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9.2 (continued)

Last time: \( f(t) \) period \( 2L \), given on \( -L < t < L \)

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

\[ a_n = \frac{1}{L} \int_{-L}^{L} f(t) \cos\left(\frac{n\pi t}{L}\right) dt \quad n = 0, 1, 2, 3, \dots \]

\[ b_n = \frac{1}{L} \int_{-L}^{L} f(t) \sin\left(\frac{n\pi t}{L}\right) dt \quad n = 1, 2, 3, \dots \]

In practice, \( -L < t < L \) is not always convenient.

Often we want \( 0 < t < 2L \).

How do the formulas above change?

\[ f(t) = \frac{1}{2} a_0 + \sum_{n=1}^{\infty} \left[ a_n \cos\left(\frac{n\pi t}{L}\right) + b_n \sin\left(\frac{n\pi t}{L}\right) \right] \]stays the same

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Periodic Function Integration Properties

If \( g(t) \) has period \( 2L \), it should look like:

A graph of a periodic function g(t) with period 2L . The horizontal axis is t and the vertical axis is f(t) . The area under the curve from -L to L is shaded in red hatching. The integral \int_{-L}^{L} g(t) dt represents this shaded area. — Figure: A graph of a periodic function \( g(t) \) with period \( 2L \). The horizontal axis is \( t \) and the vertical axis is \( f(t) \). The area under the curve from \( -L \) to \( L \) is shaded in red hatching. The integral \( \int_{-L}^{L} g(t) dt \) represents this shaded area.

\[ \int_{-L}^{L} g(t) dt \text{ is the shaded area} \]

The same periodic function g(t) as above, but the integration interval is shifted by a constant a . The shaded area now spans from -L + a to L + a . The total shaded area remains visually equivalent to the area in the first graph. — Figure: The same periodic function \( g(t) \) as above, but the integration interval is shifted by a constant \( a \). The shaded area now spans from \( -L + a \) to \( L + a \). The total shaded area remains visually equivalent to the area in the first graph.

\[ \int_{-L+a}^{L+a} g(t) dt \text{ is the shaded area} \]

Notice it's the same

The same periodic function g(t) with the integration interval shifted to span from 0 to 2L . The shaded area covers one full period of the function, demonstrating that the integral over any full period is identical. — Figure: The same periodic function \( g(t) \) with the integration interval shifted to span from \( 0 \) to \( 2L \). The shaded area covers one full period of the function, demonstrating that the integral over any full period is identical.

\[ \int_{0}^{2L} g(t) dt \text{ is still the same!} \]

Key Takeaway:

For any periodic function with period \( T \), the integral over any interval of length \( T \) is constant: \[ \int_{a}^{a+T} f(t) dt = \int_{0}^{T} f(t) dt \]

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we see

\[ \int_{-L}^{L} g(t) dt = \int_{-L+a}^{L+a} g(t) dt = \int_{0}^{2L} g(t) dt \]

must have period \( 2L \)

let's look at the coefficient formulas

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

must have period \( 2L \)

\( f(t) \) has period \( 2L \) by definition
\( \cos\left(\frac{n\pi t}{L}\right) \) has period \( \frac{2\pi}{n\pi/L} = \frac{2L}{n} \) fundamental period
but any integer multiple is also a period
so, \( \cos\left(\frac{n\pi t}{L}\right) \) also has period of \( 2L \)
and \( f(t) \cos\left(\frac{n\pi t}{L}\right) \) also has period of \( 2L \)

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So, we can shift the integration interval however we want as long as its length is \(2L\) (period)

More natural Fourier series formulas:

\(f(t)\) period \(2L\)

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

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

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

Or, since period \(P = 2L\)

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

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

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

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Fourier Series Example

Consider the function defined by:

\[ f(t) = t \quad 0 < t < \pi \quad \text{period } \pi \quad (L = \frac{\pi}{2}) \]

A graph of a periodic sawtooth function f(t) = t with period \pi . The horizontal axis is t and the vertical axis is f(t) . The function consists of repeating line segments with a slope of 1, starting at t = 0, \pi, 2\pi , etc., and ending at t = \pi, 2\pi, 3\pi , etc. with a vertical jump back to zero at each multiple of \pi . The peak value of each segment is \pi . — Figure: A graph of a periodic sawtooth function f(t) = t with period \( \pi \). The horizontal axis is \( t \) and the vertical axis is \( f(t) \). The function consists of repeating line segments with a slope of 1, starting at \( t = 0, \pi, 2\pi \), etc., and ending at \( t = \pi, 2\pi, 3\pi \), etc. with a vertical jump back to zero at each multiple of \( \pi \). The peak value of each segment is \( \pi \).

Calculating Fourier Coefficients

First, we calculate the constant term \( a_0 \):

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

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

Next, we calculate the cosine coefficients \( a_n \). Note that the frequency term is \( \frac{n\pi t}{L} = \frac{n\pi t}{\pi/2} = 2nt \):

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

Then, we calculate the sine coefficients \( b_n \):

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

Fourier Series Representation

The resulting Fourier series for \( f(t) \) is:

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

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Fourier Series Evaluation and Leibniz Series

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

evaluate at \( t = \frac{\pi}{4} \)

A Cartesian coordinate graph showing a linear function f(t) = t. The horizontal axis is labeled t and the vertical axis is labeled f(t). A straight line passes through the origin (0,0). Points are marked at t = \frac{\pi}{4} where f(t) = \frac{\pi}{4} , and t = \pi where f(t) = \pi . — Figure: A Cartesian coordinate graph showing a linear function f(t) = t. The horizontal axis is labeled t and the vertical axis is labeled f(t). A straight line passes through the origin (0,0). Points are marked at t = \( \frac{\pi}{4} \) where f(t) = \( \frac{\pi}{4} \), and t = \( \pi \) where f(t) = \( \pi \).

\[\frac{\pi}{4} = \frac{\pi}{2} - 1 + \frac{1}{3} - \frac{1}{5} + \frac{1}{7} - \dots\]

\[\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \dots \quad \text{Leibniz series}\]

Example: Piecewise Periodic Function

\[f(t) = \begin{cases} \pi - t & 0 < t < \pi \\ 0 & \pi < t < 2\pi \end{cases}\]

period \( 2\pi \) (\( L = \pi \))

A graph of a periodic sawtooth-like function f(t). The horizontal axis is t and the vertical axis is f(t). For the interval [0, \pi ], the function is a downward sloping line starting at (0, \pi ) and ending at ( \pi , 0). For the interval [ \pi , 2\pi ], the function is zero (flat on the t-axis). This pattern repeats periodically. Vertical dashed lines indicate the discontinuities at multiples of \pi . — Figure: A graph of a periodic sawtooth-like function f(t). The horizontal axis is t and the vertical axis is f(t). For the interval [0, \( \pi \)], the function is a downward sloping line starting at (0, \( \pi \)) and ending at (\( \pi \), 0). For the interval [\( \pi \), \( 2\pi \)], the function is zero (flat on the t-axis). This pattern repeats periodically. Vertical dashed lines indicate the discontinuities at multiples of \( \pi \).

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Fourier Series Coefficients and Evaluation

Calculation of Coefficients

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

\[ a_n = \frac{1}{\pi} \int_{0}^{2\pi} f(t) \cos(nt) dt = \dots = \frac{1 - (-1)^n}{n^2 \pi} = \begin{cases} 0 & n \text{ is even} \\ \frac{2}{n^2 \pi} & n \text{ is odd} \end{cases} \]

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

Fourier Series Representation

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

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

Figure: A Cartesian coordinate graph of the function f(t). The horizontal axis t ranges from -pi to pi. The vertical axis f(t) has a mark at pi. The function is zero for t between -pi and 0 (represented by a green line on the axis). At t=0, there is a jump to f(t)=pi. From t=0 to t=pi, the function is a straight line decreasing from (0, pi) to (pi, 0).

Function is discontinuous at \( t = 0 \)

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

Evaluation at \( t = 0 \)

Let's evaluate at \( t = 0 \):

\[ \frac{\pi}{2} = \frac{\pi}{4} + \sum_{n \text{ odd}}^{\infty} \frac{2}{n^2 \pi} \underbrace{\cos(n \cdot 0)}_{1} \]

\[ = \frac{\pi}{4} + \frac{2}{\pi} \left( 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \dots \right) \]

Simplify:

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

Sum of reciprocals of odd squares

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On HW, you are asked to show (Basel problem)

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