let-P-x-1-ix-n-1-ix-n-with-n-integr-decompose-the-Fraction-F-x-1-P-x-

Question Number 73486 by abdomathmax last updated on 13/Nov/19

l e t P (x) = {(1 + i x)}^{n} - {(1 - i x)}^{n} w i t h n i n t e g r

d e c o m p o s e t h e F r a c t i o n F (x) = \frac{1}{P (x)}

Commented by abdomathmax last updated on 17/Nov/19

P (x) = 0 \Leftrightarrow \frac{{(1 - i x)}^{n}}{{(1 + i x)}^{n}} = 1 \Leftrightarrow {(\frac{1 + i x}{1 - i x})}^{n} = 1

l e t z = \frac{1 + i x}{1 - i x} (e) \Rightarrow z^{n} = 1 \Rightarrow z_{k} = e^{\frac{i 2 k π}{n}}

k \in [[0, n - 1]] w e h a v e z - i z x = 1 + i x \Rightarrow

z - 1 = i x + i z x = i (z + 1) x \Rightarrow x = \frac{1}{i} \frac{z - 1}{1 + z}

= - i \frac{z - 1}{z + 1} = i \frac{1 - z}{1 + z} s o t h e r o o t s o f P a r e

x_{k} = i \frac{1 - z_{k}}{1 + z_{k}} = i \frac{1 - e^{i \frac{2 k π}{n}}}{1 + e^{i \frac{2 k π}{n}}} = i \frac{1 - c o s (\frac{2 k π}{n}) - i s i n (\frac{2 k π}{n})}{1 + c o s (\frac{2 k π}{n}) + i s i n (\frac{2 k π}{n})}

= i \frac{2 {s i n}^{2} (\frac{k π}{n}) - 2 i s i n (\frac{k π}{n}) c o s (\frac{k π}{n})}{2 {c o s}^{2} (\frac{k π}{n}) + 2 i s i n (\frac{k π}{n}) c o s (\frac{k π}{n})}

= i \frac{- i s i n (\frac{k π}{n}) e^{\frac{i k π}{n}}}{c o s (\frac{k π}{n}) e^{\frac{i k π}{n}}} = t a n (\frac{k π}{n}) \Rightarrow x_{k} = t a n (\frac{k π}{n})

k \in [[0, n - 1]] \Rightarrow P (x) = a \prod_{k = 0}^{n - 1} (x - t a n (\frac{k π}{n})}

a ? w e h a v e P (x) = {(1 + i x)}^{n} - {(1 - i x)}^{n}

= \sum_{k = 0}^{n} C_{n}^{k} {(i x)}^{k} - \sum_{k = 0}^{n} C_{n}^{k} {(- i x)}^{k}

= \sum_{k = 0}^{n} C_{n}^{k} {{(i)}^{k} - {(- i)}^{k}} x^{k}

= \sum_{p = 0}^{[\frac{n - 1}{2}]} 2 i {(- 1)}^{p} C_{n}^{2 p + 1} x^{2 p + 1}

\Rightarrow a = 2 i {(- 1)}^{[\frac{n - 1}{2}]} C_{n}^{2 [\frac{n - 1}{2}] + 1} \Rightarrow

P (x) = 2 i {(- 1)}^{[\frac{n - 1}{2}]} C_{n}^{2 [\frac{n - 1}{2}] + 1} \prod_{k = 0}^{n - 1} (x - t a n (\frac{k π}{n}))

F (x) = \frac{1}{P (x)} = \frac{1}{a \prod_{k = 0}^{n - 1} (x - x_{k})} = \sum_{k = 0}^{n - 1} \frac{λ_{k}}{x - x_{k}}

λ_{k} = \frac{1}{P^{^{'}} (x_{k})} b u t w e h a v e P (x) = {(1 + i x)}^{n} - {(1 - i x)}^{n}

\Rightarrow P^{^{'}} (x) = n i {(1 + i x)}^{n - 1} + n i {(1 - i x)}^{n - 1}

= n i {{(1 + i x)}^{n - 1} - {(1 - i x)}^{n - 1}}

= n i (2 i) I m ({(1 + i x)}^{n - 1}}

1 + i x = \sqrt{1 + x^{2}} e^{i a r c t a n x} \Rightarrow {(1 + i x)}^{n - 1} = {(1 + x^{2})}^{\frac{n - 1}{2}} e^{i (n - 1) a r c t a n (x)}

\Rightarrow I m ({(1 + i x)}^{n - 1}) = {(1 + x^{2})}^{\frac{n - 1}{2}} s i n ((n - 1) a r c t a n (x)) \Rightarrow

P^{^{'}} (x) = - 2 n {(1 + x^{2})}^{\frac{n - 1}{2}} s i n ((n - 1) a r c t a n (x)) \Rightarrow

P^{^{'}} (x_{k}) = - 2 n {(1 + {t a n}^{2} (\frac{k π}{n}))}^{\frac{n - 1}{2}} s i n ((n - 1) (\frac{k π}{n}))

= - 2 n {(\frac{1}{{c o s}^{2} (\frac{k π}{n})})}^{\frac{n - 1}{2}} s i n {k π - \frac{k π}{n}}

= \frac{- 2 n}{{(c o s (\frac{k π}{n}))}^{n - 1}} {- {(- 1)}^{k} s i n (\frac{k π}{n})

= \frac{2 n {(- 1)}^{k} s i n (\frac{k π}{n})}{{c o s (\frac{k π}{n})}^{n - 1}} = \frac{1}{λ_{k}}

Commented by mathmax by abdo last updated on 17/Nov/19

f o r g i v e P^{^{'}} (x) = - 2 n R e ({(1 + i x)}^{n - 1})

= - 2 n {(1 + x^{2})}^{\frac{n - 1}{2}} c o s {(n - 1) a r c t a n x} \Rightarrow

P^{^{'}} (x_{k}) = - 2 n {(1 + x_{k}^{2})}^{\frac{n - 1}{2}} c o s {(n - 1) a r c t a n (x_{k})}

= - 2 n {(1 + {t a n}^{2} (\frac{k π}{n}))}^{\frac{n - 1}{2}} c o s {(n - 1) \frac{k π}{n}}

= - 2 n {(\frac{1}{{c o s}^{2} (\frac{k π}{n})})}^{\frac{n - 1}{2}} c o s {k π - \frac{k π}{n}}

= - \frac{2 n}{{(c o s (\frac{k π}{n}))}^{n - 1}} {(- 1)}^{k} c o s (\frac{k π}{n}) = \frac{2 n {(- 1)}^{k + 1}}{{(c o s (\frac{k π}{n}))}^{n - 2}} = \frac{1}{λ_{k}} \Rightarrow

λ_{k} = \frac{{(- 1)}^{k + 1} {(c o s (\frac{k π}{n}))}^{n - 2}}{2 n} \Rightarrow

P (x) = \frac{1}{2 n} \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k + 1} {(c o s (\frac{k π}{n}))}^{n - 2}}{x - t a n (\frac{k π}{n})} (n > 1)