7.1 Definition A Fourier series of a 2 π 2\pi 2 π f f f
f ( x ) ∼ a 0 2 + ∑ n = 1 ∞ ( a n cos ( n x ) + b n sin ( n x ) ) f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} \left(a_n \cos(nx) + b_n \sin(nx)\right) f ( x ) ∼ 2 a 0 + ∑ n = 1 ∞ ( a n cos ( n x ) + b n sin ( n x ) )
Where
a n = 1 π ∫ − π π f ( x ) cos ( n x ) d x , b n = 1 π ∫ − π π f ( x ) sin ( n x ) d x a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos(nx)\, dx, \quad b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin(nx)\, dx a n = π 1 ∫ − π π f ( x ) cos ( n x ) d x , b n = π 1 ∫ − π π f ( x ) sin ( n x ) d x
7.2 Derivation of Fourier Coefficients The Fourier coefficients are derived using the orthogonality relations on [ − π , π ] [-\pi, \pi] [ − π , π ]
∫ − π π cos ( m x ) cos ( n x ) d x = { π m = n ≠ 0 2 π m = n = 0 0 m ≠ n \int_{-\pi}^{\pi} \cos(mx)\cos(nx)\, dx = \begin{cases} \pi & m = n \neq 0 \\ 2\pi & m = n = 0 \\ 0 & m \neq n \end{cases} ∫ − π π cos ( m x ) cos ( n x ) d x = ⎩ ⎨ ⎧ π 2 π 0 m = n = 0 m = n = 0 m = n
∫ − π π sin ( m x ) sin ( n x ) d x = { π m = n ≠ 0 0 m ≠ n \int_{-\pi}^{\pi} \sin(mx)\sin(nx)\, dx = \begin{cases} \pi & m = n \neq 0 \\ 0 & m \neq n \end{cases} ∫ − π π sin ( m x ) sin ( n x ) d x = { π 0 m = n = 0 m = n
∫ − π π cos ( m x ) sin ( n x ) d x = 0 f o r a l l m , n \int_{-\pi}^{\pi} \cos(mx)\sin(nx)\, dx = 0 \quad \mathrm{for}\; all\; m, n ∫ − π π cos ( m x ) sin ( n x ) d x = 0 for a l l m , n
To find a n a_n a n cos ( n x ) \cos(nx) cos ( n x ) [ − π , π ] [-\pi, \pi] [ − π , π ] cos ( n x ) \cos(nx) cos ( n x ) a n π = ∫ − π π f ( x ) cos ( n x ) d x a_n \pi = \int_{-\pi}^{\pi} f(x)\cos(nx)\, dx a n π = ∫ − π π f ( x ) cos ( n x ) d x b n b_n b n
7.3 Convergence Theorem 7.1 (Dirichlet”s Theorem). If f f f 2 π 2\pi 2 π
f ( x ) f(x) f ( x ) f f f f ( x + ) + f ( x − ) 2 \frac{f(x^+) + f(x^-)}{2} 2 f ( x + ) + f ( x − ) 7.4 Parseval’s Identity 1 π ∫ − π π ∣ f ( x ) ∣ 2 d x = a 0 2 2 + ∑ n = 1 ∞ ( a n 2 + b n 2 ) \frac{1}{\pi}\int_{-\pi}^{\pi} |f(x)|^2\, dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty}(a_n^2 + b_n^2) π 1 ∫ − π π ∣ f ( x ) ∣ 2 d x = 2 a 0 2 + ∑ n = 1 ∞ ( a n 2 + b n 2 )
Intuition. Parseval’s identity is the infinite-dimensional analogue of the Pythagorean theorem: The “energy” of f f f L 2 L^2 L 2
7.5 Sine and Cosine Series For functions defined on [ 0 , L ] [0, L] [ 0 , L ]
Cosine series (even extension): a n = 2 L ∫ 0 L f ( x ) cos n π x L d x a_n = \frac{2}{L}\int_0^L f(x)\cos\frac{n\pi x}{L}\, dx a n = L 2 ∫ 0 L f ( x ) cos L nπ x d x b n = 0 b_n = 0 b n = 0 Sine series (odd extension): a n = 0 a_n = 0 a n = 0 b n = 2 L ∫ 0 L f ( x ) sin n π x L d x b_n = \frac{2}{L}\int_0^L f(x)\sin\frac{n\pi x}{L}\, dx b n = L 2 ∫ 0 L f ( x ) sin L nπ x d x 7.6 Worked Example: Fourier Sine Series Problem. Find the Fourier series of f ( x ) = x f(x) = x f ( x ) = x ( − π , π ) (-\pi, \pi) ( − π , π ) 2 π 2\pi 2 π
Solution. f f f a n = 0 a_n = 0 a n = 0 n n n
b n = 1 π ∫ − π π x sin ( n x ) d x = 2 π ∫ 0 π x sin ( n x ) d x b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} x\sin(nx)\, dx = \frac{2}{\pi}\int_0^{\pi} x\sin(nx)\, dx b n = π 1 ∫ − π π x sin ( n x ) d x = π 2 ∫ 0 π x sin ( n x ) d x
Integration by parts: u = x u = x u = x d v = sin ( n x ) d x dv = \sin(nx)\, dx d v = sin ( n x ) d x
b n = 2 π [ − x cos ( n x ) n ∣ 0 π + ∫ 0 π cos ( n x ) n d x ] = 2 π [ − π cos ( n π ) n + 0 ] = − 2 cos ( n π ) n = 2 ( − 1 ) n + 1 n b_n = \frac{2}{\pi}\left[-\frac{x\cos(nx)}{n}\Big|_0^{\pi} + \int_0^{\pi} \frac{\cos(nx)}{n}\, dx\right] = \frac{2}{\pi}\left[-\frac{\pi\cos(n\pi)}{n} + 0\right] = \frac{-2\cos(n\pi)}{n} = \frac{2(-1)^{n+1}}{n} b n = π 2 [ − n x c o s ( n x ) 0 π + ∫ 0 π n c o s ( n x ) d x ] = π 2 [ − n π c o s ( nπ ) + 0 ] = n − 2 c o s ( nπ ) = n 2 ( − 1 ) n + 1
x ∼ 2 ∑ n = 1 ∞ ( − 1 ) n + 1 n sin ( n x ) x \sim 2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\sin(nx) x ∼ 2 ∑ n = 1 ∞ n ( − 1 ) n + 1 sin ( n x ) ■ \blacksquare ■
7.7 Worked Example: Fourier Cosine Series Problem. Find the Fourier cosine series of f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 [ 0 , π ] [0, \pi] [ 0 , π ]
Solution Solution. Extend f f f [ − π , π ] [-\pi, \pi] [ − π , π ] b n = 0 b_n = 0 b n = 0 n n n
a 0 = 2 π ∫ 0 π x 2 d x = 2 π ⋅ π 3 3 = 2 π 2 3 a_0 = \frac{2}{\pi}\int_0^{\pi} x^2\, dx = \frac{2}{\pi} \cdot \frac{\pi^3}{3} = \frac{2\pi^2}{3} a 0 = π 2 ∫ 0 π x 2 d x = π 2 ⋅ 3 π 3 = 3 2 π 2
For n ≥ 1 n \geq 1 n ≥ 1 a n = 2 π ∫ 0 π x 2 cos ( n x ) d x a_n = \frac{2}{\pi}\int_0^{\pi} x^2\cos(nx)\, dx a n = π 2 ∫ 0 π x 2 cos ( n x ) d x
Integrating by parts twice:
u = x 2 u = x^2 u = x 2 d v = cos ( n x ) d x dv = \cos(nx)\, dx d v = cos ( n x ) d x d u = 2 x d x du = 2x\, dx d u = 2 x d x v = sin ( n x ) / n v = \sin(nx)/n v = sin ( n x ) / n
a n = 2 π [ x 2 sin ( n x ) n ∣ 0 π − ∫ 0 π 2 x sin ( n x ) n d x ] = − 4 n π ∫ 0 π x sin ( n x ) d x a_n = \frac{2}{\pi}\left[\frac{x^2\sin(nx)}{n}\Big|_0^{\pi} - \int_0^{\pi} \frac{2x\sin(nx)}{n}\, dx\right] = -\frac{4}{n\pi}\int_0^{\pi} x\sin(nx)\, dx a n = π 2 [ n x 2 s i n ( n x ) 0 π − ∫ 0 π n 2 x s i n ( n x ) d x ] = − nπ 4 ∫ 0 π x sin ( n x ) d x
= − 4 n π [ − x cos ( n x ) n ∣ 0 π + ∫ 0 π cos ( n x ) n d x ] = -\frac{4}{n\pi}\left[-\frac{x\cos(nx)}{n}\Big|_0^{\pi} + \int_0^{\pi} \frac{\cos(nx)}{n}\, dx\right] = − nπ 4 [ − n x c o s ( n x ) 0 π + ∫ 0 π n c o s ( n x ) d x ]
= − 4 n π [ − π cos ( n π ) n + 0 ] = 4 cos ( n π ) n 2 = 4 ( − 1 ) n n 2 = -\frac{4}{n\pi}\left[-\frac{\pi\cos(n\pi)}{n} + 0\right] = \frac{4\cos(n\pi)}{n^2} = \frac{4(-1)^n}{n^2} = − nπ 4 [ − n π c o s ( nπ ) + 0 ] = n 2 4 c o s ( nπ ) = n 2 4 ( − 1 ) n
x 2 ∼ π 2 3 + 4 ∑ n = 1 ∞ ( − 1 ) n n 2 cos ( n x ) x^2 \sim \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}\cos(nx) x 2 ∼ 3 π 2 + 4 ∑ n = 1 ∞ n 2 ( − 1 ) n cos ( n x )
Setting x = 0 x = 0 x = 0 0 = π 2 3 + 4 ∑ n = 1 ∞ ( − 1 ) n n 2 0 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} 0 = 3 π 2 + 4 ∑ n = 1 ∞ n 2 ( − 1 ) n ∑ n = 1 ∞ ( − 1 ) n + 1 n 2 = π 2 12 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} = \frac{\pi^2}{12} ∑ n = 1 ∞ n 2 ( − 1 ) n + 1 = 12 π 2 ■ \blacksquare ■
7.8 Complex Fourier Series Using Euler’s formula, the Fourier series can be written in complex form:
f ( x ) ∼ ∑ n = − ∞ ∞ c n e i n x f(x) \sim \sum_{n=-\infty}^{\infty} c_n e^{inx} f ( x ) ∼ ∑ n = − ∞ ∞ c n e in x
Where c n = 1 2 π ∫ − π π f ( x ) e − i n x d x c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{-inx}\, dx c n = 2 π 1 ∫ − π π f ( x ) e − in x d x
The relationship with the real coefficients is c 0 = a 0 / 2 c_0 = a_0/2 c 0 = a 0 /2 c n = ( a n − i b n ) / 2 c_n = (a_n - ib_n)/2 c n = ( a n − i b n ) /2 n > 0 n > 0 n > 0 c − n = c n ‾ c_{-n} = \overline{c_n} c − n = c n f f f
7.9 Worked Example: Parseval’s Identity Problem. Using the Fourier series of f ( x ) = x f(x) = x f ( x ) = x ( − π , π ) (-\pi, \pi) ( − π , π ) ∑ n = 1 ∞ 1 n 2 = π 2 6 \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} ∑ n = 1 ∞ n 2 1 = 6 π 2
Solution Solution. From Section 7.6: a 0 = 0 a_0 = 0 a 0 = 0 a n = 0 a_n = 0 a n = 0 b n = 2 ( − 1 ) n + 1 n b_n = \frac{2(-1)^{n+1}}{n} b n = n 2 ( − 1 ) n + 1
Parseval: 1 π ∫ − π π x 2 d x = ∑ n = 1 ∞ b n 2 = ∑ n = 1 ∞ 4 n 2 \frac{1}{\pi}\int_{-\pi}^{\pi} x^2\, dx = \sum_{n=1}^{\infty} b_n^2 = \sum_{n=1}^{\infty} \frac{4}{n^2} π 1 ∫ − π π x 2 d x = ∑ n = 1 ∞ b n 2 = ∑ n = 1 ∞ n 2 4
1 π ⋅ 2 π 3 3 = 4 ∑ n = 1 ∞ 1 n 2 \frac{1}{\pi} \cdot \frac{2\pi^3}{3} = 4\sum_{n=1}^{\infty} \frac{1}{n^2} π 1 ⋅ 3 2 π 3 = 4 ∑ n = 1 ∞ n 2 1
2 π 2 3 = 4 ∑ n = 1 ∞ 1 n 2 \frac{2\pi^2}{3} = 4\sum_{n=1}^{\infty} \frac{1}{n^2} 3 2 π 2 = 4 ∑ n = 1 ∞ n 2 1
∑ n = 1 ∞ 1 n 2 = π 2 6 \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} ∑ n = 1 ∞ n 2 1 = 6 π 2 ■ \blacksquare ■