Question Number 73059 by mathmax by abdo last updated on 05/Nov/19
$${let}\:{P}_{{n}} \left({x}\right)=\left({x}+\mathrm{1}\right)^{{n}} −\left({x}−\mathrm{1}\right)^{{n}} \\ $$$$\left.\mathrm{1}\right)\:{fartorize}\:{inside}\:{C}\left({x}\right)\:{P}_{{n}} \left({x}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:\prod_{{k}=\mathrm{1}} ^{{p}} \:{cotan}\left(\frac{{k}\pi}{\mathrm{2}{p}+\mathrm{1}}\right) \\ $$
Commented by mathmax by abdo last updated on 06/Nov/19
![1) P_n (x)=0 ⇔(x+1)^n =(x−1)^n ⇔(((x−1)/(x+1)))^n =1 let z=((x−1)/(x+1)) (e) ⇒z^n =1 ⇒z^n =e^(i2kπ) ⇒z_k =e^((i2kπ)/n) and k∈[[0,n−1]] z=((x−1)/(x+1)) ⇒zx+z =x−1 ⇒(z−1)x=−z−1 ⇒x=((1+z)/(1−z)) ⇒ the roots of P_n (x)=0 are x_k =((1+z_k )/(1−z_k )) =((1+e^((i2kπ)/n) )/(1−e^((i2kπ)/n) )) =((1+cos(((2kπ)/n))+isin(((2kπ)/n)))/(1−cos(((2kπ)/n))−isin(((2kπ)/n)))) =((2cos^2 (((kπ)/n))+2isin(((kπ)/n))cos(((kπ)/n)))/(2sin^2 (((kπ)/n))−2isin(((kπ)/n))cos(((kπ)/n)))) =((cos(((kπ)/n))e^(i((kπ)/n)) )/(−isin(((kπ)/n))e^((ikπ)/n) )) =i cotan(((kπ)/n)) ⇒x_k =i cotan(((kπ)/n)) with k∈[[1,n−1]] and P_n (x)=aΠ_(k=1) ^(n−1) (x−icotan(((kπ)/n))) letfind a we have P_n (x)=Σ_(k=0) ^n C_n ^k x^k −Σ_(k=0) ^n C_n ^k x^k (−1)^(n−k) =Σ_(k=0) ^n { 1−(−1)^(n−k) }C_n ^k x^k ⇒a =2 C_n ^(n−1) =2n ⇒ P_n (x)=2nΠ_(k=1) ^(n−1) (x−i cotan(((kπ)/n))) ⇒P_(2n+1) (x) =2(2n+1)Π_(k=1) ^(2n) (x−icotan(((kπ)/(2n+1)))) ⇒ P_(2n+1) (0) =2(2n+1)Π_(k=1) ^(2n) (−icotan(((kπ)/(2n+1)))) =2(2n+1)(−i)^(2n) Π_(k=1) ^(2n) cotan(((kπ)/(2n+1))) and P_n (0)=1−(−1)^n ⇒P_(2n+1) (0) =2 ⇒(2n+1)(−1)^n Π_(k=1) ^(2n) cotan(((kπ)/(2n+1)))=1 ⇒ (−1)^n Π_(k=1) ^(2n) cotan(((kπ)/(2n+1)))=(1/(2n+1)) we have Π_(k=1) ^(2n) cotan(((kπ)/(2n+1)))=Π_(k=1) ^n cotan(((kπ)/(2n+1)))Π_(k=n+1) ^(2n) cotan(((kπ)/(2n+1))) =_(k−n=p) Π_(k=1) ^n cotan(((kπ)/(2n+1)))×Π_(p=1) ^n cotan((((n+p)π)/(2n+1))) rest to prove thst Π_(p=1) ^n cotan((((n+p)π)/(2n+1)))=(−1)^n Π_(p=1) ^(n ) cotan(((pπ)/(2n+1))) ⇒(Π_(k=1) ^n cotan(((kπ)/(2n+1))))^2 =(1/(2n+1)) ⇒Π_(k=1) ^n cotan(((kπ)/(2n+1)))=(1/( (√(2n+1))))](https://www.tinkutara.com/question/Q73101.png)
$$\left.\mathrm{1}\right)\:{P}_{{n}} \left({x}\right)=\mathrm{0}\:\Leftrightarrow\left({x}+\mathrm{1}\right)^{{n}} =\left({x}−\mathrm{1}\right)^{{n}} \:\Leftrightarrow\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{{n}} =\mathrm{1}\:{let}\:{z}=\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}} \\ $$$$\left({e}\right)\:\Rightarrow{z}^{{n}} =\mathrm{1}\:\Rightarrow{z}^{{n}} ={e}^{{i}\mathrm{2}{k}\pi} \:\Rightarrow{z}_{{k}} ={e}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} \:\:{and}\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right] \\ $$$${z}=\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\:\Rightarrow{zx}+{z}\:={x}−\mathrm{1}\:\Rightarrow\left({z}−\mathrm{1}\right){x}=−{z}−\mathrm{1}\:\Rightarrow{x}=\frac{\mathrm{1}+{z}}{\mathrm{1}−{z}}\:\Rightarrow \\ $$$${the}\:{roots}\:{of}\:{P}_{{n}} \left({x}\right)=\mathrm{0}\:{are}\:{x}_{{k}} =\frac{\mathrm{1}+{z}_{{k}} }{\mathrm{1}−{z}_{{k}} }\:=\frac{\mathrm{1}+{e}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} }{\mathrm{1}−{e}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} } \\ $$$$=\frac{\mathrm{1}+{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)+{isin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)}{\mathrm{1}−{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)−{isin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)}\:=\frac{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)+\mathrm{2}{isin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)−\mathrm{2}{isin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)} \\ $$$$=\frac{{cos}\left(\frac{{k}\pi}{{n}}\right){e}^{{i}\frac{{k}\pi}{{n}}} }{−{isin}\left(\frac{{k}\pi}{{n}}\right){e}^{\frac{{ik}\pi}{{n}}} }\:={i}\:{cotan}\left(\frac{{k}\pi}{{n}}\right)\:\:\Rightarrow{x}_{{k}} ={i}\:{cotan}\left(\frac{{k}\pi}{{n}}\right) \\ $$$${with}\:{k}\in\left[\left[\mathrm{1},{n}−\mathrm{1}\right]\right]\:{and}\:{P}_{{n}} \left({x}\right)={a}\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left({x}−{icotan}\left(\frac{{k}\pi}{{n}}\right)\right) \\ $$$${letfind}\:{a}\:\:{we}\:{have}\:{P}_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} −\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} {x}^{{k}} \:\left(−\mathrm{1}\right)^{{n}−{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:\left\{\:\mathrm{1}−\left(−\mathrm{1}\right)^{{n}−{k}} \right\}{C}_{{n}} ^{{k}} \:{x}^{{k}} \:\Rightarrow{a}\:=\mathrm{2}\:{C}_{{n}} ^{{n}−\mathrm{1}} =\mathrm{2}{n}\:\Rightarrow \\ $$$${P}_{{n}} \left({x}\right)=\mathrm{2}{n}\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left({x}−{i}\:{cotan}\left(\frac{{k}\pi}{{n}}\right)\right) \\ $$$$\Rightarrow{P}_{\mathrm{2}{n}+\mathrm{1}} \left({x}\right)\:=\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)\prod_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \left({x}−{icotan}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\right)\:\Rightarrow \\ $$$${P}_{\mathrm{2}{n}+\mathrm{1}} \left(\mathrm{0}\right)\:=\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)\prod_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \left(−{icotan}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\right) \\ $$$$=\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)\left(−{i}\right)^{\mathrm{2}{n}} \:\prod_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:{cotan}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:{and}\:{P}_{{n}} \left(\mathrm{0}\right)=\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \\ $$$$\Rightarrow{P}_{\mathrm{2}{n}+\mathrm{1}} \left(\mathrm{0}\right)\:=\mathrm{2}\:\Rightarrow\left(\mathrm{2}{n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} \:\prod_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:{cotan}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)=\mathrm{1}\:\Rightarrow \\ $$$$\left(−\mathrm{1}\right)^{{n}} \prod_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:{cotan}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:\:{we}\:{have} \\ $$$$\prod_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:{cotan}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)=\prod_{{k}=\mathrm{1}} ^{{n}} \:{cotan}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\prod_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:{cotan}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$$$=_{{k}−{n}={p}} \:\:\:\prod_{{k}=\mathrm{1}} ^{{n}} \:{cotan}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)×\prod_{{p}=\mathrm{1}} ^{{n}} \:{cotan}\left(\frac{\left({n}+{p}\right)\pi}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$$${rest}\:{to}\:{prove}\:{thst}\:\:\:\prod_{{p}=\mathrm{1}} ^{{n}} \:{cotan}\left(\frac{\left({n}+{p}\right)\pi}{\mathrm{2}{n}+\mathrm{1}}\right)=\left(−\mathrm{1}\right)^{{n}} \prod_{{p}=\mathrm{1}} ^{{n}\:} {cotan}\left(\frac{{p}\pi}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$$$\Rightarrow\left(\prod_{{k}=\mathrm{1}} ^{{n}} \:{cotan}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\right)^{\mathrm{2}} \:=\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:\Rightarrow\prod_{{k}=\mathrm{1}} ^{{n}} \:{cotan}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{n}+\mathrm{1}}} \\ $$
Commented by mind is power last updated on 06/Nov/19
$$\mathrm{nice} \\ $$
Commented by mathmax by abdo last updated on 06/Nov/19
$${thanks}\:{sir}. \\ $$
Answered by mind is power last updated on 06/Nov/19
![p_n (x)=0⇒((x+1)/(x−1))=e^((2ikπ)/n) ,k≤n ⇒x(1−e^((2ikπ)/n) )=−e^((2ikπ)/n) −1 ⇒k≠0 x=((e^((2ikπ)/n) +1)/(e^((2ikπ)/n) −1))=((e^(i((kπ)/n)) (2cos(((kπ)/n))))/(e^(i((kπ)/n)) (2isin(((kπ)/n)))))=−icot(((kπ)/n)) p_n (x)=aΠ_(k=1) ^(n−1) (x+icot(((kπ)/n)) let n=2p+1 p_n (x)=Π_(k=1) ^(2p) (x+icot(((kπ)/(2p+1)))) =Π_(k=1) ^p (x+icot(((kπ)/(2p+1)))).Π_(k=p+1) ^(2p) (x+icot(((kπ)/(2p+1))) =Π_(k=1) ^p (x+icot(((kπ)/(2p+1)))).Π_(k=0) ^(p−1) (x+icot(((2p−k)π)/(2p+1)))) =Π_(k=1) ^p (x+icot(((kπ)/(2p+1))))Π_(k=0) ^(p−1) (x−icot(((k+1)/(2p+1)))π)=(1+x)^n −(x−1)^n pour x=0 onaΠ_(k=1) ^p (icot(((kπ)/(2p+1))))(Π_(k=1) ^p −icot(((kπ)/(2p+1))))= ⇒{Π_(k=1) ^p cot(((kπ)/(2p+1)))}^2 =(2/a) ⇒Π_(k=1) ^p cot(((kπ)/(2p+1)))=(√(2/a))=(1/( (√(2p+1)))) a=2C_n ^1 =2n=4p+2 cause ∀k∈[1,n] 0< ((kπ)/(2p+1))<(π/2)⇒cot(((kπ)/(2p+1)))>0](https://www.tinkutara.com/question/Q73089.png)
$$\mathrm{p}_{\mathrm{n}} \left(\mathrm{x}\right)=\mathrm{0}\Rightarrow\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}−\mathrm{1}}=\mathrm{e}^{\frac{\mathrm{2ik}\pi}{\mathrm{n}}} ,\mathrm{k}\leqslant\mathrm{n} \\ $$$$ \\ $$$$\Rightarrow\mathrm{x}\left(\mathrm{1}−\mathrm{e}^{\frac{\mathrm{2ik}\pi}{\mathrm{n}}} \right)=−\mathrm{e}^{\frac{\mathrm{2ik}\pi}{\mathrm{n}}} −\mathrm{1} \\ $$$$\Rightarrow\mathrm{k}\neq\mathrm{0} \\ $$$$\mathrm{x}=\frac{\mathrm{e}^{\frac{\mathrm{2ik}\pi}{\mathrm{n}}} +\mathrm{1}}{\mathrm{e}^{\frac{\mathrm{2ik}\pi}{\mathrm{n}}} −\mathrm{1}}=\frac{\mathrm{e}^{\mathrm{i}\frac{\mathrm{k}\pi}{\mathrm{n}}} \left(\mathrm{2cos}\left(\frac{\mathrm{k}\pi}{\mathrm{n}}\right)\right)}{\mathrm{e}^{\mathrm{i}\frac{\mathrm{k}\pi}{\mathrm{n}}} \left(\mathrm{2isin}\left(\frac{\mathrm{k}\pi}{\mathrm{n}}\right)\right)}=−\mathrm{icot}\left(\frac{\mathrm{k}\pi}{\mathrm{n}}\right) \\ $$$$\mathrm{p}_{\mathrm{n}} \left(\mathrm{x}\right)=\mathrm{a}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\prod}}\left(\mathrm{x}+\mathrm{icot}\left(\frac{\mathrm{k}\pi}{\mathrm{n}}\right)\right. \\ $$$$\mathrm{let}\:\mathrm{n}=\mathrm{2p}+\mathrm{1} \\ $$$$\mathrm{p}_{\mathrm{n}} \left(\mathrm{x}\right)=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{2p}} {\prod}}\left(\mathrm{x}+\mathrm{icot}\left(\frac{\mathrm{k}\pi}{\mathrm{2p}+\mathrm{1}}\right)\right) \\ $$$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{p}} {\prod}}\left(\mathrm{x}+\mathrm{icot}\left(\frac{\mathrm{k}\pi}{\mathrm{2p}+\mathrm{1}}\right)\right).\underset{\mathrm{k}=\mathrm{p}+\mathrm{1}} {\overset{\mathrm{2p}} {\prod}}\left(\mathrm{x}+\mathrm{icot}\left(\frac{\mathrm{k}\pi}{\mathrm{2p}+\mathrm{1}}\right)\right. \\ $$$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{p}} {\prod}}\left(\mathrm{x}+\mathrm{icot}\left(\frac{\mathrm{k}\pi}{\mathrm{2p}+\mathrm{1}}\right)\right).\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{p}−\mathrm{1}} {\prod}}\left(\mathrm{x}+\mathrm{icot}\left(\frac{\left.\mathrm{2p}−\mathrm{k}\right)\pi}{\mathrm{2p}+\mathrm{1}}\right)\right) \\ $$$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{p}} {\prod}}\left(\mathrm{x}+\mathrm{icot}\left(\frac{\mathrm{k}\pi}{\mathrm{2p}+\mathrm{1}}\right)\right)\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{p}−\mathrm{1}} {\prod}}\left(\mathrm{x}−\mathrm{icot}\left(\frac{\mathrm{k}+\mathrm{1}}{\mathrm{2p}+\mathrm{1}}\right)\pi\right)=\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} −\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{n}} \\ $$$$ \\ $$$$\mathrm{pour}\:\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{ona}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{p}} {\prod}}\left(\mathrm{icot}\left(\frac{\mathrm{k}\pi}{\mathrm{2p}+\mathrm{1}}\right)\right)\left(\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{p}} {\prod}}−\mathrm{icot}\left(\frac{\mathrm{k}\pi}{\mathrm{2p}+\mathrm{1}}\right)\right)= \\ $$$$\Rightarrow\left\{\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{p}} {\prod}}\mathrm{cot}\left(\frac{\mathrm{k}\pi}{\mathrm{2p}+\mathrm{1}}\right)\right\}^{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{a}} \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{p}} {\prod}}\mathrm{cot}\left(\frac{\mathrm{k}\pi}{\mathrm{2p}+\mathrm{1}}\right)=\sqrt{\frac{\mathrm{2}}{\mathrm{a}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2p}+\mathrm{1}}} \\ $$$$\mathrm{a}=\mathrm{2C}_{\mathrm{n}} ^{\mathrm{1}} =\mathrm{2n}=\mathrm{4p}+\mathrm{2} \\ $$$$\mathrm{cause}\:\forall\mathrm{k}\in\left[\mathrm{1},\mathrm{n}\right]\:\mathrm{0}<\:\:\frac{\mathrm{k}\pi}{\mathrm{2p}+\mathrm{1}}<\frac{\pi}{\mathrm{2}}\Rightarrow\mathrm{cot}\left(\frac{\mathrm{k}\pi}{\mathrm{2p}+\mathrm{1}}\right)>\mathrm{0} \\ $$$$ \\ $$