{ ((u_(n+1) =((au_n +b)/(cu_n +d)))),((u_0 =k)) :} find u


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Question Number 207615 by mr W last updated on 21/May/24
${ ((u_(n+1) =((au_n +b)/(cu_n +d)))),((u_0 =k)) :} find u_n in terms of n.$
${\begin{cases} u_{n + 1} = \frac{{a u}_{n} + b}{{c u}_{n} + d} \\ u_{0} = k \end{cases}$ $f i n d u_{n} i n t e r m s o f n .$
Commented by mr W last updated on 20/May/24

$a n o l d q u e s t i o n Q 207477$
	Answered by mr W last updated on 21/May/24
	$u_(n+1) =((au_n +b)/(cu_n +d)) let u_n =(A_n /B_n ) (A_(n+1) /B_(n+1) )=((a×(A_n /B_n )+b)/(c×(A_n /B_n )+d))=((aA_n +bB_n )/(cA_n +dB_n )) { ((A_(n+1) =aA_n +bB_n )),((B_(n+1) =cA_n +dB_n )) :} (see also Q118849 for solution of such recurrence equation systems) let A_(n+1) +pB_(n+1) =q(A_n +pB_n ) A_(n+1) +p(cA_n +dB_n )=q(A_n +pB_n ) A_(n+1) =(q−pc)A_n +p(q−d)B_n ≡aA_n +bB_n { ((q−pc=a ...(i))),((p(q−d)=b ...(ii))) :} p(a+pc−d)=b ⇒cp^2 +(a−d)p−b=0 p=((−a+d±(√((a−d)^2 +4cb)))/(2c)) q=a+cp=((a+d±(√((a−d)^2 +4cb)))/2) (we can take any of the two pairs of p and q) A_n +pB_n =q(A_(n−1) +pB_(n−1) )=...=(A_0 +pB_0 )q^n A_(n+1) =aA_n +bB_n =aA_n +(b/p)[(A_0 +pB_0 )q^n −A_n ] ⇒A_(n+1) =(a−(b/p))A_n +(b/p)(A_0 +pB_0 )q^n let A_n =T_n +sq^n T_(n+1) +sq^(n+1) =(a−(b/p))(T_n +sq^n )+(b/p)(A_0 +pB_0 )q^n T_(n+1) =(a−(b/p))T_n +[s(a−(b/p)−q)+(b/p)(A_0 +pB_0 )]q^n set s(a−(b/p)−q)+(b/p)(A_0 +pB_0 )=0 ⇒s=((b(A_0 +pB_0 ))/(b+(q−a)p)) ⇒(s/B_0 )=((b(u_0 +p))/(b+(q−a)p)) T_(n+1) =(a−(b/p))T_n ⇒T_n =(a−(b/p))T_(n−1) =...=(a−(b/p))^n T_0 =(a−(b/p))^n (A_0 −s) ⇒A_n =(a−(b/p))^n (A_0 −s)+sq^n B_n =(((A_0 +pB_0 )q^n −A_n )/p) B_n =(((A_0 +pB_0 )q^n −(a−(b/p))^n (A_0 −s)−sq^n )/p) ⇒B_n =(((A_0 +pB_0 −s)q^n −(a−(b/p))^n (A_0 −s))/p) ⇒u_n =(A_n /B_n )=((−pB_n +(A_0 +pB_0 )q^n )/B_n )=−p+(((A_0 +pB_0 )q^n )/B_n ) ⇒u_n =−p+((p(A_0 +pB_0 )q^n )/((A_0 +pB_0 −s)q^n −(a−(b/p))^n (A_0 −s))) ⇒u_n =−p+((p(u_0 +p)q^n )/((u_0 +p−(s/B_0 ))q^n −(a−(b/p))^n (u_0 −(s/B_0 )))) ⇒u_n =−p+((p(u_0 +p)q^n )/([u_0 +p−((b(u_0 +p))/(b+(q−a)p))]q^n −(a−(b/p))^n [u_0 −((b(u_0 +p))/(b+(q−a)p))])) ⇒u_n =−p+((pq^n )/([1−(b/(b+(q−a)p))]q^n −(a−(b/p))^n [(u_0 /(u_0 +p))−(b/(b+(q−a)p))])) ⇒u_n =((pq^n )/([1−(b/(b+(q−a)p))]q^n −k(a−(b/p))^n ))−p with k=(u_0 /(u_0 +p))−(b/(b+(q−a)p))=constant example: u_(n+1) =((9u_n −2)/(−3u_n +4)) u_0 =1 a=9, b=−2, c=−3, d=4 p=((−9+4−(√((9−4)^2 +4×(−3)×(−2))))/(2×(−3)))=2 q=((9+4−(√((9−4)^2 +4×(−3)×(−2))))/2)=3 u_n =((2×3^n )/([1−((−2)/(−2+(3−9)2))]3^n −k(9−((−2)/2))^n ))−2 =((7×3^n )/( 3^(n+1) −((7k)/2)×10^n ))−2 =((7×3^n )/( 3^(n+1) +k_1 10^n ))−2=((7×3^(n+1) )/( 3^(n+2) −2×10^n ))−2 k_1 =−((7k)/2)=−(7/2)[(1/(1+2))−((−2)/(−2+(3−9)2))]=−(2/3) this is the same as Wolfram Alpha gets, see below.$
	$u_{n + 1} = \frac{{a u}_{n} + b}{{c u}_{n} + d}$ $l e t u_{n} = \frac{A_{n}}{B_{n}}$ $\frac{A_{n + 1}}{B_{n + 1}} = \frac{a \times \frac{A_{n}}{B_{n}} + b}{c \times \frac{A_{n}}{B_{n}} + d} = \frac{{a A}_{n} + {b B}_{n}}{{c A}_{n} + {d B}_{n}}$ ${\begin{cases} A_{n + 1} = {a A}_{n} + {b B}_{n} \\ B_{n + 1} = {c A}_{n} + {d B}_{n} \end{cases}$ $(s e e a l s o Q 118849 f o r s o l u t i o n o f$ $s u c h r e c u r r e n c e e q u a t i o n s y s t e m s)$ $l e t A_{n + 1} + {p B}_{n + 1} = q (A_{n} + {p B}_{n})$ $A_{n + 1} + p ({c A}_{n} + {d B}_{n}) = q (A_{n} + {p B}_{n})$ $A_{n + 1} = (q - p c) A_{n} + p (q - d) B_{n} \equiv {a A}_{n} + {b B}_{n}$ ${\begin{cases} q - p c = a . . . (i) \\ p (q - d) = b . . . (i i) \end{cases}$ $p (a + p c - d) = b$ $\Rightarrow {c p}^{2} + (a - d) p - b = 0$ $p = \frac{- a + d \pm \sqrt{{(a - d)}^{2} + 4 c b}}{2 c}$ $q = a + c p = \frac{a + d \pm \sqrt{{(a - d)}^{2} + 4 c b}}{2}$ $(w e c a n t a k e a n y o f t h e t w o p a i r s o f$ $p a n d q)$ $A_{n} + {p B}_{n} = q (A_{n - 1} + {p B}_{n - 1}) = . . . = (A_{0} + {p B}_{0}) q^{n}$ $A_{n + 1} = {a A}_{n} + {b B}_{n} = {a A}_{n} + \frac{b}{p} [(A_{0} + {p B}_{0}) q^{n} - A_{n}]$ $\Rightarrow A_{n + 1} = (a - \frac{b}{p}) A_{n} + \frac{b}{p} (A_{0} + {p B}_{0}) q^{n}$ $l e t A_{n} = T_{n} + {s q}^{n}$ $T_{n + 1} + {s q}^{n + 1} = (a - \frac{b}{p}) (T_{n} + {s q}^{n}) + \frac{b}{p} (A_{0} + {p B}_{0}) q^{n}$ $T_{n + 1} = (a - \frac{b}{p}) T_{n} + [s (a - \frac{b}{p} - q) + \frac{b}{p} (A_{0} + {p B}_{0})] q^{n}$ $s e t s (a - \frac{b}{p} - q) + \frac{b}{p} (A_{0} + {p B}_{0}) = 0$ $\Rightarrow s = \frac{b (A_{0} + {p B}_{0})}{b + (q - a) p}$ $\Rightarrow \frac{s}{B_{0}} = \frac{b (u_{0} + p)}{b + (q - a) p}$ $T_{n + 1} = (a - \frac{b}{p}) T_{n}$ $\Rightarrow T_{n} = (a - \frac{b}{p}) T_{n - 1} = . . . = {(a - \frac{b}{p})}^{n} T_{0} = {(a - \frac{b}{p})}^{n} (A_{0} - s)$ $\Rightarrow A_{n} = {(a - \frac{b}{p})}^{n} (A_{0} - s) + {s q}^{n}$ $B_{n} = \frac{(A_{0} + {p B}_{0}) q^{n} - A_{n}}{p}$ $B_{n} = \frac{(A_{0} + {p B}_{0}) q^{n} - {(a - \frac{b}{p})}^{n} (A_{0} - s) - {s q}^{n}}{p}$ $\Rightarrow B_{n} = \frac{(A_{0} + {p B}_{0} - s) q^{n} - {(a - \frac{b}{p})}^{n} (A_{0} - s)}{p}$ $\Rightarrow u_{n} = \frac{A_{n}}{B_{n}} = \frac{- {p B}_{n} + (A_{0} + {p B}_{0}) q^{n}}{B_{n}} = - p + \frac{(A_{0} + {p B}_{0}) q^{n}}{B_{n}}$ $\Rightarrow u_{n} = - p + \frac{p (A_{0} + {p B}_{0}) q^{n}}{(A_{0} + {p B}_{0} - s) q^{n} - {(a - \frac{b}{p})}^{n} (A_{0} - s)}$ $\Rightarrow u_{n} = - p + \frac{p (u_{0} + p) q^{n}}{(u_{0} + p - \frac{s}{B_{0}}) q^{n} - {(a - \frac{b}{p})}^{n} (u_{0} - \frac{s}{B_{0}})}$ $\Rightarrow u_{n} = - p + \frac{p (u_{0} + p) q^{n}}{[u_{0} + p - \frac{b (u_{0} + p)}{b + (q - a) p}] q^{n} - {(a - \frac{b}{p})}^{n} [u_{0} - \frac{b (u_{0} + p)}{b + (q - a) p}]}$ $\Rightarrow u_{n} = - p + \frac{{p q}^{n}}{[1 - \frac{b}{b + (q - a) p}] q^{n} - {(a - \frac{b}{p})}^{n} [\frac{u_{0}}{u_{0} + p} - \frac{b}{b + (q - a) p}]}$ $\Rightarrow u_{n} = \frac{{p q}^{n}}{[1 - \frac{b}{b + (q - a) p}] q^{n} - k {(a - \frac{b}{p})}^{n}} - p$ $w i t h k = \frac{u_{0}}{u_{0} + p} - \frac{b}{b + (q - a) p} = c o n s t a n t$ $\underset{―}{e x a m p l e :}$ $u_{n + 1} = \frac{9 u_{n} - 2}{- 3 u_{n} + 4}$ $u_{0} = 1$ $a = 9, b = - 2, c = - 3, d = 4$ $p = \frac{- 9 + 4 - \sqrt{{(9 - 4)}^{2} + 4 \times (- 3) \times (- 2)}}{2 \times (- 3)} = 2$ $q = \frac{9 + 4 - \sqrt{{(9 - 4)}^{2} + 4 \times (- 3) \times (- 2)}}{2} = 3$ $u_{n} = \frac{2 \times 3^{n}}{[1 - \frac{- 2}{- 2 + (3 - 9) 2}] 3^{n} - k {(9 - \frac{- 2}{2})}^{n}} - 2$ $= \frac{7 \times 3^{n}}{3^{n + 1} - \frac{7 k}{2} \times 10^{n}} - 2$ $= \frac{7 \times 3^{n}}{3^{n + 1} + k_{1} 10^{n}} - 2 = \frac{7 \times 3^{n + 1}}{3^{n + 2} - 2 \times 10^{n}} - 2$ $k_{1} = - \frac{7 k}{2} = - \frac{7}{2} [\frac{1}{1 + 2} - \frac{- 2}{- 2 + (3 - 9) 2}] = - \frac{2}{3}$ $t h i s i s t h e s a m e a s W o l f r a m A l p h a$ $g e t s, s e e b e l o w .$
	Commented by AliJumaa last updated on 21/May/24

	$i j u s t w a n t t o t e l l y o u t h a t s u p e r m a n c a n t h o l d$ $y o u r i n t e l l e g i n t$ $y o u r r e l l y a s u p e r m a t h m a t i t i o n$ $i c a n t b e l l i v e t h a t t h i n g y o u h q v e w r o t e$ $a l l t h a n k s t o y o u b e s t p r o f e s o r r$
	Commented by mr W last updated on 21/May/24

	$t h a n k s s i r s!$ $i t r i e d d i f f e r e n t w a y s, a n d t h e n$ $i c a m e t o t h e i d e a w i t h u_{n} = \frac{A_{n}}{B_{n}}$ $a n d s a w t h a t t h e q u e s t i o n b e c o m e s$ $a c u r r e n c e e q u a t i o n s y s t e m w h i c h$ $i o n c e s o l v e d i n a n o l d p o s t, t h e$ $r e s t u p o n h e r e i s t h e n c l e a r .$ ${t h a t}^{'} s t h e w a y h o w i g o t t o m y$ $s o l u t i o n .$
	Commented by Tinku Tara last updated on 21/May/24

	$Great solution . even wolfram alpha pro is$ $not able to come up with steps$
	Commented by Tawa11 last updated on 21/May/24

	$Great sir$
	Commented by mr W last updated on 21/May/24

	Commented by sniper237 last updated on 21/May/24

	$G r e a t S i r$ $O b s t a i n s t h a t p a n d q a r e t h e$ $s o l u t i o n s o f t h e e q u a t i o n \frac{a x + b}{c x + d} = x$ $A n d V_{n} = \frac{U_{n} - p}{U_{n} - q} i s g e o m e t r i c$
	Commented by mr W last updated on 23/May/24

	$a l t e r n a t i v e w a y t o s o l v e$ $u_{n} = \frac{A_{n}}{B_{n}} = \frac{{a A}_{n - 1} + {b B}_{n - 1}}{{c A}_{n - 1} + {d B}_{n - 1}}$ $[\begin{matrix} A_{n} \\ B_{n} \end{matrix}] = [\begin{matrix} a & b \\ c & d \end{matrix}] [\begin{matrix} A_{n - 1} \\ B_{n - 1} \end{matrix}]$ $= . . . = {[\begin{matrix} a & b \\ c & d \end{matrix}]}^{n} [\begin{matrix} A_{0} \\ B_{0} \end{matrix}]$ $e x a m p l e :$ $[\begin{matrix} A_{n} \\ B_{n} \end{matrix}] = {[\begin{matrix} 9 & - 2 \\ - 3 & 4 \end{matrix}]}^{n} [\begin{matrix} 1 \\ 1 \end{matrix}]$ $= \frac{1}{7} [\begin{matrix} 3^{n} + 6 \times 10^{n} & 2 \times 3^{n} - 2 \times 10^{n} \\ 3 \times 3^{n} - 3 \times 10^{n} & 6 \times 3^{n} + 10^{n} \end{matrix}] [\begin{matrix} 1 \\ 1 \end{matrix}]$ $A_{n} = \frac{1}{7} (3^{n + 1} + 4 \times 10^{n})$ $B_{n} = \frac{1}{7} (3^{n + 2} - 2 \times 10^{n})$ $\Rightarrow u_{n} = \frac{3^{n + 1} + 4 \times 10^{n}}{3^{n + 2} - 2 \times 10^{n}}$ $= \frac{7 \times 3^{n + 1}}{3^{n + 2} - 2 \times 10^{n}} - 2 ✓$
	Commented by sniper237 last updated on 21/May/24

	$i t {i s n}^{'} t t h e u s u a l p r o d u c t$ ${i t}^{'} s a n a c t i o n p r o d u c t$
	Commented by Tawa11 last updated on 21/Jun/24

	$Ohh .$
	Answered by mr W last updated on 23/May/24

	$a n o t h e r s o l u t i o n$ $(s e e a l s o Q 81871)$ $u_{n + 1} = \frac{{a u}_{n} + b}{{c u}_{n} + d}$ $u_{n + 1} + p = \frac{{a u}_{n} + b}{{c u}_{n} + d} + p = \frac{(a + c p) u_{n} + (b + d p)}{{c u}_{n} + d}$ $u_{n + 1} + q = \frac{{a u}_{n} + b}{{c u}_{n} + d} + q = \frac{(a + c q) u_{n} + (b + d q)}{{c u}_{n} + d}$ $\frac{u_{n + 1} + p}{u_{n + 1} + q} = \frac{a + c p}{a + c q} \times \frac{u_{n} + \frac{b + d p}{a + c p}}{u_{n} + \frac{b + d q}{a + c q}}$ $s e t p = \frac{b + d p}{a + c p}, \Rightarrow {c p}^{2} + (a - d) p - b = 0$ $s e t q = \frac{b + d q}{a + c q}, \Rightarrow {c q}^{2} + (a - d) q - b = 0$ $i . e . p, q a r e r o o t s o f {c x}^{2} + (a - d) x - b = 0$ $\Rightarrow p, q = \frac{- a + d \pm \sqrt{{(a - d)}^{2} + 4 b c}}{2 c}$ $\frac{u_{n + 1} + p}{u_{n + 1} + q} = \frac{a + c p}{a + c q} \times \frac{u_{n} + p}{u_{n} + q} = λ (\frac{u_{n} + p}{u_{n} + q})$ $\Rightarrow \frac{u_{n} + p}{u_{n} + q} = λ^{n} \times \frac{u_{0} + p}{u_{0} + q} = λ^{n} k$ $\Rightarrow (k λ^{n} - 1) u_{n} = p - k q λ^{n}$ $\Rightarrow u_{n} = \frac{p - k q λ^{n}}{k λ^{n} - 1} = \frac{p - q}{k λ^{n} - 1} - q$ $w i t h k = \frac{u_{0} + p}{u_{0} + q}, λ = \frac{a + c p}{a + c q}$ $e x a m p l e :$ $u_{n + 1} = \frac{9 u_{n} - 2}{- 3 u_{n} + 4}, u_{0} = 1$ $a = 9, b = - 2, c = - 3, d = 4$ $p, q = \frac{- 9 + 4 \pm \sqrt{{(9 - 4)}^{2} + 4 (- 2) (- 3)}}{2 (- 3)} = - \frac{1}{3}, 2$ $λ = \frac{a + c p}{a + c q} = \frac{9 - 3 \times (- \frac{1}{3})}{9 - 3 \times 2} = \frac{10}{3}$ $k = \frac{u_{0} + p}{u_{0} + q} = \frac{1 - \frac{1}{3}}{1 + 2} = \frac{2}{9}$ $u_{n} = \frac{- \frac{1}{3} - 2}{\frac{2}{9} \times {(\frac{10}{3})}^{n} - 1} - 2$ $= \frac{7 \times 3^{n + 1}}{3^{n + 2} - 2 \times 10^{n}} - 2$
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