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Prove that $(1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)…$ to 2n factors$=2^{2n}$ where $\omega$ is the cube root of unity

My attempt:

$(1+\omega^n+\omega^{2n})=0$ $\Rightarrow (1-\omega^n+\omega^{2n})=-2\omega^n$ $\Rightarrow \prod_{n=1 \to 2n}(-2\omega^n) =2^{2n}\omega^{1+2+3…2n}$ $\Rightarrow \prod_{n=1 \to 2n}(-2\omega^n) =2^{2n}\omega^{n(2n+1)}$

If $n$ is $3k$ type: $2^{2n}\omega^{n(2n+1)}=2^{2n}\omega^{3m}=2^{2n}$

If $n$ is $3k+1$ type: $2^{2n}\omega^{n(2n+1)}=2^{2n}\omega^{3m}=2^{2n}$

If $n$ is $3k+2$ type: $2^{2n}\omega^{n(2n+1)}=2^{2n}\omega^{(3k+2)(6k+5)} =2^{2n}\omega^{(3m+1)}=2^{2n}\omega^{3m}\omega=2^{2n}\omega$

What have I done wrong here?

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    $\begingroup$ It is not clear at all what is the general term in your product. Who said $1+ω^n+ω^{2n}=0$? Is that so for $n=3$? But wait, we don't have the term with $n=3$. What terms do we have, then? $\endgroup$ Aug 28, 2020 at 5:28
  • $\begingroup$ @Ivan Neretin I'm sorry, I forgot to mention that $\omega$ is the cube root of unity. I'm just too used to using $\omega$ exclusively for cube roots of unity $\endgroup$
    – Amadeus
    Aug 28, 2020 at 5:30
  • $\begingroup$ @WW1 I'm sorry I forgot to mention $\omega$ is the cube root of unity I have edited the question now though $\endgroup$
    – Amadeus
    Aug 28, 2020 at 5:34
  • $\begingroup$ @IvanNeretin I see my mistake now, thanks $\endgroup$
    – Amadeus
    Aug 28, 2020 at 5:44

4 Answers 4

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Since $$a^3+b^3=(a+b)(a^2-ab+b^2),$$ we obtain: $$\prod_{k=1}^{2n}(1-w^{2^{k-1}}+w^{2^{k}})=\prod_{k=1}^{2n}\frac{\left(1+w^{3\cdot2^{k-1}}\right)}{1+w^{2^{k-1}}}=\frac{2^{2n}}{((1+w)(1+w^2))^n}=2^{2n}.$$

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Hint : Evaluate first few factors and see. $$ 1st : (1-\omega+\omega^2)=-2\omega \\ 2nd : (1-\omega^2+\omega^4)=-2\omega^2 \\ 1st \times 2nd = \ldots \\ 3rd : (1-\omega^4+\omega^8)=-2\omega \\ 4th : (1-\omega^8+\omega^{16})= \ldots $$

Can you complete?

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  • $\begingroup$ I think this is a much better solution than the other one here, thanks $\endgroup$
    – Amadeus
    Aug 28, 2020 at 7:56
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Since $\omega$ is the third root of unity, and also since $2^{2n}-1\equiv 0 \mod{3}$

$$ \begin{align} 1+\omega^{2^{i}}+\omega^{2^{i+1}}&=0\\ \\ 1-\omega^{2^{i}}-\omega^{2^{i+1}}&=2\omega^{2^{i}}\\ \\ \prod_{i=0}^{2n-1}{\left(1-\omega^{2^{i}}+\omega^{2^{i+1}}\right)}&=\prod_{i=0}^{2n-1}{\left(2\omega^{2^{i}}\right)}\\ &=2^{2n}\omega^{2^{2n}-1}\\ &=2^{2n} \end{align} $$

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I know my answer is essentially the same as Michael Rozenberg's, but I couldn't help posting my answer here anyways (although the calculation is the same, my approach to this was different).

Each term is a 3-term geometric progression, and the common ratio is $-\omega^{2^{k-1}}$. The product is now $$\prod^{2n}_{k=1} \frac{(-\omega^{2^{k-1}})^3 - 1}{-\omega^{2^{k-1}}-1}$$

Using the fact that $\omega^3 = 1$, this reduces to $$\prod^{2n}_{k=1} \frac{2}{\omega^{2^{k-1}}+1}$$

As $k$ varies from $0,1,2,3,4,5,6....2n$ it can be observed that every consecutive pair of $\omega^{2^{k-1}}$ reduces to $\omega$ and $\omega^2$ by using $\omega^3 = 1$.

This gives us $n$ identical pairs of a fraction $\frac{1}{\omega+1}\cdot\frac{1}{\omega^2+1} = 1$. Product of all these fractions is $1$, which leaves the original product as $2^{2n}$.

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