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Chapter 21

Visualizing orthogonal functions.

Recall for continuous (e.g., smooth) functions f,g:[0,L]R, their dot product is defined to be f,g:=1L0Lf(x)g(x)dx, (This is not denoted as fg, to avoid confusion with usual multiplication of functions.) When f,g=0 we say f and g are orthogonal. We write f,g[0,L] rather than f,g if we want to emphasize L. (One can similarly define f,g[a,b]=1baabf(x)g(x)dx for any a<b and f,g:[a,b]R.)

One thing to keep in mind is that the dot product of two functions is not easily visualized in terms of their graphs, even for the orthogonality condition.

What actually matters about dot products of functions that they does satisfy the usual properties of the dot product that we are familiar with in Rn, and so allow us to apply to functions some of the visual intuition we have for geometry with n-vectors. For continuous functions X(x) and Y(x) from [0,L] to R, the following hold:

Moreover, if f1,f2,,fm are continuous functions on [0,L] that are not identically zero and are pairwise orthogonal (i.e., fi,fj=0 when ij) then they are linearly independent.


Example 1.

For example, consider the functions f1(x)=sin(x)/2 and f2(x)=cos(x)/2 on the interval [0,2π]. We find

So f1 and f2 are orthogonal. What's more, if we define Fθ(x):=cos(θ)f1(x)+sin(θ)f2(x) and Fθ+π/2(x):=sin(θ)f1(x)+cos(θ)f2(x), then using the linearity of the dot product we find

Fθ,Fθ+π/2=cos(θ)f1(x)+sin(θ)f2(x),sin(θ)f1(x)+cos(θ)f2(x)=cos(θ)sin(θ)f1(x),f1(x)+cos2(θ)f1(x),f2(x)sin2(θ)f2(x),f1(x)+cos(θ)sin(θ)f2(x),f2(x)=cos(θ)sin(θ)+00+cos(θ)sin(θ)=0.

So Fθ and Fθ+π/2 are orthogonal as well.

Below is a visualization of Fθ and Fθ+π/2. Choose θ by dragging P on the unit circle, and see how Fθ and Fθ+π/2 change.

Parameter
0.20.40.60.81−0.2−0.4−0.6−0.8−10.20.40.60.81−0.2−0.4−0.6−0.8−1
Drag P:
P=eiθ
θ
θ
cos
sin

θ= 0.38π radians = 68.2 degrees

sin(θ)= 0.93

cos(θ)= 0.37

Graph
0123456−0.500.5
F​θF​θ + π/2Orthogonality: Example 1x

Example 2.

Here is another exmaple. Consider the functions g1(x)=3/52(5x8) and g2(x)=78x3 on the interval [0,2]. The ugly coefficients make the gj's have length 1: we find

So g1 and g2 are orthogonal. What's more, if we define Gθ(x):=cos(θ)g1(x)+sin(θ)g2(x) and Gθ+π/2(x):=sin(θ)g1(x)+cos(θ)g2(x), then using the linearity of the dot product, again we find

Gθ,Gθ+π/2=0,

so Gθ and Gθ+π/2 are orthogonal as well.

Below is a visualization of Gθ and Gθ+π/2. Choose θ by dragging P on the unit circle, and see how Gθ and Gθ+π/2 change. Notice that it is hard to see orthogonality just by looking at the graphs.

Parameter
0.20.40.60.81−0.2−0.4−0.6−0.8−10.20.40.60.81−0.2−0.4−0.6−0.8−1
Drag P:
P=eiθ
θ
θ
cos
sin

θ= 0.38π radians = 68.2 degrees

sin(θ)= 0.93

cos(θ)= 0.37

Graph
00.511.52−3−2−10123
G​θG​θ + π/2Orthogonality: Example 2x