# Interpolation nodes for linear spline (piecewise-linear) interpolation of $x \ln x$

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.

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 $$x_{n+1}$$.

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?

• 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 $0.2869781560930248$... – 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 $$.00145$$.
• 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 $\epsilon$). – Yauhen Yakimenka Jan 29 at 21:47
• 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. – Matt F. Jan 29 at 22:11
• True, it requires solving $-k = f'(x)$ – Yauhen Yakimenka Jan 29 at 22:18