Half-Range Fourier Series/Identity Function/Cosine

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Theorem

Let $\lambda \in \R_{>0}$ be a strictly positive real number.

Let $\map f x: \openint 0 \lambda \to \R$ be the identity function on the open real interval $\openint 0 \lambda$:

$\forall x \in \openint 0 \lambda: \map f x = x$


The half-range Fourier cosine series for $\map f x$ can be expressed as:

\(\ds \map f x\) \(\sim\) \(\ds \frac \lambda 2 - \frac {4 \lambda} {\pi^2} \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \dfrac {\paren {2 n + 1} \pi x} \lambda\)
\(\ds \) \(=\) \(\ds \frac \lambda 2 - \frac {4 \lambda} {\pi^2} \paren {\cos \dfrac {\pi x} \lambda + \frac 1 {3^2} \cos \dfrac {3 \pi x} \lambda + \frac 1 {5^2} \cos \dfrac {5 \pi x} \lambda + \dotsb}\)


Proof

By definition of half-range Fourier cosine series:

$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos \dfrac {n \pi x} \lambda$


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

$a_n = \ds \frac 2 \lambda \int_0^\lambda \map f x \cos \dfrac {n \pi x} \lambda \rd x$


Thus by definition of $f$:

\(\ds a_0\) \(=\) \(\ds \frac 2 \lambda \int_0^\lambda \map f x \rd x\) Cosine of Zero is One
\(\ds \) \(=\) \(\ds \frac 2 \lambda \int_0^\lambda x \rd x\) Definition of $f$
\(\ds \) \(=\) \(\ds \frac 2 \lambda \intlimits {\frac {x^2} 2} 0 \lambda\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac 2 \lambda \paren {\frac {l^2} 2 - \frac {0^2} 2}\)
\(\ds \) \(=\) \(\ds \lambda\) simplification

$\Box$


For $n > 0$:

\(\ds a_n\) \(=\) \(\ds \frac 2 \lambda \int_0^l \map f x \cos \dfrac {n \pi x} \lambda \rd x\)
\(\ds \) \(=\) \(\ds \frac 2 \lambda \int_0^l x \cos \dfrac {n \pi x} \lambda \rd x\) Definition of $f$
\(\ds \) \(=\) \(\ds \frac 2 \lambda \intlimits {\frac {l^2} {\pi^2 n^2} \cos \dfrac {n \pi x} \lambda + \frac \lambda {\pi n} x \sin \dfrac {n \pi x} \lambda} 0 \lambda\) Primitive of $x \cos n x$
\(\ds \) \(=\) \(\ds \frac 2 \lambda \paren {\paren {\frac {\lambda^2} {\pi^2 n^2} \cos n \pi + \frac {\lambda^2} {\pi n} \sin n \pi} - \paren {\frac {\lambda^2} {\pi^2 n^2} \cos 0 + \frac \lambda {\pi n} 0 \sin 0} }\)
\(\ds \) \(=\) \(\ds \frac {2 \lambda} {\pi^2 n^2} \paren {\cos n \pi - \cos 0}\) Sine of Multiple of Pi and simplification
\(\ds \) \(=\) \(\ds \frac {2 \lambda} {\pi^2 n^2} \paren {\paren {-1}^n - 1}\) Cosine of Multiple of Pi


When $n$ is even, $\paren {-1}^n = 1$.

We can express $n = 2 r$ for $r \ge 1$.

Hence in that case:

\(\ds a_{2 r}\) \(=\) \(\ds \frac {2 \lambda} {\pi^2 n^2} \paren {\paren {-1}^n - 1}\)
\(\ds \) \(=\) \(\ds \frac {2 \lambda} {\pi^2 n^2} \paren {1 - 1}\)
\(\ds \) \(=\) \(\ds 0\)


When $n$ is odd, $\paren {-1}^n = -1$.

We can express $n = 2 r + 1$ for $r \ge 0$.

Hence in that case:

\(\ds a_{2 r - 1}\) \(=\) \(\ds \frac {2 \lambda} {\pi^2 \paren {2 r + 1}^2} \paren {-1 - 1}\)
\(\ds \) \(=\) \(\ds -\frac {4 \lambda} {\pi^2} \paren {\frac 1 {\paren {2 r + 1}^2} }\) rearranging


Finally:

\(\ds \map f x\) \(\sim\) \(\ds \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x\)
\(\ds \) \(=\) \(\ds \frac \lambda 2 - \frac {4 \lambda} {\pi^2} \sum_{r \mathop = 0}^\infty \frac 1 {\paren {2 r + 1}^2} \cos \dfrac {\paren {2 r + 1} \pi x} \lambda\) substituting for $a_0$ and $a_n$ from above
\(\ds \) \(=\) \(\ds \frac \lambda 2 - \frac {4 \lambda} {\pi^2} \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \dfrac {\paren {2 n + 1} \pi x} \lambda\) changing the name of the variable

$\blacksquare$