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Old 2007-05-02, 08:14   #1
jinydu
 
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Default 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:
<br />
(\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)<br />



<br />
\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})<br />

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.

<br />
\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}<br />

<br />
=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}<br />

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.

<br />
\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)<br />

<br />
\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)<br />

<br />
\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)<br />

<br />
\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)<br />

Finally, since the b_k's are constant with respect to n, we have:

<br />
\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)<br />


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:

<br />
\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)<br />

Subtracting the two results gives the desired equation.

Last fiddled with by jinydu on 2007-05-02 at 08:19
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Old 2007-05-02, 10:31   #2
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Problem 9b

The problem amounts to showing that if a,b\in\[-\pi,\pi\], a<b 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<b-a<2\pi, 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<c<4 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<c\leq{}cn_0+c\ (mod\ 4)<1+c\leq{}3, 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<cn_0\leq{}4, then 3<cn_0+c\ (mod\ 4)<4+c\ (mod\ 4)\leq{}6\ (mod\ 4)=2, 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<c<4. 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
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Old 2007-05-02, 12:17   #3
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You could PM it to yourself.
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