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Question Number 73227 by mathmax by abdo last updated on 08/Nov/19
![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_(k=0) ^(n−1) cos((((2k+1)π)/(2n)))](Q73227.png)
$${find}\:{the}\:{polynom}\:{T}_{{n}} \:{wich}\:{verify}\:{T}_{{n}} \left({cos}\theta\right)={cos}\left({n}\theta\right) \\ $$$$\forall{n}\:{integr}\:\forall\theta\:{real} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{T}_{\mathrm{0}} ,{T}_{\mathrm{1}} {and}\:{T}_{\mathrm{2}} {and}\:{prove}\:{that}\: \\ $$$${T}_{{n}+\mathrm{2}} =\mathrm{2}{x}\:{T}_{{n}+\mathrm{1}} −{T}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{deg}\left({T}_{{n}} \right)\:{and}\:{T}_{{n}} \left(\mathrm{1}\right)\:,{T}_{{n}} \left(−\mathrm{1}\right) \\ $$$$\left.\mathrm{4}\right)\:{find}\:{T}\:^{'} \left({cos}\theta\right)\:{for}\:\mathrm{0}<\theta<\pi\:\:{and}\:{prove}\:{that} \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right){T}_{{n}} ''−{xT}'_{{n}} \:+{n}^{\mathrm{2}} \:{T}_{{n}} =\mathrm{0} \\ $$$$\left.\mathrm{5}\right)\:{find}\:{roots}\:{of}\:{T}_{{n}} \:{and}\:{decompose}\:{T}_{{n}} {inside}\:{R}\left[{x}\right] \\ $$$$\left.\mathrm{6}\right)\:{find}\:{the}\:{value}\:{of}\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} {cos}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right) \\ $$
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) )](Q73233.png)
$$\mathrm{T}_{\mathrm{0}} \left(\mathrm{cos}\left(\theta\right)\right)=\mathrm{cos}\left(\mathrm{0}\right)=\mathrm{1}\Rightarrow\mathrm{T}_{\mathrm{0}} =\mathrm{1} \\ $$$$\mathrm{T}_{\mathrm{1}} \left(\mathrm{cos}\left(\theta\right)\right)=\mathrm{cos}\left(\theta\right)\Rightarrow\mathrm{T}_{\mathrm{1}} =\mathrm{xsin}\left(\right. \\ $$$$\mathrm{T}_{\mathrm{2}} \left(\mathrm{cos}\left(\theta\right)\right)=\mathrm{cos}\left(\mathrm{2}\theta\right)=\mathrm{2cos}^{\mathrm{2}} \left(\theta\right)−\mathrm{1}\Rightarrow\mathrm{T}_{\mathrm{2}} =\mathrm{2x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{T}_{\mathrm{n}+\mathrm{2}} \left\{\mathrm{cos}\left(\theta\right)\right\}=\mathrm{cos}\left(\mathrm{n}+\mathrm{2}\right)\theta=\mathrm{cos}\left(\theta\right)\mathrm{cos}\left(\mathrm{n}+\mathrm{1}\right)\theta−\mathrm{sin}\left(\theta\right)\mathrm{sin}\left(\mathrm{n}+\mathrm{1}\right)\theta \\ $$$$=\mathrm{2cos}\left(\theta\right)\mathrm{cos}\left(\mathrm{n}+\mathrm{1}\right)\theta\:−\left\{\mathrm{cos}\left(\theta\right)\mathrm{cos}\left(\mathrm{n}+\mathrm{1}\right)\theta+\mathrm{sin}\left(\theta\right)\mathrm{sin}\left(\mathrm{n}+\mathrm{1}\right)\theta\right\} \\ $$$$=\mathrm{2cos}\left(\theta\right)\mathrm{cos}\left(\mathrm{n}+\mathrm{1}\right)\theta−\mathrm{cos}\left(\mathrm{n}\theta\right) \\ $$$$\Rightarrow\mathrm{2cos}\left(\theta\right)\mathrm{T}_{\mathrm{n}+\mathrm{1}} \left(\mathrm{cos}\theta\right)−\mathrm{T}_{\mathrm{n}} \left(\mathrm{cos}\left(\theta\right)\right)=\mathrm{T}_{\mathrm{n}+\mathrm{1}} \left(\mathrm{cos}\left(\theta\right)\right) \\ $$$$\Rightarrow\mathrm{T}_{\mathrm{n}+\mathrm{1}} =\mathrm{2xT}_{\mathrm{n}+\mathrm{1}} −\mathrm{T}_{\mathrm{n}} \\ $$$$\mathrm{deg}\left(\mathrm{T}_{\mathrm{n}} \right)=\mathrm{n}\: \\ $$$$\mathrm{by}\:\mathrm{recursion}\:\mathrm{deg}\left(\mathrm{T}_{\mathrm{0}} \right)=\mathrm{deg}\left(\mathrm{1}\right)=\mathrm{0}\:\mathrm{true} \\ $$$$\mathrm{suppose}\:\mathrm{for}\:\mathrm{all}\:\mathrm{n}\geqslant\mathrm{2}\:\mathrm{deg}\left(\mathrm{T}_{\mathrm{n}} \right)=\mathrm{n} \\ $$$$\mathrm{T}_{\mathrm{n}+\mathrm{1}} =\mathrm{2xT}_{\mathrm{n}} −\mathrm{T}_{\mathrm{n}−\mathrm{1}} \Rightarrow\mathrm{deg}\left(\mathrm{T}_{\mathrm{n}+\mathrm{1}} \right)=\:\mathrm{deg}\left(\mathrm{T}_{\mathrm{n}} \right)+\mathrm{1}=\mathrm{n}+\mathrm{1} \\ $$$$\mathrm{T}_{\mathrm{n}} \left(\mathrm{1}\right)=\mathrm{T}_{\mathrm{n}} \left(\mathrm{cos}\left(\mathrm{0}\right)\right)=\mathrm{cos}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\mathrm{T}_{\mathrm{n}} \left(−\mathrm{1}\right)=\mathrm{T}_{\mathrm{n}} \left(\mathrm{cos}\left(\pi\right)\right)=\mathrm{cos}\left(\mathrm{n}\pi\right)=\left(−\mathrm{1}\right)^{\mathrm{n}} \\ $$$$\mathrm{T}_{\mathrm{n}} '\left(\mathrm{cos}\left(\theta\right)\right)= \\ $$$$\mathrm{T}_{\mathrm{n}} \left(\mathrm{cos}\left(\theta\right)\right)=\mathrm{cos}\left(\mathrm{n}\theta\right)\Rightarrow−\mathrm{sin}\left(\theta\right)\Gamma'\left(\mathrm{cos}\left(\theta\right)\right)=−\mathrm{nsin}\left(\mathrm{n}\theta\right) \\ $$$$\Rightarrow\Gamma'\left(\mathrm{cos}\left(\theta\right)\right)=\frac{\mathrm{nsin}\left(\mathrm{n}\theta\right)}{\mathrm{sin}\left(\theta\right)} \\ $$$$\Rightarrow\mathrm{sin}\left(\theta\right)\Gamma_{\mathrm{n}} '\left(\mathrm{cos}\left(\theta\right)\right)=\mathrm{nsin}\left(\mathrm{n}\theta\right) \\ $$$$\Rightarrow\mathrm{cos}\left(\theta\right)\Gamma'_{\mathrm{n}} \left(\mathrm{cos}\left(\theta\right)\right)−\mathrm{sin}^{\mathrm{2}} \mathrm{T}_{\mathrm{n}} ''\left(\mathrm{cos}\left(\theta\right)\right)=\mathrm{n}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{n}\theta\right) \\ $$$$\Rightarrow\mathrm{cos}\left(\theta\right)\Gamma_{\mathrm{n}} '\left(\mathrm{cos}\left(\theta\right)\right)−\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \left(\theta\right)\right)\mathrm{T}''\left(\mathrm{cos}\left(\theta\right)\right)=\mathrm{n}^{\mathrm{2}} \mathrm{T}_{\mathrm{n}} \left(\mathrm{cos}\left(\theta\right)\right) \\ $$$$\Leftrightarrow\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \left(\theta\right)\right)\mathrm{T}''_{\mathrm{n}} −\mathrm{cos}\left(\theta\right)\mathrm{T}'\left(\mathrm{cos}\left(\theta\right)\right)+\mathrm{n}^{\mathrm{2}} \mathrm{T}_{\mathrm{n}} \left(\mathrm{cos}\left(\theta\right)\right)=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\Gamma''−\mathrm{xT}^{'} +\mathrm{n}^{\mathrm{2}} \mathrm{T}=\mathrm{0} \\ $$$$\mathrm{since}\:\mathrm{T}_{\mathrm{n}} \mathrm{is}\:\mathrm{a}\:\mathrm{polonomial}\:\mathrm{of}\:\mathrm{degres}\:\mathrm{n}\:\mathrm{he}\:\mathrm{hase}\:\mathrm{n}\:\mathrm{roots}\:\mathrm{over}\:\mathrm{C} \\ $$$$\mathrm{lets}\:\mathrm{finde}\:\mathrm{roots}\:\mathrm{in}\:\left[−\mathrm{1},\mathrm{1}\right]\Rightarrow\exists\theta\in\left[\mathrm{0},\mathrm{2}\pi\right]\:\:\mathrm{x}=\mathrm{cos}\left(\theta\right) \\ $$$$\Rightarrow\mathrm{T}_{\mathrm{n}} \left(\mathrm{cos}\left(\theta\right)\right)=\mathrm{cos}\left(\mathrm{n}\theta\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{n}\theta=\frac{\pi}{\mathrm{2}}+\mathrm{k}\pi\Rightarrow\theta=\frac{\pi}{\mathrm{2n}}+\frac{\mathrm{k}\pi}{\mathrm{n}},\mathrm{k}\in\left[\mathrm{0},\mathrm{n}−\mathrm{1}\right] \\ $$$$\forall\mathrm{k},\mathrm{j}\in\left[\mathrm{0},\mathrm{n}−\mathrm{1}\left[\:\:\:\mathrm{k}\neq\mathrm{j}\Rightarrow\mathrm{cos}\left(\frac{\left(\mathrm{2k}+\mathrm{1}\right)\pi}{\mathrm{2n}}\right)\neq\mathrm{cos}\left(\frac{\left(\mathrm{2j}+\mathrm{1}\right)\pi}{\mathrm{2n}}\right)\Rightarrow\right.\right. \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{n}\:\mathrm{different}\:\mathrm{root} \\ $$$$\mathrm{witch}\:\mathrm{are}\:\mathrm{X}_{\mathrm{k}} =\mathrm{cos}\left(\frac{\left(\mathrm{2k}+\mathrm{1}\right)\pi}{\mathrm{2n}}\right),\mathrm{k}\in\left[\mathrm{0},....\mathrm{n}−\mathrm{1}\right] \\ $$$$\mathrm{T}_{\mathrm{n}} =\mathrm{a}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{x}−\mathrm{cos}\left(\frac{\left(\mathrm{2k}+\mathrm{1}\right)\pi}{\mathrm{2n}}\right)\right. \\ $$$$\mathrm{a}\:\mathrm{coeficient}\:\mathrm{of}\:\mathrm{x}^{\mathrm{n}} \\ $$$$\mathrm{T}_{\mathrm{1}} =\mathrm{x}\Rightarrow\mathrm{a}_{\mathrm{1}} =\mathrm{1} \\ $$$$\mathrm{T}_{\mathrm{2}} =\mathrm{2x}^{\mathrm{2}} −\mathrm{1}\Rightarrow\mathrm{a}_{\mathrm{2}} =\mathrm{2},\mathrm{T}_{\mathrm{n}+\mathrm{1}} =\mathrm{2xT}_{\mathrm{n}} −\mathrm{T}_{\mathrm{n}−\mathrm{1}} \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{n}+\mathrm{1}} =\mathrm{2a}_{\mathrm{n}} \Rightarrow\mathrm{a}_{\mathrm{n}} =\mathrm{a}_{\mathrm{1}} .\mathrm{2}^{\mathrm{n}−\mathrm{1}} =\mathrm{2}^{\mathrm{n}−\mathrm{1}} \:,\mathrm{n}\geqslant\mathrm{1} \\ $$$$\mathrm{T}_{\mathrm{n}} =\begin{cases}{\mathrm{1}\:\mathrm{if}\:\mathrm{n}=\mathrm{0}}\\{\mathrm{2}^{\mathrm{n}−\mathrm{1}} \underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\prod}}\left(\mathrm{x}−\mathrm{cos}\left(\frac{\left(\mathrm{2k}+\mathrm{1}\right)\pi}{\mathrm{2n}}\right)\right)}\end{cases}\:\mathrm{ifn}\geqslant\mathrm{1} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{put}\:\mathrm{x}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{T}_{\mathrm{n}} \left(\mathrm{0}\right)=\mathrm{1}=\mathrm{2}^{\mathrm{n}−\mathrm{1}} \prod_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \left(−\mathrm{cos}\left(\frac{\mathrm{2k}+\mathrm{1}}{\mathrm{2n}}\pi\right)\right)=\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{2}^{\mathrm{n}−\mathrm{1}} \underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\prod}}\left(\mathrm{cos}\left(\frac{\mathrm{2k}+\mathrm{1}}{\mathrm{n}}\pi\right)=\mathrm{T}_{\mathrm{n}} \left(\mathrm{0}\right)\right. \\ $$$$\mathrm{T}_{\mathrm{n}} \left(\mathrm{0}\right)=\mathrm{cos}\left(\frac{\mathrm{n}\pi}{\mathrm{2}}\right)= \\ $$$$ \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\prod}}\left(\mathrm{cos}\left(\frac{\mathrm{2k}+\mathrm{1}}{\mathrm{2n}}\pi\right)=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{cos}\left(\frac{\mathrm{n}\pi}{\mathrm{2}}\right)}{\mathrm{2}^{\mathrm{n}−\mathrm{1}} }\right. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 08/Nov/19
$${thank}\:{you}\:{sir}. \\ $$
Commented by mind is power last updated on 10/Nov/19
$$\mathrm{y}'\mathrm{re}\:\mathrm{welcom} \\ $$
Commented by ajfour last updated on 10/Nov/19
$${i}\:{can}\:{barely}\:{follow}\:{these},\:{but} \\ $$$${great}\:{solution},\:{you}\:{r}\:{a}\:{great}\:{solver} \\ $$$${Sir}. \\ $$
Commented by mind is power last updated on 10/Nov/19
$$\mathrm{thanx}\:\:\mathrm{sir}\:\mathrm{i}\:\mathrm{try}\:\mathrm{to}\:\mathrm{explain}\:\mathrm{most}\:\mathrm{of}\:\mathrm{my}\:\mathrm{worck}\: \\ $$