)e; thus e = 1/4!

...

and so on.

Therefore substuting the constants into the original polynomial equation,

(i) cos x = 1 - x^2/2! + x^4/4! - x^6/6! + x^8/8! + ...

You can check this by substitution a value in x, such as 1 (radians).

Similarly, you can work out an expansion for

(ii) sin x = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! + ...

(iii) e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5! + ...

Now using expansion (iii), you can expand e^(ix), where i = √(-1)

(iv) e^(ix) = 1 + ix + (ix)^2/2! + (ix)^3/3! + (ix)^4/4! + (ix)^5/5! + ...

= 1 + ix - x^2/2! - i(x)^3/3! + x^4/4! + i(x)^5/5! + ...

Separating the imaginary parts and real parts,

e^(ix) = (1 - x^2/2! + x^4/4! - x^6/6! + ...) + i(x - x^3/3! + x^5/5! - x^7/7! + ...)

= cos x + isin x

Wow!

so you get Euler's identity

e^(ix) = cos x + isin x


And a special case of it, is when x = π

e(iπ) = cosπ + isinπ = -1 + i(0) = -1

Rearrange, then you get

e^(iπ) + 1 = 0

an equation connecting the fundamental numbers i, π, e, 1, and 0 (zero), the fundamental operations +, ×, and exponentiation, the most important relation =, and nothing else.

Wow! Wow! Aren't you excited? (Nerd... if you actually read through all of the above and understood it.)

Gauss is reported to have commented that if this formula was not immediately obvious, the reader would never be a first-class mathematician.

It doesn't stop here though.

This opens a lot of branches in mathematics. One example is the hyperbolic function, which links circular and exponential functions - something you usually think as very separate branches of mathematics.

Friday, February 09, 2007

Fried rice kung fu!


Yum...

If only I could produce that fried rice everyday, I could save money at university (+ avoid eating crappy food). I could also sell my fried rice. I could take over the university canteen. Hahaha

I better start cooking now! Hai ya! Hai ya! Hai ya!



Video from averagebetty.

Get the fried rice recipe from the fried rice demon.

Thursday, February 08, 2007

Series


Can you see that 1^3 + 2^3 + 3^3 + .... + n^3 = (1 + 2 + 3 + ... + n)^2?