let p(x)=(x+i)^n −(x−i)^n with i^2 =−1 1) find p(x) at form Σ a_k x^k 2) find the roots of p(x) 3) factorize inside C[x] p(x) 4) factorize inside R[x] the polynom p(x) 5) decompose the fraction F(x)=(1/(p(x)))


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 46425 by maxmathsup by imad last updated on 25/Oct/18
$let p(x)=(x+i)^n −(x−i)^n with i^2 =−1 1) find p(x) at form Σ a_k x^k 2) find the roots of p(x) 3) factorize inside C[x] p(x) 4) factorize inside R[x] the polynom p(x) 5) decompose the fraction F(x)=(1/(p(x)))$
$l e t p (x) = {(x + i)}^{n} - {(x - i)}^{n} w i t h i^{2} = - 1$ $1) f i n d p (x) a t f o r m Σ a_{k} x^{k}$ $2) f i n d t h e r o o t s o f p (x)$ $3) f a c t o r i z e i n s i d e C [x] p (x)$ $4) f a c t o r i z e i n s i d e R [x] t h e p o l y n o m p (x)$ $5) d e c o m p o s e t h e f r a c t i o n F (x) = \frac{1}{p (x)}$
Commented by maxmathsup by imad last updated on 18/Nov/18

$1) w e h a v e P (x) = \sum_{k = 0}^{n} C_{n}^{k} i^{k} x^{n - k} - \sum_{k = 0}^{n} C_{n}^{k} {(- i)}^{k} x^{n - k}$ $= \sum_{k = 0}^{n} C_{n}^{k} (i^{k} - {(- i)}^{k}) x^{n - k} = \sum_{k = 2 p} (. . .) + \sum_{k = 2 p + 1} (. . .)$ $= \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} (i^{2 p + 1} - {(- i)}^{2 p + 1}) x^{n - 2 p - 1}$ $= \sum_{p = 0}^{[\frac{n - 1}{2}]} 2 i {(- 1)}^{p} C_{n}^{2 p + 1} x^{n - 2 p - 1} = 2 i \sum_{p = 0}^{[\frac{n - 1}{2}]} {(- 1)}^{p} C_{n}^{2 p + 1} x^{n - 2 p - 1} .$
Commented by maxmathsup by imad last updated on 18/Nov/18

$2) r o o t s o f P (x)$ $P (x) = 0 \Leftrightarrow {(x + i)}^{n} = {(x - i)}^{n} \Leftrightarrow {(\frac{x + i}{x - i})}^{n} = 1 \Rightarrow \frac{x + i}{x - i} = e^{\frac{i 2 k π}{n}} k \in [[1, n - 1]]$ $\Rightarrow x + i = (x - i) α_{k} (α_{k} = e^{\frac{i 2 k}{n}}) \Rightarrow (1 - α_{k}) x = - i - i α_{k} \Rightarrow x_{k} = - i \frac{1 + α_{k}}{1 - α_{k}}$ $x_{k} = - i \frac{1 + e^{\frac{i 2 k π}{n}}}{1 - e^{\frac{i 2 k π}{n}}} = - i \frac{1 + c o s (\frac{2 k π}{n}) + i s i n (\frac{2 k π}{n})}{1 - c o s (\frac{2 k π}{n}) - i s i n (\frac{2 k π}{n})}$ $= - i \frac{2 {c o s}^{2} (\frac{k π}{n}) + 2 i s i n (\frac{k π}{n}) c o s (\frac{k π}{n})}{2 {s i n}^{2} (\frac{k π}{n}) - 2 i s i n (\frac{k π}{n}) c o s (\frac{k π}{n})}$ $= - i \frac{c o s (\frac{k π}{n}) e^{\frac{i k π}{n}}}{- i s i n (\frac{k π}{n}) e^{\frac{i k π}{n}}} = c o t a n (\frac{k π}{n}) s o t h e r o o t s o f P (x) a r e x_{k} = c o t a n (\frac{k π}{n})$ $w i t h k \in [[1, n - 1]] .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum