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I am asked to provide an example (or prove none exists) in which $a_0 < a_1 < \cdots$. It's basically only the left side of the interval changing. I cannot think of such an example, but I also don't know how to prove it. Can anyone offer advice?

The same thing for finding an example with $a_0 = a_1 < a_2 = a_3 < a_4 = a_5< \cdots$. How do I go about finding a matching example??

Ozera
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1 Answers1

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1) scaling to the interval $[0,1]$, these inequalities would require that the solution $x^*$ be greater than $\frac12+\frac14+..+\frac1{2^n}=1-2^{-n}$ for every $n$, thus $x^*=1$ which is impossible.

2) tells us

  • $a_1=a_0$: $x^*\in[0,\frac12]$,
  • $a_1<a_2$: $x^*\in[\frac14,\frac12]$,
  • $a_2=a_3$: $x^*\in[\frac14,\frac38]$,
  • $a_3<a_4$: $x^*\in[\frac5{16},\frac38]$, and so on.

Continuing this pattern, the left boundary converges to $x^*=4^{-1}+4^{-2}+4^{-3}+...=\frac13$

Lutz Lehmann
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