
#37
Apr1808, 07:08 PM

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Do you know how to find fourier expansions?
consider (pix)/2 0<x<2pi then consider x=1 



#38
Apr1808, 11:21 PM

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You obviously know what you are doing. I was writing the answer as arctan(sin(1)/(1cos(1))), but yeah, that is more cleanly expressed as (pi1)/2. But how do you set this up using this 'summation by parts'? And I haven't figured out the how the fourier expansion works yet, but I'll look at it tomorrow. Really nice work. Most other contributions to this thread have been completely clueless. 



#39
Apr1808, 11:43 PM

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I haven't ever seen this summation by parts formula before, though I should have expected it exists, there must have been some discrete form for the continuous counterpart. As for the Fourier expansion, that seems even more out of such a course, though It is also a nice solution =]
Dick; as for the fourier expansion, since f(x) = x is an odd function, the a coefficient is automatically zero whilst [tex]b_v = \frac{2}{\pi} \int^{\pi}_0 x \sin vx dx[/tex] which some quick integration by parts yields [tex]b_v = (1)^{v+1} \frac{2}{v}[/tex], and so we get [tex] x = 2 \left( \frac{\sin x}{1}  \frac{\sin 2x}{2} + \frac{\sin 3x}{3} + ... \right) [/tex]. Rearranging and letting x=1 gives us the desired result, and putting in x= pi/2 gives us a very famous result =] EDIT: O god i just realised I found the wrong Fourier series. argh 



#40
Apr1808, 11:53 PM

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Hmm. If you did everything right then x=1 does not give the desired solution. That's an alternating sign series (1)^n*sin(n)/n. The original question doesn't have the alternating sign. Is there an extra sign that makes it work? But like I said, I didn't really look at that yet. I was trying to figure out the summation by parts thing. It makes perfect sense it such a thing would exist though. Thanks, Gib.




#41
Apr1908, 12:16 PM

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#42
Apr2008, 06:08 PM

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#43
Apr2108, 07:30 PM

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For future reference, a systematic approach to this uses the AbelDedekindDirichlet theorem. The product of a sequence with bounded partial sums (sin(n), use a trig identity) and a function of bounded variation (1/n), converges as a series. This is proved using Abel's 'summation by parts', as lurflurf intimated.




#44
Apr2208, 08:04 AM

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#45
Apr2208, 08:43 AM

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Use sin(nx)=[cos((n1/2)x)cos(n+1/2)x)]/(2*sin(x/2)). It's a telescoping series. So if A_n is the partial sum, A_n<=1/sin(x/2). Clever, huh? Wish I'd thought of it. It shows things like sin(2n)/n converge also.




#46
Apr2208, 09:14 AM

P: 352

Well that problem would be appropriate for upper level math (analysis for use of the Dirichlet test, or applied math for engineers/physicists for recognizing that it's a Fourier series). But I think it's now safe to say that was probably an inappropriate problem for freshman calculus. 



#47
Apr2208, 09:36 AM

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#48
Apr2208, 10:08 AM

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When I look at something like that, my temptation is to try to find an upper bound for:
[tex]f(n)=\sum_{j=1}^{k} \sin(jx)[/tex] using geometry, and then to rearrange things into a sum that should look like: [tex]\sum \frac{1}{n^2+n} f(n)[/tex] 



#49
Feb2710, 11:16 PM

P: 20

[tex]\sum_{n=1}^{\infty} a_n \, sin(n)[/tex]
converges whenever {an} is a decreasing sequence that tends to zero. By : Dirichlet's test 



#50
Apr1411, 12:30 PM

P: 2

Dirichlet's test will work for this series.
we can think of sin(n)/n as the product sin(n)*1/n. dirichlet's test says that if one of the sequences in the product is bounded, and the other is monotone and converges to 0 then the series of the product of sequences must converge. the sequence of partial sums of sin(n) is bounded and certainly 1/n is monotone and converges to 0, so the series must converge. 



#51
Apr1411, 05:52 PM

P: 367

nice discussion here; I know the thread is a bit old but since so much trouble was stirred over this, I would like to also point out that dirichlet's test would have worked with this ( and would have proven this fairly easily ). It uses a lemma by Abel which says that if {bn} is nondecreasing and nonnegative for each n, and terms a1,..,an are bounded: m <= a1 +... + an <= M for any n , then bk*m <= a1b1 +... + anbn <= bk*M.
Development of these theorems can be found in Spivak's Calculus book on chapters 22 (chapter on infinite series ) and 19 (chapter on integration in elementary terms ). In chapter 22, the Dirichlet test is developed in exercise 13 and in chapter 19, abel's theorem is developed in problem 35. Not saying that it is very difficult at all, but for anyone who may be curious  those are excellent sources 



#52
Apr1411, 11:55 PM

P: 2

one other technique that would work is an interesting criterion involving the partial sums. let sn be the nth partial sum of the series. If for every epsilon greater than zero there exists an N such that for all n>N we have
sn+k  sn<epsilon for all k >= 1, then the series must converge. (Knopp "Theory and Application of Infinite Series") A quick induction on k would make quick work of this series' convergence. 


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