## Linear Algebra and Its Applications, Exercise 3.4.23

Exercise 3.4.23. Given the step function $y$ with $y(x) = 1$ for $0 \le x \le \pi$ and $y(x) = 0$ for $\pi < x < 2\pi$, find the following Fourier coefficients:

$a_0 = \frac{(y, 1)}{(1, 1)} \qquad a_1 = \frac{(y, \cos x)}{(\cos x, \cos x)} \qquad b_1 = \frac{(y, \sin x)}{(\sin x, \sin x)}$

Answer: For $a_0$ the numerator is

$(y, 1) = \int_0^{2\pi} y(x) \cdot 1 \;\mathrm{d}x = \int_0^{\pi} 1 \;\mathrm{d}x + \int_{\pi}^{2\pi} 0 \;\mathrm{d}x = \pi$

and the denominator is

$(1, 1) = \int_0^{2\pi} 1^2 \;\mathrm{d}x = 2\pi$

so that $a_0 = \frac{\pi}{2\pi} = \frac{1}{2}$.

For $a_1$ the numerator is

$(y, \cos x) = \int_0^{2\pi} y(x) \cos x \;\mathrm{d}x = \int_0^{\pi} 1 \cdot \cos x \;\mathrm{d}x + \int_{\pi}^{2\pi} 0 \cdot \cos x \;\mathrm{d}x$

$= \int_0^{\pi} \cos x = \sin x \;\big|_0^{\pi} = 0 - 0 = 0$

so that $a_1 = 0$.

For $b_1$ the numerator is

$(y, \sin x) = \int_0^{2\pi} y(x) \sin x \;\mathrm{d}x = \int_0^{\pi} 1 \cdot \sin x \;\mathrm{d}x + \int_{\pi}^{2\pi} 0 \cdot \sin x \;\mathrm{d}x$

$= \int_0^{\pi} \sin x = (-\cos x) \;\big|_0^{\pi} = -(-1) - (-1) = 1 + 1 = 2$

and the denominator is

$(\sin x, \sin x) = \int_0^{2\pi} \sin^2 x \;\mathrm{d}x = \left[ \frac{1}{2}x - \frac{1}{4} \sin 2x \right] \;\big|_0^{2\pi}$

$= \left[ \frac{1}{2}\cdot(2\pi) - \frac{1}{4} \sin 2\pi \right] - \left[ \frac{1}{2} \cdot 0 - \frac{1}{4} \sin 2 \cdot 0 \right] = \pi - \frac{1}{4} \cdot 0 - 0 + \frac{1}{4} \cdot 0 = \pi$

so that $b_1 = \frac{2}{\pi}$.

So we have $a_0 = \frac{1}{2}$, $a_1 = 0$, and $b_1 = \frac{2}{\pi}$.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fifth Edition and the accompanying free online course, and Dr Strang’s other books.

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