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Thread: Power series

  1. #1
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    Power series




    These two problems have stumped me. I doubt they're particularly difficult. I'm just having trouble grasping the concepts.....
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by osiron23 View Post



    These two problems have stumped me. I doubt they're particularly difficult. I'm just having trouble grasping the concepts.....
    I will help with the first....but...for the second...physics ::shivers::.

    a) So first we must find an expression for $\displaystyle \cos\left(x^2\right)$

    We first note (I assume you know this) that $\displaystyle \forall{x}\in\mathbb{R}~\cos(x)=\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n}}{(2n)!}$

    So now if we substitute in $\displaystyle x=\psi^2$ we get

    $\displaystyle \forall\psi^2\in\mathbb{R}~\cos\left(\psi^2\right) =\sum_{n=0}^{\infty}\frac{(-1)^n\psi^{4n}}{(2n)!}$

    And noting that $\displaystyle x\in\mathbb{R}\implies\psi^2\in\mathbb{R}$ and that since a variable is a variable let us just replace $\displaystyle \psi$ with x[/tex]. So

    $\displaystyle \forall{x}\in\mathbb{R}~~\cos\left(x^2\right)=\sum _{n=0}^{\infty}\frac{(-1)^nx^{4n}}{(2n)!}$

    b) So since this series converges for all real values we can say it converges on any $\displaystyle [0,\xi]\subset\mathbb{R}$. So now let us consider

    $\displaystyle \begin{aligned}\int_0^{\xi}\cos\left(x^2\right)dx= &\int_0^{\xi}\sum_{n=0}^{\infty}\frac{(-1)^nx^{4n}}{(2n)!}\\
    &=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}\int_0^{\xi}x^{4n}dx\\
    &=\sum_{n=0}^{\infty}\frac{(-1)^n\xi^{4n+1}}{(4n+1)(2n)!}\end{aligned}$

    And since integrating a power series changes the endpoints of its interval of convergence we are done (because the set of real numbers does not have endpoints)

    c) So you must just calculate $\displaystyle \sum_{n=0}^{N}\frac{(-1)^n}{(4n+1)(2n)!}$ where $\displaystyle N$ is given by the first integer that makes $\displaystyle \frac{(-1)^n}{(4n+1)(2n)!}<.001$


    Ok...lets do this

    a) Just multiply by $\displaystyle \frac{\frac{1}{R^2}}{\frac{1}{R^2}}$

    b) Why dont you take a stab at this

    Hint: Consider $\displaystyle \left(\frac{1}{1+x}\right)'$
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  3. #3
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    Thanks

    Thanks a lot!!! Is number 5 really a physics problem though? I'm in trouble then.
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