May 5th, 2018, 01:11 AM  #1 
Newbie Joined: Dec 2010 Posts: 19 Thanks: 0  Does Σ cos x exist?
Hi, all. Recently I calculate some Σ. The idea of Σ cos x just popped up. I have googled a lot but none is about this kind of question. Does Σ cos x exist? That is, is it meaningful? To speak the Σ more precisely, For example. It's that kind of Σ. My guess: Σ cos x = 0 or it doesn't have meaning? Thanks in advance. Last edited by Kavesat; May 5th, 2018 at 02:05 AM. 
May 5th, 2018, 03:48 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
Frankly I am not clear what you are asking! As for $\displaystyle \sum_{n=1}^N \cos(n)$, as long as N is a finite number, yes, of course it exists. And, yes, $\displaystyle \sum_{n=1}^N n= \frac{N(N+1)}{2}$. For example, taking N= 4, $\displaystyle \sum_{n=1}^4 \cos(n)= \cos(1)+ \cos(2)+ \cos(3)+ \cos(4)$ and $\displaystyle \sum_{n=1}^4= 1+ 2+ 3+ 4= 10= \frac{4(5)}{2}$ But $\displaystyle \sum x$ is incomplete. There is no information, as there is with "n=1" below the $\displaystyle \sum$ or "N" above it, about what is to change in the terms of the sum or how it changes. Is "x" intended to be a positive integer or a real number? Last edited by skipjack; May 5th, 2018 at 08:38 AM. 
May 5th, 2018, 03:56 AM  #3 
Newbie Joined: Dec 2010 Posts: 19 Thanks: 0 
Sorry, my bad. My math is often in a mess. I shouldn't use different variables. The question I asked should be the following. And thank you for reply. Because 1 ≦ cos(k) ≦1...hmm, I am not sure it is right or wrong. I guess the sum will finally be 0. And what I thought Σ cos might not exist? Because I can't find exact example from google search and I don't have this knowledge. I can only guess or assume. Hmm...I can't shrink the picture smaller, sorry again. Well, if anyone tells me more or correct all my saying, I will be thankful. I was recently interested in Complex number like (1+i)^(1i) but the education of my math only taught me some basic Complex number. And I can't resolve it. So, I search lots google and youtube. Finally I understand more and enough to resolve (1+i)^(1i). Well, I am glad to know math can be learnt by one's own if there's enough resource. I type this only to describe I am serious in math. Last edited by skipjack; May 5th, 2018 at 08:45 AM. 
May 5th, 2018, 06:15 AM  #4 
Senior Member Joined: Sep 2016 From: USA Posts: 443 Thanks: 254 Math Focus: Dynamical systems, analytic function theory, numerics 
That cleared things up!

May 5th, 2018, 06:45 AM  #5 
Newbie Joined: Dec 2010 Posts: 19 Thanks: 0  What do you mean? Did I say too much? I just thought if I said more about myself, not directly putting a question coldly then gone, more replies tend to come. Or I did wrong? But people sometimes tell the askers should describe how they try to or think the question before making a thread. Ok, if I do wrong, next time I will put only a question, no more addon or other descriptions. Last edited by skipjack; May 5th, 2018 at 08:46 AM. 
May 5th, 2018, 06:45 AM  #6  
Senior Member Joined: May 2016 From: USA Posts: 1,126 Thanks: 468  Quote:
$\displaystyle x = \sum_{k=j}^n f(k)$ normally implies that j and n are real numbers, n  k is a nonnegative integer, f(k) is defined for every number j + i in the interval [j, n], where i is a nonnegative integer, and that x equals the sum of f(j) + f(j + 1) + f(j + 2) ... f(j + n). Is that how you understand it? $\cos(k)$ is defined for every real number so it certainly is defined for a finite number of integers. Therefore $\displaystyle \sum_{k=j}^n \cos(k)$ is defined and calculable. You can't find anything about it because it is so trivial a result. HOWEVER $\displaystyle x = \sum_{k=j}^{\infty} f(k)$ is a completely different thing. Infinity is NOT a real number. We are not talking about a finite number of summands that can in fact just be added up, but an infinite number of summands. What is really meant here is $\displaystyle x = \lim_{n \rightarrow \infty} \sum_{k=j}^n f(k).$ Have you studied infinite series? Last edited by skipjack; May 5th, 2018 at 08:47 AM.  
May 5th, 2018, 07:12 AM  #7 
Newbie Joined: Dec 2010 Posts: 19 Thanks: 0  Thanks for reply, sir. Yes, I have learned some lim like this. Sorry, I can't totally understand what you mean about Σ. If the following Σ is right, why can't you directly replace the k with cos(k) and then calculate? if you can replace, the value will go down  then go up + between 1 and 1...They will neutralize each other. cos1=0.54030230586 cos2=0.41614683654 cos3=0.9899924966 cos4=0.65364362086 cos5=0.28366218546 cos7=0.75390225434 Last edited by Kavesat; May 5th, 2018 at 07:46 AM. 
May 5th, 2018, 07:43 AM  #8 
Senior Member Joined: Jun 2015 From: England Posts: 853 Thanks: 258 
Just thinking about your question, The cos (or sin) of successive integers is not very interesting because they are periodic functions with period 2pi. However the sums of series with succussive integers times an angle, ie double, triple quadruple etc is very interesting and useful. These are called fourier series. Here are a few, both using sin and cos Perhaps your sums were part of something like this? 
May 5th, 2018, 09:09 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 19,532 Thanks: 1750 
$\displaystyle \sum_{k=1}^n \cos(k) = \frac{\sin(n + 1/2)}{2\sin(1/2)}  \frac12$

May 5th, 2018, 09:29 AM  #10  
Newbie Joined: Dec 2010 Posts: 19 Thanks: 0  Quote:
There is net pal helping me with drawing the sum for each point k=1 ~150 Value is from【1.6， 0.6】. I'm still interested in Σ(cosk). Quote:
Last edited by Kavesat; May 5th, 2018 at 10:16 AM.  

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