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Question Number 143368 by mr W last updated on 13/Jun/21

Commented by mr W last updated on 13/Jun/21

$T h i s i s a s o l v e d o l d q u e s t i o n$ $(Q 142207) . T h e a n s w e r i s$ $a_{n} = \frac{\sqrt{2} [{(\sqrt{2} + 1)}^{2 (2 n - 1)} + 1]}{{(\sqrt{2} + 1)}^{2 (2 n - 1)} - 1}$ $A n y o t h e r w a y s t o s o l v e ?$
	Answered by mr W last updated on 13/Jun/21
	$here a new way. a_(n+1) =((a_n +(4/3))/(1+(2/3)a_n )) λa_(n+1) =((λa_n +((4λ)/3))/(1+(2/(3λ))×λa_n )) let tan θ_n =λa_n and ((4λ)/3)=−(2/(3λ))=tan α ...(I) tan θ_(n+1) =((tan θ_n +tan α)/(1−tan θ_n tan α))=tan (θ_n +α) ⇒θ_(n+1) =θ_n +α ⇒θ_n =θ_1 +(n−1)α tan θ_n =((tan θ_1 +tan (n−1)α)/(1−tan θ_1 tan (n−1)α)) λa_n =((λa_1 +tan (n−1)α)/(1−λa_1 tan (n−1)α)) ⇒a_n =((a_1 +(1/λ)tan (n−1)α)/(1−a_1 λtan (n−1)α)) ...(II) from (I): ((4λ)/3)=−(2/(3λ)) λ^2 =−(1/2) ⇒λ=±(i/( (√2))) we take λ=(i/( (√2))) tan α=((4λ)/3)=((2(√2)i)/3) ⇒α=(tanh^(−1) ((2(√2))/3))i tan (n−1)α=[tanh {(n−1)tanh^(−1) ((2(√2))/3)}]i put into (II): a_n =((2+((√2)/i)×[tanh {(n−1)tanh^(−1) ((2(√2))/3)}]i)/(1−2×(i/( (√2)))×[tanh {(n−1)tanh^(−1) ((2(√2))/3)}]i)) a_n =((2+(√2)[tanh {(n−1)tanh^(−1) ((2(√2))/3)}])/(1+(√2)[tanh {(n−1)tanh^(−1) ((2(√2))/3)}])) tanh^(−1) ((2(√2))/3)=(1/2)ln ((1+((2(√2))/3))/(1−((2(√2))/3)))=2ln ((√2)+1) tanh {(n−1)tanh^(−1) ((2(√2))/3)} =tanh {ln ((√2)+1)^(2(n−1)) } =((((√2)+1)^(4(n−1)) −1)/(((√2)+1)^(4(n−1)) +1)) =1−(2/(((√2)+1)^(4(n−1)) +1)) a_n =((2+(√2)[1−(2/(((√2)+1)^(4(n−1)) +1))])/(1+(√2)[1−(2/(((√2)+1)^(4(n−1)) +1))])) a_n =(((√2)((√2)+1)−((2(√2))/(((√2)+1)^(4(n−1)) +1)))/( (√2)+1−((2(√2))/(((√2)+1)^(4(n−1)) +1)))) a_n =(((√2)((√2)+1)+(√2)((√2)+1)^(4(n−1)+1) −2(√2))/( (√2)+1+((√2)+1)^(4(n−1)+1) −2(√2))) a_n =(((√2)[((√2)−1)+((√2)+1)^(4(n−1)+1) ])/( ((√2)+1)^(4(n−1)+1) −((√2)−1))) a_n =(((√2)[((√2)+1)^(2(2n−1)) +1])/( ((√2)+1)^(2(2n−1)) −1))$
	$h e r e a n e w w a y .$ $a_{n + 1} = \frac{a_{n} + \frac{4}{3}}{1 + \frac{2}{3} a_{n}}$ $λ a_{n + 1} = \frac{λ a_{n} + \frac{4 λ}{3}}{1 + \frac{2}{3 λ} \times λ a_{n}}$ $l e t \tan θ_{n} = λ a_{n} a n d$ $\frac{4 λ}{3} = - \frac{2}{3 λ} = \tan α . . . (I)$ $\tan θ_{n + 1} = \frac{\tan θ_{n} + \tan α}{1 - \tan θ_{n} \tan α} = \tan (θ_{n} + α)$ $\Rightarrow θ_{n + 1} = θ_{n} + α$ $\Rightarrow θ_{n} = θ_{1} + (n - 1) α$ $\tan θ_{n} = \frac{\tan θ_{1} + \tan (n - 1) α}{1 - \tan θ_{1} \tan (n - 1) α}$ $λ a_{n} = \frac{λ a_{1} + \tan (n - 1) α}{1 - λ a_{1} \tan (n - 1) α}$ $\Rightarrow a_{n} = \frac{a_{1} + \frac{1}{λ} \tan (n - 1) α}{1 - a_{1} λ \tan (n - 1) α} . . . (I I)$ $f r o m (I) :$ $\frac{4 λ}{3} = - \frac{2}{3 λ}$ $λ^{2} = - \frac{1}{2}$ $\Rightarrow λ = \pm \frac{i}{\sqrt{2}}$ $w e t a k e λ = \frac{i}{\sqrt{2}}$ $\tan α = \frac{4 λ}{3} = \frac{2 \sqrt{2} i}{3}$ $\Rightarrow α = (\tanh^{- 1} \frac{2 \sqrt{2}}{3}) i$ $\tan (n - 1) α = [\tanh {(n - 1) \tanh^{- 1} \frac{2 \sqrt{2}}{3}}] i$ $p u t i n t o (I I) :$ $a_{n} = \frac{2 + \frac{\sqrt{2}}{i} \times [\tanh {(n - 1) \tanh^{- 1} \frac{2 \sqrt{2}}{3}}] i}{1 - 2 \times \frac{i}{\sqrt{2}} \times [\tanh {(n - 1) \tanh^{- 1} \frac{2 \sqrt{2}}{3}}] i}$ $a_{n} = \frac{2 + \sqrt{2} [\tanh {(n - 1) \tanh^{- 1} \frac{2 \sqrt{2}}{3}}]}{1 + \sqrt{2} [\tanh {(n - 1) \tanh^{- 1} \frac{2 \sqrt{2}}{3}}]}$ $\tanh^{- 1} \frac{2 \sqrt{2}}{3} = \frac{1}{2} \ln \frac{1 + \frac{2 \sqrt{2}}{3}}{1 - \frac{2 \sqrt{2}}{3}} = 2 \ln (\sqrt{2} + 1)$ $\tanh {(n - 1) \tanh^{- 1} \frac{2 \sqrt{2}}{3}}$ $= \tanh {\ln {(\sqrt{2} + 1)}^{2 (n - 1)}}$ $= \frac{{(\sqrt{2} + 1)}^{4 (n - 1)} - 1}{{(\sqrt{2} + 1)}^{4 (n - 1)} + 1}$ $= 1 - \frac{2}{{(\sqrt{2} + 1)}^{4 (n - 1)} + 1}$ $a_{n} = \frac{2 + \sqrt{2} [1 - \frac{2}{{(\sqrt{2} + 1)}^{4 (n - 1)} + 1}]}{1 + \sqrt{2} [1 - \frac{2}{{(\sqrt{2} + 1)}^{4 (n - 1)} + 1}]}$ $a_{n} = \frac{\sqrt{2} (\sqrt{2} + 1) - \frac{2 \sqrt{2}}{{(\sqrt{2} + 1)}^{4 (n - 1)} + 1}}{\sqrt{2} + 1 - \frac{2 \sqrt{2}}{{(\sqrt{2} + 1)}^{4 (n - 1)} + 1}}$ $a_{n} = \frac{\sqrt{2} (\sqrt{2} + 1) + \sqrt{2} {(\sqrt{2} + 1)}^{4 (n - 1) + 1} - 2 \sqrt{2}}{\sqrt{2} + 1 + {(\sqrt{2} + 1)}^{4 (n - 1) + 1} - 2 \sqrt{2}}$ $a_{n} = \frac{\sqrt{2} [(\sqrt{2} - 1) + {(\sqrt{2} + 1)}^{4 (n - 1) + 1}]}{{(\sqrt{2} + 1)}^{4 (n - 1) + 1} - (\sqrt{2} - 1)}$ $a_{n} = \frac{\sqrt{2} [{(\sqrt{2} + 1)}^{2 (2 n - 1)} + 1]}{{(\sqrt{2} + 1)}^{2 (2 n - 1)} - 1}$
	Answered by mindispower last updated on 14/Jun/21
	$let f(x)=((3x+4)/(2x+3)),f^n (x)=fofo.....f(x) n composition of f f^n (x)=((a_n x+b_n )/(c_n x+d_n )) A= (((3 2)),((4 3)) ),A^n = (((a_n c_n )),((b_n d_n )) )...claimt A^1 =A true A^(n+1) =A.A^n = (((3a_n +2b_n 2d_n +3c_n )),((4b_n +3d_n 4c_n +3d_n )) ) f^(n+1) (x)=f^n of=((a_n .((3x+4)/(2x+3))+b_n )/(c_n .((3x+4)/(2x+3))+d_n )) =(((3a_n +2b_n )x+4a_n +3b_n )/((3c_n +2d_n )x+4c_n +3d_n ))...True det(A−xI_2 )=0 ⇔(3−x)^2 −8=0⇒x∈{2(√2)+3;2(√2)−3} (A−(3+2(√2))I_2 )u=0 ⇒u(x,y)∣−2(√2)x+2y=0⇒u(1,(√2)) (A−(2(√2)−3))u=0⇒(6−2(√2))x+2y=0 u(1,3−(√2)) P= (((1 1)),(((√2) 3−(√2))) ) P^− =(1/(3−2(√2))) (((3−(√2) −1 )),((−(√2) 1)) ) A^n =(1/(3−2(√2))) (((1 1)),(((√2) 3−(√2))) ). ((((2(√2)+3)^n 0)),((0 (2(√2)−3)^n )) ). (((3−(√2) −1)),((−(√2) 1)) ) this give us a_n ,b_n ,c_n ,d_n then f^n (x) a_(n+1) =f^n (2),f^o =Id,f(x)=x$
	$l e t f (x) = \frac{3 x + 4}{2 x + 3}, f^{n} (x) = f o f o . . . . . f (x)$ $n c o m p o s i t i o n o f f$ $f^{n} (x) = \frac{a_{n} x + b_{n}}{c_{n} x + d_{n}}$ $A = (\begin{matrix} 3 2 \\ 4 3 \end{matrix}), A^{n} = (\begin{matrix} a_{n} c_{n} \\ b_{n} d_{n} \end{matrix}) . . . c l a i m t$ $A^{1} = A t r u e$ $A^{n + 1} = A . A^{n} = (\begin{matrix} 3 a_{n} + 2 b_{n} 2 d_{n} + 3 c_{n} \\ 4 b_{n} + 3 d_{n} 4 c_{n} + 3 d_{n} \end{matrix})$ $f^{n + 1} (x) = f^{n} o f = \frac{a_{n} . \frac{3 x + 4}{2 x + 3} + b_{n}}{c_{n} . \frac{3 x + 4}{2 x + 3} + d_{n}}$ $= \frac{(3 a_{n} + 2 b_{n}) x + 4 a_{n} + 3 b_{n}}{(3 c_{n} + 2 d_{n}) x + 4 c_{n} + 3 d_{n}} . . . T r u e$ $d e t (A - {x I}_{2}) = 0$ $\Leftrightarrow {(3 - x)}^{2} - 8 = 0 \Rightarrow x \in {2 \sqrt{2} + 3; 2 \sqrt{2} - 3}$ $(A - (3 + 2 \sqrt{2}) I_{2}) u = 0$ $\Rightarrow u (x, y) ∣ - 2 \sqrt{2} x + 2 y = 0 \Rightarrow u (1, \sqrt{2})$ $(A - (2 \sqrt{2} - 3)) u = 0 \Rightarrow (6 - 2 \sqrt{2}) x + 2 y = 0$ $u (1, 3 - \sqrt{2})$ $P = (\begin{matrix} 1 1 \\ \sqrt{2} 3 - \sqrt{2} \end{matrix})$ $P^{-} = \frac{1}{3 - 2 \sqrt{2}} (\begin{matrix} 3 - \sqrt{2} - 1 \\ - \sqrt{2} 1 \end{matrix})$ $A^{n} = \frac{1}{3 - 2 \sqrt{2}} (\begin{matrix} 1 1 \\ \sqrt{2} 3 - \sqrt{2} \end{matrix}) . (\begin{matrix} {(2 \sqrt{2} + 3)}^{n} 0 \\ 0 {(2 \sqrt{2} - 3)}^{n} \end{matrix}) . (\begin{matrix} 3 - \sqrt{2} - 1 \\ - \sqrt{2} 1 \end{matrix})$ $t h i s g i v e u s a_{n}, b_{n}, c_{n}, d_{n}$ $t h e n f^{n} (x)$ $a_{n + 1} = f^{n} (2), f^{o} = I d, f (x) = x$
	Commented by mr W last updated on 14/Jun/21

	$t h a n k s s i r!$
	Commented by mindispower last updated on 17/Jun/21

	$p l e a s u r s i r$
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