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anonymous

  • 5 years ago

I wanted to know why I can't apply the alternating series test to this series sigma n=1 to infinity (-1)^n-1 sin n

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  1. anonymous
    • 5 years ago
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    Can you make the expression clearer?

  2. anonymous
    • 5 years ago
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    Is it,\[(-)^{n-1}\sin n\]?

  3. anonymous
    • 5 years ago
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    \[\sum_{n=1}^{\infty}(-1)^{n-1} \sin( n )\] maybe that's what he meant?

  4. anonymous
    • 5 years ago
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    lol, that's what I've read :)

  5. anonymous
    • 5 years ago
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    yah thats what it is ty

  6. anonymous
    • 5 years ago
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    The alternating series test says you need the following for it to apply:

  7. anonymous
    • 5 years ago
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    Your terms need to be expressible in the form\[a_n=(-)^nb_n\]and the b_n have to be monotonic decreasing to zero as n approaches infinity.

  8. anonymous
    • 5 years ago
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    sin(n) does *not* monotonically decrease to zero.

  9. anonymous
    • 5 years ago
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    It oscillates.

  10. anonymous
    • 5 years ago
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    You can't apply the test here.

  11. anonymous
    • 5 years ago
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    u mean when we plug in values in the series

  12. anonymous
    • 5 years ago
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    Yes. A you increase n, you should get a decrease in b_n...and this shouldn't stop.

  13. anonymous
    • 5 years ago
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    But sin(n) will decrease and then start to increase again, and then start to decrease and then...as n increases.

  14. anonymous
    • 5 years ago
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    but when i plug in values it seems like its decreasing 0,.909,.141,..756

  15. anonymous
    • 5 years ago
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    This is what sine does for increasing n: http://www.wolframalpha.com/input/?i=sin%28n%29 Compare this with e^(-n) which *is* monotonic decreasing: http://www.wolframalpha.com/input/?i=e^%28-n%29

  16. anonymous
    • 5 years ago
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    i understand it now but this is the the behavior of sin, so we can never use the alternative test bc its not always decreasing

  17. anonymous
    • 5 years ago
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    Well, if it's like this, you can't use alternating series test. But don't be tricked into then thinking you can NEVER apply it whenever you see sine or cosine, because sometimes you can have series like,\[\sum_{n=1}^{\infty}\frac{\cos n \pi}{n^2}\]

  18. anonymous
    • 5 years ago
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    right it depends

  19. anonymous
    • 5 years ago
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    Here, you *can* use the test because, if you look at what happens as you increase n from 1 to 2 to 3 to...you see for cos(n pi):\[\cos(\pi)=-1\]\[\cos(2\pi)=1\]\[\cos(3 \pi)= -1\]in which case, the series can be written\[\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\]This is alternating, and the term 1/n^2 is monotonic decreasing...so the test will apply.

  20. anonymous
    • 5 years ago
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    i understand what u r saying , i have to see what i am given to determine whether ia m able to use it or not

  21. anonymous
    • 5 years ago
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    Yes, always, and it's the same for any other test. Theorems in mathematics are derived from statements assumed at the beginning. These are your conditions. If the conditions don't exist, the theorem doesn't apply.

  22. anonymous
    • 5 years ago
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    in the example above are we able to use it ∑n=1∞cosnπ/n2

  23. anonymous
    • 5 years ago
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    When you embark on solving these problems, read through the assumptions for the test and check them off like a recipe..."Do I have monotone decreasing? Yes. Do I have blah blah? No, so can't use this theorem...have to look for something else."

  24. anonymous
    • 5 years ago
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    Oh yeah, you can use it.

  25. anonymous
    • 5 years ago
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    okay ty

  26. anonymous
    • 5 years ago
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    The bit that isn't alternating needs to do two things: 1. go to zero as n goes to infinity 2. be monotonic decreasing

  27. anonymous
    • 5 years ago
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    *monotone

  28. anonymous
    • 5 years ago
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    k ty

  29. anonymous
    • 5 years ago
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    There's a lot of information lying around. These notes are pretty good http://tutorial.math.lamar.edu/Classes/CalcII/AlternatingSeries.aspx and there's stuff from MIT and the like either in open course ware or YouTube.

  30. anonymous
    • 5 years ago
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    cool ty for info

  31. anonymous
    • 5 years ago
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    ok. good luck.

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