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Iosif Pinelis
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In general, it is not true at all that $x^*_h\to x_1$ as $h\downarrow0$. Indeed, suppose that $n=3$, $x_1=2$, $x_2=0$, $x_3=-1$.

Then for small $h>0$ there will be a critical point $x^*_h$ close to the midpoint $1$ between $x_1=2$ and $x_2=0$, with $x_1=2$ being the closest to $x^*_h$ among the points $x_1,x_2,x_3$ -- because $f'(1)=-2Ce^{-2/h^2}/h^2<0$. However, $x^*_h-x_1$ will be close to $1-2=-1$ and not to $0$.


For an illustration, here isare the graphgraphs $\{(x,f(x))\colon x\in[-1.5,2.5]\}$ and $\{(x,f(x))\colon x\in[1-\frac12\,10^{-7},1+\frac12\,10^{-7}]\}$ for $h=3/10$:

enter image description here

enter image description here

In general, it is not true at all that $x^*_h\to x_1$ as $h\downarrow0$. Indeed, suppose that $n=3$, $x_1=2$, $x_2=0$, $x_3=-1$.

Then for small $h>0$ there will be a critical point $x^*_h$ close to the midpoint $1$ between $x_1=2$ and $x_2=0$, with $x_1=2$ being the closest to $x^*_h$ among the points $x_1,x_2,x_3$ -- because $f'(1)=-2Ce^{-2/h^2}/h^2<0$. However, $x^*_h-x_1$ will be close to $1-2=-1$ and not to $0$.


For an illustration, here is the graph $\{(x,f(x))\colon x\in[-1.5,2.5]\}$ for $h=3/10$:

enter image description here

In general, it is not true at all that $x^*_h\to x_1$ as $h\downarrow0$. Indeed, suppose that $n=3$, $x_1=2$, $x_2=0$, $x_3=-1$.

Then for small $h>0$ there will be a critical point $x^*_h$ close to the midpoint $1$ between $x_1=2$ and $x_2=0$, with $x_1=2$ being the closest to $x^*_h$ among the points $x_1,x_2,x_3$ -- because $f'(1)=-2Ce^{-2/h^2}/h^2<0$. However, $x^*_h-x_1$ will be close to $1-2=-1$ and not to $0$.


For an illustration, here are the graphs $\{(x,f(x))\colon x\in[-1.5,2.5]\}$ and $\{(x,f(x))\colon x\in[1-\frac12\,10^{-7},1+\frac12\,10^{-7}]\}$ for $h=3/10$:

enter image description here

enter image description here

Source Link
Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229

In general, it is not true at all that $x^*_h\to x_1$ as $h\downarrow0$. Indeed, suppose that $n=3$, $x_1=2$, $x_2=0$, $x_3=-1$.

Then for small $h>0$ there will be a critical point $x^*_h$ close to the midpoint $1$ between $x_1=2$ and $x_2=0$, with $x_1=2$ being the closest to $x^*_h$ among the points $x_1,x_2,x_3$ -- because $f'(1)=-2Ce^{-2/h^2}/h^2<0$. However, $x^*_h-x_1$ will be close to $1-2=-1$ and not to $0$.


For an illustration, here is the graph $\{(x,f(x))\colon x\in[-1.5,2.5]\}$ for $h=3/10$:

enter image description here