# power series representation in terms of another

Hi how to do the following: Given $f(z) = \sum c_n z^n$

How to express $\sum c_{3n} z^{3n}$ in terms of $f(z)$

Thanks a lot!

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The most natural approach involves a detour through complex numbers. Are they familiar to you? –  André Nicolas Jun 10 at 3:40
unfortunately no :( but please give an answer anyway.. i am suspicious whether there is a purely calculus manipulations based answer as well ? thank you –  Salih Ucan Jun 10 at 3:46

If you allow the use of complex numbers (numbers of the form $a + bi$ where $i=\sqrt{-1}$ and $a$ and $b$ are ordinary real numbers) then there are three cube roots of 1: $$\begin{array}{l} 1 \\ \frac{-1+i\sqrt{3}}{2} \equiv \omega \\ \frac{-1-i\sqrt{3}}{2} = \omega^2 \\ \end{array}$$ You can easily verify that $$\omega^2 = \left(\frac{-1+i\sqrt{3}}{2}\right)^2 = \frac{1 -2i\sqrt{3} - 3}{4} = \frac{-1-i\sqrt{3}}{2}$$ and that $$\omega^3 = \frac{-1+i\sqrt{3}}{2} \frac{-1-i\sqrt{3}}{2} = \frac{1+3}{4} = 1$$ Now let's look at what $f(z\omega)$ would be: When the power is a multiple of 3, the term will remain $c_{3n}z^{3n}$ because $\omega^{3n} = 1$. But when the power is $3n+1$ the term is $\omega c_{3n+1}z^{3n+1}$ and when the power is $3n+2$ the term is $\omega^2 c_{3n+2}z^{3n+2}$. $$f(\omega z) = \sum c_{3n} z^{3n} + \omega \sum c_{3n+1}z^{3n+1} + \omega^2 \sum c_{3n+2}z^{3n+2}$$ Similarly, $$f(\omega^2 z) = \sum c_{3n} z^{3n} + \omega^2 \sum c_{3n+1}z^{3n+1} + \omega \sum c_{3n+2}z^{3n+2}$$ because $\omega^4 = \omega$.

Now here is the cute step: Notice that $\omega + \omega^2 = -1$. So $$f(\omega z) + f(\omega^2 z)= 2 \sum c_{3n} z^{3n} -\sum c_{3n+1}z^{3n+1} -\sum c_{3n+2}z^{3n+2}$$ and the answer to your question is obtained by adding $f(z)$ to get rid of all the $3n+1$ and $3n+2$ powers: $$\sum c_{3n} z^{3n} = \frac{f(z)+f(\omega z) + f(\omega^2 z)}{3}$$

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There's a $2$ missing at the end. –  G.T.R Jun 10 at 10:28
Qite right, I will try to edit that in –  Mark Fischler Jun 10 at 15:56

Hint: We give an approach using complex numbers. Let $\omega_1=\frac{-1+\sqrt{-3}}{2}$ and $\omega_2=\frac{-1-\sqrt{-3}}{2}$. If $k$ is divisible by $3$, then $1^k+\omega_1^k+\omega_2^k=3$. In all other cases, $1^k+\omega_1^k+\omega_2^k=0$. This is because if $k\equiv 1\pmod{3}$, then $\omega_1^k=\omega_1$ and $\omega_2^k=\omega_2$, while if $k\equiv 2\pmod{3}$, then $\omega_1^k=\omega_2$ and $\omega_2^k=\omega_1$.

Now consider $$\frac{f(z)+f(\omega_1 z)+f(\omega_2 z)}{3}.$$ Substitute in the power series, and add. The terms involving powers not divisible by $3$ will vanish.

Remark: A similar approach, using fourth roots of unity instead of cube roots, will give an expression for $\sum c_{4n} z^{4n}$. At a simpler level, using the square roots of unity, we obtain an expression for $\sum c_{2n}z^{2n}$.

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