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Math Help - Taylor Series

  1. #1
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    Taylor Series

    Use a known Taylor Series to find the Taylor series about c=0 for the given function and find its radius of convergence.

    f(x)= cos x^3

    So, the existing Taylor series for cos(x) is
    \sum \frac{-1^k}{2k!}(x)^{2k}

    So the new Taylor Series for cos x^3 would be
    \sum \frac{-1^k}{2k!}(x)^{3k}

    Am I on the right track? And how do I find its radius of convergence?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by kl.twilleger View Post
    Use a known Taylor Series to find the Taylor series about c=0 for the given function and find its radius of convergence.

    f(x)= cos x^3

    So, the existing Taylor series for cos(x) is
    \sum \frac{-1^k}{2k!}(x)^{2k}

    So the new Taylor Series for cos x^3 would be
    \sum \frac{-1^k}{2k!}(x)^{3k}

    Am I on the right track? And how do I find its radius of convergence?
    \begin{aligned}\cos\left(x\right)&=\sum_{n=0}^{\in  fty}\frac{(-1)^nx^{2n}}{(2n)!}\\<br />
&\implies\cos\left(x^3\right)=\sum_{n=0}^{\infty}\  frac{(-1)^n\left(x^3\right)^{2n}}{(2n)!}\\<br />
&=\sum_{n=0}^{\infty}\frac{(-1)^nx^{6n}}{(2n)!}\end{aligned}

    And x\in\mathbb{R}\implies{x^3\in\mathbb{R}} because the reals are a field. Therefore this converges \forall{x}\in\mathbb{R}
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