let j =e^((i2π)/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 ∫


Question and Answers Forum

All Questions Topic List
Relation and Functions Questions
Previous in All Question Next in All Question
Previous in Relation and Functions Next in Relation and Functions
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
$1) we have j =cos(((2π)/3))+isin(((2π)/3)) =−(1/2)+i((√3)/2) ⇒ P(x)=(1+(−(1/2)+i((√3)/2))x)^n −(1−(−(1/2) +i((√3)/2))x)^n =(1−(x/2) +i((√3)/2)x)^n −(1+(x/2)−i((√3)/2)x)^n =A^n (x)−B^n (x) ∣A(x)∣ =(√((1−(x/2))^2 +(3/4)x^2 )) =(√(((2−x)^2 +3x^2 )/4))=(1/2)(√((x−2)^2 +3x^2 )) we know that x+iy =(√(x^2 +y^2 )) e^(i arctan((y/x))) ⇒ A(x) =(1/2)(√((x−2)^2 +3x^2 )) e^(i arctan((((√3)x)/(2−x)))) ⇒A^n (x) =(1/2^n ){(x−2)^2 +3x^2 }^(n/2) e^(i n arctan((((√3)x)/(2−x)))) ∣B(x)∣ =(1/2)(√((x+2)^2 +3x^2 )) ⇒B(x) =(1/2)(√((x+2)^2 +3x^2 ))e^(i arctan(((−(√3)x)/(2+x)))) ⇒ B^n (x) =(1/2^n ){ (x+2)^2 +3x^2 }^(n/2) e^(−in arctan((((√3)x)/(2+x)))) ⇒ P(x) =(1/2^n ){ (x−2)^2 +3x^2 }^(n/2) e^(in arctan((((√3)x)/(2−x)))) −(1/2^n ){(x+2)^2 +3x^2 }^(n/2) e^(−in arctan((((√3)x)/(2+x)))) and we see that B(x) =A(−x).$
$1) w e h a v e j = c o s (\frac{2 π}{3}) + i s i n (\frac{2 π}{3}) = - \frac{1}{2} + i \frac{\sqrt{3}}{2} \Rightarrow$ $P (x) = {(1 + (- \frac{1}{2} + i \frac{\sqrt{3}}{2}) x)}^{n} - {(1 - (- \frac{1}{2} + i \frac{\sqrt{3}}{2}) x)}^{n}$ $= {(1 - \frac{x}{2} + i \frac{\sqrt{3}}{2} x)}^{n} - {(1 + \frac{x}{2} - i \frac{\sqrt{3}}{2} x)}^{n} = A^{n} (x) - B^{n} (x)$ $∣ A (x) ∣ = \sqrt{{(1 - \frac{x}{2})}^{2} + \frac{3}{4} x^{2}} = \sqrt{\frac{{(2 - x)}^{2} + 3 x^{2}}{4}} = \frac{1}{2} \sqrt{{(x - 2)}^{2} + 3 x^{2}}$ $w e k n o w t h a t x + i y = \sqrt{x^{2} + y^{2}} e^{i a r c t a n (\frac{y}{x})} \Rightarrow$ $A (x) = \frac{1}{2} \sqrt{{(x - 2)}^{2} + 3 x^{2}} e^{i a r c t a n (\frac{\sqrt{3} x}{2 - x})} \Rightarrow A^{n} (x) = \frac{1}{2^{n}} {{(x - 2)}^{2} + 3 x^{2}}^{\frac{n}{2}} e^{i n a r c t a n (\frac{\sqrt{3} x}{2 - x})}$ $∣ B (x) ∣ = \frac{1}{2} \sqrt{{(x + 2)}^{2} + 3 x^{2}} \Rightarrow B (x) = \frac{1}{2} \sqrt{{(x + 2)}^{2} + 3 x^{2}} e^{i a r c t a n (\frac{- \sqrt{3} x}{2 + x})} \Rightarrow$ $B^{n} (x) = \frac{1}{2^{n}} {{(x + 2)}^{2} + 3 x^{2}}^{\frac{n}{2}} e^{- i n a r c t a n (\frac{\sqrt{3} x}{2 + x})} \Rightarrow$ $P (x) = \frac{1}{2^{n}} {{(x - 2)}^{2} + 3 x^{2}}^{\frac{n}{2}} e^{i n a r c t a n (\frac{\sqrt{3} x}{2 - x})} - \frac{1}{2^{n}} {{(x + 2)}^{2} + 3 x^{2}}^{\frac{n}{2}} e^{- i n a r c t a n (\frac{\sqrt{3} x}{2 + x})}$ $a n d w e s e e t h a t B (x) = A (- x) .$
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
$4) we have P(x) =Σ_(k=0) ^n C_n ^k j^k x^k −Σ_(k=0) ^n C_n ^k (−j)^k x^k =Σ_(k=0) ^n C_n ^k (j^k −(−j)^k )x^k but j^k −(−j)^k =e^(i((2kπ)/3)) −(−e^(i((2π)/3)) )^k ={1−(−1)^k } e^(i((2kπ)/3)) =0 if k=2p and =2 e^(i((2(2p+1)π)/3)) if k =2p+1 ⇒ P(x) =Σ_(p=0) ^([((n−1)/2)]) 2 C_n ^(2p+1) e^(i((2(2p+1)π)/3)) x^(2p+1) ⇒ ∫_0 ^1 P(x)dx =2Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) e^(i((4p+2)/3)π) (1/(2p+2)) =Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) (e^(i((4p+2)/3)π) /(p+1)) .$
$4) w e h a v e P (x) = \sum_{k = 0}^{n} C_{n}^{k} j^{k} x^{k} - \sum_{k = 0}^{n} C_{n}^{k} {(- j)}^{k} x^{k}$ $= \sum_{k = 0}^{n} C_{n}^{k} (j^{k} - {(- j)}^{k}) x^{k} b u t j^{k} - {(- j)}^{k} = e^{i \frac{2 k π}{3}} - {(- e^{i \frac{2 π}{3}})}^{k}$ $= {1 - {(- 1)}^{k}} e^{i \frac{2 k π}{3}} = 0 i f k = 2 p a n d = 2 e^{i \frac{2 (2 p + 1) π}{3}} i f k = 2 p + 1 \Rightarrow$ $P (x) = \sum_{p = 0}^{[\frac{n - 1}{2}]} 2 C_{n}^{2 p + 1} e^{i \frac{2 (2 p + 1) π}{3}} x^{2 p + 1} \Rightarrow$ $\int_{0}^{1} P (x) d x = 2 \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} e^{i \frac{4 p + 2}{3} π} \frac{1}{2 p + 2}$ $= \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} \frac{e^{i \frac{4 p + 2}{3} π}}{p + 1} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum