Download original notes
PAGE 1

9.2 (continued)

last time: f(t) period 2L given on -L < t < L

\[ f(t) \sim \frac{1}{2} a_0 + \sum_{n=1}^{\infty} \left[ a_n \cos\left(\frac{n\pi t}{L}\right) + b_n \sin\left(\frac{n\pi t}{L}\right) \right] \]
\[ a_n = \frac{1}{L} \int_{-L}^{L} f(t) \cos\left(\frac{n\pi t}{L}\right) dt \]
\[ b_n = \frac{1}{L} \int_{-L}^{L} f(t) \sin\left(\frac{n\pi t}{L}\right) dt \]

usually, we prefer f(t) period 2L but given on 0 < t < 2L

  • first formula stays the same — no assumption on the interval of t
  • the other two may be affected by the change of integration interval
PAGE 2

Periodic Functions and Integration

Consider a function \( g(t) \) with period \( 2L \). It looks like this:

A Cartesian coordinate system showing a periodic function g(t) with period 2L . The horizontal axis is labeled t and the vertical axis is labeled g(t) . The graph shows several cycles of a wave-like function. The region between t = -L and t = L is shaded with diagonal lines, representing the integral of the function over one full period.
Figure 1: Periodic function \( g(t) \) with shaded area from \( -L \) to \( L \).
\[ \int_{-L}^{L} g(t) dt \text{ is the area of the shaded region.} \]
A Cartesian coordinate system showing the same periodic function g(t) . The horizontal axis is labeled with points -2L, -L, -L+a, 0, L, L+a, 2L, 3L . The shaded region is now shifted, starting at t = -L + a and ending at t = L + a . Visually, the total area of the shaded regions remains constant despite the shift.
Figure 2: Periodic function \( g(t) \) with shifted shaded area from \( -L + a \) to \( L + a \).
\[ \int_{-L+a}^{L+a} g(t) dt \text{ is the area of the shaded region.} \]

Notice they are the same.

If we make \( a = L \) then we get:

PAGE 3
A graph of a periodic function g(t) on a Cartesian coordinate system. The horizontal axis is labeled t and the vertical axis is labeled g(t) . The function g(t) oscillates with a period of 2L . Key points on the t -axis are marked at -2L, -L, 0, L, 2L, and 3L . Shaded regions under the curve illustrate the area over one full period, specifically between 0 and L , and between L and 2L .
\[ \int_{0}^{2L} g(t) dt = \int_{-L}^{L} g(t) dt = \int_{-L+a}^{L+a} g(t) dt \]

(more natural interval)

the same as long as the integration interval is \( 2L \) (one period)

\( \rightarrow g(t) \) MUST be periodic w/ period \( 2L \)

look at
\[ a_n = \frac{1}{L} \int_{-L}^{L} f(t) \cos\left( \frac{n \pi t}{L} \right) dt \]
collectively must have \( 2L \) as period
  • \( f(t) \) has period \( 2L \) by definition
  • \( \cos\left( \frac{n \pi t}{L} \right) \) has period of \( \frac{2\pi}{n\pi/L} = \frac{2L}{n} \) fundamental period

\( \rightarrow \) but any integer multiple is also a period

\( \rightarrow \) so \( 2L \) is a period of \( \cos\left( \frac{n \pi t}{L} \right) \)

PAGE 4

So, \( f(t) \cos\left( \frac{n \pi t}{L} \right) \) also has period \( 2L \).

Therefore,

\[ a_n = \frac{1}{L} \int_{0}^{2L} f(t) \cos\left( \frac{n \pi t}{L} \right) dt \]
\[ b_n = \frac{1}{L} \int_{0}^{2L} f(t) \sin\left( \frac{n \pi t}{L} \right) dt \]

Alternative Formulation

Alternatively, we can choose to work with period \( P = 2L \).

Change \( L = P/2 \)

\[ f(t) \approx \frac{1}{2} a_0 + \sum_{n=1}^{\infty} \left[ a_n \cos\left( \frac{2n \pi}{P} t \right) + b_n \sin\left( \frac{2n \pi}{P} t \right) \right] \]
\[ a_n = \frac{2}{P} \int_{0}^{P} f(t) \cos\left( \frac{2n \pi}{P} t \right) dt \]
\[ b_n = \frac{2}{P} \int_{0}^{P} f(t) \sin\left( \frac{2n \pi}{P} t \right) dt \]
PAGE 5

Fourier Series Example

Consider the function defined by:

\[ f(t) = t \quad \text{for } 0 < t < \pi \]

Period \( T = \pi \) (which implies \( L = \pi/2 \))

A graph of a periodic sawtooth wave function f(t) = t. The horizontal axis is labeled t and the vertical axis is labeled f(t). The function consists of repeating diagonal line segments. Each segment starts at a multiple of pi on the t-axis (e.g., -2 pi, -pi, 0, pi, 2 pi) and rises linearly with a slope of 1 until it reaches a height of pi at the next multiple of pi, where it then drops vertically back to zero. The peaks are marked with a dashed vertical line down to the t-axis at pi, 2 pi, and 3 pi. The vertical axis has a tick mark at pi.
Figure 1: Periodic sawtooth function \( f(t) = t \) with period \( \pi \).

Calculating Fourier Coefficients

We calculate the coefficients for the Fourier series expansion:

\[ a_0 = \frac{1}{\pi/2} \int_{0}^{\pi} t \, dt = \frac{2}{\pi} \left( \frac{1}{2} \cdot \pi \cdot \pi \right) = \pi \]

Note: The integral \( \int_{0}^{\pi} t \, dt \) represents the area under the curve from 0 to \( \pi \).

\[ a_n = \frac{1}{\pi/2} \int_{0}^{\pi} t \cos(2nt) \, dt = \dots = 0 \]

\[ b_n = \frac{1}{\pi/2} \int_{0}^{\pi} t \sin(2nt) \, dt = \dots = -\frac{1}{n} \]

Fourier Series Expansion

Substituting the coefficients into the general form (where the constant term is \( \frac{a_0}{2} \)):

\[ t \sim \frac{\pi}{2} + \sum_{n=1}^{\infty} -\frac{1}{n} \sin(2nt) \]

\[ t \sim \frac{\pi}{2} - \sin(2t) - \frac{1}{2} \sin(4t) - \frac{1}{3} \sin(6t) - \frac{1}{4} \sin(8t) - \dots \]

Evaluation at a Point

Let's evaluate at \( t = \frac{\pi}{4} \):

Since \( t = \frac{\pi}{4} \) is where \( f(t) \) is continuous, the Fourier series converges to the value of \( f(t) \).

PAGE 6
\[ \frac{\pi}{4} = \frac{\pi}{2} - 1 + \frac{1}{3} - \frac{1}{5} + \frac{1}{7} - \frac{1}{9} + \dots \]
\[ -\frac{\pi}{4} = -1 + \frac{1}{3} - \frac{1}{5} + \frac{1}{7} - \frac{1}{9} + \dots \]
\[ \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \dots \]Leibniz series

Example

\[ f(t) = \begin{cases} \pi - t & 0 < t < \pi \\ 0 & \pi < t < 2\pi \end{cases} \]
period \( 2\pi \) (\( L = \pi \))
A Cartesian coordinate system showing a periodic function f(t). The horizontal axis t ranges from -3\pi to 3\pi. The vertical axis f(t) has a maximum value of \pi. The function is defined as a sawtooth wave: it is a downward sloping line from (0, \pi) to (\pi, 0), and then it is zero from \pi to 2\pi. This pattern repeats every 2\pi units. Vertical dashed lines indicate the discontinuities at multiples of 2\pi.
Graph of the periodic function \( f(t) \) with period \( 2\pi \).
\[ a_0 = \frac{1}{\pi} \int_{0}^{2\pi} f(t) \, dt = \frac{1}{\pi} \left( \frac{1}{2} \cdot \pi \cdot \pi \right) = \frac{\pi}{2} \]
\[ a_n = \frac{1}{\pi} \int_{0}^{2\pi} f(t) \cos(nt) \, dt = \dots = \frac{1 - (-1)^n}{n^2 \pi} = \begin{cases} 0 & n \text{ is even} \\ \frac{2}{n^2 \pi} & n \text{ is odd} \end{cases} \]
\[ b_n = \frac{1}{\pi} \int_{0}^{2\pi} f(t) \sin(nt) \, dt = \dots = \frac{1}{n} \]
PAGE 7

Fourier Series Convergence and the Basel Problem

The Fourier series for the function \( f(t) \) is given by:

\[ f(t) \sim \frac{\pi}{4} + \sum_{n \text{ odd}}^{\infty} \frac{2}{n^2 \pi} \cos(nt) + \sum_{n=1}^{\infty} \frac{1}{n} \sin(nt) \]

At \( t = 0 \), the function is NOT continuous.

The Fourier series converges to the average of the left and right limits:

\[ \frac{f(0^-) + f(0^+)}{2} = \frac{\pi}{2} \]

Evaluating the series at \( t = 0 \):

\[ \frac{\pi}{2} = \frac{\pi}{4} + \sum_{n \text{ odd}}^{\infty} \frac{2}{n^2 \pi} \]
\[ = \frac{\pi}{4} + \frac{2}{\pi} \left( 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \dots \right) \]

Simplify

\[ \frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \dots \]

This represents the sum of reciprocals of odd squares.

On HW, you are asked to show:

\[ \frac{\pi^2}{6} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \dots \]

(Basel problem)