Equivalence of Definitions of Trigonometric Series

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Theorem

The following definitions of the concept of Trigonometric Series are equivalent:

Conventional Form

A trigonometric series is a series of the type:

$\map S x = \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$

where:

the domain of $x$ is the set of real numbers $\R$
the coefficients $a_0, a_1, a_2, \ldots, a_n, \ldots; b_1, b_2, \ldots, b_n, \ldots$ are real numbers independent of $x$.

Complex Form

A trigonometric series can be expressed in a form using complex functions as follows:

$\map S x = \ds \sum_{n \mathop = -\infty}^\infty c_n e^{i n x}$

where:

the domain of $x$ is the set of real numbers $\R$
the coefficients $\ldots, c_{-n}, \ldots, c_{-2}, c_{-1}, c_0, c_1, c_2, \ldots, c_n, \ldots$ are real numbers independent of $x$
$c_{-n} = \overline {c_n}$ where $\overline {c_n}$ is the complex conjugate of $c_n$.


Proof

Let $\map S x = \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$.

Set:

\(\ds \cos n x\) \(=\) \(\ds \frac {e^{i n x} + e^{- i n x} } 2\) Euler's Cosine Identity
\(\ds \sin n x\) \(=\) \(\ds \frac {e^{i n x} - e^{- i n x} } {2 i}\) Euler's Sine Identity


Then:

\(\ds \map S x\) \(=\) \(\ds \dfrac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \frac {e^{i n x} + e^{- i n x} } 2 + b_n \frac {e^{i n x} - e^{- i n x} } {2 i} }\)
\(\ds \) \(=\) \(\ds \dfrac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {\paren {\frac {a_n} 2 + \frac {b_n} {2 i} } e^{i n x} + \paren {\frac {a_n} 2 - \frac {b_n} {2 i} } e^{- i n x} }\)
\(\ds \) \(=\) \(\ds \dfrac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {\paren {\frac {a_n - i b_n} 2} e^{i n x} + \paren {\frac {a_n + i b_n} 2} e^{- i n x} }\)
\(\ds \) \(=\) \(\ds \dfrac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {\frac {a_n - i b_n} 2} e^{i n x} + \sum_{n \mathop = 1}^\infty \paren {\frac {a_n + i b_n} 2} e^{- i n x}\)


Let $b_0 := 0$.

Then let:

$\forall n \in \N: c_n = \dfrac {a_n - i b_n} 2$

Thus it follows that:

$c_0 = \dfrac {a_0 - 0 \times i} 2 = \dfrac {a_0} 2$


Then let:

$c_{-n} = \dfrac {a_n + i b_n} 2$

from which it follows that:

$c_{-n} = \overline {\paren {\dfrac {a_n - i b_n} 2} }$

and again:

$c_0 = \dfrac {a_0 + 0 \times i} 2 = \dfrac {a_0} 2$


Then it is noted that:

$e^{i \times 0 \times x} = e^0 = 1$

and it follows that:

\(\ds \map S x\) \(=\) \(\ds \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {\frac {a_n - i b_n} 2} e^{i n x} + \sum_{n \mathop = 1}^\infty \paren {\frac {a_n + i b_n} 2} e^{- i n x}\) from above
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {\frac {a_n - i b_n} 2} e^{i n x} + \sum_{n \mathop = 1}^\infty \paren {\frac {a_n + i b_n} 2} e^{- i n x}\) subsuming $a_0$ into first summation
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty c_n e^{i n x} + \sum_{n \mathop = 1}^\infty c_{-n} e^{- i n x}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty c_n e^{i n x} + \sum_{n \mathop = -\infty}^{-1} c_n e^{i n x}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = -\infty}^\infty c_n e^{i n x}\)

where it is noted that $c_{-n} = \overline {c_n}$ as required.

$\blacksquare$


Sources

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