Exercise 3.4.21. Given the function on the interval , what is the closest function to ? What is the closest line to ?

Answer: To find the closest function to the function we first project onto the function on the given interval to obtain , and then project onto to obtain .

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

The numerator is

where we used the trigonometric identity .

To integrate we substitute the variable so that . We then have

We then have

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

(Note that we do not need to calculate the denominator in the expression for . 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 onto by taking the dot product of with and then normalizing by dividing by the dot product of with itself:

The numerator is

where we used the trigonometric identity .

To integrate we substitute the variable so that . We then have

We then have

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

So the closest function to is .

To find the closest function to the function we first project onto the constant function with the value 1 on the given interval to obtain , and then project onto the function to obtain .

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

The numerator is

To integrate we substitute the variable so that . We then have

We then have

Since the numerator in the expression for is zero we have . (Recall that the denominator is guaranteed to be positive.)

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

The numerator is

To integrate this we use integration by parts, taking advantage of the formula . (The following is adapted from a post on socratic.org.) We let and . Then is simply , and (the integrand of , as discussed above).

We then have

The second integral we can evaluate by substituting and so that

Substituting for the second integral above we then have

We then have

The denominator in the expression for is

We then have

The straight line closest to the function is thus the line .

ADDENDUM: Suppose that is a continuous function defined on the interval and for some . Then we want to show that the inner product .

The basic idea of the proof is as follows: The function is always nonnegative, and thus its integral over the interval is nonnegative as well. If is nonzero for some then since is continuous will also be nonzero for some interval that includes , with . This implies that the integral of over that subinterval will be positive.

But we also have since and is contained within . So if then we also have and the inner product 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.