Euler's Formula

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

e^ix = 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) + (x²/2!)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: e^x = 1 + (x/1) + (x²/2!) + (x³/3!) + ....

(1) Let f(x) = e^x

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

(3) We know that e^x 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:
e^x = 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:
e^ix = 1 + ix + (ix)²/2! + (ix)³/3! + ...

(2) Since i² = -1 and i⁴ = 1, this gives us: (for details on i, see here)
e^ix = (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:
e^ix = cos x + i sin x.

QED

Corollary: De Moivre's Formula

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

Proof:

(1) (e^ix)ⁿ = e^inx

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

QED