power series in closed form.

**Showcase_22** · October 3rd 2008, 02:07 PM

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

Express the power series:

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

in closed form.

(Hint: Use the fact $1+\zeta+\zeta^2=0$ for the cube root of unity $\zeta=exp(\frac{2\pi i}{3}.))$

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 $z^n=1$ ?

**NonCommAlg** · October 4th 2008, 12:22 AM

let $\zeta_1, \zeta_2,$ be the roots of $\zeta^2 + \zeta + 1=0.$ then since $\zeta_1^3=\zeta_2^3=1,$ we will clearly have: $\zeta_1^{3n}=\zeta_2^{3n}=1, \ \zeta_1^{3n+1}+\zeta_2^{3n+1}=\zeta_1+\zeta_2=-1, \ \zeta_1^{3n+2}+ \zeta_2^{3n+2}=\zeta_1^2+\zeta_2^2=-1, \ \ n \geq 0.$ thus:

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

$f(x)=\frac{1}{3}(e^x + e^{\zeta_1 x} + e^{\zeta_2x}).$ but: $e^{\zeta_1 x} + e^{\zeta_2 x}=e^{\frac{-x}{2}} \left(e^{\frac{\sqrt{3}}{2}xi} + e^{\frac{-\sqrt{3}}{2}xi} \right)=2e^{-\frac{x}{2}} \cos \left(\frac{\sqrt{3}}{2}x \right).$ thus the closed form of your series is: $f(x)=\frac{1}{3}\left(e^x + 2e^{-\frac{x}{2}} \cos \left(\frac{\sqrt{3}}{2}x \right) \right).$

**Showcase_22** · October 4th 2008, 01:50 AM

Thanks for that post! I'm understanding parts of it =D

$\zeta_{1}^3+\zeta_{2}^3=1$

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

$\zeta_{1}+\zeta_{2}=-1$

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

$\zeta_{1}^2+\zeta_{2}^2=-1$

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.

**NonCommAlg** · October 4th 2008, 01:17 PM

$\zeta$ is just a variable like $x$ and $\zeta_1, \zeta_2$ are the roots of the quadratic equation $\zeta^2+\zeta+1=0.$ you know that the sum and product of the roots of a quadratic equation $x^2+ax+b=0$ are $-a$ and $b$

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

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