Finding $a\in(0,10)$ so that the linear spline for $f(x)=e^{-x^2}$ is sufficiently accurate and $n_1+n_2\le 12$

Question

Suppose we're given a function $f(x)=e^{-x^2}$ and we want to make a piecewise linear interpolation on the interval $[0,10]$ under the following constraints:

1. We divide the interval $[0,10]$ into two intervals $[0,a]$ and $[a,10]$ and make equidistant partitions into $n_1$ and $n_2$ intervals respectively.
2. We're looking for a given (global) accuracy of $\varepsilon=10^{-2}.$
3. $n_1+n_2\le 12.$

My thoughts:

I took a look at the function $f(x)=e^{-x^2}$ restricted to $[0,10].$ I computed the first three derivatives $$\begin{aligned}f'(x)&=-2xe^{-x^2}\\f''(x)&=(4x^2-2)e^{-x^2}\\f'''(x)&=-4x(2x^2-3)e^{-x^2}\end{aligned}$$ $f$ is decreasing on $[0,10]$ and has a saddle point at $x=\frac1{\sqrt 2}.$ From the lectures, I know, if we're interpolating a function $f\in C^2([a,b]),$ with a linear spline $\varphi,$ there is the following estimation: $$|f(x)-\varphi(x)|\le\frac18h^2M_2,$$ where $h$ is the maximum distance of the nodes and $M_2:=\max\limits_{[a,b]}|f''(x)|.$ In our case, $$|f(x)-\varphi(x)|\le\frac18\max\left\{\frac{a}{n_1},\frac{10-a}{n_2}\right\}^2\underbrace{\left|f''(\sqrt{3/2})\right|}_{\frac4{e^{3/2}}}\le\varepsilon=10^{-2}.$$

I'm not sure, though, what $a$ should be. If we look at $f$ on $\Bbb R,$ it has a horizontal asymptote $y=0$ (since $\lim\limits_{x\to\infty}f'(x)=0$ and $\lim\limits_{x\to\infty}f(x)=0$) so, if we choose some $a\ge\frac1{\sqrt 2},$ maybe we would need less nodes to more accurately interpolate $f_{\Big|[a,10]}$ as it seems to approach the $x$-axis, but I don't quite know how to argue that.

Any advice or guidance would be appreciated.

Answer 1

Look at the graph of $f’’$ over $[0,10]$. Choose short intervals (large values of $n_1$ and $n_2$) in regions where $f’’$ is large, and longer ones where it’s small. The idea is to make the product $h^2M_2$ roughly constant.

Other than the fuzzy advice I gave above, I don’t know how to approach this mathematically. But it should be easy to investigate experimentally using a program like Mathematica.

If you really want to be rigorous, you can set up an optimization problem where $a$ and $n_1$ and $n_2$ are the independent variables, and the approximation error is the objective function to be minimized. Then, just throw the problem to your favorite optimization procedure.

Answer 2

Attached a MATHEMATICA script which computes the maximum deviation absolute value, for each semi-segment considering the practical possible configurations of $n_1, n_2, a$. The first plot shows the behavior for the set and the last plot shows the needed solution.

Clear[fun]
fun[a_, n_?IntegerQ] := 
 Module[{dif = {}, a1, a2, n1, n2, dx1, dx2, xm, ym, k},
  g[x_] := Exp[-x^2];
  a1 = a;
  a2 = 10 - a;
  n1 = n;
  n2 = 12 - n;
  dx1 = a1/n1;
  dx2 = a2/n2;
  X1 = Table[{dx1 k, g[dx1 k]}, {k, 0, n1}];
  X2 = Table[{a + dx2 k, g[a + dx2 k]}, {k, 0, n2}];
  For[k = 1, k < n1, k++,
   {xm, ym} = (X1[[k]] + X1[[k + 1]])/2;
   AppendTo[dif, Abs[g[xm] - ym]]
   ];
  For[k = 1, k < n2, k++,
   {xm, ym} = (X2[[k]] + X2[[k + 1]])/2;
   AppendTo[dif, Abs[g[xm] - ym]]
   ];
  Return[Max[dif]]
  ]

grs = Table[Plot[fun[a, n], {a, 0.5, 9.5}], {n, 3, 10}];
Show[AppendTo[grs, Plot[0.01, {a, 0.05, 9.5}, PlotStyle -> Red]], PlotRange -> {0, 0.03}]

Show[{grs[[8]], Plot[0.01, {a, 0.05, 9.5}, PlotStyle -> Red]}, PlotRange -> {0, 0.01}]