11.9k views
1 vote

\: \: \: \: \: \: \: \: \: \: \: \: \: \: \:

\: \: \: \: \: \: \: \: \: \: \: \: \: \: \:​-example-1
User Binyomin
by
7.7k points

1 Answer

1 vote


\omega is a 2020th root of unity, so
\omega^(2020)=1 and we have a kind of reflection identity of
\omega^(2020-n) = \omega^(-n) for
n\in\Bbb Z.

Let's evaluate the product first. Denote it by
P_k. We split the product where the
j=k factor would belong, and pull out powers of
\omega^k.


\displaystyle P_k = \prod_(j=1, j\\eq k) (\omega^k - \omega^j) \\\\ ~~~~ = \prod_(a=1)^(k-1) (\omega^k - \omega^a) \prod_(b=k+1)^(2019) (\omega^k - \omega^b) \\\\ ~~~~ = (\omega^k)^(k-1) \prod_(a=1)^(k-1) (1-\omega^(a-k)) \cdot (\omega^k)^(2019-k) \prod_(b=k+1)^(2019) (1 - \omega^(b-k)) \\\\ ~~~~ = (\omega^k)^(2018) \prod_(a=1)^(k-1) (1 - \omega^(-a)) \prod_(b=1)^(2019-k) (1 - \omega^b) \\\\ ~~~~ = \omega^(-2k) \prod_(a=1)^(k-1) (1-\omega^(-a)) \prod_(b=1)^(2019-k) (1-\omega^b)

Now introduce some factors to "complete" the
b-product and have it contain 2019 factors.


\displaystyle P_k = \omega^(-2k) \prod_(a=1)^(k-1) (1-\omega^(-a)) (\displaystyle \prod_(b=1)^(2019) (1 - \omega^b))/(\displaystyle \prod_(b=2020-k)^(2019) (1 - \omega^b))

It's relatively straightforward to show that if
\zeta is an
n-th root of unity, then


\displaystyle \sum_(m=1)^(n-1) (1-\zeta^m) = n

which gives


\displaystyle P_k = 2020 \omega^(-2k) (\displaystyle \prod_(a=1)^(k-1) (1-\omega^(-a)))/(\displaystyle \prod_(b=2020-k)^(2019) (1 - \omega^b))

Shifting the index in the denominator and again using the reflection property eliminates all but one factor.


\displaystyle P_k = 2020 \omega^(-2k) (\displaystyle \prod_(a=1)^(k-1) (1-\omega^(-a)))/(\displaystyle \prod_(b=1)^k (1 - \omega^(-b))) \\\\ ~~~~ = (2020 \omega^(-2k))/(1 - \omega^(-k))

Now evaluate the sum. We can exploit symmetry. Split the sum at the 1010th term, so that


\displaystyle - \sum_(k=1)^(2019) P_k = -2020 \left(\sum_(k=1)^(1009) (\omega^(-2k))/(1 - \omega^(-k)) + (\omega^(-2020))/(1-\omega^(-1010)) + \sum_(k=1011)^(2019) (\omega^(-2k))/(1-\omega^(-k))\right)

The middle terms reduces to 1/2. Shifting the index in the second sum, we can condense it to


\displaystyle -\sum_(k=1)^(2019) P_k = -2020 \left(\frac12 + \sum_(k=1)^(1009) \left((\omega^(-2k))/(1-\omega^(-k)) + (\omega^k)/(1-\omega^k)\right)\right)

Join the fractions.


\displaystyle (\omega^(-2k))/(1-\omega^(-k)) + (\omega^k)/(1-\omega^k) = 1 - (2-\omega^(2k)-\omega^(-2k))/(2-\omega^k-\omega^(-k)) = -(1+\omega^k + \omega^(-k))

The remaining sums are trivial.


-\displaystyle \sum_(k=1)^(2019) P_k = -2020 \left(\frac12 - 1009 - (1-\omega^(1010))/(1-\omega) - (1-(-\omega)^(1010))/(1+\omega)\right) \\\\ ~~~~ = 2020\cdot1009 - 1010 \\\\ ~~~~ = (2000+20)(1000+9) - 1000 - 10 \\\\ ~~~~ = 2\cdot1000^2 + 37\cdot1000 + 170

Taking this last result mod 1000, we find the last 3 digits to be 170.

User Enzokie
by
7.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories