Fourier Series/Exponential of x over Minus Pi to Pi

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Theorem

Let $\map f x$ be the real function defined on $\R$ as:

$\map f x$ and its $7$th approximation
$\map f x = \begin{cases}

e^x & : -\pi < x \le \pi \\ \map f {x + 2 \pi} & : \text{everywhere} \end{cases}$


Then its Fourier series can be expressed as:

\(\ds \map f x\) \(\sim\) \(\ds \frac {\sinh \pi} \pi \paren {1 + 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n} {1 + n^2} \paren {\cos n x - n \sin n x} }\)
\(\ds \) \(=\) \(\ds \frac {\sinh \pi} \pi \paren {1 + 2 \paren {-\dfrac {\cos x - \sin x} 2 + \dfrac {\cos 2 x - 2 \sin 2 x} 5 - \dfrac {\cos 3 x - 3 \sin 3 x} {10} + \dotsb} }\)


Proof

By definition of Fourier series:

$\displaystyle \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$


where for all $n \in \Z_{> 0}$:

\(\ds a_n\) \(=\) \(\ds \dfrac 1 \pi \int_{-\pi}^\pi \map f x \cos n x \rd x\)
\(\ds b_n\) \(=\) \(\ds \dfrac 1 \pi \int_{-\pi}^\pi \map f x \sin n x \rd x\)


Thus by definition of $f$:

\(\ds a_0\) \(=\) \(\ds \frac 1 \pi \int_{-\pi}^\pi \map f x \rd x\) Cosine of Zero is One
\(\ds \) \(=\) \(\ds \frac 1 \pi \int_{-\pi}^\pi e^x \rd x\) Definition of $f$
\(\ds \) \(=\) \(\ds \frac 1 \pi \bigintlimits {e^x} {-\pi} \pi\) Primitive of Exponential Function
\(\ds \) \(=\) \(\ds \frac 1 \pi \paren {e^\pi - e^{-\pi} }\)
\(\ds \) \(=\) \(\ds \frac 2 \pi \dfrac {\paren {e^\pi - e^{-\pi} } } 2\)
\(\ds \) \(=\) \(\ds \frac 2 \pi \sinh \pi\) Definition of Hyperbolic Sine

$\Box$


For $n > 0$:

\(\ds a_n\) \(=\) \(\ds \dfrac 1 \pi \int_{-\pi}^\pi \map f x \cos n x \rd x\)
\(\ds \) \(=\) \(\ds \dfrac 1 \pi \int_{-\pi}^\pi e^x \cos n x \rd x\) Definition of $f$
\(\ds \) \(=\) \(\ds \frac 1 \pi \intlimits {\frac {e^x \paren {\cos n x + n \sin n x} } {1 + n^2} } {-\pi} \pi\) Primitive of $e^x \cos n x$
\(\ds \) \(=\) \(\ds \frac 1 \pi \paren {\frac {e^\pi \paren {\cos n \pi + n \sin n \pi} } {1 + n^2} - \frac {e^{-\pi} \paren {\cos n \paren {-\pi} + n \sin n \paren {-\pi} } } {1 + n^2} }\)
\(\ds \) \(=\) \(\ds \frac 1 \pi \paren {\frac {e^\pi \cos n \pi} {1 + n^2} - \frac {e^{-\pi} \cos n \paren {-\pi} } {1 + n^2} }\) Sine of Multiple of Pi
\(\ds \) \(=\) \(\ds \frac 1 \pi \paren {\frac {e^\pi \paren {-1}^n - e^{-\pi} \paren {-1}^n} {1 + n^2} }\) Cosine of Multiple of Pi
\(\ds \) \(=\) \(\ds \frac 2 \pi \frac {\paren {-1}^n} {1 + n^2} \frac {e^\pi - e^{-\pi} } 2\) manipulation
\(\ds \) \(=\) \(\ds \frac {2 \paren {-1}^n} {\paren {1 + n^2} \pi} \sinh \pi\) Definition of Hyperbolic Sine

$\Box$


Now for the $\sin n x$ terms:

\(\ds b_n\) \(=\) \(\ds \frac 1 \pi \int_{-\pi}^\pi \map f x \sin n x \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 \pi \int_{-\pi}^\pi e^x \sin n x \rd x\) Definition of $f$
\(\ds \) \(=\) \(\ds \frac 1 \pi \intlimits {\frac {e^x \paren {\sin n x - n \cos n x} } {1 + n^2} } {-\pi} \pi\) Primitive of $e^x \sin n x$
\(\ds \) \(=\) \(\ds \frac 1 \pi \paren {\frac {e^\pi \paren {\sin n \pi - n \cos n \pi} } {1 + n^2} - \frac {e^{-\pi} \paren {\sin n \paren {-\pi} - n \cos n \paren {-\pi} } } {1 + n^2} }\)
\(\ds \) \(=\) \(\ds \frac 1 \pi \paren {\frac {-e^\pi n \cos n \pi} {1 + n^2} - \frac {-e^{-\pi} n \cos n \paren {-\pi} } {1 + n^2} }\) Sine of Multiple of Pi
\(\ds \) \(=\) \(\ds -\frac 1 \pi \paren {\frac {e^\pi n \paren {-1}^n - e^{-\pi} n \paren {-1}^n} {1 + n^2} }\) Cosine of Multiple of Pi
\(\ds \) \(=\) \(\ds -\frac {2 n} \pi \frac {\paren {-1}^n} {1 + n^2} \frac {e^\pi - e^{-\pi} } 2\) manipulation
\(\ds \) \(=\) \(\ds -\frac {2 n \paren {-1}^n} {\paren {1 + n^2} \pi} \sinh \pi\) Definition of Hyperbolic Sine

$\Box$


Finally:

\(\ds \map f x\) \(\sim\) \(\ds \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}\)
\(\ds \) \(=\) \(\ds \frac 1 2 \frac 2 \pi \sinh \pi + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \paren {-1}^n} {\paren {1 + n^2} \pi} \sinh \pi \cos n x - \frac {2 n \paren {-1}^n} {\paren {1 + n^2} \pi} \sinh \pi \sin n x}\) substituting for $a_0$, $a_n$ and $b_n$ from above
\(\ds \) \(=\) \(\ds \frac {\sinh \pi} \pi \paren {1 + 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n} {1 + n^2} \paren {\cos n x - n \sin n x} }\) simplifying

$\blacksquare$


Sources

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