Infinite limit of cos sum

  • Thread starter flyerpower
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Homework Statement


ScreenShot038.jpg


I have no idea how to start. Any hints?

Thanks in advance.
 

Answers and Replies

  • #2
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Hi flyerpower! :smile:

The trick is to write this sum in a Riemann sum. So, write the sum as something of the form

[tex]\sum_{k=1}^n{f(x_k)\Delta x}[/tex]

The limit of such a sum is an integral. So if you can write your sum as a Riemann sum, then you can find integrals to calculate the limit.
 
  • #3
I like Serena
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Hmm, nice problem! :smile:

@MM: I tried it your way, but couldn't find such a function. Do you have a suggestion how to get there?

What I did, was to write the cosine function as a Taylor expansion with only 1 term and an upper estimate for the remainder term. From there I could find the limit....
 
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  • #4
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I've already tried finding a function suitable for Riemann but, unfortunately, i couldn't come up with a result. I'll keep trying if you say it can be solved using riemann sums:).

@serena what value did you come up with by using taylor series?
 
  • #5
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I've already tried finding a function suitable for Riemann but, unfortunately, i couldn't come up with a result. I'll keep trying if you say it can be solved using riemann sums:).

@serena what value did you come up with by using taylor series?
I'd try ILSerena's method, it is much easier. :smile:
It should work with riemann sums, but I think it will get a bit complicated...
 
  • #6
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@serena what value did you come up with by using taylor series?
1/3.

(You're not doing some online homework test for extra credit, for which you only need the answer I hope?)
 
  • #7
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(You're not doing some online homework test for extra credit, for which you only need the answer I hope?)
No :), actually i'm preparing for an important math exam and i'm just practicing. I wanted to know your result and try with taylor series.
 
  • #8
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So, i kinda guessed it using my intuition.

So that sum expands like this : cos PI/(2n+1) + cos PI/(2n+2) + ... cos PI/(2n+n)

then if we apply the limit n -> infinity :

PI/(2n+1) -> 0, so does PI/(2n+2) .... PI/(2n+n), so every term converges to cos(0) = 1

We will have 1+1+1+1+1.....+1 = n

And then if we plug it into the initial limit we have n/(3n+1) which converges to 1/3.
Is my intuition correct ?:)


Also if you may write here the Taylor series method i would really appreciate. I don't know how to write that cosine function with only one term if Taylor series are infinite polynomial series :).
 
  • #9
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Yes, your intuition is correct! :smile:

The main thing the Taylor series remainder term provides, is to show that the remainder vanishes.

For reference, the Taylor series is explained on wiki:
http://en.wikipedia.org/wiki/Taylor_series

In particular on this page you can find that the expansion for the cosine is:
[tex]\cos x = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - ...[/tex]

Sadly the wiki article does not explain about the remainder term, but it is explained in this wiki article:
http://en.wikipedia.org/wiki/Taylor's_theorem

Long story short, we have:
[tex]1 - \frac {x^2} {2!} \le \cos x \le 1[/tex]

If you fill that in, you should find your limit (although it's still not trivial :wink:).
 
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  • #10
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Thank you for the explanation, i got it.
Taylor series seem to be very useful in many situations, i'll take a deeper look on them even though i didn't study that at school.
 
  • #11
I like Serena
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Well, you wouldn't.
That is, not in high school (assuming that is what you mean).
It's college/university material, which is not taught in high school (afaik).
This problem has university in mathematics or physics written over it.
How did you get by it?
 
  • #12
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I just finished high school and now i'm preparing for college admission.
As for this problem, i found it in the math book that i use for practice.
 

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