#1




Problem 2.14(c)
For Problem 2.14(c), to determine the min value, the way I think would be try to solve the equation in (b) and get L. Maybe L is the second part of the min. However, how to solve the equation is a really hard question. Thus, could anyone tell me how to solve the equation or give me a hint on how to get the right answer?

#2




Re: Problem 2.14(c)
Yes, solving the equation is really hard. It is simpler to show that if takes on the value in the second part of the min, the condition in (b) is satisfied.
Quote:
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#3




Re: Problem 2.14(c)
Quote:
(e.g. when , then and ). I believe the right thing to do would be to assume that because the min bound will still hold and I believe the condition in (b) is then satisfied? But how do I prove that? I appreciate any help. 
#4




Re: Problem 2.14(c)
There is a typo in the equation, sorry.
The second term in the minimum should be . Rather than solve the inequality in (b) to get this bound, you may rather just verify that this is a bound by showing that if , then the inequality in (b) is satisfied, namely . Quote:
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#5




Re: Problem 2.14(c)
Thank you for your reply!

#6




Re: Problem 2.14(c)

#7




Re: Problem 2.14(c)

#8




Re: Problem 2.14(c)
Yes, the problem should state that K>1, otherwise the problem is trivial.
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Have faith in probability 
#9




Re: Problem 2.14(c)
I'm pretty stuck on this one  any hints?

#10




Re: Problem 2.14(c)
Hi, I'm also stuck on this one. I don't know if I'm missing an algebraic argument (in verifying that 2^l > 2Kl^d) or if I'm missing something more important. Any hint would be appreciated.

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