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?

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. 
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:

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

Re: Problem 2.14(c)
Quote:

Re: Problem 2.14(c)

Re: Problem 2.14(c)

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

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.

Re: Problem 2.14(c)
I´ve been struggling with this problem too. Essentialiy we have to prove that the second expression in the min expression
. is a valid as explains magdon in Quote:
(1) must be satisfied. I have been finding upper bounds to the right hand side of (1), using the following tricks if (the case must be proved apart). , , because (this is not the seven in the exponent) and . Then we arrive at an expression that can be compared easily with the left hand side of (1) proving that this inequality is valid. 
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