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Math Help - series

  1. #1
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    series

    \sum_{k\ =\ 0}^n (cos 1)^k

    i'm supposed to tell if this converges or diverges, & if it converges, evaluate the sum

    i know that cosine oscillates..but because it's cosine of a specific value, i'm a little confused. since cos 1 is about 1, will this just go up to infinity? and if so, how would i show it?
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  2. #2
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    Hi

    \sum_{k\ =\ 0}^n q^k = \frac{1-q^{n+1}}{1-q}

    When |q| < 1 the series converge to \frac{1}{1-q}

    This is the case since |cos 1| < 1
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    Quote Originally Posted by buttonbear View Post
    \sum_{k\ =\ 0}^n (cos 1)^k

    i'm supposed to tell if this converges or diverges, & if it converges, evaluate the sum

    i know that cosine oscillates..but because it's cosine of a specific value, i'm a little confused. since cos 1 is about 1, will this just go up to infinity? and if so, how would i show it?
    At first glance I would replace cos1 with its Taylor series and you could make a few quick observations as to whether the series converges or diverges.
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  4. #4
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    Running-Gag's method is a lot better.
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  5. #5
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    i haven't actually learned taylor series yet..

    is the first method using the definition of geometric series?
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    Quote Originally Posted by buttonbear View Post
    i haven't actually learned taylor series yet..

    is the first method using the definition of geometric series?
    Did you really read running-gag's post ?

    \sum_{k= 0}^n q^k = \frac{1-q^{n+1}}{1-q}
    When |q| < 1 the series converges to \frac{1}{1-q}

    This is the case since |cos 1| < 1
    what is this ?
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  7. #7
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    there's really no reason to be rude. it looks similar to the definition for the geometric series, where the sum converges to a/1-r, but not identical. i haven't seen the other form that running-gag posted, so i was just asking. i'm here to understand, if you don't want to help me, you don't have to.
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  8. #8
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    Quote Originally Posted by buttonbear View Post
    there's really no reason to be rude. it looks similar to the definition for the geometric series, where the sum converges to a/1-r, but not identical. i haven't seen the other form that running-gag posted, so i was just asking. i'm here to understand, if you don't want to help me, you don't have to.
    Did I look rude ?? It wasn't my intention. And I tried to help you, as I didn't give immediately the solution, but made you think about it...

    And what is a in your formula ? It's the first term of the series ! (it should be better with a smilie ><)
    Your series starts at k=0, and when k=0, [\cos(1)]^k=1
    So a=1.
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  9. #9
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    i'm sorry if you weren't trying to be, it just seemed like it because you were like "did you actually read it??" >.< i've had a couple of bad experiences on here, so, i just wanted to make sure. i appreciate your help, and, i agree that it's better when you don't give the answer right away. thanks again, sorry for the misunderstanding.

    edit:
    the series actually started at k=1, so, that's probably where i'm getting hung up..
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    Quote Originally Posted by buttonbear View Post
    edit:
    the series actually started at k=1, so, that's probably where i'm getting hung up..
    Do you mean that the series is actually \sum_{k\ =\ 1}^n (cos1)^k ?

    \sum_{k\ =\ 1}^n q^k = \left(\sum_{k\ =\ 0}^n q^k\right) - 1 = \frac{1-q^{n+1}}{1-q} - 1 = \frac{q-q^{n+1}}{1-q}
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  11. #11
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    yes, that's what i mean

    it's actually \sum_{k\ =\ 1}^{\infty} (cos1)^k, though i don't know how much that infinity affects it

    so, you're saying you can just subtract 1 from the rest of the series to reconcile the difference in starting value of k?
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  12. #12
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    Quote Originally Posted by buttonbear View Post
    yes, that's what i mean

    it's actually \sum_{k\ =\ 1}^{\infty} (cos1)^k, though i don't know how much that infinity affects it

    so, you're saying you can just subtract 1 from the rest of the series to reconcile the difference in starting value of k?
    Or you can factorise :
    \sum_{k=1}^\infty (\cos(1))^k=\cos(1) \sum_{k=1}^\infty (\cos(1))^{k-1}
    if you let n=k-1 (for more convenience), you get :
    \cos(1) \sum_{n=0}^\infty (\cos(1))^n

    Anyway, you can just apply the formula \frac{a}{1-q}
    If the series starts at k=1, then the first term a=\cos(1)^1=\cos(1)
    that's all
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    i'm sorry that i'm having so much trouble getting my head around this, but, what is q, in this case? i guess i'm confused by the whole n=k-1 thing.. i'm not sure how it directly converts to the formula a/1-r
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  14. #14
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    a is the first term and q is the ratio (2nd term is aq, third term is aq² etc)

    \sum_{k\ =\ 0}^{\infty} aq^k = a\:\sum_{k\ =\ 0}^{\infty} q^k = \frac{a}{1-q}

    When taking the sum from 1 and not from 0 you can
    either take away the first term (a)
    \sum_{k\ =\ 1}^{\infty} aq^k = \left(\sum_{k\ =\ 0}^{\infty} aq^k\right) - a = \frac{a}{1-q} - a = \frac{aq}{1-q}

    or factor q
    \sum_{k\ =\ 1}^{\infty} aq^k = q \sum_{k\ =\ 1}^{\infty} aq^{k-1} = q \sum_{j\ =\ 0}^{\infty} aq^j = q\:\frac{a}{1-q} = \frac{aq}{1-q}

    Fortunately the result is the same
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  15. #15
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    that's a really helpful explanation, thank you!

    so, for this problem, i ended up with \frac{cos1}{1-cos1} for the sum, because i said a= cos1 and then to get to the next term you multiply it by itself, so r=cos1 also?

    is this correct?
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