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 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: $ (\sum_{n=M}^{N-1}\sum_{k=1}^{n}a_{n+1}b_k)-(\sum_{n=M}^{N-1}\sum_{k=1}^{n}a_nb_k)=(\sum_{k=1}^{M}\sum_{n=M}^{N-1}a_{n+1}b_k)+(\sum_{k=M+1}^{N-1}\sum_{n=k}^{N-1}a_{n+1}b_k)-(\sum_{k=1}^{M}\sum_{n=M}^{N-1}a_nb_k)-(\sum_{k=M+1}^{N-1}\sum_{n=k}^{N-1}a_nb_k)$ $ \sum_{n=M}^{N-1}\sum_{k=1}^{n}a_nb_k=(a_Mb_1+a_Mb_2+...+a_Mb_M)+(a_{M+1}b_1+a_{M+1}b_2+...+a_{M+1}b_M+a_{M+1}b_{M+1})+...+(a_{N-1}b_1+...+a_{N-1}b_M+...+a_{N-1}b_{N-1})$ Now group the terms by $b$'s instead of $a$'s. Every one of the terms in parentheses contains $b_1,...,b_M$. After that, every time $k$ increases by one, there is one less parenthesis containing a $b$ of that index until $k=N-1$, which only the last term has. $ \sum_{n=M}^{N-1}\sum_{k=1}^{n}a_nb_k=(a_Mb_1+a_{M+1}b_1+...+a_{N-1}b_1)+...+(a_Mb_M+a_{M+1}b_M+...+a_{N-1}b_M)+(a_{M+1}b_{M+1}+a_{M+2}b_{M+1}+...+a_{N-1}b_{M+1})+...+a_{N-1}b_{N-1}$ $ =b_1(a_M+a_{M+1}+...+a_{N-1})+...+b_M(a_M+a_{M+1}+...+a_{N-1})+b_{M+1}(a_{M+1}+a_{M+2}+...+a_{N-1})+...+b_{N-1}a_{N-1}$ Now we see that the sums are of two types. For $1\leq{}k\leq{}M$, we sum the $a$'s from $n=M$ to $n=N-1$. For $M\leq{}k\leq{}N-1$, we sum the $a$'s from the term that matches the $b$, namely $k$, to $N-1$. $ \sum_{n=M}^{N-1}\sum_{k=1}^{n}a_nb_k=b_1(\sum_{n=M}^{N-1}a_n)+...+b_M(\sum_{n=M}^{N-1}a_n)+b_{M+1}(\sum_{n=M+1}^{N-1}a_n)+...+b_{N-1}(\sum_{n=N-1}^{N-1}a_n)$ $ \sum_{n=M}^{N-1}\sum_{k=1}^{n}a_nb_k=(b_1+...+b_M)\sum_{n=M}^{N-1}a_n+b_{M+1}(\sum_{n=M+1}^{N-1}a_n)+...+b_{N-1}(\sum_{n=N-1}^{N-1}a_n)$ $ \sum_{n=M}^{N-1}\sum_{k=1}^{n}a_nb_k=(\sum_{k=1}^{M}b_k\sum_{n=M}^{N-1}a_n)+b_{M+1}(\sum_{n=M+1}^{N-1}a_n)+...+b_{N-1}(\sum_{n=N-1}^{N-1}a_n)$ $ \sum_{n=M}^{N-1}\sum_{k=1}^{n}a_nb_k=(\sum_{k=1}^{M}b_k\sum_{n=M}^{N-1}a_n)+(\sum_{k=M+1}^{N-1}b_k\sum_{n=k}^{N-1}a_n)$ Finally, since the $b_k$'s are constant with respect to $n$, we have: $ \sum_{n=M}^{N-1}\sum_{k=1}^{n}a_nb_k=(\sum_{k=1}^{M}\sum_{n=M}^{N-1}a_nb_k)+(\sum_{k=M+1}^{N-1}\sum_{n=k}^{N-1}a_nb_k)$ Now do the same for $\sum_{n=M}^{N-1}\sum_{k=1}^{n}a_{n+1}b_k$, replacing with $a_n$ with $a_{n+1}$. Using identical reasoning, the result will be: $ \sum_{n=M}^{N-1}\sum_{k=1}^{n}a_{n+1}b_k=(\sum_{k=1}^{M}\sum_{n=M}^{N-1}a_{n+1}b_k)+(\sum_{k=M+1}^{N-1}\sum_{n=k}^{N-1}a_{n+1}b_k)$ 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 $a,b\in$-\pi,\pi$$, $a and either $a\neq{}-\pi$ or $b\neq{}\pi$, then there exists a $N\in\aleph$ such that there is at least one integer $n$ in each of the intervals $$1,N$,$-N,-1$,$N+1,2N$,$-2N,-N-1$,$2N+1,3N$,$-3N,-2N-1$,...$ such that $\frac{2k-1}{2(b-a)}\pi\leq{}n\leq\frac{2k+1}{2(b-a)}\pi$ for some odd integer $k$. First note that by the conditions on $a$ and $b$, we must have $0, which implies $\frac{1}{b-a}>\frac{1}{2\pi}$ and so $\frac{\pi}{2(b-a)}>\frac{1}{4}$. Also, $k=2m+1$ for some integer $m$ because $k$ is odd. So $2k-1=4m+2-1=4m+1$ and $2k+1=4m+2+1=4m+3$. As a result, it is sufficient to find $n$ such that: $(4m+1)\frac{\pi}{2(b-a)}\leq{}n\leq{}(4m+3)\frac{\pi}{2(b-a)}$ Letting $c=\frac{2(b-a)}{\pi}$, we have that $0 and our desired inequality becomes: $4m+1\leq{}cn\leq{}4m+3$ $1\leq{}cn\ mod\ 4\leq{}3$ Now suppose that we start at any fixed $n_0$ and increase $n$ by 1 successively. Every time we do this, we find that: $c(n+1)\ (mod\ 4)\ -\ cn\ (mod\ 4)=c(n+1)-cn\ (mod\ 4)=c\ (mod\ 4)=c$ It suffices to show that no matter which $n_0$, there exists a fixed number of steps $M$, independent of $n_0$ such that we increment $n$ by 1 repeatedly, we will have $1\leq{}cn\ mod\ 4\leq{}3$ within $M$. Once we do this, we can repeat the same argument indefinitely to find that every $M$ consecutive integers contains at least one integer that satisfies $1\leq{}cn\ mod\ 4\leq{}3$. Consider the case where $c\leq{}2$. If $0\leq{}cn_0\ mod\ 4<1$, then $0, so that $n$ cannot "jump over" [1,3]. If $cn_0+c$ is not now in [1,3], it will have moved a distance $c$ closer. Since the original distance was at most 1 (obtained when $cn_0=0\ mod\ 4$), the number of steps required is at most $\frac{1}{c}$. If $1\leq{}cn_0\ mod\ 4\leq{}3$, then we are already done. If $3, then $3, so that $n$ is in either (3, 4) or [0,1]. If $cn_0+c$ is not now in [0,1], it will have moved a distance $c$ closer. Since the original distance was less than 1 (nearly obtained when $cn_0$ is slightly larger than $3\ mod\ 4$), the number of steps required is at most $\frac{1}{c}$. It could take up to $\frac{1}{c}$ to get inside [1,3] so in the worst cast, the number of steps required to reach [1,3] is at most $\frac{2}{c}$. The remaining case is $2. But in $mod\ 4$ arithmetic, adding $c$ is the same as subtracting $4-c$ and $0<4-c<2$. 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 $\frac{1}{c}$ steps while those in [0,1) may required $\frac{2}{c}$ steps. In all cases, the total number of steps is bounded by $\frac{2}{c}$. So if we choose $N>\frac{2}{c}$, then $$1,N$,$-N,-1$,$N+1,2N$,$-2N,-N-1$,$2N+1,3N$,$-3N,-2N-1$,...$ each contain $N$ successive integers and hence at least one integer $n$ that satisfies the condition $\frac{2k-1}{2(b-a)}\pi\leq{}n\leq\frac{2k+1}{2(b-a)}\pi$ for some odd integer $k$. 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.

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