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Math Help - power series

  1. #1
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    power series

    hi
    I know the power series representation of sin x is Sum [ (-1)^n*(x)^(2n+1)/(2n+1)!] but what about (sinx)^2?

    another quick question :
    give an example of set A that has no supremum and why does it not violate the Completeness Axiom.
    is {tanx : 0 < x < pi/2} correct?
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  2. #2
    Senior Member vincisonfire's Avatar
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    To get the power series expansion of  sin^2(x) , you can do 2 things :
    1. Compute and evaluate the derivative up to some order and try to find a general expression for the coefficients;
    2.  sin^2(x) = \sum_{i=0}^\infty c_nx^n = \left( \sum_{i=0}^\infty (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} \right)^2
    Equate coefficient on both sides using this formula
    <br />
\left(\sum_{n=0}^\infty a_n x^n\right)\left(\sum_{n=0}^\infty b_n x^n\right)
    = \sum_{i=0}^\infty \sum_{j=0}^\infty  a_i b_j x^{i+j}
    = \sum_{n=0}^\infty \left(\sum_{i=0}^n a_i b_{n-i}\right) x^n
    I suggest the second method.
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  3. #3
    MHF Contributor chisigma's Avatar
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    A third way is the use of identity...

    \sin^{2} x = \frac{1}{2} - \frac{1}{2}\cdot \cos 2x (1)

    ... so that is...

    \sin^{2} x = \frac{1}{2} \cdot \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(2x)^{2n}}{(2n)!} (2)

    Kind regards

    \chi \sigma
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  4. #4
    Senior Member vincisonfire's Avatar
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    chisigma's trick is even better
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  5. #5
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    lol why didn't I think of that!
    thank you!!!
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