Fourier Series/4 minus x squared over Range of 2

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Theorem

Let $\map f x$ be the real function defined on $\openint 0 2$ as:

$\map f x$ and its $7$th approximation
$\map f x = 4 - x^2$


Then its Fourier series can be expressed as:

$\map f x \sim \ds \frac 8 3 - \frac 4 {\pi^2} \sum_{n \mathop = 1}^\infty \frac {\cos n \pi x} {n^2} + \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\sin n \pi x} n$


Proof

By definition of Fourier series:

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

where:

\(\ds a_n\) \(=\) \(\ds \int_0^2 \map f x \cos n \pi x \rd x\)
\(\ds b_n\) \(=\) \(\ds \int_0^2 \map f x \sin n \pi x \rd x\)

for all $n \in \Z_{>0}$.


Thus:

\(\ds a_0\) \(=\) \(\ds \int_0^2 \map f x \rd x\) Cosine of Zero is One
\(\ds \) \(=\) \(\ds \int_0^2 \paren {4 - x^2} \rd x\)
\(\ds \) \(=\) \(\ds \intlimits {4 x - \frac {x^3} 3} 0 2\) Primitive of Power
\(\ds \) \(=\) \(\ds \paren {4 \times 2 - \frac {2^3} 3} - \paren {0 - 0}\)
\(\ds \) \(=\) \(\ds 8 - \frac 8 3\)
\(\ds \) \(=\) \(\ds \frac {16} 3\)

$\Box$


For $n > 0$:

\(\ds a_n\) \(=\) \(\ds \int_0^2 \map f x \cos n \pi x \rd x\)
\(\ds \) \(=\) \(\ds \int_0^2 \paren {4 - x^2} \cos n \pi x \rd x\) Definition of $f$
\(\ds \) \(=\) \(\ds 4 \int_0^2 \cos n \pi x \rd x - \int_0^2 x^2 \cos n \pi x \rd x\) Linear Combination of Definite Integrals


Splitting this up into bits:


\(\ds \) \(\) \(\ds 4 \int_0^2 \cos n \pi x \rd x\)
\(\ds \) \(=\) \(\ds 4 \intlimits {\frac {\sin n \pi x} {n \pi} } 0 2\) Primitive of $\cos n \pi x$
\(\ds \) \(=\) \(\ds 4 \paren {\frac {\sin 2 n \pi} {n \pi} } - 4 \paren {\frac {\sin 0} {n \pi} }\)
\(\ds \) \(=\) \(\ds 0\) Sine of Multiple of Pi


\(\ds \) \(\) \(\ds \int_0^2 x^2 \cos n \pi x \rd x\)
\(\ds \) \(=\) \(\ds \intlimits {\frac {2 x \cos n \pi x} {\paren {n \pi}^2} + \paren {\frac {x^2} {n \pi} - \frac 2 {\paren {n \pi}^3} } \sin n \pi x} 0 2\) Primitive of $x^2 \cos n \pi x$
\(\ds \) \(=\) \(\ds \paren {\frac {4 \cos 2 n \pi} {n^2 \pi^2} + \paren {\frac 4 {n \pi} - \frac 2 {\paren {n \pi}^3} } \sin 2 n \pi} - \paren {\frac {0 \cos 0} {n^2 \pi^2} + \paren {\frac 0 {n \pi} - \frac 2 {\paren {n \pi}^3} } \sin 0}\)
\(\ds \) \(=\) \(\ds \frac {4 \cos 2 n \pi} {n^2 \pi^2}\) Sine of Multiple of Pi and removal of vanishing terms
\(\ds \) \(=\) \(\ds \frac 4 {n^2 \pi^2}\) Cosine of Multiple of Pi


Reassembling $a_n$ from the remaining non-vanishing terms:

\(\ds a_n\) \(=\) \(\ds 0 - \frac 4 {n^2 \pi^2}\)
\(\ds \) \(=\) \(\ds -\frac 4 {n^2 \pi^2}\)

$\Box$


Now for the $\sin n \pi x$ terms:

\(\ds b_n\) \(=\) \(\ds \int_0^2 \map f x \sin n \pi x \rd x\)
\(\ds \) \(=\) \(\ds \int_0^2 \paren {4 - x^2} \sin n \pi x \rd x\) Definition of $f$
\(\ds \) \(=\) \(\ds 4 \int_0^2 \sin n \pi x \rd x - \int_0^2 x^2 \sin n \pi x \rd x\) Linear Combination of Definite Integrals


Splitting this up into bits:


\(\ds \) \(\) \(\ds 4 \int_0^2 \sin n \pi x \rd x\)
\(\ds \) \(=\) \(\ds 4 \intlimits {\frac {-\cos n \pi x} {n \pi} } 0 2\) Primitive of $\sin n \pi x$
\(\ds \) \(=\) \(\ds 4 \paren {\frac {-\cos 2 n \pi} {n \pi} } - 4 \paren {\frac {-\cos 0} {n \pi} }\)
\(\ds \) \(=\) \(\ds 4 \paren {\frac {-1} {n \pi} } - 4 \paren {\frac {-1} {n \pi} }\) Cosine of Multiple of Pi
\(\ds \) \(=\) \(\ds 0\)


\(\ds \) \(\) \(\ds \int_0^2 x^2 \sin n x \rd x\)
\(\ds \) \(=\) \(\ds \intlimits {\frac {2 x \sin n \pi x} {\paren {n \pi}^2} + \paren {\frac 2 {\paren {n \pi}^3} - \frac {x^2} {n \pi} } \cos n \pi x} 0 2\) Primitive of $x^2 \sin n \pi x$
\(\ds \) \(=\) \(\ds \paren {\frac {4 \sin 2 n \pi} {n^2 \pi^2} + \paren {\frac 2 {\paren {n \pi}^3} - \frac 4 {n \pi} } \cos 2 n \pi} - \paren {\frac {0 \sin 0} {n^2 \pi^2} + \paren {\frac 2 {\paren {n \pi}^3} - \frac 0 {n \pi} } \cos 0}\)
\(\ds \) \(=\) \(\ds \paren {\frac 2 {\paren {n \pi}^3} - \frac 4 {n \pi} } \cos 2 n \pi - \frac 2 {\paren {n \pi}^3} \cos 0\) Sine of Multiple of Pi and removal of vanishing terms
\(\ds \) \(=\) \(\ds -\frac 4 {n \pi}\) Cosine of Multiple of Pi and simplifying


Reassembling $b_n$ from the remaining non-vanishing terms:

\(\ds b_n\) \(=\) \(\ds 0 - \paren {-\frac 4 {n \pi} }\)
\(\ds \) \(=\) \(\ds \frac 4 {n \pi}\)

$\Box$


Finally:

\(\ds \map f x\) \(\sim\) \(\ds \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \paren {\frac {16} 3} + \sum_{n \mathop = 1}^\infty \paren {-\frac 4 {n^2 \pi^2} \cos n \pi x + \frac 4 {n \pi} \sin n \pi x}\) substituting for $a_0$, $a_n$ and $b_n$ from above
\(\ds \) \(=\) \(\ds \frac 8 3 - \frac 4 {\pi^2} \sum_{n \mathop = 1}^\infty \frac {\cos n \pi x} {n^2} + \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\sin n \pi x} n\) rearranging

$\blacksquare$


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