Close curves on an Andrews plot On the Andrews plot, close points (that is inputs) appear as close functions (the curves), but the reverse does not necessarily hold. That is there can be two curves on the Andrews plot that are close, but the original points are far away. Is this correct? If yes, and as an aid to understanding, how to mathematically create input points that are distant but whose Andrews' curves are close?
 A: If you take cross correlation as measure for distance between curves
For the curves in the Andrews plot
$$f_x(t) = x_1 \frac{1}{\sqrt{2}} + x_2 \sin(t) + x_3 \cos(t) + x_4 \sin(2t)+ ...$$
The cross covariance
$$\int_{-\pi}^{\pi} f_x(t) f_y(t) dt$$
can be simplified as separate products (because the trigonometric functions are orthogonal)
$$ x_1 y_1 \int_{-\pi}^{\pi} \frac{1}{\sqrt{2}} dt + x_2 y _2 \int_{-\pi}^{\pi} \sin(t) \sin(t) dt + x_3 y_3 \int_{-\pi}^{\pi} \cos(t) \cos(t) dt + ...$$
and this is the covariance between the vectors 
$$  x_1y_1 \pi + x_2y_2 \pi + x_3y_3 \pi + ... = \pi \vec{x}\cdot \vec{y}$$
See also Parseval's theorem and orthonormal functions

It is a bit tricky, the definitions of cross variance and cross correlation. Of course you need to see the difference relative to the size of the vectors. If the vectors are larger then the difference and the covariance will be larger. So you need to normalize the functions and then you can make the same argument/computations for correlation as for covariance above.
Thus if the vectors/points have a high correlation (are close together) then the curves will have a high cross correlation as well (close together).
This works in both ways.

If you take the mean absolute difference as measure.
In this case it will be more or less the same
$$\small\begin{array}{}
\overbrace{\frac{1}{2\pi} \int_{-\pi}^{\pi}|f_y(t)-f_x(t)|dt}^{\text{mean absolute difference}}&\leq&\overbrace{\sqrt{\frac{1}{2\pi}\int_{-\pi}^{\pi}(f_y(t)-f_x(t))^2dt}}^{\text{root mean squared difference}} =\\
&\leq&\sqrt{\frac{1}{2\pi}\int_{-\pi}^{\pi}f_x(t)^2dt+\frac{1}{2\pi}\int_{-\pi}^{\pi}f_y(t)^2dt-\frac{2}{2\pi}\int_{-\pi}^{\pi}f_x(t)f_y(t)dt} \end{array}$$
but in this case it goes in one way and due to the inequality it is possible to have a small mean absolute difference while the root mean squared difference is large. 
The inequality is stronger when the difference $f_y(t)-f_x(t)$ is less homogeneously spaced.


*

*The equality occurs when only $x_1 \neq y_1$ and for all other $i \neq 1$ we have $x_i = y_i$. (because then the difference $f_y(t)-f_x(t)$ is constant and the mean absolute difference equals the root mean squared distance)

*In case for some $j>1$ we have $x_j \neq y_j$ and all other $i \neq j$ we have $x_i = y_i$ then we have for the $\sin(kt)$ case with integer values of $k$ (and the same for the $\cos(kt)$ case)
$$\small\frac{1}{2\pi}\int_{-\pi}^{\pi}|f_y(t)-f_x(t)|dt = (y_j-x_j) \frac{1}{2\pi} \int_{-\pi}^{\pi} \vert \sin(kt) \vert dt = (y_j-x_j) \frac{2}{\pi} \approx 0.637$$
whereas
$$\small\sqrt{\frac{1}{2\pi}\int_{-\pi}^{\pi}(f_y(t)-f_x(t))^2dt} = (y_j-x_j) \sqrt{ \frac{1}{2\pi}\int_{-\pi}^{\pi}  \sin(kt)^2 dt} = (y_j-x_j) \frac{1}{\sqrt{2}} \approx 0.707$$ 
This is independent from from the particular coordinate. It is the same for all $sin(kt)$ or $cos(kt)$

*For combinations of coordinates that are different you will have to work it out a bit. I am not sure whether you can get extremely large differences between the mean absolute difference and the root mean squared difference, but in general the inequality may be somewhat larger than the two cases above. For example:
$$\small\frac{1}{2\pi}\int_{-\pi}^{\pi} \vert 0.5+ 0.5\sin(t) \vert dt = \frac{1}{2} = 0.500$$
$$\small\sqrt{\frac{1}{2\pi}\int_{-\pi}^{\pi} (0.5+ 0.5\sin(t))^2 dt} = \sqrt{\frac{3}{8}} \approx 0.612$$
