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.

5 comments:

Angel said...

Well, it wasn't "immediately obvious", but I actually struggle with the IB Maths HL syllabus, so I don't really seek to become a 'first-class mathematician'. However, I did understand it and found it real nice. I'd always used it direct from my formula booklet without the slightest idea of where it came from.

Gosh, I hope not everybody at Oxford is as good at maths as you, or I'll be in trouble :P

A fellow Oxford acceptee seeking to fulfill a (nice) 39 IB offer for Economics and Management at Merton.

Angel (from Mexico City)

Anonymous said...

It is quite scary how excited my maths teacher was when he showed us this!

Anonymous said...

That was a very nice derivation for Euler's equation.

Hope to see you all in October!!

Christian (Denmark/Sweden/Mexico)

momes said...

Ah you can also derive Euler's identity from the taylor polynomial for e^x

e ^ (x+iy) = e^x + e^(iy)
= e ^(x) ( 1 + iy + (iy)^2/2 + (iy)^3/3! ......)

then using sinx and cosx taylor polynomials , the above expression simplifies to cos y + isin y!

A cambridge acceptee looking to fulfill a 39 IB offer as well =)

momes said...

You can also derive this relation through the taylor polynomial of e^x!