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Cauchy sequences, induction, telescoping property

  • Thread starter Meggle
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  • #1
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Homework Statement



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Homework Equations



I am guessing a combination of induction and the telescoping property.

The Attempt at a Solution



I'm studying this extramurally, and I've just hit a wall with this last chunk of the sequences section, so if someone can suggest a good text book that would be really good too. My googlefu is not helping me this week.

Ok. So prove the first chunk of a). The only way I can think of to prove that chunk is by mathematical induction again, and I think from the fact that it doesn't explicitly say to use that that there ought to be another way. But I can't see it. So I think I've proved it using mathematical induction. Although that did involve saying |-[tex]\alpha[/tex](b-a)| = [tex]\alpha[/tex]|(b-a)| which I'm a bit dubious about.

Second section of a) let n=1, then:
|s1+1 - s1| = |b - a| = [tex]\alpha[/tex] 0|b - a|
so the result holds for n=1.
Assume the result holds for some positive integer k, i.e. assume:
|sk+1 - sk| = [tex]\alpha[/tex]k-1|b - a|
so I think the first part shows the case where k=2 and [tex]\alpha[/tex]k-1=[tex]\alpha[/tex]1=[tex]\alpha[/tex] but I don't know how to make use of this.
??

So moving on to b) sn+1 +[tex]\alpha[/tex]sn where n=1 becomes s2 +[tex]\alpha[/tex]s1 = b + [tex]\alpha[/tex]a ok for n=1. But I don't know how to prove the general, as the method my readings have relies on incorporating the definition of <sn> into the equation to be proved, and I don't have a definition for sk+1, or any examples in my readings of a proof involving sk+2 instead of sk+1.

Hence why I feel like I've just not got the jist of this section of the course material, but I've been back and forth through my readings and it's not becoming any clearer. :cry: Can anyone suggest the fundamental thinking I'm obviously not getting? Even some worked examples somewhere would be brilliant.

And I don't know why all my alphas are floating up so high, sorry. :yuck:
 

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Answers and Replies

  • #2
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You don't need to do induction on the first part of (a). It's just substitution of the formula they provide. If you get stuck at a spot, try going in the reverse direction and finding a spot where it meets the forward direction.

Although that did involve saying |-[tex]
\alpha
[/tex](b-a)| = [tex]
\alpha
[/tex]|(b-a)| which I'm a bit dubious about.
That's perfectly fine here since they tell you [tex]
\alpha
[/tex] > 0.

For the 2nd part of (a), you were correct in assuming it works for k. Now you have to prove it works for k+1. To do this, you'll first have to substitute |s_(k+2) - s_(k+1)| with their respective values using the formula from the first part of (a). Next, after a bit of simplification and rearranging, you should be able to use your induction hypothesis to prove it works for k+1.
 

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