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

Euler's Identity

One of the simplest and most elegant equations is Euler's Identity:
e^πi + 1 = 0.

It states that Euler's number e (which is equal to roughly 2.718...) to the power of i*π (taken at roughly 3.14...) is equal to -1. This equation is used to derive cyclotomic integers which are used in Kummer's proof for Fermat's Last Theorem for regular primes.

To be fair, for many people, xⁱ does not have a clear value. What does it mean to put an exponent to an imaginary power? In mathematics, it is ok if the details are not intuitive as long as they are logically consistent. The Maclaurin Series, for example, can be used to define exponents to a complex power (see here). Newton did something similar when he generalized the Binomial Theorem to include complex powers (see here for details)

Theorem: Euler's Identity is the equation: e^πi = -1

This equation derives directly from Euler's Formula:
e^ix = cos x - isin x [See here for proof]

We get:
e^πi = cos(π) + isin(π) = -1 + i(0) = -1. [Review of e (see here), sin and cos (see here), i (see here), and π (see here)]

QED

Corollary: e^πi + 1 = 0

This directly follows from above.

QED