Complex Form of Fourier Series

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Let the function $f (x)$ be defined on the interval $[- π, π] .$ Using the well-known Euler's formulas

\cos φ = \frac{e^{i φ} + e^{- i φ}}{2}, \sin φ = \frac{e^{i φ} - e^{- i φ}}{2 i},

we can write the Fourier series of the function in complex form:

f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} (a_{n} \cos n x + b_{n} \sin n x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} (a_{n} \frac{e^{i n x} + e^{- i n x}}{2} + b_{n} \frac{e^{i n x} - e^{- i n x}}{2 i}) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} \frac{a_{n} - i b_{n}}{2} e^{i n x} + \sum_{n = 1}^{\infty} \frac{a_{n} + i b_{n}}{2} e^{- i n x} = \sum_{n = - \infty}^{\infty} c_{n} e^{i n x} .

Here we have used the following notations:

c_{0} = \frac{a_{0}}{2}, c_{n} = \frac{a_{n} - i b_{n}}{2}, c_{- n} = \frac{a_{n} + i b_{n}}{2} .

The coefficients $c_{n}$ are called complex Fourier coefficients. They are defined by the formulas

c_{n} = \frac{1}{2 π} \int_{- π}^{π} f (x) e^{- i n x} d x, n = 0, \pm 1, \pm 2, \dots

If necessary to expand a function $f (x)$ of period $2 L,$ we can use the following expressions:

f (x) = \sum_{n = - \infty}^{\infty} c_{n} e^{\frac{i n π x}{L}},

where

c_{n} = \frac{1}{2 L} \int_{- L}^{L} f (x) e^{- \frac{i n π x}{L}} d x, n = 0, \pm 1, \pm 2, \dots

The complex form of Fourier series is algebraically simpler and more symmetric. Therefore, it is often used in physics and other sciences.

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Using complex form, find the Fourier series of the function

f (x) = sign x = {\begin{cases} - 1, & - π \leq x \leq 0 \\ 1, & 0 < x \leq π \end{cases} .

Using complex form, find the Fourier series of the function

f (x) = sign x = {\begin{cases} - 1, & - π \leq x \leq 0 \\ 1, & 0 < x \leq π \end{cases} .

Solution.

We calculate the coefficients $c_{0}$ and $c_{n}$ for $n \neq 0 :$

c_{0} = \frac{1}{2 π} \int_{- π}^{π} f (x) d x = \frac{1}{2 π} [\int_{- π}^{0} (- 1) d x + \int_{0}^{π} d x] = \frac{1}{2 π} [{(- x) |}_{- π}^{0} + {x |}_{0}^{π}] = \frac{1}{2 π} (- π + π) = 0,

c_{n} = \frac{1}{2 π} \int_{- π}^{π} f (x) e^{- i n x} d x = \frac{1}{2 π} [\int_{- π}^{0} (- 1) e^{- i n x} d x + \int_{0}^{π} e^{- i n x} d x] = \frac{1}{2 π} [- \frac{{(e^{- i n x}) |}_{- π}^{0}}{- i n} + \frac{{(e^{- i n x}) |}_{0}^{π}}{- i n}] = \frac{i}{2 π n} [- (1 - e^{i n π}) + e^{- i n π} - 1] = \frac{i}{2 π n} [e^{i n π} + e^{- i n π} - 2] = \frac{i}{π n} [\frac{e^{i n π} + e^{- i n π}}{2} - 1] = \frac{i}{π n} [\cos n π - 1] = \frac{i}{π n} [{(- 1)}^{n} - 1] .

If $n = 2 k,$ then $c_{2 k} = 0.$ If $n = 2 k - 1,$ then $c_{2 k - 1} = - \frac{2 i}{(2 k - 1) π} .$

Hence, the Fourier series of the function in complex form is

f (x) = sign x = - \frac{2 i}{π} \sum_{k = - \infty}^{\infty} \frac{1}{2 k - 1} e^{i (2 k - 1) x} .

We can transform the series and write it in the real form. Rename: $n = 2 k - 1,$ $n = \pm 1, \pm 2, \pm 3, \dots$ Then

f (x) = sign x = - \frac{2 i}{π} \sum_{k = - \infty}^{\infty} \frac{1}{2 k - 1} e^{i (2 k - 1) x} = - \frac{2 i}{π} \sum_{n = - \infty}^{\infty} \frac{e^{i n x}}{n} = - \frac{2 i}{π} \sum_{n = 1}^{\infty} (\frac{e^{- i n x}}{- n} + \frac{e^{i n x}}{n}) = \frac{4}{π} \sum_{n = 1}^{\infty} \frac{e^{i n x} - e^{- i n x}}{2 i n} = \frac{4}{π} \sum_{n = 1}^{\infty} \frac{\sin n x}{n} = \frac{4}{π} \sum_{k = 1}^{\infty} \frac{\sin (2 k - 1) x}{2 k - 1} .

Graph of the function and its Fourier approximation for $n = 5$ and $n = 50$ are shown in Figure $1.$