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Thread: power series in closed form.

  1. #1
    Super Member Showcase_22's Avatar
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    power series in closed form.

    This is a really confusing question that I haven't been able to solve:

    Express the power series:

    $\displaystyle 1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac {x^9}{9!}+......$

    in closed form.

    (Hint: Use the fact $\displaystyle 1+\zeta+\zeta^2=0$ for the cube root of unity $\displaystyle \zeta=exp(\frac{2\pi i}{3}.))$
    Normally when I post questions, I write down the working that i've done so far (I hate the people who post on here purely for answers and never intend to solve the questions themselves!) but I really have no idea how to start this.

    Since it wants me to use [tex]z^3=1[tex] this will only give me the first three terms. However, since there's an n number of terms would I need the $\displaystyle z^n=1$?
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  2. #2
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    let $\displaystyle \zeta_1, \zeta_2,$ be the roots of $\displaystyle \zeta^2 + \zeta + 1=0.$ then since $\displaystyle \zeta_1^3=\zeta_2^3=1,$ we will clearly have: $\displaystyle \zeta_1^{3n}=\zeta_2^{3n}=1, \ \zeta_1^{3n+1}+\zeta_2^{3n+1}=\zeta_1+\zeta_2=-1, \ \zeta_1^{3n+2}+ \zeta_2^{3n+2}=\zeta_1^2+\zeta_2^2=-1, \ \ n \geq 0.$ thus:

    $\displaystyle 1+\zeta_1^{3n+k} + \zeta_2^{3n+k}=\begin{cases}3 \ \ \ \text{if} \ \ k=0 \\ 0 \ \ \ \text{if} \ \ k=1,2 \end{cases} .$ now let $\displaystyle f(x)=\sum_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}.$ then: $\displaystyle e^x + e^{\zeta_1 x} + e^{\zeta_2x}=\sum_{n=0}^{\infty}(1+\zeta_1^n+\zeta _2^n)\frac{x^n}{n!}=\sum_{k=0}^2 \sum_{n=0}^{\infty}(1+\zeta_1^{3n+k}+\zeta_2^{3n+k })\frac{x^{3n+k}}{(3n+k)!}=3f(x),$ which gives us:

    $\displaystyle f(x)=\frac{1}{3}(e^x + e^{\zeta_1 x} + e^{\zeta_2x}).$ but: $\displaystyle e^{\zeta_1 x} + e^{\zeta_2 x}=e^{\frac{-x}{2}} \left(e^{\frac{\sqrt{3}}{2}xi} + e^{\frac{-\sqrt{3}}{2}xi} \right)=2e^{-\frac{x}{2}} \cos \left(\frac{\sqrt{3}}{2}x \right).$ thus the closed form of your series is: $\displaystyle f(x)=\frac{1}{3}\left(e^x + 2e^{-\frac{x}{2}} \cos \left(\frac{\sqrt{3}}{2}x \right) \right).$
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  3. #3
    Super Member Showcase_22's Avatar
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    Thanks for that post! I'm understanding parts of it =D

    $\displaystyle \zeta_{1}^3+\zeta_{2}^3=1$
    Is this because $\displaystyle \zeta_{1}$ and $\displaystyle \zeta_{2}$ are the cube roots of unity? I thought that just $\displaystyle \zeta$ was the cube root of unity?

    $\displaystyle \zeta_{1}+\zeta_{2}=-1$
    Assuming $\displaystyle \zeta_{1}$ and $\displaystyle \zeta_{2}$ are the cube roots of unity, you would also have to include 1 (ie. the cube roots of unity are 1, $\displaystyle \zeta_{1}$ and $\displaystyle \zeta_{2}$. Since the sum of these roots must equal 0 then $\displaystyle \zeta_{1}+\zeta_{2}+1=0$ so $\displaystyle \zeta_{1}+\zeta_{2}=-1$?

    $\displaystyle \zeta_{1}^2+\zeta_{2}^2=-1$
    How did you end up with this? I've tried working it through on my whiteboard but i'm not ending up with anything like it.

    As soon as I know that i'm thinking of it correctly I will probably be able to understand the rest of it.
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  4. #4
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    $\displaystyle \zeta$ is just a variable like $\displaystyle x$ and $\displaystyle \zeta_1, \zeta_2$ are the roots of the quadratic equation $\displaystyle \zeta^2+\zeta+1=0.$ you know that the sum and product of the roots of a quadratic equation $\displaystyle x^2+ax+b=0$ are $\displaystyle -a$ and $\displaystyle b$

    respectively. so $\displaystyle \zeta_1 + \zeta_2=-1$ and $\displaystyle \zeta_1 \zeta_2 = 1.$ hence $\displaystyle \zeta_1^2 + \zeta_2^2 = (\zeta_1 + \zeta_2)^2 - 2 \zeta_1 \zeta_2=1-2=-1.$ also if you multiply both sides of $\displaystyle \zeta^2 + \zeta + 1 = 0$ by $\displaystyle \zeta - 1,$ you get $\displaystyle \zeta^3 - 1=0.$ thus $\displaystyle \zeta_1^3=\zeta_2^3=1.$
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