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## Homework Statement

I have no idea how to start. Any hints?

Thanks in advance.

- Thread starter flyerpower
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- #1

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I have no idea how to start. Any hints?

Thanks in advance.

- #2

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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

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Homework Helper

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Hmm, nice problem!

@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....

@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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@serena what value did you come up with by using taylor series?

- #5

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I'd try ILSerena's method, it is much easier.

@serena what value did you come up with by using taylor series?

It should work with riemann sums, but I think it will get a bit complicated...

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Homework Helper

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1/3.@serena what value did you come up with by using taylor series?

(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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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.(You're not doing some online homework test for extra credit, for which you only need the answer I hope?)

- #8

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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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Homework Helper

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Yes, your intuition is correct!

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 ).

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 ).

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- #10

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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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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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As for this problem, i found it in the math book that i use for practice.

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