Prove the required claim by induction on m. This requires bounding from below and from above, as per the hint.
Hey there!
I'm really stuck on a question and i can't figure out how I should solve it.
I have a sum given by
Given the special case where
I have to prove that is logarithmically bounded by
And here's a *HINT* : Look at the terms
in
Well, I really don't get it...if someone could help me I would greatly appreciate!
ps. I'm glad I found thoses forums, I'll certainly become an active member, being a studient in computer sciences :P.
THANKS!
Thank you for your reply!
I kinda guessed it was by induction but didn't think about the bounding from below. (I know... i had a rough day)
But still I'm confused about the induction hypothesis :S
Normally, inductions are all about a base case ( ) and an inductive step (do we need to show for ?)
I'm really confused! Thank you!
Yes.
To be more explicit, prove P(m) for all by induction on m where P(m) is .
In the induction step, when you are proving P(m+1) from P(m), you know the lower and upper bounds on . Also, it is easy to get lower and upper bounds on by replacing the tail of the sum with its smallest or biggest term, respectively.
Ok so i give it a try. Given the special case , the sum is given by
Lets prove, by induction, the following logarithmic bounds
*edited ^
First, lets start with the base case
???? ????
well from there I get stuck, I think I did something wrong ...
Thank you for your help so far!!
Look again at post #5. You assume the induction hypothesis P(m), i.e., . From this assumption, you need to prove P(m+1), i.e., . Now, . The induction hypothesis gives you an upper and a lower bound on the first term of this sum, i.e., . As for the tail of , replace all terms with the smallest term to get a lower bound and with the largest term to get an upper bound. By adding the obtained lower bounds on and the tail you get a lower bound on , and by adding the obtained upper bounds on and the tail, you get an upper bound on . The two resulting inequalities are precisely P(m+1).
And one last thing!!!
when you say replace all terms you mean replacing all the terms of the tail right?
I can't figure out how it can equal the lower and upper bounds when added to bounds.
For example, for lower bound:
where T is the modified tail you're talking about. would T be like
this is what I understood..
Yes, there are exactly in the tail of .
You are correct. When we replace all terms with the smallest term , we get , and .
For the upper bound, we won't get the exact equality, but if we replace all terms with the largest term and add the upper bound of , i.e., m + 2, we get something that is ≤ (rather than =) the required upper bound m + 3.