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nth roots of unity

sbhatnagar

Active member
Jan 27, 2012
95
Challenge Problem $1,a_1,a_2,a_3, \cdots ,a_{n-1}$ are the $n^{\text{th}}$ roots of unity.

Find the value of

i) $(1-a_1)(1-a_2)(1-a_3) \cdots (1-a_{n-1})$

ii)$\displaystyle \frac{1}{2-a_1}+\frac{1}{2-a_2}+\frac{1}{2-a_3}+\cdots +\frac{1}{2-a_{n-1}}$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Challenge Problem $1,a_1,a_2,a_3, \cdots ,a_{n-1}$ are the $n^{\text{th}}$ roots of unity.

Find the value of

i) $(1-a_1)(1-a_2)(1-a_3) \cdots (1-a_{n-1})$

ii)$\displaystyle \frac{1}{2-a_1}+\frac{1}{2-a_2}+\frac{1}{2-a_3}+\cdots +\frac{1}{2-a_{n-1}}$
Is...

$\displaystyle f(x)=\frac{x^{n}-1}{x-1}= 1 + x + x^{2} + ... + x^{n-1} = \prod_{k=1}^{n-1} (x-a_{k})$ (1)


... so that the answer to i) is $f(1)=n$. The answer to ii) will be given in a successive post...

Kind regards

$\chi$ $\sigma$
 
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Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Challenge Problem $1,a_1,a_2,a_3, \cdots ,a_{n-1}$ are the $n^{\text{th}}$ roots of unity.

Find the value of

i) $(1-a_1)(1-a_2)(1-a_3) \cdots (1-a_{n-1})$

ii)$\displaystyle \frac{1}{2-a_1}+\frac{1}{2-a_2}+\frac{1}{2-a_3}+\cdots +\frac{1}{2-a_{n-1}}$
As chisigma has noted, the roots of $f(x)=\dfrac{x^{n}-1}{x-1}$ are $a_1,a_2,a_3, \ldots ,a_{n-1}$. The roots of $$f(2-x)=\frac{(2-x)^{n}-1}{(2-x)-1} = \bigl((2-x)^{n}-1\bigr)(1-x)^{-1} = \bigl((2-x)^{n}-1\bigr)(1+x+x^2+\ldots)$$ are $(2-a_1),(2-a_2),(2-a_3), \ldots ,(2-a_{n-1})$. The sum of the reciprocals of these roots will be the negative of the coefficient of $x$ in that last expression, divided by the constant term. Therefore $$ \frac{1}{2-a_1}+\frac{1}{2-a_2}+\frac{1}{2-a_3}+\ldots +\frac{1}{2-a_{n-1}} = \boxed{\dfrac{n2^{n-1} -2^n+1}{2^n-1}}.$$

You can check that result by working out the value of the sum when $n=4$. In that case, $a_1,a_2,a_3$ are $-1,i,-i$, and the sum is $\dfrac13+\dfrac1{2-i}+\dfrac1{2+i}$, which simplifies to $\dfrac{17}{15}$, in accordance with the boxed formula.
 

sbhatnagar

Active member
Jan 27, 2012
95
Here is another solution for 2!

We have
$$ \frac{x^n-1}{x-1}=(x-a_1)(x-a_2)(x-a_3)\cdots (x-a_{n-1}) $$
Let $f(x)=\log\frac{x^n -1}{x-1}$
$$f(x)=\log\dfrac{x^n-1}{x-1}=\log(x-a_1)+\log(x-a_2)+\log(x-a_3)+\cdots +\log(x-a_{n-1})$$
The derivative of $f(x)$ will be
$$f'(x)=\dfrac{ \left[n x^{-1+n}-\dfrac{-1+x^n}{(-1+x)}\right]}{-1+x^n}=\frac{1}{x-a_1}+\frac{1}{x-a_2}+\frac{1}{x-a_3}+\cdots +\frac{1}{x-a_{n-1}}$$

putting $x=2$

$$f'(2)=\frac{n 2^{n-1}-2^n+1}{2^n-1}=\frac{1}{2-a_1}+\frac{1}{2-a_2}+\frac{1}{2-a_3}+\cdots +\frac{1}{2-a_{n-1}}$$
 
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