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Question Number 107500 by Ar Brandon last updated on 11/Aug/20

Given a \in R - {\pm 1}

1 . Show that \forall x \in R 1 - 2 acos (x) + a^{2} > 0

2 . Show that;

\underset{k = 1}{\prod^{n}} (1 - 2 acos (\frac{2 k π}{n}) + a^{2}) = \underset{k = 1}{\prod^{n}} (a - e^{2 ik π / n}) (a - e^{- 2 ik π / n})

3 . Deduce that;

\underset{k = 1}{\prod^{n}} (1 - 2 acos (\frac{2 k π}{n}) + a^{2}) = {(a^{n} - 1)}^{2}

4 . Using {Reimann}^{'} s sum, calculate

I = \int_{0}^{2 π} \ln (1 - 2 acos (x) + a^{2}) dx

Answered by Ar Brandon last updated on 11/Aug/20

1 . 1 - 2 acos (x) + a^{2}

= \cos^{2} - 2 acos (x) + a^{2} + \sin^{2} (x)

= {(\cos (x) - a)}^{2} + \sin^{2} (x) > 0

2 . 1 - 2 a \cos \frac{2 k π}{n} + a^{2}

= {(\cos \frac{2 k π}{n} - a)}^{2} + \sin^{2} \frac{2 k π}{n}

= {(\cos \frac{2 k π}{n} - a)}^{2} - {(isin \frac{2 k π}{n})}^{2}

= (\cos \frac{2 k π}{n} - isin \frac{2 k π}{n} - a) (\cos \frac{2 k π}{n} + isin \frac{2 k π}{n} - a)

= (a - e^{2 ik π / n}) (a - e^{- 2 ik π / n})

3 . S e e M r 1549442205 P V T^{'} s e x p l a n a t i o n o n Q 107498

4 . I = \int_{0}^{2 π} \ln (1 - 2 a \cos (x) + a^{2}) dx

= \lim_{n \to \infty} \frac{2 π}{n} \underset{k = 1}{\sum^{n}} \ln (1 - 2 a \cos (\frac{2 π k}{n}) + a^{2})

= \lim_{n \to \infty} \frac{2 π}{n} \ln \underset{k = 1}{\prod^{n}} (1 - 2 a \cos (\frac{2 π k}{n}) + a^{2})

= \lim_{n \to \infty} \frac{2 π}{n} \ln {(a^{n} - 1)}^{2} = \lim_{n \to \infty} \frac{4 π}{n} \ln (a^{n} - 1)

= 4 π \lim_{n \to \infty} [\frac{\ln (a^{n} - 1)}{n}] = 4 π \lim_{n \to \infty} [\frac{a^{n} lna}{a^{n} - 1}]

= 4 π \ln (a)