Linear Algebra and Its Applications, Exercise 3.4.21

Exercise 3.4.21. Given the function f(x) = \sin 2x on the interval -\pi \le x \le \pi, what is the closest function a \cos x + b \sin x to f? What is the closest line c + dx to f?

Answer: To find the closest function a \cos x + b \sin x to the function f(x) = \sin 2x we first project f onto the function \cos x on the given interval to obtain a, and then project f onto \sin x to obtain b.

We project f onto \cos x by taking the dot product of f with \cos x and then normalizing by dividing by the dot product of \cos x with itself:

a = (f, \cos x)/(\cos x, \cos x)

The numerator is

(f, \cos x) = \int_{-\pi}^{\pi} f(x) \cos x \;\mathrm{d}x = \int_{-\pi}^{\pi} \sin 2x \cos x \;\mathrm{d}x

= 2 \int_{-\pi}^{\pi} \sin x \cos^2 x \;\mathrm{d}x

where we used the trigonometric identity \sin 2\theta = \sin \theta \cos \theta.

To integrate we substitute the variable u = \cos x so that \mathrm{d}u = -\sin x \;\mathrm{d}x. We then have

\int \sin x \cos^2 x \;\mathrm{d}x = -\int \cos^2 x (-\sin x) \;\mathrm{d}x

-\int u^2 \;\mathrm{d}u = -\frac{1}{3}u^3 = -\frac{1}{3} \cos^3 x

We then have

(f, \cos x) = 2 \int_{-\pi}^{\pi} \sin x \cos^2 x \;\mathrm{d}x = 2 (-\frac{1}{3} \cos^3 x) \;\big|_{-\pi}^{\pi}

= -\frac{2}{3} \cos^3 \pi - [-\frac{2}{3} \cos^3 (-\pi)] = -\frac{2}{3} (-1)^3 - [-\frac{2}{3} (-1)^3]

= -\frac{2}{3} \cdot (-1) - [-\frac{2}{3} \cdot (-1)] = \frac{2}{3} - \frac{2}{3} = 0

Since the numerator in the expression for a is zero, we have a = 0.

(Note that we do not need to calculate the denominator in the expression for a. We know it must be positive, and thus the quotient is defined. See below for a sketch of a proof of this.)

We next project f onto \sin x by taking the dot product of f with \sin x and then normalizing by dividing by the dot product of \sin x with itself:

a = (f, \sin x)/(\sin x, \sin x)

The numerator is

(f, \sin x) = \int_{-\pi}^{\pi} f(x) \sin x \;\mathrm{d}x = \int_{-\pi}^{\pi} \sin 2x \sin x \;\mathrm{d}x

= 2 \int_{-\pi}^{\pi} \sin^2 x \cos x \;\mathrm{d}x

where we used the trigonometric identity \sin 2\theta = \sin \theta \cos \theta.

To integrate we substitute the variable u = \sin x so that \mathrm{d}u = \cos x \;\mathrm{d}x. We then have

\int \sin^2 x \cos x \;\mathrm{d}x = \int u^2 \;\mathrm{d}u = \frac{1}{3}u^3 = \frac{1}{3} \sin^3 x

We then have

(f, \sin x) = 2 \int_{-\pi}^{\pi} \sin^2 x \cos x \;\mathrm{d}x = 2 (\frac{1}{3} \sin^3 x) \;\big|_{-\pi}^{\pi}

= \frac{2}{3} \sin^3 \pi - \frac{2}{3} \sin^3 (-\pi) = \frac{2}{3} (0)^3 - \frac{2}{3} (0)^3

= 0 - 0 = 0

Since the numerator in the expression for b is zero, we have b = 0. (Again, we are guaranteed that the denominator is positive and the quotient defined.)

So the closest function a \cos x + b \sin x to f(x) = \sin 2x is 0 \cdot \cos x + 0 \cdot \sin x = 0.

To find the closest function c + dx to the function f(x) = \sin 2x we first project f onto the constant function with the value 1 on the given interval to obtain c, and then project f onto the function x to obtain d.

We project f onto the constant function with value 1 by taking the dot product of f with 1 and then normalizing by dividing by the dot product of 1 with itself:

c = (f, 1)/(1, 1)

The numerator is

(f, 1) = \int_{-\pi}^{\pi} f(x) \cdot 1 \;\mathrm{d}x = \int_{-\pi}^{\pi} \sin 2x \;\mathrm{d}x

To integrate we substitute the variable u = 2x so that \mathrm{d}u = 2 \;\mathrm{d}x. We then have

\int \sin 2x \;\mathrm{d}x = \int \frac{1}{2} \sin 2x \cdot 2 \;\mathrm{d}x

= \frac{1}{2} \int \sin u \;\mathrm{d}u = \frac{1}{2}(-\cos u) = -\frac{1}{2} \cos 2x

We then have

(f, 1) = \int_{-\pi}^{\pi} \sin 2x \;\mathrm{d}x = -\frac{1}{2} \cos 2x \;\big|_{-\pi}^{\pi}

= -\frac{1}{2} \cos 2\pi - (-\frac{1}{2} \cos (-2\pi) = -\frac{1}{2} (1)^3 - (-\frac{1}{2} (1)^3

= -\frac{1}{2} + \frac{1}{2}= 0

Since the numerator in the expression for c = (f, 1)/(1, 1) is zero we have c = 0. (Recall that the denominator is guaranteed to be positive.)

We project f onto the function x by taking the dot product of f with x and then normalizing by dividing by the dot product of x with itself:

d = (f, x)/(x, x)

The numerator is

(f, x) = \int_{-\pi}^{\pi} f(x) \cdot x \;\mathrm{d}x = \int_{-\pi}^{\pi} x \sin 2x \;\mathrm{d}x

To integrate this we use integration by parts, taking advantage of the formula \int u \;\mathrm{d}v = uv - \int v \;\mathrm{d}u. (The following is adapted from a post on socratic.org.) We let \mathrm{d}v = \sin 2x \;\mathrm{d}x and u = x. Then \mathrm{d}u is simply \mathrm{d}x, and v = -\frac{1}{2} \cos 2x (the integrand of \sin 2x, as discussed above).

We then have

\int x \sin x \;\mathrm{d}x = \int u \;\mathrm{d}v = uv - \int v \;\mathrm{d}u

= x (-\frac{1}{2} \cos 2x) - \int (-\frac{1}{2} \cos 2x) \;\mathrm{d}x

= -\frac{1}{2} x \cos 2x + \frac{1}{2} \int \cos 2x \;\mathrm{d}x

The second integral we can evaluate by substituting w = 2x and \mathrm{d}w = 2 \;\mathrm{d}x so that

\int \cos 2x \;\mathrm{d}x = \frac{1}{2} \int \cos w \;\mathrm{d}w = \frac{1}{2} \sin w = \frac{1}{2} \sin 2x

Substituting for the second integral above we then have

\int x \sin x \;\mathrm{d}x = -\frac{1}{2} x \cos 2x + \frac{1}{2} \int \cos 2x \;\mathrm{d}x = -\frac{1}{2} x \cos 2x + \frac{1}{2} (\frac{1}{2} \sin 2x)

= -\frac{1}{2} x \cos 2x + \frac{1}{4} \sin 2x

We then have

(f, x) = \int_{-\pi}^{\pi} x \sin 2x \;\mathrm{d}x = -\frac{1}{2} x \cos 2x \;\big|_{-\pi}^{\pi} + \frac{1}{4} \sin 2x \;\big|_{-\pi}^{\pi}

= -\frac{1}{2} \pi \cos 2\pi - (-\frac{1}{2} (-\pi) \cos (-2\pi) + \frac{1}{4} \sin 2\pi - \frac{1}{4} \sin 2(-\pi)

= -\frac{1}{2} \pi \cdot 1 + \frac{1}{2} (-\pi) \cdot 1 + \frac{1}{4} \cdot 0 - \frac{1}{4} \cdot 0 = -\frac{\pi}{2} - \frac{\pi}{2}= -\pi

The denominator in the expression for d is

(x, x) = \int_{-\pi}^{\pi} x^2 \;\mathrm{d}x = \frac{1}{3} x^3 \;\big|_{-\pi}^{\pi}

= \frac{1}{3} \pi^3 - \frac{1}{3} (-\pi)^3 = \frac{2}{3} \pi^3

We then have

d = (f, x)/(x, x) = -\pi / (\frac{2}{3} \pi^3) = -\frac{3}{2\pi^2}

The straight line c + dx closest to the function \sin 2x is thus the line -\frac{3}{2\pi^2} x.

ADDENDUM: Suppose that g is a continuous function defined on the interval [a, b] and g(t) \ne 0 for some a \le t \le b. Then we want to show that the inner product (g, g) > 0.

The basic idea of the proof is as follows: The function g^2 is always nonnegative, and thus its integral over the interval [a, b] is nonnegative as well. If g(t) is nonzero for some a \le t \le b then since g is continuous g will also be nonzero for some interval [c, d] that includes t, with a \le c < d \le b. This implies that the integral of g^2 over that subinterval [c, d] will be positive.

But we also have \int_a^b g(x)^2 \;\mathrm{d}x \ge \int_c^d g(x)^2 \;\mathrm{d}x since g(x)^2 \ge 0 and [c, d] is contained within [a, b]. So if \int_c^d g(x)^2 \;\mathrm{d}x > 0 then we also have \int_a^b g(x)^2 \;\mathrm{d}x > 0 and the inner product (g, g) is positive.

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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