Given a ∈R−{±1} 1. Show that ∀x∈R 1−2acos(x)+a^2 >0 2. Show that; Π_(k=1) ^n (1−2acos(((2kπ)/n))+a^2 )=Π_(k=1) ^n (a−e^(2ikπ/n) )(a−e^(−2ikπ/n) ) 3. Deduce that; Π_(k=1) ^n (1−2acos(((2kπ)/n))+a^2 )=(a^n −1)^2 4. Using Reimann′s sum, calculate I=∫


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Question Number 107500 by Ar Brandon last updated on 11/Aug/20
$Given a ∈R−{±1} 1. Show that ∀x∈R 1−2acos(x)+a^2 >0 2. Show that; Π_(k=1) ^n (1−2acos(((2kπ)/n))+a^2 )=Π_(k=1) ^n (a−e^(2ikπ/n) )(a−e^(−2ikπ/n) ) 3. Deduce that; Π_(k=1) ^n (1−2acos(((2kπ)/n))+a^2 )=(a^n −1)^2 4. Using Reimann′s sum, calculate I=∫_0 ^(2π) ln(1−2acos(x)+a^2 )dx$
$Given a \in R - {\pm 1}$ $1 . Show that \forall x \in R 1 - 2 acos (x) + a^{2} > 0$ $2 . Show that;$ $\underset{k = 1}{\prod^{n}} (1 - 2 acos (\frac{2 k π}{n}) + a^{2}) = \underset{k = 1}{\prod^{n}} (a - e^{2 ik π / n}) (a - e^{- 2 ik π / n})$ $3 . Deduce that;$ $\underset{k = 1}{\prod^{n}} (1 - 2 acos (\frac{2 k π}{n}) + a^{2}) = {(a^{n} - 1)}^{2}$ $4 . Using {Reimann}^{'} s sum, calculate$ $I = \int_{0}^{2 π} \ln (1 - 2 acos (x) + a^{2}) dx$
	Answered by Ar Brandon last updated on 11/Aug/20

	$1 . 1 - 2 acos (x) + a^{2}$ $= \cos^{2} - 2 acos (x) + a^{2} + \sin^{2} (x)$ $= {(\cos (x) - a)}^{2} + \sin^{2} (x) > 0$ $2 . 1 - 2 a \cos \frac{2 k π}{n} + a^{2}$ $= {(\cos \frac{2 k π}{n} - a)}^{2} + \sin^{2} \frac{2 k π}{n}$ $= {(\cos \frac{2 k π}{n} - a)}^{2} - {(isin \frac{2 k π}{n})}^{2}$ $= (\cos \frac{2 k π}{n} - isin \frac{2 k π}{n} - a) (\cos \frac{2 k π}{n} + isin \frac{2 k π}{n} - a)$ $= (a - e^{2 ik π / n}) (a - e^{- 2 ik π / n})$ $3 . S e e M r 1549442205 P V T^{'} s e x p l a n a t i o n o n Q 107498$ $4 . I = \int_{0}^{2 π} \ln (1 - 2 a \cos (x) + a^{2}) dx$ $= \lim_{n \to \infty} \frac{2 π}{n} \underset{k = 1}{\sum^{n}} \ln (1 - 2 a \cos (\frac{2 π k}{n}) + a^{2})$ $= \lim_{n \to \infty} \frac{2 π}{n} \ln \underset{k = 1}{\prod^{n}} (1 - 2 a \cos (\frac{2 π k}{n}) + a^{2})$ $= \lim_{n \to \infty} \frac{2 π}{n} \ln {(a^{n} - 1)}^{2} = \lim_{n \to \infty} \frac{4 π}{n} \ln (a^{n} - 1)$ $= 4 π \lim_{n \to \infty} [\frac{\ln (a^{n} - 1)}{n}] = 4 π \lim_{n \to \infty} [\frac{a^{n} lna}{a^{n} - 1}]$ $= 4 π \ln (a)$
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