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Math Help - Maclaurin Series Expansion

  1. #1
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    Maclaurin Series Expansion

    How can I calculate it for \frac{1}{1+cos^2(x)} by using the fact that \frac{1}{1+x^2} = 1 - x^2 + x^4 - ...?

    I tried letting u = cos(x), then

    \frac{1}{1+cos^2(x)} = \frac{1}{1+u^2} = 1 - u^2 + u^4 - ... = 1 - cos^2(x) + cos^4(x) - ...

    But I don't think this is right because the first term should be 0.5, not 1... and I don't see how a -0.5 might pop out of this series of cos terms....and even if it somehow does, I think this question is not meant to be that difficult...

    Any ideas?

    Thanks
    Last edited by HD09; October 10th 2009 at 05:58 AM.
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  2. #2
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    Quote Originally Posted by HD09 View Post
    How can I calculate it for \frac{1}{1+cos^2(x)} by using the fact that \frac{1}{1+x^2} = 1 - x^2 + x^4 - ...?

    I tried letting u = cos(x), then

    \frac{1}{1+cos^2(x)} = \frac{1}{1+u^2} = 1 - u^2 + u^4 - ... = 1 - cos^2(x) + cos^4(x) - ...

    But I don't think this is right because the first term should be 0.5, not 1... and I don't see how a -0.5 might pop out of this series of cos terms....and even if it somehow does, I think this question is not meant to be that difficult...

    Any ideas?

    Thanks
    What about the identity \cos^2 x = \frac {1 + \cos 2x}{2}? There's a convenient "divide by 2" in there which should get you your 0.5 you're looking for.
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  3. #3
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    Quote Originally Posted by Matt Westwood View Post
    What about the identity \cos^2 x = \frac {1 + \cos 2x}{2}? There's a convenient "divide by 2" in there which should get you your 0.5 you're looking for.

    Hmmm... but then cos^4(x) = \frac{1+2cos(2x)+cos^2(2x)}{2} puts it right back in again

    The answer I'm trying to get to is \frac{1}{2} + \frac{1}{4}x^2 + ...
    Last edited by mr fantastic; October 10th 2009 at 03:21 PM. Reason: Merged posts
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  4. #4
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    Maclaurin Series

    I tried letting u = cos(x), then

    \frac{1}{1+cos^2(x)} = \frac{1}{1+u^2} = 1 - u^2 + u^4 - ... = 1 - cos^2(x) + cos^4(x) - ...

    But I don't think this is right because the first term should be 0.5, not 1... and I don't see how a -0.5 might pop out of this series of cos terms....and even if it somehow does, I think this question is not meant to be that difficult...

    The answer I am trying to work towards is \frac{1}{2} + \frac{1}{4}x^2+...
    Last edited by mr fantastic; October 10th 2009 at 03:24 PM. Reason: Moved from another thread.
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  5. #5
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    Quote Originally Posted by HD09 View Post
    How can I calculate it for \frac{1}{1+cos^2(x)} by using the fact that \frac{1}{1+x^2} = 1 - x^2 + x^4 - ...?

    I tried letting u = cos(x), then

    \frac{1}{1+cos^2(x)} = \frac{1}{1+u^2} = 1 - u^2 + u^4 - ... = 1 - cos^2(x) + cos^4(x) - ...

    But I don't think this is right because the first term should be 0.5, not 1... and I don't see how a -0.5 might pop out of this series of cos terms....and even if it somehow does, I think this question is not meant to be that difficult...

    Any ideas?

    Thanks
    I don't see any simple way of doing this by using the result from an infinite geometric series. Is this how you were told to do it?
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  6. #6
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    Well, I was told to use my previous answer... I guess I can just use the derivatives I calculated for it and the ones for u(x) to get the derivatives for f(x) if there's no simpler way to do it. Thanks for replying.
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  7. #7
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    Quote Originally Posted by HD09 View Post
    Well, I was told to use my previous answer... I guess I can just use the derivatives I calculated for it and the ones for u(x) to get the derivatives for f(x) if there's no simpler way to do it. Thanks for replying.

    Well, of course that 1 - cos^2(x) + cos^4(x)-.... cannot be a valid answer since you're not giving a McClaurin series IN X but in cos x, and that's not, apparently, what you were asked to do.

    Sometimes , in some maths forum, there had been discussions as whether the infinite series 1 -1 + 1 - 1 +.... """converges""" to 1/2, and one of the reasons is precisely this developement: if you use the usual McClaurin series for cos(x), you get:

    == cos x = 1 - x^2/2 + x^4/4 -....

    == cox^2(x) = (1 - x^2/2 +...)^2 = 1 - x^2 +...

    == cox^4(x) = (1 - x^2/2 +...)^4 = 1 - 2x^2 +....

    etc.

    Now, if you want to know the value of f(x) = 1/[1 + cox^2(x)] at x = 0, which clearly is 1/2, using the above result with

    1/[1 + cosx^2(x)] = 1 - cos^2(x) + cos^4(x) -...

    and you look at the constant coefficient in the RH, you get 1 - 1 + 1 - 1 +..., so "it must be" that its sum is 1/2.
    You can read about this Grandi's Series here Grandi's series - Wikipedia, the free encyclopedia
    and if you know something about Cesaro's sums then it may be make a little more sense.

    Tonio
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