Friday, February 24, 2006

Euler's Formula

Today's proof for Euler's Formula is based on the Taylor's Series. Euler's Formula is the equation:

eix = cosx + isinx

In a previous blog, I spoke about Euler's Identity which is derived from Euler's Formula. Richard Cotes was the first person to provide a proof but the great popularizer of this result was Leonhard Euler. Euler's Identity is used in the construction of cyclotomic integers which are used in Kummer's proof of Fermat's Last Theorem for regular primes.

Lemma 1: Maclaurin's Series

f(x) = f(0) + (x/1!)f'(0) + (x
2/2!)f''(0) + (x3/3!)f'''(0) + ... + (xn/n!)fn(0)

(1) Taylor's series gives us:
f(x) = f(a) + f'(a)(x-a) + [f''(a)/2!](x-a)2 + .... + [f(n)(a)/n!](x-a)n + ....
[See here for its proof]

(2) Now, if a=0, then we have:
f(x) = f(0) + (x/1!)f'(0) + (x2/2!)f''(0) + (x3/3!)f'''(0) + ... + (xn/n!)fn(0).

QED

Lemma 2: ex = 1 + (x/1) + (x2/2!) + (x3/3!) + ....

(1) Let f(x) = ex

(2) From the properties of ex [See here for details]:
f(x) = ex → f(0) = e0 = 1
f'(x) = ex → f'(0) =e0 = 1
fn(x) = ex → fn(0) = e0 = 1

(3) We know that ex is continuous since it has a derivative at each point. [See here for details of why this is true]

(4) By Lemma 1 above, we have:
ex = 1 + (x/1!)(1) + (x2/2!)(1) + ... (xn/n!)(1)

QED

Lemma 3: sinx = x - (x3/3!) + (x5/5!) - (x7/7!) + ...

(1) Let f(x) = sin x

(2) From the properties of sin, we know:

f(0) = sin(0) = 0 [See here for details if needed]

f'(x) = cos x → f'(0) = 1 [See here for proof if needed]

f''(x) = -sin(x) → f'(0) = 0 [See here for proof if needed]

f'''(x) = -cos x → f'(0) = -1

(3) From this, we see that:

fn(0) = 0 if n is even.
fn(0) = 1 if (n-1)/2 is even
fn(0) = -1 if (n-1)/2 is odd


(4) Putting this all together gives us:
sin x = (x/1)(1) + (x2/2!)(0) + (x3/3!)(-1) + ...

QED

Lemma 4: cos x = 1 - (x2)/2! + (x4/4!) - (x6/6!) + ...

(1) Let f(x) = cos x

(2) From the properties of cos, we know:

f(0) = cos(0) = 1 [Details if needed are found here]

f'(x) = -sin x → f'(0) = 0 [Details if needed are found here]

f''(x) = -cos(x) → f'(0) = -1

f'''(x) = sin x → f'(0) = 0

(3) From this, we see that:

fn(0) = 0 if n is odd.
fn(0) = 1 if (n/2) is even
fn(0) = -1 if (n/2) is odd

(4) Putting this all together gives us:
cos x = 1 + (x2/2!)(-1) + (x3/3!)(0) + (x4/4!)(1) + ...

QED

Theorem: Euler's Formula

e
ix = cos x + isin x

(1) From Lemma 2, we have:
eix = 1 + ix + (ix)2/2! + (ix)3/3! + ...

(2) Since i2 = -1 and i4 = 1, this gives us: (for details on i, see here)
eix = (1 - x2/2! + x4/4! + ...) + i(x - x3/3! + x5/5! + ...)

(3) From Lemma 4 above, we see that:
cos(x) = (1 - x2/2! + x4/4! + ...)

(4) From Lemma 3 above, we see that:
isin(x) = i(x - x3/3! + x5/5! + ...)

(5) Combining step #2 with step #3 and step #4 gives us:
eix = cos x + i sin x.

QED

Corollary: De Moivre's Formula

(cos x + isin x)n = cos(nx) + isin(nx)

Proof:

(1) (eix)n = einx

(2) (cos x + i sin x)n = cos(nx) + isin(nx) [Applying Euler's formula above]

QED

4 comments:

Mahndisa S. Rigmaiden said...

12 04 06

What a great blog you have here!!! I did a similar derivation of the Inverse tangent function using these same principles. But the explicit representation for Tan^-1 came out algebraically. See here. Meanwhile I will blogroll you:)

batuj said...

When compared to other silly blogs, yours seems to me like "God among insects". Keep the good work bro.

Michael Ejercito said...

This theorem resulted in a breakthrough in mathematics. Before, imaginary exponents were undefined. This theorem introduced imaginary exponents, as well as logarithms of numbers that are not positive and real.

Sanju said...

Though this proves the desired result, it doesn't come out of intuition like the previous simpler proof..