I need to approximate $x \ln x$ on $[0,1]$ as a piecewise-linear function. If $P(x)$ is a piecewise-linear approximation, I want to minimize $$ \max_{0 \le x \le 1} |P(x) - x \ln x| \rightarrow \min_P. $$

I tried interpolation over evenly spread points $x_i = \frac{i}{n}$, for $i=0,1,\dotsc,n$. I also tried Chebyshev nodes together with additional nodes $x=0$ and $x=1$. The latter works much better. For my numerical problem, it was enough but I am interested in optimal solution (mathematical mind, you know :)

Is there a way to optimally spread points $x_0, x_1, \dotsc, x_n$ over $[0,1]$ for piecewise-linear interpolation/approximation?

Update To illustrate the difference between uniform and Chebyshev nodes, here is the error of approximation with $n=20$ points.

Uniform vs Chebyshev nodes

My guess is that for optimal nodes, the "bumps" should be of the same height.

Update 2

I have also the following idea of a numerical method.

First, let us see what is the maximum error of approximation between points $x_n$ and $x_{n+1}$.

error of approximation

We can find equation of an approximating (red) line $y = kx + b$, where $$ k = \frac{x_{n+1} \ln x_{n+1} - x_n \ln x_n}{x_{n+1} - x_n}, \quad b = -k x_n+x_n \ln x_n \,. $$

Next, for $x_n \le x \le x_{n+1}$, we can define the error of approximation at point $x$: $$ \delta(x) = k x + b - x \ln x \,. $$ To find the maximum error, we just solve $\delta'(x) = 0$: $$ 0 = (kx + b - x \ln x)' = k - \ln x - 1. $$ Therefore, the maximum error on $[x_n,x_{n+1}]$ is achieved at $x_c=e^{k-1}$ and it is equal to $$ d(x_n, x_{n+1}) = k x_c + b - x_c \ln x_c \,, $$ where $k$, $b$, and $x_c$ depend on $x_n$ and $x_{n+1}$ as described above. Note also that if we fix $x_n$, the $d(x_n, x_{n+1})$ is increasing function in $x_{n+1}$.

Now, the method itself. Assume we want to ensure that approximation error is not larger than $\epsilon$. Then the following iterative procedure can be applied to construct the list of points.

  1. Set $x_0 = 0$.
  2. For $n=0,1,2,\dotsc$ find (e.g. by binary search) $x_{n+1}$ that ensures $d(x_n, x_{n+1}) = \epsilon$.
  3. Stop when some $x_{n+1} \ge 1$. Change $x_{n+1}$ to $x_{n+1} = 1$ (not necessary but to make things nicer).

But is there an analytical solution?

  • $\begingroup$ For example, if we set three points $x_0=0, x_1, x_2=1$, then it seems the optimal position for $x_1$ is $0.2869781560930248$... $\endgroup$ – Yauhen Yakimenka Jan 29 at 10:25

Each piece of deviation from a linear interpolation resembles a parabola. So if we interpolate a function $f$ over a segment of width $w$, then the maximum deviation of the curve from the segment is roughly $(w/2)^2 f''/2$. To get a maximum deviation of $\epsilon$, we should choose $w=\sqrt{8\epsilon/f''}$. To be more precise, we should use a value for $f''$ in the middle of the parabola, whose location we can estimate from the previous two points of interpolation.

This leads to the following procedure: Let \begin{align} x_0 &=1 + 1/N\\ x_1 &= 1\\ x_{n+1} &= x_n - \sqrt{8\epsilon\big/ f''\!\left(x_n+\frac{x_n - x_{n-1}}2\right)}\\ \end{align} Then we interpolate with the points $\{0,x_{N-1},\ldots,x_1\}$.

Here $N=20$, $f(x)=x \log x$, and by trial and error we choose $\epsilon=.00141$. This gives the points $$\{0.0000, 0.0039, 0.0112, 0.0242, 0.0427, 0.0670, 0.0968, 0.1323, 0.1735, 0.2203, \\ \ \ 0.2727, 0.3308, 0.3945, 0.4638, 0.5388, 0.6195, 0.7057, 0.7976, 0.8951, 1.0000\}.$$ Linear interpolation on those points approximates $f$ with a maximum deviation of only $.00145$.

  • $\begingroup$ Hmm... actually, I hoped for the analytical solution :) I think I have an iterative method that is exact... I will test it and post it here if it works $\endgroup$ – Yauhen Yakimenka Jan 29 at 20:25
  • $\begingroup$ See Update 2 in the question. What do you think? $\endgroup$ – Yauhen Yakimenka Jan 29 at 21:34
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    $\begingroup$ If I set $\epsilon=0.00145$ in my numerical method, I also get 20 points: {0.000000,0.003942,0.013735,0.029329,0.050725,0.077920,0.110915,0.149711,0.194307,0.244702,0.300898,0.362893,0.430689,0.504284,0.583680,0.668875,0.759871,0.856666,0.959261,1}. So your method is rather good (at least, for this $\epsilon$). $\endgroup$ – Yauhen Yakimenka Jan 29 at 21:47
  • $\begingroup$ For most $f$, there is no closed form for the maximum error on an interval, so I like having a method that doesn't depend on numerical maximization at each of those 20 steps. $\endgroup$ – Matt F. Jan 29 at 22:11
  • $\begingroup$ True, it requires solving $-k = f'(x)$ $\endgroup$ – Yauhen Yakimenka Jan 29 at 22:18

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