An ODE, Orthogonal Functions, and the Chebyshev Polynomials

Orthogonal functions are very useful and rather easy to use.  The most famous ones are sin(n x) and cos(n x) on the interval [$-\pi$, $\pi$].  It is extremely easy to do linear regression with orthogonal functions.

Linear regression involves approximating a known function $f(x)$ as the weighted sum of other functions.  This shows up all the time in various disguises in machine learning:

  • When using AdaBoosting we perform a weighted sum of weak classifiers. 
  • Support vector machines find a linear combination of the features that separates the data into two classes.  
  • Each node in a neural net typically computes a weighted sum of the nodes in the prior layer.

For orthogonal functions, computing the linear regression is simple.  Suppose that the features are $f_1, f_2, \ldots, f_n$ and the functions $f_i$ are orthonormal, then the best approximation of the function $g(x)$ is the same as the solution to the linear regression

$$ g(x) \approx \sum_{i=1}^n \alpha_i f_i(x)$$

where $\alpha_i$ is the “inner product” of $f_i$ and $g$ denoted

$\alpha_i= \langle\;g, f_i\rangle = \int g(x) f(x) w(x) dx.$  (Two function are orthogonal if their inner product (or dot product) is zero.  The set of orthogonal functions $\{f_i\}$ is called orthonormal if $<f_i, f_i>=1$ for all $i$.)  The function $w(x)$ is a positive weight function which is typically just $w(x)=1$.

For example, if we try to approximate the function $g(x)=x^2$ on the interval [-1,1] by combining the features $f_1(x) = 1/\sqrt(2)$ and $f_2(x) = x \sqrt{3 \over 2} $, then

$$ g(x) \approx \sum_{i=1}^n \alpha_i f_i(x)$$


$$\alpha_1 = \langle g, f_1\rangle = \int_{-1}^1 g(x) f_1(x) dx =  \int_{-1}^1 x^2 1/\sqrt2 dx = \sqrt{2}/3,$$

$$\alpha_2 = \langle g, f_2\rangle = \int_{-1}^1 g(x) f_2(x)  dx=  \int_{-1}^1 x^2  \sqrt{3/2}\; x dx =0,$$

and the best approximation is then $\sum_{i=1}^n \alpha_i f_i(x) = \alpha_1 f_1(x) = 1/3$.


The other day I needed to do a linear regression and I wanted to use orthogonal functions, so I started by using the trigonometric functions to approximate my function $g(x)$.  Using trig functions to approximate another function on a closed interval like [-1,1] is the same as computing the Fourier Series.  The resulting approximations were not good enough at x=-1 and x=1, so I wanted to change the weight function $w(x)$.  Now assume that $u_1(x), u_2(x), \ldots, u_n(x)$ are orthonormal with the weight function w(x) on [-1,1].  Then

$$ \int_{-1}^1 u_i(x) u_j(x) w(x) dx = \delta_{i,j}$$

where $\delta_{i,j}$ is one if $i=j$ and zero otherwise.  If we make a change of variable $x=y(t)$, then

$$ \int_{y^{-1}(-1)}^{y^{-1}(1)} u_i(y(t)) u_j(y(t)) w(y(t)) y'(t) dx = \delta_{i,j}.$$

It would be nice if

$$(1)\quad w(y(t)) y'(t)=1,$$

because then I could just set $u_i(y(t))$ equal to a trig function and the resulting $u_i(x)$ would be orthonormal with respect to the weight function $w(x)$. Condition (1) on $y$ is really an ordinary differential equation

$$y’ = {1\over{w(y)}}.$$

In order to get a better fit at $x=\pm 1$, I tried two weight functions $w(x) = {1\over{(1-x)(1+x)}}$ and $w(x) = {1\over{\sqrt{(1-x)(1+x)}}}$.  For the former, the differential equation has solutions that look similar to the hyperbolic tangent function. The latter $w(x)$ yields the nice solution

$$y(t) = \sin( c + t).$$

So now if I choose $c=\pi/2$, then $y(t) = \cos(t)$, $y^{-1}(-1) = \pi$, $y^{-1}(1)=0$, so I need functions that are orthonormal on the interval $[-\pi, 0]$.  The functions $\cos(n x)$ are orthogonal on that interval, but I need to divide by $\sqrt{\pi/2}$ to make them orthornormal.  So, I set $u_n(y(t)) = \sqrt{2/\pi}\cos(n\;t)$ and now

$$ \int_{-1}^1 {{u_i(x) u_j(x)}\over {\sqrt{(1-x)(1+x)}}}  dx = \delta_{i,j}.$$

The $u_n(x)$ functions are orthonormal with respect to the weight $w(x)$ which is exactly what I wanted.

Solving for $u(x)$ yields

$$u_n(x) = \sqrt{2/\pi} \cos( n \;\arccos(x)),$$

$$u_0(x) = \sqrt{2/\pi}, u_1(x) = \sqrt{2/\pi} x,$$

$$u_2(x) = \sqrt{2/\pi}(2x^2-1), u_3(x) = \sqrt{2/\pi}(4x^3-3x),\ldots.$$


If you happen to be an approximation theorist, you would probably notice that these are the Chebyshev polynomials multiplied by $\sqrt{2/\pi}$.   I was rather surprised when I stumbled across them, but when I showed this to my friend Ludmil, he assured me that this accidental derivation was obvious.  (But still fun :) )


  1. ThePig’s avatar

    This post reminds me of a friend who used to chastise me when I used “orthogonal” when I meant “independent.” For sin x and cos x are orthogonal but they are clearly not independent.

  2. Leo’s avatar

    seems that Mr Tchebychev followed the same thinking path..

    1. hundalhh’s avatar

      I wonder if there is some way to know ahead of time or intuit that the result is going to be polynomial.

      I guess for any weight function, it’s possible to get orthogonal polynomials, but the Chebyshev polynomials are so nice—integer coefficients and all.

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