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Thread: Differentiation of Power Series

  1. #1
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    Differentiation of Power Series

    Let $\displaystyle s(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... $and $\displaystyle c(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} -$ ... 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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    Quote Originally Posted by tbyou87 View Post
    Let $\displaystyle s(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... $and $\displaystyle c(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} -$ ... 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 $\displaystyle (s^2+c^2)' = 0$

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

    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 $\displaystyle s(x) = \sin x$ and $\displaystyle c(x) = \cos x$

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

    so, for instance:

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

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

    $\displaystyle = c(x)$
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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 $\displaystyle \left( s^2(x) + c^2(x) \right)' = 0$ it means that $\displaystyle s^2(x) + c^2(x) = k$ for $\displaystyle k$ some constant.

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