Question-162825

Question Number 162825 by KONE last updated on 01/Jan/22

Answered by mr W last updated on 01/Jan/22

2. P_(n+2) −2XP_(n+1) +P_n =0 t^2 −2Xt+1=0 (char. eqn.) t=X±(√(X^2 −1)) P_n =A(X+(√(X^2 −1)))^n +B(X−(√(X^2 −1)))^n P_0 =A+B=1 P_1 =A(X+(√(X^2 −1)))+B(X−(√(X^2 −1)))=X ⇒A=B=(1/2) P_n =(((X+(√(X^2 −1)))^n +(X−(√(X^2 −1)))^n )/2) P_n =(((X+(√(X^2 −1)))^n +(X+(√(X^2 −1)))^(−n) )/2) P_n =((e^(nln (X+(√(X^2 −1)))) +e^(−nln (X+(√(X^2 −1)))) )/2) P_n =cosh [n ln (X+(√(X^2 −1)))] P_n =cosh (n cosh^(−1) X)

$$\mathrm{2}. \\ $$$${P}_{{n}+\mathrm{2}} −\mathrm{2}{XP}_{{n}+\mathrm{1}} +{P}_{{n}} =\mathrm{0} \\ $$$${t}^{\mathrm{2}} −\mathrm{2}{Xt}+\mathrm{1}=\mathrm{0}\:\left({char}.\:{eqn}.\right) \\ $$$${t}={X}\pm\sqrt{{X}^{\mathrm{2}} −\mathrm{1}} \\ $$$${P}_{{n}} ={A}\left({X}+\sqrt{{X}^{\mathrm{2}} −\mathrm{1}}\right)^{{n}} +{B}\left({X}−\sqrt{{X}^{\mathrm{2}} −\mathrm{1}}\right)^{{n}} \\ $$$${P}_{\mathrm{0}} ={A}+{B}=\mathrm{1} \\ $$$${P}_{\mathrm{1}} ={A}\left({X}+\sqrt{{X}^{\mathrm{2}} −\mathrm{1}}\right)+{B}\left({X}−\sqrt{{X}^{\mathrm{2}} −\mathrm{1}}\right)={X} \\ $$$$\Rightarrow{A}={B}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${P}_{{n}} =\frac{\left({X}+\sqrt{{X}^{\mathrm{2}} −\mathrm{1}}\right)^{{n}} +\left({X}−\sqrt{{X}^{\mathrm{2}} −\mathrm{1}}\right)^{{n}} }{\mathrm{2}} \\ $$$${P}_{{n}} =\frac{\left({X}+\sqrt{{X}^{\mathrm{2}} −\mathrm{1}}\right)^{{n}} +\left({X}+\sqrt{{X}^{\mathrm{2}} −\mathrm{1}}\right)^{−{n}} }{\mathrm{2}} \\ $$$${P}_{{n}} =\frac{{e}^{{n}\mathrm{ln}\:\left({X}+\sqrt{{X}^{\mathrm{2}} −\mathrm{1}}\right)} +{e}^{−{n}\mathrm{ln}\:\left({X}+\sqrt{{X}^{\mathrm{2}} −\mathrm{1}}\right)} }{\mathrm{2}} \\ $$$${P}_{{n}} =\mathrm{cosh}\:\left[{n}\:\mathrm{ln}\:\left({X}+\sqrt{{X}^{\mathrm{2}} −\mathrm{1}}\right)\right] \\ $$$${P}_{{n}} =\mathrm{cosh}\:\left({n}\:\mathrm{cosh}^{−\mathrm{1}} {X}\right) \\ $$

Commented by KONE last updated on 02/Jan/22