All primitives

Algebra Level 4

$\large\prod\limits_{n=1}^{\infty}\frac{ \zeta_{2n-1}}{\zeta_{2n}}$

The fraction above evaluates to $A^{\frac{i \pi}{b}}$ , where $A$ and $b$ are integers and that $A$ cannot be reduced to another integer when raised to the exponent $\frac{1}{b}$ .

Find $A+b$ .

Note: $\zeta_{k}$ denotes the first primitive $k^\text{th}$ root of unity.

The answer is 5.

1 solution

Efren Medallo
Apr 14, 2017

Knowing that $\zeta_n$ evaluates to $e^{\frac{2 \pi i }{n}}$ , we can reimagine the expression as

$\large \frac{e^{[(2\pi i)(\frac{1}{1} + \frac{1}{3}+ \frac{1}{5}...)]}}{e^{[(2\pi i)(\frac{1}{2} + \frac{1}{4} + \frac{1}{6}...)]}}$

and since these series have common factors, we can combine them and get

$\large e^{[(2 \pi i)(\frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4}...)]}$

which later simplifies to

$e^{[(2 \pi i )(\ln 2)]}$

Going further

$e^{i \pi \ln 4 }$

$4^{i \pi}$

and that's how we get $4+1 = \boxed{5}$ .

you shall change the latex from huge to large which will look like this $\large \frac{\prod\limits_{n=1}^{\infty} \zeta_{2n-1}}{\prod\limits_{n=1}^{\infty} \zeta_{2n}}$ .

Isnt it fine?

Where as huge will look like

$\huge\frac{\prod\limits_{n=1}^{\infty} \zeta_{2n-1}}{\prod\limits_{n=1}^{\infty} \zeta_{2n}}$

Md Zuhair - 4 years, 1 month ago

Yes, you're right. \huge is too huge for the expression. Thanks for that!

Efren Medallo - 4 years, 1 month ago

All primitives

The answer is 5.

1 solution

Where as huge will look like

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