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Question Number 46425 by maxmathsup by imad last updated on 25/Oct/18

l e t p (x) = {(x + i)}^{n} - {(x - i)}^{n} w i t h i^{2} = - 1

1) f i n d p (x) a t f o r m Σ a_{k} x^{k}

2) f i n d t h e r o o t s o f p (x)

3) f a c t o r i z e i n s i d e C [x] p (x)

4) f a c t o r i z e i n s i d e R [x] t h e p o l y n o m p (x)

5) 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 maxmathsup by imad last updated on 18/Nov/18

1) w e h a v e P (x) = \sum_{k = 0}^{n} C_{n}^{k} i^{k} x^{n - k} - \sum_{k = 0}^{n} C_{n}^{k} {(- i)}^{k} x^{n - k}

= \sum_{k = 0}^{n} C_{n}^{k} (i^{k} - {(- i)}^{k}) x^{n - k} = \sum_{k = 2 p} (\dots) + \sum_{k = 2 p + 1} (\dots)

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

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

Commented by maxmathsup by imad last updated on 18/Nov/18

2) r o o t s o f P (x)

P (x) = 0 \Leftrightarrow {(x + i)}^{n} = {(x - i)}^{n} \Leftrightarrow {(\frac{x + i}{x - i})}^{n} = 1 \Rightarrow \frac{x + i}{x - i} = e^{\frac{i 2 k π}{n}} k \in [[1, n - 1]]

\Rightarrow x + i = (x - i) α_{k} (α_{k} = e^{\frac{i 2 k}{n}}) \Rightarrow (1 - α_{k}) x = - i - i α_{k} \Rightarrow x_{k} = - i \frac{1 + α_{k}}{1 - α_{k}}

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

= - i \frac{c o s (\frac{k π}{n}) e^{\frac{i k π}{n}}}{- i s i n (\frac{k π}{n}) e^{\frac{i k π}{n}}} = c o t a n (\frac{k π}{n}) s o t h e r o o t s o f P (x) a r e x_{k} = c o t a n (\frac{k π}{n})

w i t h k \in [[1, n - 1]] .