find in terms of n the value of A_n = ∫_0 ^1 Π_(k=1) ^(n−1) (x^2 −2xcos(((kπ)/n)) +1)dx with n from N^★ .


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Question Number 31460 by abdo imad last updated on 08/Mar/18

$f i n d i n t e r m s o f n t h e v a l u e o f$ $A_{n} = \int_{0}^{1} \prod_{k = 1}^{n - 1} (x^{2} - 2 x c o s (\frac{k π}{n}) + 1) d x w i t h n f r o m N^{★} .$
Commented by abdo imad last updated on 10/Mar/18

$l e t d e c o m p o s e i n s i d e C [x] p (x) = x^{2 n} - 1 t h e r o o t s o f$ $p (x) a r e z_{k} = e^{i \frac{k π}{n}} w i t h k \in [0, 2 n - 1]] w e h a v e$ $z_{0} = 1, z_{1} = e^{i \frac{π}{n}}, z_{2} = e^{i \frac{2 π}{n}}, . . . z_{n - 1} = e^{i \frac{(n - 1) π}{n}}, z_{n} = - 1,$ $z_{n + 1} = e^{i \frac{(n + 1) π}{n}} = {\bar{z}}_{n - 1}, z_{n + 2} = {\bar{z}}_{n - 2}, z_{2 n - 1} = {\bar{z}}_{1} \Rightarrow$ $p (x) = \prod_{k = 0}^{2 n - 1} (x - z_{k}) = (x^{2} - 1) \prod_{k = 1}^{n - 1} (x - z_{k}) (x - {\bar{z}}_{k})$ $= (x^{2} - 1) \prod_{k = 1}^{n - 1} (x^{2} - 2 R e (z_{k}) x + ∣ z_{k} ∣^{2})$ $= (x^{2} - 1) \prod_{k = 1}^{n - 1} (x^{2} - 2 c o s (\frac{k π}{n}) x + 1) \Rightarrow$ $\prod_{k = 1}^{n - 1} (x^{2} - 2 c o s (\frac{k π}{n}) x + 1) = \frac{x^{2 n} - 1}{x^{2} - 1} \Rightarrow$ $A_{n} = \int_{0}^{1} \frac{x^{2 n} - 1}{x^{2} - 1} d x = \int_{0}^{1} (1 + x^{2} + x^{4} + . . . + x^{2 n - 2}) d x$ $A_{n} = {[x + \frac{1}{3} x^{3} + \frac{1}{5} x^{5} + . . . . + \frac{1}{2 n - 1} x^{2 n - 1}]}_{0}^{1}$ $= 1 + \frac{1}{3} + \frac{1}{5} + . . . . . + \frac{1}{2 n - 1} l e t f i n d A_{n} i n t e r m s o f H_{n}$ $A_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + . . . + \frac{1}{2 n - 1} + \frac{1}{2 n} - \frac{1}{2} - \frac{1}{4} - . . . - \frac{1}{2 n}$ $A_{n} = H_{2 n} - \frac{1}{2} H_{n} w i t h H_{n} = \sum_{k = 1}^{n} \frac{1}{k} .$
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