- #26

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X0 isn't a it's alpha α

That explains something. Imagine my misreading it.

Okay, why is ##U_n## bounded below by a? By zero yes.

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- Thread starter mtayab1994
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- #26

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X0 isn't a it's alpha α

That explains something. Imagine my misreading it.

Okay, why is ##U_n## bounded below by a? By zero yes.

- #27

pasmith

Homework Helper

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From part three we can see that ##g## is strictly decreasing so that means ##(U_{n})## is monotonic decreasing and bounded below by a>0 so therefore it converges to a number greater than 0.

No. In the case that [itex]x_0 = \alpha = s[/itex] you have [itex]U_n = 0[/itex] for all [itex]n[/itex].

It is enough that [itex]U_n[/itex] is bounded below by zero, and you need to show your working in support of the conclusion that [itex]U_n[/itex] is monotonic decreasing.

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- #29

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You want to use what you learned about g being monotonic in working through why U is.

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