Question Number 207477 by AliJumaa last updated on 16/May/24
$${is}\:{there}\:{any}\:{generale}\:{form}\:{for}\:{this}\:{sequense}\: \\ $$$$\begin{cases}{{u}_{{n}+\mathrm{1}} =\frac{{au}_{{n}} +{b}}{{cu}_{{n}} +{d}}}\\{{u}_{{m}} ={k}}\end{cases} \\ $$$${I}\:{need}\:{u}_{{n}} \:{in}\:{terms}\:{of}\:{n}\:{i}\:{have}\:{try}\:{to}\:{derrive}\:{it}\:{for}\:{a}\:{long}\:{time}\:{but}\:{i}\:{cant} \\ $$
Commented by Frix last updated on 17/May/24
$${u}_{{n}+\mathrm{1}} =\frac{{a}}{{c}}−\frac{\frac{{ad}−{bc}}{{c}^{\mathrm{2}} }}{{u}_{{n}} +\frac{{d}}{{c}}} \\ $$$${A}=\frac{{a}}{{c}}\wedge{B}=\frac{{ad}−{bc}}{{c}^{\mathrm{2}} }\wedge{C}=\frac{{d}}{{c}} \\ $$$${u}_{{n}+\mathrm{1}} ={A}−\frac{{B}}{{u}_{{n}} +{C}} \\ $$$$\mathrm{Still}\:\mathrm{this}\:\mathrm{looks}\:\mathrm{rather}\:\mathrm{nasty}… \\ $$
Commented by Tinku Tara last updated on 17/May/24
$${assuming}\:{a},{c}\neq\mathrm{0} \\ $$$${b}/{a}={j} \\ $$$${d}/{c}={k} \\ $$$${f}\left({n}+\mathrm{1}\right)=\frac{{f}\left({n}\right)+{j}}{{f}\left({n}\right)+{k}} \\ $$$$\mathrm{Wolfram}\:\mathrm{alpha}\:\mathrm{gives}\:\mathrm{a}\:\mathrm{solution} \\ $$
Commented by Tinku Tara last updated on 17/May/24
Commented by Tinku Tara last updated on 17/May/24
https://m.wolframalpha.com/input?i=f%28n%2B1%29%3D%28%28f%28n%29%2Bj%29%2F%28f%28n%29%2Bk%29%29&lang=en
Commented by AliJumaa last updated on 17/May/24
$${All}\:{thanks}\:{to}\:{you} \\ $$
Commented by Tinku Tara last updated on 17/May/24
$$\mathrm{Maybe}\:\mathrm{someone}\:\mathrm{can}\:\mathrm{find}\:\mathrm{steps} \\ $$$$\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{given}\:\mathrm{by}\:\mathrm{wolfram} \\ $$$$\mathrm{alpha} \\ $$
Commented by AliJumaa last updated on 17/May/24
$${i}\:{want}\:{to}\:{thank}\:{you}\:{for}\:{the}\:{link}\: \\ $$$${i}'{ve}\:{tried}\:{this}\:{before}\:{and}\:{it}\:{was}\:{so}\:{complecated} \\ $$
Commented by mr W last updated on 20/May/24
$${i}\:{have}\:{solved}\:{this}\:{in}\:{my}\:{own}\:{way}, \\ $$$${see}\:{Q}\mathrm{207615} \\ $$