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Math Help - Differentiation of Power Series

  1. #1
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    Differentiation of Power Series

    Let  s(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... and c(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ... for x in R.

    Prove that s' = c and c' = s.

    I think you are supposed to differentiate term by term but i'm not sure exactly how that is done.

    Thanks
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  2. #2
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    Quote Originally Posted by tbyou87 View Post
    Let  s(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... and c(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ... for x in R.

    Prove that s' = c and c' = s.

    I think you are supposed to differentiate term by term but i'm not sure exactly how that is done.

    Thanks
    Differenciate term by terms.

    (b)Show that (s^2+c^2)' = 0

    (c)Conclude that s^2+c^2 = 1.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tbyou87 View Post
    Let  s(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... and c(x) = 1 - /frac{x^2}{2!} + /frac{x^4}{4!} - ... for x in R.

    Prove that s' = c and c' = s.

    I think you are supposed to differentiate term by term but i'm not sure exactly how that is done.

    Thanks
    i suppose we are not to assume that s(x) = \sin x and c(x) = \cos x

    just differentiate term by term. meaning, differentiate each term separately

    so, for instance:

    s(x) = x - \frac {x^3}{3!} + \frac {x^5}{5!} + \cdots

    \Rightarrow s'(x) = 1 - \frac {x^2}{2!} + \frac {x^4}{4!} - \cdots

    = c(x)
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    So the first one makes sense to me.

    I'm guessing for part b and c we are still using differentiation term by term. I guess i'm not seeing how to group them together so that they equal 0 for b and 1 for c.

    Thanks
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tbyou87 View Post
    So the first one makes sense to me.

    I'm guessing for part b and c we are still using differentiation term by term. I guess i'm not seeing how to group them together so that they equal 0 for b and 1 for c.

    Thanks
    for b you can do pretty easily by the chain rule. no term by term stuff.

    for c, i forgot the short way, try doing it the loooonnnggg tteeedddiiiooouuusss way of actually multiplying out the power series and summing them
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    is up to his old tricks again! Jhevon's Avatar
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    ah! i remember the short way

    since you (should have, by now) proved that \left( s^2(x) + c^2(x) \right)' = 0 it means that s^2(x) + c^2(x) = k for k some constant.

    indeed, since the series s(x) and c(x) work for all x, we have that the constant k is the same for all x. in particular, try x = 0 and the conclusion is immediate
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