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Thread: Power series

  1. #1
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    Power series

    Hi I'm having troubles with a question and was wondering if someone could guide me through it:
    Given two mystery functions called anna and bob. Both have power series representations
    that work for all x $\displaystyle \epsilon$ R
    Anna(x) = $\displaystyle \sum_{n=0}^{\infty }a_{n}x^{n}$, bob(x)= $\displaystyle \sum_{n=0}^{\infty }b_{n}x^{n}$
    Furthermore, anna(0) = 0; bob(0) = 1 and we know that deriving bob gives anna and deriving
    anna gives bob.
    a) Find anna and bob, i.e. nd the coefficients $\displaystyle a_{n}$ and $\displaystyle b_{n}$ of their power series.
    b) Express the exponential function $\displaystyle e^{x}$ and $\displaystyle e^{-x}$ as a combination of anna and bob. Conversely,
    express anna and bob as a combination of $\displaystyle e^{x}$ and $\displaystyle e^{-x}$
    Thanks in advance
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  2. #2
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    Re: Power series

    Quote Originally Posted by Dragonkiller View Post
    Hi I'm having troubles with a question and was wondering if someone could guide me through it:
    Given two mystery functions called anna and bob. Both have power series representations
    that work for all x $\displaystyle \epsilon$ R
    Anna(x) = $\displaystyle \sum_{n=0}^{\infty }a_{n}x^{n}$, bob(x)= $\displaystyle \sum_{n=0}^{\infty }b_{n}x^{n}$
    Furthermore, anna(0) = 0; bob(0) = 1 and we know that deriving bob gives anna and deriving
    anna gives bob.
    a) Find anna and bob, i.e. nd the coefficients $\displaystyle a_{n}$ and $\displaystyle b_{n}$ of their power series.
    b) Express the exponential function $\displaystyle e^{x}$ and $\displaystyle e^{-x}$ as a combination of anna and bob. Conversely,
    express anna and bob as a combination of $\displaystyle e^{x}$ and $\displaystyle e^{-x}$
    Thanks in advance
    Here is the important information: You know that differentiating Anna will give Bob, and differentiating Bob will give Anna. So differentiating Anna TWICE will get back to Anna. This gives us the differential equation

    $\displaystyle \displaystyle \begin{align*} A'' &= A \\ A'' - A &= 0 \\ \textrm{Characteristic Equation: } m^2 - 1 &= 0 \\ m^2 &= 1 \\ m &= \pm 1 \\ \textrm{Therefore } A &= C_1 e^x + C_2 e^{-x} \end{align*}$

    We also know that $\displaystyle \displaystyle \begin{align*} A(0) = 0 \end{align*}$ and $\displaystyle \displaystyle \begin{align*} A'(0) = 1 \end{align*}$, so we get

    $\displaystyle \displaystyle \begin{align*} 0 &= C_1 e^0 + C_2 e^{-0} \\ 0 &= C_1 + C_2 \\ \\ 1 &= C_1 e^0 - C_2 e^{-0} \\ 1 &= C_1 - C_2 \\ \\ C_1 = \frac{1}{2}, C_2 = -\frac{1}{2} \end{align*}$

    So therefore

    $\displaystyle \displaystyle \begin{align*} A = \frac{1}{2}e^x - \frac{1}{2}e^{-x} \end{align*}$ and $\displaystyle \displaystyle \begin{align*} B = \frac{1}{2}e^x + \frac{1}{2}e^{-x} \end{align*}$

    Go from here.
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