Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
1  
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
add comment

2 Answers 2

up vote 2 down vote accepted

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: $$ \sum c_{3n} z^{3n} = \frac{f(z)+f(\omega z) + f(\omega^2 z)}{3} $$

share|improve this answer
    
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
add comment

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 $\sum c_{2n}z^{2n}$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.