20070502, 08:14  #1 
Dec 2003
Hopefully Near M48
1758_{10} Posts 
Display Sums with Latex
I'm typing this here because I have homework due on Thursday and I can't learn TeXnicCenter that fast. I'll print this out Wednesday, please don't delete.
Problem 7a Justification for: Now group the terms by 's instead of 's. Every one of the terms in parentheses contains . After that, every time increases by one, there is one less parenthesis containing a of that index until , which only the last term has. Now we see that the sums are of two types. For , we sum the 's from to . For , we sum the 's from the term that matches the , namely , to . Finally, since the 's are constant with respect to , we have: Now do the same for , replacing with with . Using identical reasoning, the result will be: Subtracting the two results gives the desired equation. Last fiddled with by jinydu on 20070502 at 08:19 
20070502, 10:31  #2 
Dec 2003
Hopefully Near M48
1758_{10} Posts 
Problem 9b
The problem amounts to showing that if , and either or , then there exists a such that there is at least one integer in each of the intervals such that for some odd integer . First note that by the conditions on and , we must have , which implies and so . Also, for some integer because is odd. So and . As a result, it is sufficient to find such that: Letting , we have that and our desired inequality becomes: Now suppose that we start at any fixed and increase by 1 successively. Every time we do this, we find that: It suffices to show that no matter which , there exists a fixed number of steps , independent of such that we increment by 1 repeatedly, we will have within . Once we do this, we can repeat the same argument indefinitely to find that every consecutive integers contains at least one integer that satisfies . Consider the case where . If , then , so that cannot "jump over" [1,3]. If is not now in [1,3], it will have moved a distance closer. Since the original distance was at most 1 (obtained when ), the number of steps required is at most . If , then we are already done. If , then , so that is in either (3, 4) or [0,1]. If is not now in [0,1], it will have moved a distance closer. Since the original distance was less than 1 (nearly obtained when is slightly larger than ), the number of steps required is at most . It could take up to to get inside [1,3] so in the worst cast, the number of steps required to reach [1,3] is at most . The remaining case is . But in arithmetic, adding is the same as subtracting and . The reasoning is essentially the same as in the previous paragraph, except we must now subtract instead of add and the roles of [0, 1) and (3, 4) are interchanged; starting points in (3,4) always reach [1,3] within steps while those in [0,1) may required steps. In all cases, the total number of steps is bounded by . So if we choose , then each contain successive integers and hence at least one integer that satisfies the condition for some odd integer . Last fiddled with by jinydu on 20070502 at 10:33 
20070502, 12:17  #3 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10254_{8} Posts 
You could PM it to yourself.

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