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May 5th, 2018, 02:11 AM   #1
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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 03:05 AM.
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May 5th, 2018, 04:48 AM   #2
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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 09:38 AM.
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May 5th, 2018, 04:56 AM   #3
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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)^(1-i) 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)^(1-i).
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 09:45 AM.
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May 5th, 2018, 07:15 AM   #4
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That cleared things up!
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May 5th, 2018, 07:45 AM   #5
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Quote:
Originally Posted by SDK View Post
That cleared things up!
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 add-on or other descriptions.

Last edited by skipjack; May 5th, 2018 at 09:46 AM.
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May 5th, 2018, 07:45 AM   #6
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Quote:
Originally Posted by Kavesat View Post
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.
Let's try to straighten this out, one step at a time.

$\displaystyle x = \sum_{k=j}^n f(k)$

normally implies that j and n are real numbers, n - k is a non-negative integer, f(k) is defined for every number j + i in the interval [j, n], where i is a non-negative 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 09:47 AM.
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May 5th, 2018, 08:12 AM   #7
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Quote:
Originally Posted by JeffM1 View Post
Have you studied infinite series?
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 08:46 AM.
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May 5th, 2018, 08:43 AM   #8
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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?
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May 5th, 2018, 10:09 AM   #9
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$\displaystyle \sum_{k=1}^n \cos(k) = \frac{\sin(n + 1/2)}{2\sin(1/2)} - \frac12$
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May 5th, 2018, 10:29 AM   #10
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Quote:
Originally Posted by skipjack View Post
$\displaystyle \sum_{k=1}^n \cos(k) = \frac{\sin(n + 1/2)}{2\sin(1/2)} - \frac12$
Thank you a lot, sir. Your formula is very useful to me.
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:
Originally Posted by studiot View Post
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?
Thanks, sir. It is too hard for me currently. I will search more to try to understand its math.

Last edited by Kavesat; May 5th, 2018 at 11:16 AM.
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