let-j-e-i2pi-3-and-P-x-1-jx-n-1-jx-n-1-find-P-x-at-form-of-arctan-2-find-the-roots-of-P-x-3-factorize-inside-C-x-the-polynome-P-x-4-calculate-0-1-P-x-dx-

Question Number 62225 by maxmathsup by imad last updated on 17/Jun/19

l e t j = e^{\frac{i 2 π}{3}} a n d P (x) = {(1 + j x)}^{n} - {(1 - j x)}^{n}

1) f i n d P (x) a t f o r m o f a r c t a n

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] t h e p o l y n o m e P (x)

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

Commented by mathmax by abdo last updated on 24/Jun/19

1) w e h a v e j = c o s (\frac{2 π}{3}) + i s i n (\frac{2 π}{3}) = - \frac{1}{2} + i \frac{\sqrt{3}}{2} \Rightarrow

P (x) = {(1 + (- \frac{1}{2} + i \frac{\sqrt{3}}{2}) x)}^{n} - {(1 - (- \frac{1}{2} + i \frac{\sqrt{3}}{2}) x)}^{n}

= {(1 - \frac{x}{2} + i \frac{\sqrt{3}}{2} x)}^{n} - {(1 + \frac{x}{2} - i \frac{\sqrt{3}}{2} x)}^{n} = A^{n} (x) - B^{n} (x)

∣ A (x) ∣ = \sqrt{{(1 - \frac{x}{2})}^{2} + \frac{3}{4} x^{2}} = \sqrt{\frac{{(2 - x)}^{2} + 3 x^{2}}{4}} = \frac{1}{2} \sqrt{{(x - 2)}^{2} + 3 x^{2}}

w e k n o w t h a t x + i y = \sqrt{x^{2} + y^{2}} e^{i a r c t a n (\frac{y}{x})} \Rightarrow

A (x) = \frac{1}{2} \sqrt{{(x - 2)}^{2} + 3 x^{2}} e^{i a r c t a n (\frac{\sqrt{3} x}{2 - x})} \Rightarrow A^{n} (x) = \frac{1}{2^{n}} {{(x - 2)}^{2} + 3 x^{2}}^{\frac{n}{2}} e^{i n a r c t a n (\frac{\sqrt{3} x}{2 - x})}

∣ B (x) ∣ = \frac{1}{2} \sqrt{{(x + 2)}^{2} + 3 x^{2}} \Rightarrow B (x) = \frac{1}{2} \sqrt{{(x + 2)}^{2} + 3 x^{2}} e^{i a r c t a n (\frac{- \sqrt{3} x}{2 + x})} \Rightarrow

B^{n} (x) = \frac{1}{2^{n}} {{(x + 2)}^{2} + 3 x^{2}}^{\frac{n}{2}} e^{- i n a r c t a n (\frac{\sqrt{3} x}{2 + x})} \Rightarrow

P (x) = \frac{1}{2^{n}} {{(x - 2)}^{2} + 3 x^{2}}^{\frac{n}{2}} e^{i n a r c t a n (\frac{\sqrt{3} x}{2 - x})} - \frac{1}{2^{n}} {{(x + 2)}^{2} + 3 x^{2}}^{\frac{n}{2}} e^{- i n a r c t a n (\frac{\sqrt{3} x}{2 + x})}

a n d w e s e e t h a t B (x) = A (- x) .

Commented by mathmax by abdo last updated on 24/Jun/19

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

t h e r o o t s o f Z^{n} = 1 a r e Z_{k} = e^{i \frac{2 k π}{n}} w i t h k \in [[0, n - 1]]

Z = \frac{1 + j x}{1 - j x} \Rightarrow Z - j Z x = 1 + j x \Rightarrow Z - 1 = j x (1 + Z) \Rightarrow x = \frac{Z - 1}{j (1 + Z)} s o t h e r o o t s

o f P (x) = 0 a r e x_{k} = - \frac{1}{j} \frac{1 - Z k}{1 + Z_{k}}

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

= - \frac{1}{j} \frac{- i s i n (\frac{k π}{n}) e^{i \frac{k π}{n}}}{c o s (\frac{k π}{n}) e^{\frac{i k π}{n}}} = \frac{i}{j} t a n (\frac{k π}{n}) \Rightarrow x_{k} = \frac{i}{j} t a n (\frac{k π}{n}) a n d k \in [[0, n - 1]] .

Commented by mathmax by abdo last updated on 24/Jun/19

3) P (x) = λ \prod_{k = 0}^{n - 1} (x - \frac{i}{j} t a n (\frac{k π}{n})) w i t h 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 mathmax by abdo last updated on 24/Jun/19

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

= \sum_{k = 0}^{n} C_{n}^{k} (j^{k} - {(- j)}^{k}) x^{k} b u t j^{k} - {(- j)}^{k} = e^{i \frac{2 k π}{3}} - {(- e^{i \frac{2 π}{3}})}^{k}

= {1 - {(- 1)}^{k}} e^{i \frac{2 k π}{3}} = 0 i f k = 2 p a n d = 2 e^{i \frac{2 (2 p + 1) π}{3}} i f k = 2 p + 1 \Rightarrow

P (x) = \sum_{p = 0}^{[\frac{n - 1}{2}]} 2 C_{n}^{2 p + 1} e^{i \frac{2 (2 p + 1) π}{3}} x^{2 p + 1} \Rightarrow

\int_{0}^{1} P (x) d x = 2 \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} e^{i \frac{4 p + 2}{3} π} \frac{1}{2 p + 2}

= \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} \frac{e^{i \frac{4 p + 2}{3} π}}{p + 1} .