Half-Range Fourier Cosine Series/Identity Function/0 to Pi

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Theorem

Let $\map f x$ be the real function defined on $\openint 0 \pi$ as:

$\map f x$ and its $4$th approximation
$\map f x = x$


Then its half-range Fourier cosine series can be expressed as:

\(\ds x\) \(\sim\) \(\ds \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\cos \paren {2 n - 1} x} {\paren {2 n - 1}^2}\)
\(\ds \) \(=\) \(\ds \frac \pi 2 - \frac 4 \pi \paren {\cos x + \frac {\cos 3 x} {3^2} + \frac {\cos 5 x} {5^2} + \cdots}\)


Proof 1

By definition of half-range Fourier cosine series:

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


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

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


Thus by definition of $f$:

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

$\Box$


For $n > 0$:

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

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


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

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

Hence in that case:

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


Finally:

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

$\blacksquare$


Proof 2

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

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


From Half-Range Fourier Cosine Series for Identity Function, the half-range Fourier cosine series for $\map f x$ can be expressed as:

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


The result follows by setting $\lambda = \pi$.

$\blacksquare$


Sources