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 |
$\blacksquare$