Fourier Series/x over 0 to 2, x-2 over 2 to 4

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Theorem

Let $\map f x$ be the real function defined on $\openint 0 4$ as:

$\map f x = \begin{cases}

x & : 0 < x \le 2 \\ x - 2 & : 2 < x < 4 \end{cases}$


Then its Fourier series can be expressed as:

$\ds \map f x \sim 1 + \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{r - 1} } {2 r - 1} \paren {1 + \frac {4 \paren {-1}^r} {\paren {2 r - 1} \pi} } \cos \frac {\paren {2 r - 1} \pi x} 4$


Proof

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

$\forall x \in \openint 0 4: \begin{cases}

x & : 0 < x \le 2 \\ x - 2 & : 2 < x < 4 \end{cases}$


Let $f$ be expressed by a half-range Fourier cosine series:

$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos \frac {n \pi x} 4$

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

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


In this context, $l = 4$ and so this can be expressed as:

\(\ds a_n\) \(=\) \(\ds \frac 2 4 \paren {\int_0^2 x \cos \frac {n \pi x} 4 \rd x + \int_2^4 \paren {x - 2} \cos \frac {n \pi x} 4 \rd x}\)
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\int_0^2 x \cos \frac {n \pi x} 4 \rd x + \int_2^4 \paren {x - 2} \cos \frac {n \pi x} 4 \rd x}\)


First the case when $n = 0$:

\(\ds a_0\) \(=\) \(\ds \frac 1 2 \paren {\int_0^2 x \rd x + \int_2^4 \paren {x - 2} \rd x}\)
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\intlimits {\frac {x^2} 2} 0 2 + \intlimits {\frac {x^2} 2 - 2 x} 2 4}\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\paren {\frac {2^2} 2 - \frac {0^2} 2} + \paren {\paren {\frac {4^2} 2 - 2 \times 4} - \paren {\frac {2^2} 2 - 2 \times 2} } }\)
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\paren {\frac 4 2 - 0} + \paren {\paren {\frac {16} 2 - 8} - \paren {\frac 4 2 - 4} } }\)
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {2 + 0 + 2}\)
\(\ds \) \(=\) \(\ds 2\)


When $n \ne 0$:

\(\ds a_n\) \(=\) \(\ds \frac 1 2 \paren {\int_0^2 x \cos \frac {n \pi x} 4 \rd x + \int_2^4 \paren {x - 2} \cos \frac {n \pi x} 4 \rd x}\)
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\int_0^2 x \cos \frac {n \pi x} 4 \rd x + \int_2^4 x \cos \frac {n \pi x} 4 \rd x - 2 \int_2^4 \cos \frac {n \pi x} 4 \rd x}\) Linear Combination of Definite Integrals
\(\ds \) \(=\) \(\ds \frac 1 2 \int_0^4 x \cos \frac {n \pi x} 4 \rd x - \int_2^4 \cos \frac {n \pi x} 4 \rd x\) Sum of Integrals on Adjacent Intervals for Integrable Functions


Splitting it up into two:

\(\ds \) \(\) \(\ds \frac 1 2 \int_0^4 x \cos \frac {n \pi x} 4 \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 2 \intlimits {\frac {16} {n^2 \pi^2} \cos \frac {n \pi x} 4 + \frac 4 {n \pi} x \sin \frac {n \pi x} 4 } 0 4\) Primitive of $x \cos a x$
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\paren {\frac {16} {n^2 \pi^2} \cos n \pi + \frac {16} {n \pi} \sin n \pi} - \paren {\frac {16} {n^2 \pi^2} \cos 0 + \frac 4 {n \pi} \times 0 \sin 0} }\)
\(\ds \) \(=\) \(\ds \frac 8 {n^2 \pi^2} \paren {\cos n \pi - \cos 0}\) Sine of Multiple of Pi and simplification
\(\ds \) \(=\) \(\ds \frac {8 \paren {\paren {-1}^n - 1} } {n^2 \pi^2}\) Cosine of Multiple of Pi
\(\ds \) \(=\) \(\ds \begin {cases} 0 & : \text {$n$ even} \\ \dfrac {-16} {n^2 \pi^2} & : \text {$n$ odd} \end {cases}\)
\(\ds \) \(=\) \(\ds \dfrac {-16} {\paren {2 r - 1}^2 \pi^2}\) substituting $n = 2 r - 1$


On the other hand:

\(\ds \) \(\) \(\ds \int_2^4 \cos \frac {n \pi x} 4 \rd x\)
\(\ds \) \(=\) \(\ds \intlimits {\frac 4 {n \pi} \sin \frac {n \pi x} 4} 2 4\) Primitive of $\cos a x$
\(\ds \) \(=\) \(\ds \frac 4 {n \pi} \paren {\sin \frac {4 n \pi} 4 - \sin \frac {2 n \pi} 4 }\)
\(\ds \) \(=\) \(\ds \frac 4 {n \pi} \paren {\sin n \pi - \sin \frac {n \pi} 2}\)
\(\ds \) \(=\) \(\ds -\frac 4 {n \pi} \sin \frac {n \pi} 2\) Sine of Multiple of Pi

The above expression is $0$ when $n$ is even, by Sine of Multiple of Pi.

Thus we may substitute $n = 2 r - 1$.

Then:

\(\ds \) \(\) \(\ds \int_2^4 \cos \frac {n \pi x} 4 \rd x\)
\(\ds \) \(=\) \(\ds -\frac 4 {n \pi} \sin \frac {n \pi} 2\)
\(\ds \) \(=\) \(\ds -\frac 4 {\paren {2 r - 1} \pi} \sin \frac {\paren {2 r - 1} \pi} 2\)
\(\ds \) \(=\) \(\ds -\frac 4 {\paren {2 r - 1} \pi} \sin \paren {\paren {r - 1} + \frac 1 2} \pi\)
\(\ds \) \(=\) \(\ds -\frac 4 {\paren {2 r - 1} \pi} \paren {-1}^{r - 1}\) Sine of Half-Integer Multiple of Pi


Combining the results:

\(\ds \map f x\) \(\sim\) \(\ds \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos \frac {n \pi x} 4\)
\(\ds \) \(=\) \(\ds \frac {a_0} 2 + \sum_{n \text{ odd} } a_n \cos \frac {n \pi x} 4\) as both integrals vanish for even $n$
\(\ds \) \(=\) \(\ds \frac {a_0} 2 + \sum_{r \mathop = 1}^\infty a_{2 r - 1} \cos \frac {\paren{2 r - 1} \pi x} 4\) using our substitution $n = 2 r - 1$
\(\ds \) \(=\) \(\ds \frac 2 2 + \sum_{r \mathop = 1}^\infty \paren{\dfrac {-16} {\paren {2 r - 1}^2 \pi^2} + \frac 4 {\paren {2 r - 1} \pi} \paren {-1}^{r - 1} } \cos \frac {\paren{2 r - 1} \pi x} 4\) from above
\(\ds \) \(=\) \(\ds 1 + \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{r - 1} } {2 r - 1} \paren {\frac {- 4 \paren {-1}^{r - 1} } {\paren {2 r - 1} \pi} + 1} \cos \frac {\paren {2 r - 1} \pi x} 4\) factorisation
\(\ds \) \(=\) \(\ds 1 + \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{r - 1} } {2 r - 1} \paren {1 + \frac {4 \paren {-1}^r} {\paren {2 r - 1} \pi} } \cos \frac {\paren {2 r - 1} \pi x} 4\)

$\blacksquare$


Sources

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