Chapter 21

Visualizing orthogonal functions.

Recall for continuous (e.g., smooth) functions $f,g:[0,L]\to \mathbf{R}$, their dot product is defined to be $$⟨f,g⟩:=\frac{1}{L}{\int }_0^{L}f(x)g(x)dx,$$ (This is not denoted as $f\cdot g$, 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 $⟨f,g{⟩}_{[a,b]}=\frac{1}{b-a}{\int }_a^{b}f(x)g(x)dx$ for any $a<b$ and $f,g:[a,b]\to \mathbf{R}$.)

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 ${\mathbf{R}}^n$, 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 $\mathbf{R}$, the following hold:

• $⟨c_1{X}_1+c_2{X}_2,Y⟩=c_1⟨X_1,Y⟩+c_2⟨X_2,Y⟩$,
• $⟨X,c_1{Y}_1+c_2{Y}_2⟩=c_1⟨X,Y_1⟩+c_2⟨X,Y_2⟩$,
• $⟨Y,X⟩=⟨X,Y⟩$,
• $⟨X,X⟩={\int }_0^{L}X(x{)}^{2}dx\ge 0$, and $⟨X,X⟩=0$ precisely when $X$ is the zero function.

Moreover, if $f_1,f_2,\dots ,f_m$ are continuous functions on $[0,L]$ that are not identically zero and are pairwise orthogonal (i.e., $⟨f_i,f_j⟩=0$ when $i\ne j$) then they are linearly independent.

Example 1.

For example, consider the functions $f_1(x)=\sin (x)/\sqrt{2}$ and $f_2(x)=\cos (x)/\sqrt{2}$ on the interval $[0,2\pi ]$. We find

• $⟨f_1,f_1⟩=1$,
• $⟨f_2,f_2⟩=1$,
• $⟨f_1,f_2⟩=0$.

So $f_1$ and $f_2$ are orthogonal. What's more, if we define $F_{\theta }(x):=\cos (\theta )f_1(x)+\sin (\theta )f_2(x)$ and $F_{\theta +\pi /2}(x):=-\sin (\theta )f_1(x)+\cos (\theta )f_2(x)$, then using the linearity of the dot product we find

$$\begin{array}{rl}⟨F_{\theta },F_{\theta +\pi /2}⟩& =⟨\cos (\theta )f_1(x)+\sin (\theta )f_2(x),-\sin (\theta )f_1(x)+\cos (\theta )f_2(x)⟩\\ & =-\cos (\theta )\sin (\theta )⟨f_1(x),f_1(x)⟩+{\cos }^2(\theta )⟨f_1(x),f_2(x)⟩-{\sin }^2(\theta )⟨f_2(x),f_1(x)⟩+\cos (\theta )\sin (\theta )⟨f_2(x),f_2(x)⟩\\ & =-\cos (\theta )\sin (\theta )+0-0+\cos (\theta )\sin (\theta )\\ & =0.\end{array}$$

So $F_{\theta }$ and $F_{\theta +\pi /2}$ are orthogonal as well.

Below is a visualization of $F_{\theta }$ and $F_{\theta +\pi /2}$. Choose $\theta$ by dragging $P$ on the unit circle, and see how $F_{\theta }$ and $F_{\theta +\pi /2}$ change.

Example 2.

Here is another exmaple. Consider the functions $g_1(x)=\sqrt{3/52}(5x-8)$ and $g_2(x)={\displaystyle \frac{\sqrt{7}}{8}}{x}^3$ on the interval $[0,2]$. The ugly coefficients make the $g_j$'s have length 1: we find

• $⟨g_1,g_1⟩=1$,
• $⟨g_2,g_2⟩=1$,
• $⟨g_1,g_2⟩=0$.

So $g_1$ and $g_2$ are orthogonal. What's more, if we define $G_{\theta }(x):=\cos (\theta )g_1(x)+\sin (\theta )g_2(x)$ and $G_{\theta +\pi /2}(x):=-\sin (\theta )g_1(x)+\cos (\theta )g_2(x)$, then using the linearity of the dot product, again we find

$$⟨G_{\theta },G_{\theta +\pi /2}⟩=0,$$

so $G_{\theta }$ and $G_{\theta +\pi /2}$ are orthogonal as well.

Below is a visualization of $G_{\theta }$ and $G_{\theta +\pi /2}$. Choose $\theta$ by dragging $P$ on the unit circle, and see how $G_{\theta }$ and $G_{\theta +\pi /2}$ change. Notice that it is hard to see orthogonality just by looking at the graphs.