## Linear Algebra and Its Applications, Exercise 3.4.22

Exercise 3.4.22. Given an arbitrary function $y$ find the coefficient $b_1$ that minimizes the quantity $\|b_1\sin x - y\|^2 = \int_0^{2\pi} (b_1\sin x - y(x))^2 \;\mathrm{d}x$

(Use the method of setting the derivative to zero.) How does this value of $b_1$ compare with the Fourier coefficient $b_1$? What is $b_1$ if $y(x) = \cos x$?

Answer: We are looking for a value of $b_1$ that minimizes the expression on the right, so we need to differentiate with respect to $b_1$. Expanding the right-hand side of the equation above, we have $\int_0^{2\pi} (b_1\sin x - y(x))^2 \;\mathrm{d}x = \int_0^{2\pi} [b_1^2\sin^2 x - 2b_1y(x)\sin x + y(x)^2] \;\mathrm{d}x$ $= \int_0^{2\pi} b_1^2\sin^2 x \;\mathrm{d}x - 2 \int_0^{2\pi} b_1y(x)\sin x \;\mathrm{d}x + \int_0^{2\pi} y(x)^2 \;\mathrm{d}x$

Since $b_1$ is not dependent on $x$ we can pull it out of the integral, so that $\int_0^{2\pi} (b_1\sin x - y(x))^2 \;\mathrm{d}x = b_1^2 \int_0^{2\pi} \sin^2 x \;\mathrm{d}x - 2b_1 \int_0^{2\pi} y(x) \sin x \;\mathrm{d}x + \int_0^{2\pi} y(x)^2 \;\mathrm{d}x$

Differentiating with respect to $b_1$ we have $\frac{\mathrm{d}}{\mathrm{d}b_1} \int_0^{2\pi} (b_1\sin x - y(x))^2 \;\mathrm{d}x$ $\frac{\mathrm{d}}{\mathrm{d}b_1} \left[ b_1^2 \int_0^{2\pi} \sin^2 x \;\mathrm{d}x - 2b_1 \int_0^{2\pi} y(x) \sin x \;\mathrm{d}x + \int_0^{2\pi} y(x)^2 \;\mathrm{d}x \right]$ $= 2b_1 \int_0^{2\pi} \sin^2 x \;\mathrm{d}x - 2 \int_0^{2\pi} y(x) \sin x \;\mathrm{d}x$

Equating the derivative to zero gives us $2b_1 \int_0^{2\pi} \sin^2 x \;\mathrm{d}x = 2 \int_0^{2\pi} y(x) \sin x \;\mathrm{d}x$

or $b_1 = \left( \int_0^{2\pi} y(x) \sin x \;\mathrm{d}x \right) / \left( \int_0^{2\pi} \sin^2 x \;\mathrm{d}x \right)$

Note that this is identical to the expression for the Fourier coefficient $b_1$ on page 178; the numerator is the dot product of $y(x)$ with $\sin x$ and the denominator is the dot product of $\sin x$ with itself.

If $y(x) = \cos x$ then the numerator of $b_1$ becomes $\int_0^{2\pi} \cos x \sin x \;\mathrm{d}x = 0$

since $\cos x$ and $\sin x$ are orthogonal, and we therefore have $b_1 = 0$.

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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