 mersenneforum.org Display Sums with Latex
 Register FAQ Search Today's Posts Mark Forums Read 2007-05-02, 08:14 #1 jinydu   Dec 2003 Hopefully Near M48 2×3×293 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 2007-05-02 at 08:19   2007-05-02, 10:31 #2 jinydu   Dec 2003 Hopefully Near M48 2·3·293 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 2007-05-02 at 10:33   2007-05-02, 12:17 #3 Mini-Geek Account Deleted   "Tim Sorbera" Aug 2006 San Antonio, TX USA 2·3·23·31 Posts You could PM it to yourself.   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post science_man_88 science_man_88 2 2018-07-21 17:11 wildrabbitt Software 7 2018-01-12 19:46 wildrabbitt Math 2 2017-10-30 12:47 ProximaCentauri Miscellaneous Math 7 2014-12-05 22:28 jinydu Forum Feedback 2 2007-05-02 11:44

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