The Orthogonality

Appendix to A Radical Approach to Real Analysis 2nd edition c 2006 David M. Bressoud

March 16, 2006

The term orthogonal means “at right .” The concept comes from , but it is not too much of a stretch to see how it comes to be applied to functions.

An easy way to determine whether or not two vectors are orthogonal is to take their inner product (also called the ). The inner product has two equivalent definitions. On the one hand, it is the product of the lengths of the vectors multiplied by the cosine of the between them,

~v · ~w = k~vk k~wk cos θ.

On the other hand, if we know the decomposition of these vectors into the unit vectors, then the dot product is the sum of the products of the corresponding coefficients,     v1~ı + v2~ + v3~k · w1~ı + w2~ + w3~k = v1w1 + v2w2 + v3w3.

The first definition gives meaning to the the inner product. The second makes it easy to calculate. From the first definition, we can use the inner product to decide whether or not two vectors are orthogonal: They are orthogonal if and only if their inner product is 0. We can also use it to find the or length of any vector. Since the angle between any vector and itself is 0, we have that √ k~vk = ~v · ~v.

Functions are like vectors in that the sum of two functions is another , and any constant multiple of a function is a function. If we can define a natural inner product on functions, then we can use it to find the norm of a function (analogous to the length of a vector), and we can use it to define orthonality of functions.

A natural inner product for functions is given by the of their ordinary product. This is the limit of a sum of products, and so really is analogous to the inner product of vectors. We denote this inner product by Z 1 hf, gi = f(x)g(x) dx. −1

1 The Orthogonality Relation 2

It follows that the norm or “length” of a function is

Z 1 1/2 kfk = phf, fi = f 2(x) dx . −1

Two functions are orthogonal if and only if

Z 1 hf, gi = f(x)g(x) dx = 0. −1

The functions  cos(πx/2), cos(3πx/2), cos(5πx/2), cos(7πx/2),... are pairwise orthogonal—any two distinct functions from this list are orthogonal—and their norms are all 1:  (2m − 1)πx (2n − 1)πx cos , cos 2 2 Z 1 (2m − 1)πx (2n − 1)πx = cos cos dx −1 2 2  0, if m 6= n = (1) 1, if m = n.

To prove equation (1), we start with the trigonometric identity, 1 cos(A) cos(B) = cos(A + B) + cos(A − B) . 2 When we use this to simplify our integral, we get that

Z 1 (2m − 1)πx (2n − 1)πx cos cos dx −1 2 2 1 Z 1 = cos (m + n − 1)πx + cos (m − n)πx dx. 2 −1 If m = n, this is

" #1 1 Z 1 1 sin (2n + 1)πx cos (2n + 1)πx + 1 dx = + x = 1. 2 −1 2 (2n − 1)π −1 If m 6= n, this is " #1 1 sin (m + n − 1)πx sin (m − n)πx + = 0. 2 (m + n − 1)π (m − n)π −1

Appendix to A Radical Approach to Real Analysis 2nd edition. c 2006 David M. Bressoud