Thursday, July 14, 2011

Euler's Gem 2

Here is a sketch of a second derivation of Euler's famous formula:

e = cosθ + i sinθ

as presented by William Dunham in his book Euler, The Master of Us All. First post here.

The first step is to recall a standard trigonometric substitution in calculus:


y = sin x
x = sin-1 y

√(1 - y2) = cos x

We're interested in the integral:

∫ dy / √(1 - y2)

Substituting with x we see that:

dy = cos x dx

And the integral is

∫ (1/cos x) cos x dx = ∫ dx = x
x = ∫ dy / √(1 - y2)

Now Euler makes a complex change of variable:

y = iz
x = ∫ dy / √(1 - y2)
= ∫ i dz / √(1-(iz)2)
= i ∫ dz / √(1 + z2)
= i ln [√(1 + z2) + z]

The last step is another standard result from calculus which I will assume for the time being (more here).

Undo the substitution:

z = y/i = sin x / i
z2 = -sin2 x
√(1 + z2) = √(1 - sin2 x)
= cos x

x = i ln (cos x + sin x / i)

We will use two identities involving i:

u / i = - i u
1 / (cos u - i sin u) = (cos u + i sin u)

(For the second one, see the previous post). Now:

x = i ln (cos x + sin x / i)
x = i ln (cos x - i sin x)
ix = - ln (cos x - i sin x)
= ln [ 1 / (cos x - i sin x) ]
= ln (cos x + i sin x)

Just eponentiate:

eix = cos x + i sin x

Wow, again!

No comments: