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Successive substitution:

Why is the successive substitution proces $x=2(1-e^{-x})$ monotone convergent for start value $x_0=0,5$ while in the tabel the value for $f'(x_0)=1,2121>1$? So we would think it is monotone divergent.

We took $f(x)=2(1-e^{-x})$ in the table.

enter image description here

WinstonCherf
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1 Answers1

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What actually happens is that the iteration moves away from $x$ and towards the fixed point between $1.5$ and $2$. Note that $f([0,\infty))=[0,2)$ with $f$ monotonically increasing and $f'(x)<1$ for $x>\ln 0.5$. As $e^x\ge 1+x$ one gets $$ 2(1-e^{-x})\ge\frac{2x}{1+x} > x $$ for $x<1$ so that the iteration of $f$ is increasing for $x_0\in (0,1)$ until it reaches the contractivity region $x_k>\ln 0.5$ of the fixed point close to $1.59$

  k        x[k]           f'(x[k])

  0   0.5             1.21306131943
  1   0.786938680575  0.910472575971
  2   1.08952742403   0.672750838184
  3   1.32724916182   0.530411594063
  4   1.46958840594   0.46004028126
  5   1.53995971874   0.428779474283
  6   1.57122052572   0.415582825519
  7   1.58441717448   0.410134553473
  8   1.58986544653   0.407906104959
  9   1.59209389504   0.406998119281
 10   1.59300188072   0.40662873854
 11   1.59337126146   0.406478565452
 12   1.59352143455   0.406417527894
 13   1.59358247211   0.406392721917
 14   1.59360727808   0.406382641074
 15   1.59361735893   0.406378544415

The dynamic of the iteration can be illustrated in a phase portrait

enter image description here

Lutz Lehmann
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  • Thnx for your answer, but unfortunately, it's still not really clear... Since f'(x) > 1 for x = 0,5 I still do not undersand why I can say that the function is monotone divergent for x_0 < 0 and monoton convergent for x_0 > 0? This is the answer my book gives... (I only got a basic introduction to numerical methods) – WinstonCherf Dec 14 '17 at 18:29
  • This has to be interpreted that for $x_0>x^$ where $x^=2+W_0(-2e^{-2})=1.59..$ (using the Lambert-W function) is the positive fixed point, you get a monotonically falling sequence. Try to demonstrate that between the two fixed points, $f(x)-x$ is positive, and outside the middle interval it is always negative. – Lutz Lehmann Dec 14 '17 at 18:57
  • So it is possible that in the picture for start value x_0=0,5 the successive proces is monotone convergent while theoreticly f'(0,5)>1 and so for start value x_0=0,5 the proces would be monotonic divergent? The best way is to sketch some start values and give conclusions from the picture? – WinstonCherf Dec 14 '17 at 19:04
  • You can prove everything by starting with the observation that $f$ is concave. – Lutz Lehmann Dec 14 '17 at 19:42
  • Ok, can you tell me how? – WinstonCherf Dec 14 '17 at 19:46