Fourier Series/Pi Squared minus x Squared over Minus Pi to Pi
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Theorem
Let $\map f x$ be the real function defined on $\openint {-\pi} \pi$ as:
- $\map f x = \pi^2 - x^2$
$f$ can be expressed as a half-range Fourier cosine series thus:
\(\ds \map f x\) | \(\sim\) | \(\ds \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {\cos n x} {n^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \pi^2} 3 + 4 \paren {\cos x - \frac 1 4 \cos 2 x + \frac 1 9 \cos 3 x - \cdots}\) |
Proof
We have that:
- $\pi^2 - \paren {-x}^2 = \pi^2 - x^2$
and so $\map f x$ is even on $\openint {-\pi} \pi$.
It follows from Fourier Series for Even Function over Symmetric Range:
- $\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x$
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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 \pi \int_0^\pi \paren {\pi^2 - x^2} \rd x\) | Definition of $f$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 \pi \intlimits {\pi^2 x - \frac {x^3} 3} 0 \pi\) | Primitive of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 \pi \paren {\paren {\pi^3 - \frac {\pi^3} 3} - \paren {0 - \frac 0 3} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 \pi \paren {\frac {2 \pi^3} 3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 \pi^2} 3\) |
$\Box$
Then for $n > 0$:
\(\ds a_n\) | \(=\) | \(\ds \frac 2 \pi \int_0^\pi \map f x \cos n x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 \pi \int_0^\pi \paren {\pi^2 - x^2} \cos n x \rd x\) | Definition of $f$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \pi \int_0^\pi \cos n x \rd x - \frac 2 \pi \int_0^\pi x^2 \cos n x \rd x\) | Linear Combination of Definite Integrals |
Splitting this up into two:
\(\ds \) | \(\) | \(\ds 2 \pi \int_0^\pi \cos n x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \pi \intlimits {\frac {\sin n x} n} 0 \pi\) | Primitive of $\cos a x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \pi \paren {\frac {\sin n \pi} n - \frac {\sin 0} n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \pi \paren {0 - 0}\) | Sine of Multiple of Pi | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
\(\ds \) | \(\) | \(\ds -\frac 2 \pi \int_0^\pi x^2 \cos n x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 2 \pi \intlimits {\frac {2 x \cos n x} {n^2} + \paren {\frac {x^2} n - \frac 2 {n^3} } \sin n x} 0 \pi\) | Primitive of $x^2 \cos n x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 2 \pi \paren {\paren {\frac {2 \pi \cos n \pi} {n^2} + \paren {\frac {\pi^2} n - \frac 2 {n^3} } \sin n \pi} - \paren {\frac {0 \cos 0} {n^2} + \paren {\frac {0^2} n - \frac 2 {n^3} } \sin 0} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 2 \pi \paren {\frac {2 \pi \cos n \pi} {n^2} }\) | Sine of Multiple of Pi and removing vanishing terms | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac {4 \cos n \pi} {n^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac {4 \paren {-1}^n} {n^2}\) | Cosine of Multiple of Pi | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{n - 1} \frac 4 {n^2}\) | simplification |
$\Box$
Finally:
\(\ds \map f x\) | \(\sim\) | \(\ds \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \frac {4 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac 4 {n^2} \cos n x\) | substituting for $a_0$ and $a_n$ from above | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {\cos n x} {n^2}\) | simplifying |
$\blacksquare$
Sources
- 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Exercises on Chapter $\text I$: $6 \, \text{(i)}$.