Half-Range Fourier Cosine Series/Identity Function/0 to Pi
Theorem
Let $\map f x$ be the real function defined on $\openint 0 \pi$ as:
- $\map f x = x$
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
- 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Exercises on Chapter $\text I$: $6 \, \text{(ii)}$.