find-the-polynom-T-n-wich-verify-T-n-cos-cos-n-n-integr-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-cos-for

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 \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, \dots . 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