let-p-x-x-i-3-n-x-i-3-n-with-x-real-1-simlify-p-x-2-find-the-roots-of-P-x-3-decompose-inside-C-x-p-x-4-calculate-0-1-p-x-dx-

Question Number 61651 by maxmathsup by imad last updated on 05/Jun/19

l e t p (x) = {(x + i \sqrt{3})}^{n} + {(x - i \sqrt{3})}^{n} w i t h x r e a l

1) s i m l i f y p (x)

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

3) d e c o m p o s e i n s i d e C [x] p (x)

4) c a l c u l a t e \int_{0}^{1} p (x) d x

Commented by maxmathsup by imad last updated on 06/Jun/19

1) w e h a v e P (x) = \sum_{k = 0}^{n} C_{n}^{k} {(i \sqrt{3})}^{k} x^{n - k} + \sum_{k = 0}^{n} C_{n}^{k} {(- i \sqrt{3})}^{k} x^{n - k}

= \sum_{k = 0}^{n} C_{n}^{k} {i^{k} + {(- i)}^{k}) 3^{\frac{k}{2}} x^{n - k} = \sum_{k = 2 p} (\dots) + \sum_{k = 2 p + 1} (\dots)

= 2 \sum_{p = 0}^{[\frac{n}{2}]} C_{n}^{2 p} {(- 1)}^{p} 3^{p} x^{n - 2 p} = 2 \sum_{p = 0}^{[\frac{n}{2}]} {(- 3)}^{p} C_{n}^{2 p} x^{n - 2 p}

a n o t h e r w a y w i t h a r c t a n

w e h a v e P (x) = 2 R e ({(x + i \sqrt{3})}^{n}) b u t ∣ x + i \sqrt{3} ∣ = \sqrt{x^{2} + 3} \Rightarrow

x + i \sqrt{3} = \sqrt{x^{2} + 3} (\frac{x}{\sqrt{x^{2} + 3}} + i \frac{\sqrt{3}}{\sqrt{x^{2} + 3}}) = r e^{i θ} \Rightarrow r = \sqrt{x^{2} + 3} a n d t a n θ = \frac{\sqrt{3}}{x}

(w e s u p p o s e x \neq 0) \Rightarrow θ a r c t a n (\frac{\sqrt{3}}{x}) \Rightarrow x + i \sqrt{3} = \sqrt{x^{2} + 3} e^{i a r c t a n (\frac{\sqrt{3}}{x})} \Rightarrow

{(x + i \sqrt{3})}^{n} = {(x^{2} + 3)}^{\frac{n}{2}} e^{i n a r c t a n (\frac{\sqrt{3}}{x})} \Rightarrow P (x) = 2 {(x^{2} + 3)}^{\frac{n}{2}} c o s (n a r c t a n (\frac{\sqrt{3}}{x}))

Commented by maxmathsup by imad last updated on 06/Jun/19

2) P (x) = 0 \Leftrightarrow {(x - i \sqrt{3})}^{n} = - {(x + i \sqrt{3})}^{n} \Leftrightarrow {(\frac{x - i \sqrt{3}}{x + i \sqrt{3}})}^{n} = - 1 \Leftrightarrow Z^{n} = - 1 w i t h

Z = \frac{x - i \sqrt{3}}{x + i \sqrt{3}} \Rightarrow Z x + i \sqrt{3} Z = x - i \sqrt{3} \Rightarrow (Z - 1) x = - i \sqrt{3} Z - i \sqrt{3} \Rightarrow

(Z - 1) x = - i \sqrt{3} (Z + 1) \Rightarrow x = i \sqrt{3} \frac{1 + Z}{1 - Z}

Z^{n} = - 1 \Rightarrow Z^{n} = e^{i (2 k + 1) π} \Rightarrow Z_{k} = e^{i \frac{(2 k + 1) π}{n}} a n d k \in [[0, n - 1]] \Rightarrow t h e r o o t s o f P (x)

a r e x_{k} = i \sqrt{3} \frac{1 + e^{\frac{i (2 k + 1) π}{n}}}{1 - e^{\frac{i (2 k + 1) π}{n}}} l e t s k m p l i f y x_{k}

x_{k} = i \sqrt{3} \frac{1 + c o s ((2 k + 1) \frac{π}{n}) + 2 i s i n ((2 k + 1) \frac{π}{2 n}) c o s ((2 k + 1) \frac{π}{2 n})}{1 - c o s ((2 k + 1) \frac{π}{n}) - 2 i s i n (2 k + 1) \frac{π}{2 n}) c o s ((2 k + 1) \frac{π}{2 n})}

= i \sqrt{3} \frac{2 {c o s}^{2} ((2 k + 1) \frac{π}{2 n}) + 2 i s i n ((2 k + 1) \frac{π}{2 n}) c o s ((2 k + 1) \frac{π}{2 n})}{2 {s i n}^{2} (l 2 k + 1) \frac{π}{2 n} - 2 i s i n (2 k + 1) \frac{π}{2 n}) c o s (2 k + 1) \frac{π}{2 n}}

= i \sqrt{3} \frac{c o s (2 k + 1) \frac{π}{2 n} (e^{i (2 k + 1) \frac{π}{2 n}})}{- i s i n (2 k + 1) \frac{π}{2 n} (e^{i (2 k + 1) \frac{π}{2 n}})} = - \sqrt{3} c o t a n ((2 k + 1) \frac{π}{2 n}) k \in [[0, n - 1]]

\times

Commented by maxmathsup by imad last updated on 06/Jun/19

3) P (x) = λ \prod_{k = 0}^{n - 1} (x - Z_{k}) = λ \prod_{k = 0}^{n - 1} (x + \sqrt{3} c o t a n ((2 k + 1) \frac{π}{2 n}))

λ i s t h e d o m i n e n t c o e f f i c i e n t o f P (x) .

Commented by maxmathsup by imad last updated on 06/Jun/19

4) w e h a v e P (x) = 2 \sum_{p = 0}^{[\frac{n}{2}]} {(- 3)}^{p} C_{n}^{2 p} x^{n - 2 p} \Rightarrow

\int_{0}^{1} P (x) d x = 2 \sum_{p = 0}^{[\frac{n}{2}]} {(- 3)}^{p} C_{n}^{2 p} {[\frac{1}{n - 2 p + 1} x^{n - 2 p + 1}]}_{0}^{1}

= 2 \sum_{p = 0}^{[\frac{n}{2}]} \frac{{(- 3)}^{p} C_{n}^{2 p}}{n - 2 p + 1} .