find the polynom T_n wich verify T_n (cosθ)=cos(nθ) ∀n integr ∀θ real 1) find T_0 ,T_1 and T_2 and prove that T_(n+2) =2x T_(n+1) −T_n 3) find deg(T_n ) and T_n (1) ,T_n (−1) 4) find T^′ (cosθ) for 0<θ<π and prove that (1−x^2 )T_n ′′−xT′_n +n^2 T_n =0 5) find roots of T_n and decompose T_n inside R[x] 6) find the value of Π


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Question Number 73227 by mathmax by abdo last updated on 08/Nov/19

$f i n d t h e p o l y n o m T_{n} w i c h v e r i f y T_{n} (c o s θ) = c o s (n θ)$ $\forall n i n t e g r \forall θ r e a l$ $1) f i n d T_{0}, T_{1} a n d T_{2} a n d p r o v e t h a t$ $T_{n + 2} = 2 x T_{n + 1} - T_{n}$ $3) f i n d d e g (T_{n}) a n d T_{n} (1), T_{n} (- 1)$ $4) f i n d T^{^{'}} (c o s θ) f o r 0 < θ < π a n d p r o v e t h a t$ $(1 - x^{2}) T_{n}^{″} - {x T}_{n}^{'} + n^{2} T_{n} = 0$ $5) f i n d r o o t s o f T_{n} a n d d e c o m p o s e T_{n} i n s i d e R [x]$ $6) f i n d t h e v a l u e o f \prod_{k = 0}^{n - 1} c o s (\frac{(2 k + 1) π}{2 n})$
	Answered by mind is power last updated on 08/Nov/19
	$T_0 (cos(θ))=cos(0)=1⇒T_0 =1 T_1 (cos(θ))=cos(θ)⇒T_1 =xsin( T_2 (cos(θ))=cos(2θ)=2cos^2 (θ)−1⇒T_2 =2x^2 −1 T_(n+2) {cos(θ)}=cos(n+2)θ=cos(θ)cos(n+1)θ−sin(θ)sin(n+1)θ =2cos(θ)cos(n+1)θ −{cos(θ)cos(n+1)θ+sin(θ)sin(n+1)θ} =2cos(θ)cos(n+1)θ−cos(nθ) ⇒2cos(θ)T_(n+1) (cosθ)−T_n (cos(θ))=T_(n+1) (cos(θ)) ⇒T_(n+1) =2xT_(n+1) −T_n deg(T_n )=n by recursion deg(T_0 )=deg(1)=0 true suppose for all n≥2 deg(T_n )=n T_(n+1) =2xT_n −T_(n−1) ⇒deg(T_(n+1) )= deg(T_n )+1=n+1 T_n (1)=T_n (cos(0))=cos(0)=1 T_n (−1)=T_n (cos(π))=cos(nπ)=(−1)^n T_n ′(cos(θ))= T_n (cos(θ))=cos(nθ)⇒−sin(θ)Γ′(cos(θ))=−nsin(nθ) ⇒Γ′(cos(θ))=((nsin(nθ))/(sin(θ))) ⇒sin(θ)Γ_n ′(cos(θ))=nsin(nθ) ⇒cos(θ)Γ′_n (cos(θ))−sin^2 T_n ′′(cos(θ))=n^2 cos(nθ) ⇒cos(θ)Γ_n ′(cos(θ))−(1−cos^2 (θ))T′′(cos(θ))=n^2 T_n (cos(θ)) ⇔(1−cos^2 (θ))T′′_n −cos(θ)T′(cos(θ))+n^2 T_n (cos(θ))=0 ⇒(1−x^2 )Γ′′−xT^′ +n^2 T=0 since T_n is a polonomial of degres n he hase n roots over C lets finde roots in [−1,1]⇒∃θ∈[0,2π] x=cos(θ) ⇒T_n (cos(θ))=cos(nθ)=0 ⇒nθ=(π/2)+kπ⇒θ=(π/(2n))+((kπ)/n),k∈[0,n−1] ∀k,j∈[0,n−1[ k≠j⇒cos((((2k+1)π)/(2n)))≠cos((((2j+1)π)/(2n)))⇒ we get n different root witch are X_k =cos((((2k+1)π)/(2n))),k∈[0,....n−1] T_n =aΠ_(k=1) ^n (x−cos((((2k+1)π)/(2n))) a coeficient of x^n T_1 =x⇒a_1 =1 T_2 =2x^2 −1⇒a_2 =2,T_(n+1) =2xT_n −T_(n−1) ⇒a_(n+1) =2a_n ⇒a_n =a_1 .2^(n−1) =2^(n−1) ,n≥1 T_n = { ((1 if n=0)),((2^(n−1) Π_(k=0) ^(n−1) (x−cos((((2k+1)π)/(2n)))))) :} ifn≥1 if we put x=0 ⇒T_n (0)=1=2^(n−1) Π_(k=0) ^(n−1) (−cos(((2k+1)/(2n))π))=(−1)^n 2^(n−1) Π_(k=0) ^(n−1) (cos(((2k+1)/n)π)=T_n (0) T_n (0)=cos(((nπ)/2))= ⇒Π_(k=0) ^(n−1) (cos(((2k+1)/(2n))π)=(((−1)^n cos(((nπ)/2)))/2^(n−1) )$
	$T_{0} (\cos (θ)) = \cos (0) = 1 \Rightarrow T_{0} = 1$ $T_{1} (\cos (θ)) = \cos (θ) \Rightarrow T_{1} = xsin ($ $T_{2} (\cos (θ)) = \cos (2 θ) = {2 \cos}^{2} (θ) - 1 \Rightarrow T_{2} = {2 x}^{2} - 1$ $T_{n + 2} {\cos (θ)} = \cos (n + 2) θ = \cos (θ) \cos (n + 1) θ - \sin (θ) \sin (n + 1) θ$ $= 2 \cos (θ) \cos (n + 1) θ - {\cos (θ) \cos (n + 1) θ + \sin (θ) \sin (n + 1) θ}$ $= 2 \cos (θ) \cos (n + 1) θ - \cos (n θ)$ $\Rightarrow 2 \cos (θ) T_{n + 1} (\cos θ) - T_{n} (\cos (θ)) = T_{n + 1} (\cos (θ))$ $\Rightarrow T_{n + 1} = {2 xT}_{n + 1} - T_{n}$ $\deg (T_{n}) = n$ $by recursion \deg (T_{0}) = \deg (1) = 0 true$ $suppose for all n ⩾ 2 \deg (T_{n}) = n$ $T_{n + 1} = {2 xT}_{n} - T_{n - 1} \Rightarrow \deg (T_{n + 1}) = \deg (T_{n}) + 1 = n + 1$ $T_{n} (1) = T_{n} (\cos (0)) = \cos (0) = 1$ $T_{n} (- 1) = T_{n} (\cos (π)) = \cos (n π) = {(- 1)}^{n}$ $T_{n}^{'} (\cos (θ)) =$ $T_{n} (\cos (θ)) = \cos (n θ) \Rightarrow - \sin (θ) Γ^{'} (\cos (θ)) = - nsin (n θ)$ $\Rightarrow Γ^{'} (\cos (θ)) = \frac{nsin (n θ)}{\sin (θ)}$ $\Rightarrow \sin (θ) Γ_{n}^{'} (\cos (θ)) = nsin (n θ)$ $\Rightarrow \cos (θ) Γ_{n}^{'} (\cos (θ)) - \sin^{2} T_{n}^{″} (\cos (θ)) = n^{2} \cos (n θ)$ $\Rightarrow \cos (θ) Γ_{n}^{'} (\cos (θ)) - (1 - \cos^{2} (θ)) T^{″} (\cos (θ)) = n^{2} T_{n} (\cos (θ))$ $\Leftrightarrow (1 - \cos^{2} (θ)) T_{n}^{″} - \cos (θ) T^{'} (\cos (θ)) + n^{2} T_{n} (\cos (θ)) = 0$ $\Rightarrow (1 - x^{2}) Γ^{″} - {xT}^{^{'}} + n^{2} T = 0$ $since T_{n} is a polonomial of degres n he hase n roots over C$ $lets finde roots in [- 1, 1] \Rightarrow \exists θ \in [0, 2 π] x = \cos (θ)$ $\Rightarrow T_{n} (\cos (θ)) = \cos (n θ) = 0$ $\Rightarrow n θ = \frac{π}{2} + k π \Rightarrow θ = \frac{π}{2 n} + \frac{k π}{n}, k \in [0, n - 1]$ $\forall k, j \in [0, n - 1 [k \neq j \Rightarrow \cos (\frac{(2 k + 1) π}{2 n}) \neq \cos (\frac{(2 j + 1) π}{2 n}) \Rightarrow$ $we get n different root$ $witch are X_{k} = \cos (\frac{(2 k + 1) π}{2 n}), k \in [0, . . . . n - 1]$ $T_{n} = a \underset{k = 1}{\prod^{n}} (x - \cos (\frac{(2 k + 1) π}{2 n})$ $a coeficient of x^{n}$ $T_{1} = x \Rightarrow a_{1} = 1$ $T_{2} = {2 x}^{2} - 1 \Rightarrow a_{2} = 2, T_{n + 1} = {2 xT}_{n} - T_{n - 1}$ $\Rightarrow a_{n + 1} = {2 a}_{n} \Rightarrow a_{n} = a_{1} . 2^{n - 1} = 2^{n - 1}, n ⩾ 1$ $T_{n} = {\begin{cases} 1 if n = 0 \\ 2^{n - 1} \underset{k = 0}{\prod^{n - 1}} (x - \cos (\frac{(2 k + 1) π}{2 n})) \end{cases} ifn ⩾ 1$ $if we put x = 0$ $\Rightarrow T_{n} (0) = 1 = 2^{n - 1} \prod_{k = 0}^{n - 1} (- \cos (\frac{2 k + 1}{2 n} π)) = {(- 1)}^{n} 2^{n - 1} \underset{k = 0}{\prod^{n - 1}} (\cos (\frac{2 k + 1}{n} π) = T_{n} (0)$ $T_{n} (0) = \cos (\frac{n π}{2}) =$ $\Rightarrow \underset{k = 0}{\prod^{n - 1}} (\cos (\frac{2 k + 1}{2 n} π) = \frac{{(- 1)}^{n} \cos (\frac{n π}{2})}{2^{n - 1}}$
	Commented by mathmax by abdo last updated on 08/Nov/19

	$t h a n k y o u s i r .$
	Commented by mind is power last updated on 10/Nov/19

	$y^{'} re welcom$
	Commented by ajfour last updated on 10/Nov/19

	$i c a n b a r e l y f o l l o w t h e s e, b u t$ $g r e a t s o l u t i o n, y o u r a g r e a t s o l v e r$ $S i r .$
	Commented by mind is power last updated on 10/Nov/19

	$thanx sir i try to explain most of my worck$
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