We can define trigonometric functions as power series, using Taylor’s Theorem.

Euler’s Number

e^{x} = n = 0 \sum \infty \frac{x ^{n}}{n !} = 1 + x + \frac{x ^{2}}{2} + \frac{x ^{3}}{6} + \dots

Cosine

cos (x) = n = 0 \sum \infty \frac{( - 1 ) ^{n}}{( 2 n )!} x^{2 n} = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots

sin (x) = n = 0 \sum \infty (- 1)^{n} \frac{x ^{2 n + 1}}{( 2 n + 1 )!} = x - \frac{x ^{3}}{6} + \frac{x ^{5}}{120} - \dots

e^{i x} = cos (x) + i sin (x) \forall x \in C

A trigonometric polynomial of degree $n$ can be written in real form

f (x) = a_{0} + k = 1 \sum n (a_{k} cos (k x) + b_{k} sin (k x))

where

{a_{0}, a_{1}, \dots, a_{n}} {b_{0}, b_{1}, \dots, b_{n}} \in C \in C

and in complex exponential form:

f (x) = k = 1 \sum n c_{k} e^{ik x} c_{k} \in C

where equivalence is shown by Euler’s Formula. Indeed,

{e^{i 0 x} = e^{0} = 1} {e^{i x}, e^{- i x}} {e^{2 i x}, e^{- 2 i x}} ⟷ {1} ⟷ {cos (x), sin (x)} ⟷ {cos (2 x), sin (2 x)}

If $a_{0} = c_{0}$ , and

a_{k} = c_{k} + c_{- k} b_{k} = c_{k} - c_{- k}

then

c_{k} e^{ik x} + c_{- k} e^{- ik x} = c_{k} (cos (k x) + i sin (k x)) + c_{- k} (cos (- k x) + i sin (- k x)) = (c_{k} + c_{- k}) cos (k x) + (c_{k} + c_{- k}) sin (k x)

and the proof is equivalent.

In the space of integrable functions over $[- π, π]$ with inner product given by $⟨ f, g ⟩ = \int_{- π}^{π} f \cdot \overline{g} d x$

Two functions $f, g$ are orthonormal if

$⟨ f, f ⟩$ is not always equal to 1. If

⟨ f, g ⟩ = {01 f \neq = g f = g

Then $\int_{- π}^{π} ∣ f ∣^{2} = 0$

The set

{1/ 2 π, cos (n x) / π, sin (n x) / π}

is orthonormal.

Proof: Trivial.

The set

{\frac{1}{2 π} e^{in x}}_{n \in Z}

form an orthonormal list.

Proof: Show that $\overline{e^{in x}} = e^{- in x}$ . The rest is trivial.