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Maclaurin series for sin^2(x)

  1. May 23, 2013 #1
    since the maclaurin series for sin x is alternating in sign (EQ1) so when you square it to get sin[itex]^{2}[/itex](x) (EQ2) the (-1)[itex]^{n}[/itex] should become (-1)[itex]^{2n}[/itex] (EQ3) which can be simplified down to (EQ4), but when i checked that series at wolframalpha the series was still alternating like: Why is that? So when we square it do we ignore squaring the (-1)[itex]^{n}[/itex] and put that after we're done squaring the series?


    In the above post when I say EQ#, I'm referring to these equations
    140jd7c.jpg
     
    Last edited by a moderator: Apr 18, 2017
  2. jcsd
  3. May 23, 2013 #2

    micromass

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    You seem to think that ##\left(\sum \alpha_n\right)^2 = \sum \alpha_n^2##. This is of course false. It's the same thing as saying that ##(x + y)^2 = x^2 + y^2##.
     
  4. May 23, 2013 #3

    I can now see why sin[itex]^{2}[/itex](x) should be alternating, but how would I then simplify this algerbracially? how would I simplify ##\left(\sum \alpha_n\right)^2## ? Is using the half angle property for sin[itex]^{2}[/itex](x) my only method?
     
    Last edited: May 24, 2013
  5. May 24, 2013 #4

    Mark44

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    It's probably the easiest way to go.
     
  6. May 24, 2013 #5

    micromass

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    The half-angle property is indeed the easiest way. But there also is a explicit way to multiply two series. This is called the Cauchy product: http://en.wikipedia.org/wiki/Cauchy_product So you can solve it with this too, but this is a lot more complicated.

    Another way is to explicitely find the derivatives of ##\sin^2(x)## and see if you can find a pattern. But this is also not a very simple way to go.
     
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