Question Number 207615 by mr W last updated on 21/May/24
$$\begin{cases}{{u}_{{n}+\mathrm{1}} =\frac{{au}_{{n}} +{b}}{{cu}_{{n}} +{d}}}\\{{u}_{\mathrm{0}} ={k}}\end{cases} \\ $$$${find}\:{u}_{{n}} \:{in}\:{terms}\:{of}\:{n}. \\ $$
Commented by mr W last updated on 20/May/24
$${an}\:{old}\:{question}\:{Q}\mathrm{207477} \\ $$
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.](https://www.tinkutara.com/question/Q207618.png)
$${u}_{{n}+\mathrm{1}} =\frac{{au}_{{n}} +{b}}{{cu}_{{n}} +{d}} \\ $$$${let}\:{u}_{{n}} =\frac{{A}_{{n}} }{{B}_{{n}} } \\ $$$$\frac{{A}_{{n}+\mathrm{1}} }{{B}_{{n}+\mathrm{1}} }=\frac{{a}×\frac{{A}_{{n}} }{{B}_{{n}} }+{b}}{{c}×\frac{{A}_{{n}} }{{B}_{{n}} }+{d}}=\frac{{aA}_{{n}} +{bB}_{{n}} }{{cA}_{{n}} +{dB}_{{n}} } \\ $$$$\begin{cases}{{A}_{{n}+\mathrm{1}} ={aA}_{{n}} +{bB}_{{n}} }\\{{B}_{{n}+\mathrm{1}} ={cA}_{{n}} +{dB}_{{n}} }\end{cases} \\ $$$$ \\ $$$$\left({see}\:{also}\:\:{Q}\mathrm{118849}\:{for}\:{solution}\:{of}\right. \\ $$$$\left.{such}\:{recurrence}\:{equation}\:{systems}\right) \\ $$$$ \\ $$$${let}\:{A}_{{n}+\mathrm{1}} +{pB}_{{n}+\mathrm{1}} ={q}\left({A}_{{n}} +{pB}_{{n}} \right) \\ $$$${A}_{{n}+\mathrm{1}} +{p}\left({cA}_{{n}} +{dB}_{{n}} \right)={q}\left({A}_{{n}} +{pB}_{{n}} \right) \\ $$$${A}_{{n}+\mathrm{1}} =\left({q}−{pc}\right){A}_{{n}} +{p}\left({q}−{d}\right){B}_{{n}} \equiv{aA}_{{n}} +{bB}_{{n}} \\ $$$$\begin{cases}{{q}−{pc}={a}\:\:\:…\left({i}\right)}\\{{p}\left({q}−{d}\right)={b}\:\:\:…\left({ii}\right)}\end{cases} \\ $$$${p}\left({a}+{pc}−{d}\right)={b} \\ $$$$\Rightarrow{cp}^{\mathrm{2}} +\left({a}−{d}\right){p}−{b}=\mathrm{0} \\ $$$${p}=\frac{−{a}+{d}\pm\sqrt{\left({a}−{d}\right)^{\mathrm{2}} +\mathrm{4}{cb}}}{\mathrm{2}{c}} \\ $$$${q}={a}+{cp}=\frac{{a}+{d}\pm\sqrt{\left({a}−{d}\right)^{\mathrm{2}} +\mathrm{4}{cb}}}{\mathrm{2}} \\ $$$$\left({we}\:{can}\:{take}\:{any}\:{of}\:{the}\:{two}\:{pairs}\:{of}\right. \\ $$$$\left.{p}\:{and}\:{q}\right) \\ $$$$ \\ $$$${A}_{{n}} +{pB}_{{n}} ={q}\left({A}_{{n}−\mathrm{1}} +{pB}_{{n}−\mathrm{1}} \right)=…=\left({A}_{\mathrm{0}} +{pB}_{\mathrm{0}} \right){q}^{{n}} \\ $$$${A}_{{n}+\mathrm{1}} ={aA}_{{n}} +{bB}_{{n}} ={aA}_{{n}} +\frac{{b}}{{p}}\left[\left({A}_{\mathrm{0}} +{pB}_{\mathrm{0}} \right){q}^{{n}} −{A}_{{n}} \right] \\ $$$$\Rightarrow{A}_{{n}+\mathrm{1}} =\left({a}−\frac{{b}}{{p}}\right){A}_{{n}} +\frac{{b}}{{p}}\left({A}_{\mathrm{0}} +{pB}_{\mathrm{0}} \right){q}^{{n}} \\ $$$${let}\:{A}_{{n}} ={T}_{{n}} +{sq}^{{n}} \\ $$$${T}_{{n}+\mathrm{1}} +{sq}^{{n}+\mathrm{1}} =\left({a}−\frac{{b}}{{p}}\right)\left({T}_{{n}} +{sq}^{{n}} \right)+\frac{{b}}{{p}}\left({A}_{\mathrm{0}} +{pB}_{\mathrm{0}} \right){q}^{{n}} \\ $$$${T}_{{n}+\mathrm{1}} =\left({a}−\frac{{b}}{{p}}\right){T}_{{n}} +\left[{s}\left({a}−\frac{{b}}{{p}}−{q}\right)+\frac{{b}}{{p}}\left({A}_{\mathrm{0}} +{pB}_{\mathrm{0}} \right)\right]{q}^{{n}} \\ $$$${set}\:{s}\left({a}−\frac{{b}}{{p}}−{q}\right)+\frac{{b}}{{p}}\left({A}_{\mathrm{0}} +{pB}_{\mathrm{0}} \right)=\mathrm{0} \\ $$$$\Rightarrow{s}=\frac{{b}\left({A}_{\mathrm{0}} +{pB}_{\mathrm{0}} \right)}{{b}+\left({q}−{a}\right){p}} \\ $$$$\Rightarrow\frac{{s}}{{B}_{\mathrm{0}} }=\frac{{b}\left({u}_{\mathrm{0}} +{p}\right)}{{b}+\left({q}−{a}\right){p}} \\ $$$${T}_{{n}+\mathrm{1}} =\left({a}−\frac{{b}}{{p}}\right){T}_{{n}} \\ $$$$\Rightarrow{T}_{{n}} =\left({a}−\frac{{b}}{{p}}\right){T}_{{n}−\mathrm{1}} =…=\left({a}−\frac{{b}}{{p}}\right)^{{n}} {T}_{\mathrm{0}} =\left({a}−\frac{{b}}{{p}}\right)^{{n}} \left({A}_{\mathrm{0}} −{s}\right) \\ $$$$\Rightarrow{A}_{{n}} =\left({a}−\frac{{b}}{{p}}\right)^{{n}} \left({A}_{\mathrm{0}} −{s}\right)+{sq}^{{n}} \\ $$$${B}_{{n}} =\frac{\left({A}_{\mathrm{0}} +{pB}_{\mathrm{0}} \right){q}^{{n}} −{A}_{{n}} }{{p}} \\ $$$${B}_{{n}} =\frac{\left({A}_{\mathrm{0}} +{pB}_{\mathrm{0}} \right){q}^{{n}} −\left({a}−\frac{{b}}{{p}}\right)^{{n}} \left({A}_{\mathrm{0}} −{s}\right)−{sq}^{{n}} }{{p}} \\ $$$$\Rightarrow{B}_{{n}} =\frac{\left({A}_{\mathrm{0}} +{pB}_{\mathrm{0}} −{s}\right){q}^{{n}} −\left({a}−\frac{{b}}{{p}}\right)^{{n}} \left({A}_{\mathrm{0}} −{s}\right)}{{p}} \\ $$$$\Rightarrow{u}_{{n}} =\frac{{A}_{{n}} }{{B}_{{n}} }=\frac{−{pB}_{{n}} +\left({A}_{\mathrm{0}} +{pB}_{\mathrm{0}} \right){q}^{{n}} }{{B}_{{n}} }=−{p}+\frac{\left({A}_{\mathrm{0}} +{pB}_{\mathrm{0}} \right){q}^{{n}} }{{B}_{{n}} } \\ $$$$\Rightarrow{u}_{{n}} =−{p}+\frac{{p}\left({A}_{\mathrm{0}} +{pB}_{\mathrm{0}} \right){q}^{{n}} }{\left({A}_{\mathrm{0}} +{pB}_{\mathrm{0}} −{s}\right){q}^{{n}} −\left({a}−\frac{{b}}{{p}}\right)^{{n}} \left({A}_{\mathrm{0}} −{s}\right)} \\ $$$$\Rightarrow{u}_{{n}} =−{p}+\frac{{p}\left({u}_{\mathrm{0}} +{p}\right){q}^{{n}} }{\left({u}_{\mathrm{0}} +{p}−\frac{{s}}{{B}_{\mathrm{0}} }\right){q}^{{n}} −\left({a}−\frac{{b}}{{p}}\right)^{{n}} \left({u}_{\mathrm{0}} −\frac{{s}}{{B}_{\mathrm{0}} }\right)} \\ $$$$\Rightarrow{u}_{{n}} =−{p}+\frac{{p}\left({u}_{\mathrm{0}} +{p}\right){q}^{{n}} }{\left[{u}_{\mathrm{0}} +{p}−\frac{{b}\left({u}_{\mathrm{0}} +{p}\right)}{{b}+\left({q}−{a}\right){p}}\right]{q}^{{n}} −\left({a}−\frac{{b}}{{p}}\right)^{{n}} \left[{u}_{\mathrm{0}} −\frac{{b}\left({u}_{\mathrm{0}} +{p}\right)}{{b}+\left({q}−{a}\right){p}}\right]} \\ $$$$\Rightarrow{u}_{{n}} =−{p}+\frac{{pq}^{{n}} }{\left[\mathrm{1}−\frac{{b}}{{b}+\left({q}−{a}\right){p}}\right]{q}^{{n}} −\left({a}−\frac{{b}}{{p}}\right)^{{n}} \left[\frac{{u}_{\mathrm{0}} }{{u}_{\mathrm{0}} +{p}}−\frac{{b}}{{b}+\left({q}−{a}\right){p}}\right]} \\ $$$$\Rightarrow{u}_{{n}} =\frac{{pq}^{{n}} }{\left[\mathrm{1}−\frac{{b}}{{b}+\left({q}−{a}\right){p}}\right]{q}^{{n}} −{k}\left({a}−\frac{{b}}{{p}}\right)^{{n}} }−{p} \\ $$$${with}\:{k}=\frac{{u}_{\mathrm{0}} }{{u}_{\mathrm{0}} +{p}}−\frac{{b}}{{b}+\left({q}−{a}\right){p}}={constant} \\ $$$$ \\ $$$$\underline{{example}:} \\ $$$${u}_{{n}+\mathrm{1}} =\frac{\mathrm{9}{u}_{{n}} −\mathrm{2}}{−\mathrm{3}{u}_{{n}} +\mathrm{4}} \\ $$$${u}_{\mathrm{0}} =\mathrm{1} \\ $$$${a}=\mathrm{9},\:{b}=−\mathrm{2},\:{c}=−\mathrm{3},\:{d}=\mathrm{4} \\ $$$${p}=\frac{−\mathrm{9}+\mathrm{4}−\sqrt{\left(\mathrm{9}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{4}×\left(−\mathrm{3}\right)×\left(−\mathrm{2}\right)}}{\mathrm{2}×\left(−\mathrm{3}\right)}=\mathrm{2} \\ $$$${q}=\frac{\mathrm{9}+\mathrm{4}−\sqrt{\left(\mathrm{9}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{4}×\left(−\mathrm{3}\right)×\left(−\mathrm{2}\right)}}{\mathrm{2}}=\mathrm{3} \\ $$$${u}_{{n}} =\frac{\mathrm{2}×\mathrm{3}^{{n}} }{\left[\mathrm{1}−\frac{−\mathrm{2}}{−\mathrm{2}+\left(\mathrm{3}−\mathrm{9}\right)\mathrm{2}}\right]\mathrm{3}^{{n}} −{k}\left(\mathrm{9}−\frac{−\mathrm{2}}{\mathrm{2}}\right)^{{n}} }−\mathrm{2} \\ $$$$\:\:\:\:\:=\frac{\mathrm{7}×\mathrm{3}^{{n}} }{\:\mathrm{3}^{{n}+\mathrm{1}} −\frac{\mathrm{7}{k}}{\mathrm{2}}×\mathrm{10}^{{n}} }−\mathrm{2} \\ $$$$\:\:\:\:=\frac{\mathrm{7}×\mathrm{3}^{{n}} }{\:\mathrm{3}^{{n}+\mathrm{1}} +{k}_{\mathrm{1}} \mathrm{10}^{{n}} }−\mathrm{2}=\frac{\mathrm{7}×\mathrm{3}^{{n}+\mathrm{1}} }{\:\mathrm{3}^{{n}+\mathrm{2}} −\mathrm{2}×\mathrm{10}^{{n}} }−\mathrm{2} \\ $$$${k}_{\mathrm{1}} =−\frac{\mathrm{7}{k}}{\mathrm{2}}=−\frac{\mathrm{7}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}−\frac{−\mathrm{2}}{−\mathrm{2}+\left(\mathrm{3}−\mathrm{9}\right)\mathrm{2}}\right]=−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${this}\:{is}\:{the}\:{same}\:{as}\:{Wolfram}\:{Alpha} \\ $$$${gets},\:{see}\:{below}. \\ $$
Commented by AliJumaa last updated on 21/May/24
$${i}\:{just}\:{want}\:{to}\:{tell}\:{you}\:{that}\:{super}\:{man}\:{cant}\:{hold} \\ $$$${your}\:{intellegint}\: \\ $$$${your}\:{relly}\:{a}\:{super}\:{mathmatition}\: \\ $$$${i}\:{cant}\:{bellive}\:{that}\:{thing}\:{you}\:{hqve}\:{wrote}\: \\ $$$${all}\:{thanks}\:{to}\:{you}\:{best}\:{profesorr} \\ $$
Commented by mr W last updated on 21/May/24
$${thanks}\:{sirs}! \\ $$$${i}\:{tried}\:{different}\:{ways},\:{and}\:{then} \\ $$$${i}\:{came}\:{to}\:{the}\:{idea}\:{with}\:{u}_{{n}} =\frac{{A}_{{n}} }{{B}_{{n}} } \\ $$$${and}\:{saw}\:{that}\:{the}\:{question}\:{becomes} \\ $$$${a}\:{currence}\:{equation}\:{system}\:{which} \\ $$$${i}\:{once}\:{solved}\:{in}\:{an}\:{old}\:{post},\:{the} \\ $$$${rest}\:{upon}\:{here}\:{is}\:{then}\:{clear}. \\ $$$${that}'{s}\:{the}\:{way}\:{how}\:{i}\:{got}\:{to}\:{my} \\ $$$${solution}. \\ $$
Commented by Tinku Tara last updated on 21/May/24
$$\mathrm{Great}\:\mathrm{solution}.\:\mathrm{even}\:\mathrm{wolfram}\:\mathrm{alpha}\:\mathrm{pro}\:\mathrm{is} \\ $$$$\mathrm{not}\:\mathrm{able}\:\mathrm{to}\:\mathrm{come}\:\mathrm{up}\:\mathrm{with}\:\mathrm{steps} \\ $$
Commented by Tawa11 last updated on 21/May/24
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 21/May/24
Commented by sniper237 last updated on 21/May/24
$${Great}\:{Sir} \\ $$$${Obstains}\:\:{that}\:{p}\:{and}\:{q}\:{are}\:{the}\: \\ $$$${solutions}\:{of}\:\:{the}\:{equation}\:\:\frac{{ax}+{b}}{{cx}+{d}}={x}\: \\ $$$${And}\:{V}_{{n}} =\frac{{U}_{{n}} −{p}}{{U}_{{n}} −{q}}\:{is}\:{geometric} \\ $$
Commented by mr W last updated on 23/May/24
![alternative way to solve u_n =(A_n /B_n )=((aA_(n−1) +bB_(n−1) )/(cA_(n−1) +dB_(n−1) )) [(A_n ),(B_n ) ]= [(a,b),(c,d) ] [(A_(n−1) ),(B_(n−1) ) ] =...= [(a,b),(c,d) ]^n [(A_0 ),(B_0 ) ] example: [(A_n ),(B_n ) ]= [(9,(−2)),((−3),4) ]^n [(1),(1) ] =(1/7) [((3^n +6×10^n ),(2×3^n −2×10^n )),((3×3^n −3×10^n ),(6×3^n +10^n )) ] [(1),(1) ] A_n =(1/7)(3^(n+1) +4×10^n ) B_n =(1/7)(3^(n+2) −2×10^n ) ⇒u_n =((3^(n+1) +4×10^n )/(3^(n+2) −2×10^n )) =((7×3^(n+1) )/(3^(n+2) −2×10^n ))−2 ✓](https://www.tinkutara.com/question/Q207636.png)
$${alternative}\:{way}\:{to}\:{solve} \\ $$$${u}_{{n}} =\frac{{A}_{{n}} }{{B}_{{n}} }=\frac{{aA}_{{n}−\mathrm{1}} +{bB}_{{n}−\mathrm{1}} }{{cA}_{{n}−\mathrm{1}} +{dB}_{{n}−\mathrm{1}} } \\ $$$$\begin{bmatrix}{{A}_{{n}} }\\{{B}_{{n}} }\end{bmatrix}=\begin{bmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{bmatrix}\begin{bmatrix}{{A}_{{n}−\mathrm{1}} }\\{{B}_{{n}−\mathrm{1}} }\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=…=\begin{bmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{bmatrix}^{{n}} \begin{bmatrix}{{A}_{\mathrm{0}} }\\{{B}_{\mathrm{0}} }\end{bmatrix} \\ $$$${example}: \\ $$$$\begin{bmatrix}{{A}_{{n}} }\\{{B}_{{n}} }\end{bmatrix}=\begin{bmatrix}{\mathrm{9}}&{−\mathrm{2}}\\{−\mathrm{3}}&{\mathrm{4}}\end{bmatrix}^{{n}} \begin{bmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{bmatrix} \\ $$$$=\frac{\mathrm{1}}{\mathrm{7}}\begin{bmatrix}{\mathrm{3}^{{n}} +\mathrm{6}×\mathrm{10}^{{n}} }&{\mathrm{2}×\mathrm{3}^{{n}} −\mathrm{2}×\mathrm{10}^{{n}} }\\{\mathrm{3}×\mathrm{3}^{{n}} −\mathrm{3}×\mathrm{10}^{{n}} }&{\mathrm{6}×\mathrm{3}^{{n}} +\mathrm{10}^{{n}} }\end{bmatrix}\begin{bmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{bmatrix} \\ $$$${A}_{{n}} =\frac{\mathrm{1}}{\mathrm{7}}\left(\mathrm{3}^{{n}+\mathrm{1}} +\mathrm{4}×\mathrm{10}^{{n}} \right) \\ $$$${B}_{{n}} =\frac{\mathrm{1}}{\mathrm{7}}\left(\mathrm{3}^{{n}+\mathrm{2}} −\mathrm{2}×\mathrm{10}^{{n}} \right) \\ $$$$\Rightarrow{u}_{{n}} =\frac{\mathrm{3}^{{n}+\mathrm{1}} +\mathrm{4}×\mathrm{10}^{{n}} }{\mathrm{3}^{{n}+\mathrm{2}} −\mathrm{2}×\mathrm{10}^{{n}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{7}×\mathrm{3}^{{n}+\mathrm{1}} }{\mathrm{3}^{{n}+\mathrm{2}} −\mathrm{2}×\mathrm{10}^{{n}} }−\mathrm{2}\:\checkmark \\ $$
Commented by sniper237 last updated on 21/May/24
$${it}\:{isn}'{t}\:{the}\:\:{usual}\:{product}\: \\ $$$${it}'{s}\:\:{an}\:{action}\:{product} \\ $$
Commented by Tawa11 last updated on 21/Jun/24
$$\mathrm{Ohh}. \\ $$
Answered by mr W last updated on 23/May/24
$$\boldsymbol{{an}}\:\boldsymbol{{other}}\:\boldsymbol{{solution}} \\ $$$$\left({see}\:{also}\:{Q}\mathrm{81871}\right) \\ $$$$ \\ $$$${u}_{{n}+\mathrm{1}} =\frac{{au}_{{n}} +{b}}{{cu}_{{n}} +{d}} \\ $$$${u}_{{n}+\mathrm{1}} +{p}=\frac{{au}_{{n}} +{b}}{{cu}_{{n}} +{d}}+{p}=\frac{\left({a}+{cp}\right){u}_{{n}} +\left({b}+{dp}\right)}{{cu}_{{n}} +{d}} \\ $$$${u}_{{n}+\mathrm{1}} +{q}=\frac{{au}_{{n}} +{b}}{{cu}_{{n}} +{d}}+{q}=\frac{\left({a}+{cq}\right){u}_{{n}} +\left({b}+{dq}\right)}{{cu}_{{n}} +{d}} \\ $$$$\frac{{u}_{{n}+\mathrm{1}} +{p}}{{u}_{{n}+\mathrm{1}} +{q}}=\frac{{a}+{cp}}{{a}+{cq}}×\frac{{u}_{{n}} +\frac{{b}+{dp}}{{a}+{cp}}}{{u}_{{n}} +\frac{{b}+{dq}}{{a}+{cq}}} \\ $$$${set}\:{p}=\frac{{b}+{dp}}{{a}+{cp}},\:\Rightarrow\:{cp}^{\mathrm{2}} +\left({a}−{d}\right){p}−{b}=\mathrm{0} \\ $$$${set}\:{q}=\frac{{b}+{dq}}{{a}+{cq}},\:\Rightarrow\:{cq}^{\mathrm{2}} +\left({a}−{d}\right){q}−{b}=\mathrm{0} \\ $$$${i}.{e}.\:{p},\:{q}\:{are}\:{roots}\:{of}\:{cx}^{\mathrm{2}} +\left({a}−{d}\right){x}−{b}=\mathrm{0} \\ $$$$\Rightarrow{p},\:{q}=\frac{−{a}+{d}\pm\sqrt{\left({a}−{d}\right)^{\mathrm{2}} +\mathrm{4}{bc}}}{\mathrm{2}{c}} \\ $$$$\frac{{u}_{{n}+\mathrm{1}} +{p}}{{u}_{{n}+\mathrm{1}} +{q}}=\frac{{a}+{cp}}{{a}+{cq}}×\frac{{u}_{{n}} +{p}}{{u}_{{n}} +{q}}=\lambda\left(\frac{{u}_{{n}} +{p}}{{u}_{{n}} +{q}}\right) \\ $$$$\Rightarrow\frac{{u}_{{n}} +{p}}{{u}_{{n}} +{q}}=\lambda^{{n}} ×\frac{{u}_{\mathrm{0}} +{p}}{{u}_{\mathrm{0}} +{q}}=\lambda^{{n}} {k} \\ $$$$\Rightarrow\left({k}\lambda^{{n}} −\mathrm{1}\right){u}_{{n}} ={p}−{kq}\lambda^{{n}} \\ $$$$\Rightarrow{u}_{{n}} =\frac{{p}−{kq}\lambda^{{n}} }{{k}\lambda^{{n}} −\mathrm{1}}=\frac{{p}−{q}}{{k}\lambda^{{n}} −\mathrm{1}}−{q} \\ $$$${with}\:{k}=\frac{{u}_{\mathrm{0}} +{p}}{{u}_{\mathrm{0}} +{q}},\:\lambda=\frac{{a}+{cp}}{{a}+{cq}} \\ $$$$ \\ $$$${example}: \\ $$$${u}_{{n}+\mathrm{1}} =\frac{\mathrm{9}{u}_{{n}} −\mathrm{2}}{−\mathrm{3}{u}_{{n}} +\mathrm{4}},\:{u}_{\mathrm{0}} =\mathrm{1} \\ $$$${a}=\mathrm{9},\:{b}=−\mathrm{2},\:{c}=−\mathrm{3},\:{d}=\mathrm{4} \\ $$$${p},\:{q}=\frac{−\mathrm{9}+\mathrm{4}\pm\sqrt{\left(\mathrm{9}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{4}\left(−\mathrm{2}\right)\left(−\mathrm{3}\right)}}{\mathrm{2}\left(−\mathrm{3}\right)}=−\frac{\mathrm{1}}{\mathrm{3}},\:\mathrm{2} \\ $$$$\lambda=\frac{{a}+{cp}}{{a}+{cq}}=\frac{\mathrm{9}−\mathrm{3}×\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{9}−\mathrm{3}×\mathrm{2}}=\frac{\mathrm{10}}{\mathrm{3}} \\ $$$${k}=\frac{{u}_{\mathrm{0}} +{p}}{{u}_{\mathrm{0}} +{q}}=\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}+\mathrm{2}}=\frac{\mathrm{2}}{\mathrm{9}} \\ $$$${u}_{{n}} =\frac{−\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{2}}{\frac{\mathrm{2}}{\mathrm{9}}×\left(\frac{\mathrm{10}}{\mathrm{3}}\right)^{{n}} −\mathrm{1}}−\mathrm{2} \\ $$$$\:\:\:\:=\frac{\mathrm{7}×\mathrm{3}^{{n}+\mathrm{1}} }{\mathrm{3}^{{n}+\mathrm{2}} −\mathrm{2}×\mathrm{10}^{{n}} }−\mathrm{2} \\ $$